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1038cf1a34c4fe095deb8979b4b0af74 | https://artofproblemsolving.com/wiki/index.php/2016_AIME_I_Problems/Problem_6 | In $\triangle ABC$ let $I$ be the center of the inscribed circle, and let the bisector of $\angle ACB$ intersect $AB$ at $L$ . The line through $C$ and $L$ intersects the circumscribed circle of $\triangle ABC$ at the two points $C$ and $D$ . If $LI=2$ and $LD=3$ , then $IC=\tfrac{m}{n}$ , where $m$ and $n$ are relatively prime positive integers. Find $m+n$ | WLOG assume $\triangle ABC$ is isosceles. Then, $L$ is the midpoint of $AB$ , and $\angle CLB=\angle CLA=90^\circ$ . Draw the perpendicular from $I$ to $CB$ , and let it meet $CB$ at $E$ . Since $IL=2$ $IE$ is also $2$ (they are both inradii). Set $BD$ as $x$ . Then, triangles $BLD$ and $CEI$ are similar, and $\tfrac{2}{3}=\tfrac{CI}{x}$ . Thus, $CI=\tfrac{2x}{3}$ $\triangle CBD \sim \triangle CEI$ , so $\tfrac{IE}{DB}=\tfrac{CI}{CD}$ . Thus $\tfrac{2}{x}=\tfrac{(2x/3)}{(2x/3+5)}$ . Solving for $x$ , we have: $x^2-2x-15=0$ , or $x=5, -3$ $x$ is positive, so $x=5$ . As a result, $CI=\tfrac{2x}{3}=\tfrac{10}{3}$ and the answer is $\boxed{013}$ | null | 013 |
1038cf1a34c4fe095deb8979b4b0af74 | https://artofproblemsolving.com/wiki/index.php/2016_AIME_I_Problems/Problem_6 | In $\triangle ABC$ let $I$ be the center of the inscribed circle, and let the bisector of $\angle ACB$ intersect $AB$ at $L$ . The line through $C$ and $L$ intersects the circumscribed circle of $\triangle ABC$ at the two points $C$ and $D$ . If $LI=2$ and $LD=3$ , then $IC=\tfrac{m}{n}$ , where $m$ and $n$ are relatively prime positive integers. Find $m+n$ | WLOG assume $\triangle ABC$ is isosceles (with vertex $C$ ). Let $O$ be the center of the circumcircle, $R$ the circumradius, and $r$ the inradius. A simple sketch will reveal that $\triangle ABC$ must be obtuse (as an acute triangle will result in $LI$ being greater than $DL$ ) and that $O$ and $I$ are collinear. Next, if $OI=d$ $DO+OI=R+d$ and $R+d=DL+LI=5$ . Euler gives us that $d^{2}=R(R-2r)$ , and in this case, $r=LI=2$ . Thus, $d=\sqrt{R^{2}-4R}$ . Solving for $d$ , we have $R+\sqrt{R^{2}-4R}=5$ , then $R^{2}-4R=25-10R+R^{2}$ , yielding $R=\frac{25}{6}$ . Next, $R+d=5$ so $d=\frac{5}{6}$ . Finally, $OC=OI+IC$ gives us $R=d+IC$ , and $IC=\frac{25}{6}-\frac{5}{6}=\frac{10}{3}$ . Our answer is then $\boxed{013}$ | null | 013 |
1038cf1a34c4fe095deb8979b4b0af74 | https://artofproblemsolving.com/wiki/index.php/2016_AIME_I_Problems/Problem_6 | In $\triangle ABC$ let $I$ be the center of the inscribed circle, and let the bisector of $\angle ACB$ intersect $AB$ at $L$ . The line through $C$ and $L$ intersects the circumscribed circle of $\triangle ABC$ at the two points $C$ and $D$ . If $LI=2$ and $LD=3$ , then $IC=\tfrac{m}{n}$ , where $m$ and $n$ are relatively prime positive integers. Find $m+n$ | Since $\angle{LAD} = \angle{BDC}$ and $\angle{DLA}=\angle{DCB}$ $\triangle{DLA}\sim\triangle{DBC}$ . Also, $\angle{DAC}=\angle{BLC}$ and $\angle{ACD}=\angle{LCB}$ so $\triangle{DAC}\sim\triangle{BLC}$ . Now we can call $AC$ $b$ and $BC$ $a$ . By angle bisector theorem, $\frac{AD}{DB}=\frac{AC}{BC}$ . So let $AD=bk$ and $DB=ak$ for some value of $k$ . Now call $IC=x$ . By the similar triangles we found earlier, $\frac{3}{ak}=\frac{bk}{x+2}$ and $\frac{b}{x+5}=\frac{x+2}{a}$ . We can simplify this to $abk^2=3x+6$ and $ab=(x+5)(x+2)$ . So we can plug the $ab$ into the first equation and get $(x+5)(x+2)k^2=3(x+2) \rightarrow k^2(x+5)=3$ . We can now draw a line through $A$ and $I$ that intersects $BC$ at $E$ . By mass points, we can assign a mass of $a$ to $A$ $b$ to $B$ , and $a+b$ to $D$ . We can also assign a mass of $(a+b)k$ to $C$ by angle bisector theorem. So the ratio of $\frac{DI}{IC}=\frac{(a+b)k}{a+b}=k=\frac{2}{x}$ . So since $k=\frac{2}{x}$ , we can plug this back into the original equation to get $\left(\frac{2}{x}\right)^2(x+5)=3$ . This means that $\frac{3x^2}{4}-x-5=0$ which has roots -2 and $\frac{10}{3}$ which means our $CI=\frac{10}{3}$ and our answer is $\boxed{013}$ | null | 013 |
1038cf1a34c4fe095deb8979b4b0af74 | https://artofproblemsolving.com/wiki/index.php/2016_AIME_I_Problems/Problem_6 | In $\triangle ABC$ let $I$ be the center of the inscribed circle, and let the bisector of $\angle ACB$ intersect $AB$ at $L$ . The line through $C$ and $L$ intersects the circumscribed circle of $\triangle ABC$ at the two points $C$ and $D$ . If $LI=2$ and $LD=3$ , then $IC=\tfrac{m}{n}$ , where $m$ and $n$ are relatively prime positive integers. Find $m+n$ | Since $\angle BCD$ and $\angle BAD$ both intercept arc $BD$ , it follows that $\angle BAD=\gamma$ . Note that $\angle AID=\alpha+\gamma$ by the external angle theorem. It follows that $\angle DAI=\angle AID=\alpha+\gamma$ , so we must have that $\triangle AID$ is isosceles, yielding $AD=ID=5$ . Note that $\triangle DLA \sim \triangle DAC$ , so $\frac{DA}{DL} = \frac{DC}{DA}$ . This yields $DC = \frac{25}{3}$ . It follows that $CI = DC - DI = \frac{10}{3}$ , giving a final answer of $\boxed{013}$ | null | 013 |
1038cf1a34c4fe095deb8979b4b0af74 | https://artofproblemsolving.com/wiki/index.php/2016_AIME_I_Problems/Problem_6 | In $\triangle ABC$ let $I$ be the center of the inscribed circle, and let the bisector of $\angle ACB$ intersect $AB$ at $L$ . The line through $C$ and $L$ intersects the circumscribed circle of $\triangle ABC$ at the two points $C$ and $D$ . If $LI=2$ and $LD=3$ , then $IC=\tfrac{m}{n}$ , where $m$ and $n$ are relatively prime positive integers. Find $m+n$ | We can just say that quadrilateral $ADBC$ is a right kite with right angles at $A$ and $B$ . Let us construct another similar right kite with the points of tangency on $AC$ and $BC$ called $E$ and $F$ respectively, point $I$ , and point $C$ . Note that we only have to look at one half of the circle since the diagram is symmetrical. Let us call $CI$ $x$ for simplicity's sake. Based on the fact that $\triangle BCD$ is similar to $\triangle FCI$ we can use triangle proportionality to say that $BD$ is $2\frac{x+5}{x}$ . Using geometric mean theorem we can show that $BL$ must be $\sqrt{3x+6}$ . With Pythagorean Theorem we can say that $3x+6+9=4{(\frac{x+5}{x})}^2$ . Multiplying both sides by $x^2$ and moving everything to LHS will give you $3{x}^3+11{x}^2-40x-100=0$ Since $x$ must be in the form $\frac{p}{q}$ we can assume that $x$ is most likely a positive fraction in the form $\frac{p}{3}$ where $p$ is a factor of $100$ . Testing the factors in synthetic division would lead $x = \frac{10}{3}$ , giving us our desired answer $\boxed{013}$ . ~Lopkiloinm | null | 013 |
1038cf1a34c4fe095deb8979b4b0af74 | https://artofproblemsolving.com/wiki/index.php/2016_AIME_I_Problems/Problem_6 | In $\triangle ABC$ let $I$ be the center of the inscribed circle, and let the bisector of $\angle ACB$ intersect $AB$ at $L$ . The line through $C$ and $L$ intersects the circumscribed circle of $\triangle ABC$ at the two points $C$ and $D$ . If $LI=2$ and $LD=3$ , then $IC=\tfrac{m}{n}$ , where $m$ and $n$ are relatively prime positive integers. Find $m+n$ | [asy] size(150); import olympiad; real c=8.1,a=5*(c+sqrt(c^2-64))/6,b=5*(c-sqrt(c^2-64))/6; pair A=(0,0),B=(c,0),D=(c/2,-sqrt(25-(c/2)^2)); pair C=intersectionpoints(circle(A,b),circle(B,a))[0]; pair I=incenter(A,B,C); pair L=extension(C,D,A,B); dot(I^^A^^B^^C^^D); draw(C--D); path midangle(pair d,pair e,pair f) {return e--e+((f-e)/length(f-e)+(d-e)/length(d-e))/2;} draw(A--B--D--cycle); draw(circumcircle(A,B,D)); draw(A--C--B); draw(A--I--B^^C--I); draw(incircle(A,B,C)); label("$A$",A,SW,fontsize(8)); label("$B$",B,SE,fontsize(8)); label("$C$",C,N,fontsize(8)); label("$D$",D,S,fontsize(8)); label("$I$",I,NE,fontsize(8)); label("$L$",L,SW,fontsize(8)); label("$\alpha$",A,5*dir(midangle(C,A,I)),fontsize(8)); label("$\alpha$",A,5*dir(midangle(I,A,B)),fontsize(8)); label("$\beta$",B,12*dir(midangle(A,B,I)),fontsize(8)); label("$\beta$",B,12*dir(midangle(I,B,C)),fontsize(8)); label("$\gamma$",C,5*dir(midangle(A,C,I)),fontsize(8)); label("$\gamma$",C,5*dir(midangle(I,C,B)),fontsize(8)); [/asy] Connect $D$ to $A$ and $D$ to $B$ to form quadrilateral $ACBD$ . Since quadrilateral $ACBD$ is cyclic, we can apply Ptolemy's Theorem on the quadrilateral.
Denote the length of $BD$ and $AD$ as $z$ (they must be congruent, as $\angle ABD$ and $\angle DAB$ are both inscribed in arcs that have the same degree measure due to the angle bisector intersecting the circumcircle at $D$ ), and the lengths of $BC$ $AC$ $AB$ , and $CI$ as $a,b,c, x$ , respectively.
After applying Ptolemy's, one will get that:
\[z(a+b)=c(x+5)\]
Next, since $ACBD$ is cyclic, triangles $ALD$ and $CLB$ are similar, yielding the following equation once simplifications are made to the equation $\frac{AD}{CB}=\frac{AL}{BL}$ , with the length of $BL$ written in terms of $a,b,c$ using the angle bisector theorem on triangle $ABC$
\[zc=3(a+b)\]
Next, drawing in the bisector of $\angle BAC$ to the incenter $I$ , and applying the angle bisector theorem, we have that:
\[cx=2(a+b)\]
Now, solving for $z$ in the second equation, and $x$ in the third equation and plugging them both back into the first equation, and making the substitution $w=\frac{a+b}{c}$ , we get the quadratic equation:
\[3w^2-2w-5=0\]
Solving, we get $w=5/3$ , which gives $z=5$ and $x=10/3$ , when we rewrite the above equations in terms of $w$ . Thus, our answer is $\boxed{013}$ and we're done. | null | 013 |
1038cf1a34c4fe095deb8979b4b0af74 | https://artofproblemsolving.com/wiki/index.php/2016_AIME_I_Problems/Problem_6 | In $\triangle ABC$ let $I$ be the center of the inscribed circle, and let the bisector of $\angle ACB$ intersect $AB$ at $L$ . The line through $C$ and $L$ intersects the circumscribed circle of $\triangle ABC$ at the two points $C$ and $D$ . If $LI=2$ and $LD=3$ , then $IC=\tfrac{m}{n}$ , where $m$ and $n$ are relatively prime positive integers. Find $m+n$ | Let $AB=c,BC=a,CA=b$ , and $x=\tfrac{a+b}{c}$ . Then, notice that $\tfrac{CI}{IL}=\tfrac{a+b}{c}=x$ , so $CI=IL\cdot{}x=2x$ . Also, by the incenter-excenter lemma, $AD=BD=ID=IL+LD=5$ . Therefore, by Ptolemy's Theorem on cyclic quadrilateral $ABCD$ $5a+5b=c(2x+5)$ , so $5\left(\tfrac{a+b}{c}\right)=2x+5$ , so $5x=2x+5$ . Solving, we get that $x=\tfrac{5}{3}$ , so $CI=\tfrac{10}{3}$ and the answer is $10+3=\boxed{013}$ | null | 013 |
1038cf1a34c4fe095deb8979b4b0af74 | https://artofproblemsolving.com/wiki/index.php/2016_AIME_I_Problems/Problem_6 | In $\triangle ABC$ let $I$ be the center of the inscribed circle, and let the bisector of $\angle ACB$ intersect $AB$ at $L$ . The line through $C$ and $L$ intersects the circumscribed circle of $\triangle ABC$ at the two points $C$ and $D$ . If $LI=2$ and $LD=3$ , then $IC=\tfrac{m}{n}$ , where $m$ and $n$ are relatively prime positive integers. Find $m+n$ | Without loss of generality, let $\triangle ABC$ be isosceles. Note that by the incenter-excenter lemma, $DI = DA = DB.$ Hence, $DA=DB=5.$ Let the point of tangency of the incircle and $\overline{BC}$ be $F$ and the point of tangency of the incircle and $\overline{AC}$ be $E.$ We note that $\angle ALC = \angle BLC = 90^\circ$ and $LA=LB=4,$ which immediately gives $AE=BF=4.$ Applying the Pythagorean Theorem on $\triangle ALC$ and $\triangle IEC$ gives $2^2+x^2=y^2$ and $4^2+(2+y)^2 = (4+x)^2.$ Solving for $y$ gives us $y=\frac{10}{3}.$ Therefore, $IC = \frac{10}{3}$ so the answer is $\boxed{13}.$ | null | 13 |
df530a4ba8fc4da75b93e64f11f0472e | https://artofproblemsolving.com/wiki/index.php/2016_AIME_I_Problems/Problem_7 | For integers $a$ and $b$ consider the complex number \[\frac{\sqrt{ab+2016}}{ab+100}-\left({\frac{\sqrt{|a+b|}}{ab+100}}\right)i\]
Find the number of ordered pairs of integers $(a,b)$ such that this complex number is a real number. | We consider two cases:
Case 1: $ab \ge -2016$
In this case, if \[0 = \text{Im}\left({\frac{\sqrt{ab+2016}}{ab+100}-\left({\frac{\sqrt{|a+b|}}{ab+100}}\right)i}\right) = -\frac{\sqrt{|a+b|}}{ab+100}\] then $ab \ne -100$ and $|a + b| = 0 = a + b$ . Thus $ab = -a^2$ so $a^2 < 2016$ . Thus $a = -44,-43, ... , -1, 0, 1, ..., 43, 44$ , yielding $89$ values. However since $ab = -a^2 \ne -100$ , we have $a \ne \pm 10$ . Thus there are $87$ allowed tuples $(a,b)$ in this case.
Case 2: $ab < -2016$
In this case, we want \[0 = \text{Im}\left({\frac{\sqrt{ab+2016}}{ab+100}-\left({\frac{\sqrt{|a+b|}}{ab+100}}\right)i}\right) = \frac{\sqrt{-ab-2016} - \sqrt{|a+b|}}{ab+100}\] Squaring, we have the equations $ab \ne -100$ (which always holds in this case) and \[-(ab + 2016)= |a + b|.\] Then if $a > 0$ and $b < 0$ , let $c = -b$ . If $c > a$ \[ac - 2016 = c - a \Rightarrow (a - 1)(c + 1) = 2015 \Rightarrow (a,b) = (2, -2014), (6, -402), (14, -154), (32, -64).\] Note that $ab < -2016$ for every one of these solutions. If $c < a$ , then \[ac - 2016 = a - c \Rightarrow (a + 1)(c - 1) = 2015 \Rightarrow (a,b) = (2014, -2), (402, -6), (154, -14), (64, -32).\] Again, $ab < -2016$ for every one of the above solutions. This yields $8$ solutions. Similarly, if $a < 0$ and $b > 0$ , there are $8$ solutions. Thus, there are a total of $16$ solutions in this case.
Thus, the answer is $87 + 16 = \boxed{103}$ | null | 103 |
10b89ea3f13aa2c628bafff65bf48904 | https://artofproblemsolving.com/wiki/index.php/2016_AIME_I_Problems/Problem_9 | problem_id
10b89ea3f13aa2c628bafff65bf48904 Triangle $ABC$ has $AB=40,AC=31,$ and $\sin{A}...
10b89ea3f13aa2c628bafff65bf48904 It has been noted that this answer won't actua...
Name: Text, dtype: object | We start by drawing a diagram;
[asy] size(400); import olympiad; import geometry; pair A = (0, 20) ,B=(30,10) ,C=(15,0), Q=(30,20) ,R=(30,0), S=(0,0); draw(A--B--C--cycle); draw(A--Q); draw(Q--R); draw(R--S); draw(S--A); label("$A$", A, W); label("$B$", B, E); label("$C$", C, N); label("$Q$", Q, E); label("$R$", R, E); label("$S$", S, W); label("$w$", (-1,10)); label("$l$", (15,21)); label("$y$", (7.5,-1)); label("$x$", (31,15)); label("$31$",(7.5,10), E); label("$40$",(15,15), N); markangle(Label("$\alpha$", Relative(0.5)), n=1, C, A, B); markangle(Label("$\beta$", Relative(0.5)), n=1, B, A, Q); markangle(Label("$\gamma$", Relative(0.5)), n=1, S, A, C); [/asy]
We know that $\sin \alpha = \frac{1}{5}$ . Since $\sin \alpha = \cos (90- \alpha)$ \[\cos (90- \alpha) = \frac{1}{5} \implies \cos (\beta + \gamma) = \frac{1}{5}\]
Using our angle sum identities, we expand this to $\cos \beta \cdot \cos \gamma - \sin \beta \cdot \sin \gamma = \frac{1}{5}$ .
We can now use the right triangle definition of cosine and sine to rewrite this equation as;
\[\frac{l}{40} \cdot \frac{w}{31} - \frac{x}{40} \cdot \frac{y}{31} = \frac{1}{5} \implies lw- xy = 8 \cdot 31 \implies lw = xy + 31 \cdot 8\]
Hang on; $lw$ is the area we want to maximize! Therefore, to maximize this area we must maximize $xy = 40 \sin \beta \cdot 31 \sin \gamma = 31 \cdot 40 \cdot \frac{1}{2}( \cos (\beta - \gamma) - \cos ( \beta + \gamma)) = 31 \cdot 20 \cdot (\cos(\beta-\gamma)-\frac{1}{5})$ .
Since $\cos(\beta-\gamma)$ is the only variable component of this expression, to maximize the expression we must maximize $\cos(\beta-\gamma)$ . The cosine function has a maximum value of 1, so our equation evaluates to $xy = 31 \cdot 20 \cdot (1-\frac{1}{5}) = 31 \cdot 20 \cdot \frac{4}{5} = 31 \cdot 16$ (Note that at this max value, since $\beta$ and $\gamma$ are both acute, $\beta-\gamma=0 \implies \beta=\gamma$ ).
Finally, $lw = xy + 31 \cdot 8 = 31 \cdot 16 + 31\cdot 8 = 31 \cdot 24 = \boxed{744}$ | null | 744 |
10b89ea3f13aa2c628bafff65bf48904 | https://artofproblemsolving.com/wiki/index.php/2016_AIME_I_Problems/Problem_9 | problem_id
10b89ea3f13aa2c628bafff65bf48904 Triangle $ABC$ has $AB=40,AC=31,$ and $\sin{A}...
10b89ea3f13aa2c628bafff65bf48904 It has been noted that this answer won't actua...
Name: Text, dtype: object | As above, we note that angle $A$ must be acute. Therefore, let $A$ be the origin, and suppose that $Q$ is on the positive $x$ axis and $S$ is on the positive $y$ axis. We approach this using complex numbers. Let $w=\text{cis} A$ , and let $z$ be a complex number with $|z|=1$ $\text{Arg}(z)\ge 0^\circ$ and $\text{Arg}(zw)\le90^\circ$ . Then we represent $B$ by $40z$ and $C$ by $31zw$ . The coordinates of $Q$ and $S$ depend on the real part of $40z$ and the imaginary part of $31zw$ . Thus \[[AQRS]=\Re(40z)\cdot \Im(31zw)=1240\left(\frac{z+\overline{z}}{2}\right)\left(\frac{zw-\overline{zw}}{2i}\right).\] We can expand this, using the fact that $z\overline{z}=|z|^2$ , finding \[[AQRS]=620\left(\frac{z^2w-\overline{z^2w}+w-\overline{w}}{2i}\right)=620(\Im(z^2w)+\Im(w)).\] Now as $w=\text{cis}A$ , we know that $\Im(w)=\frac15$ . Also, $|z^2w|=1$ , so the maximum possible imaginary part of $z^2w$ is $1$ . This is clearly achievable under our conditions on $z$ . Therefore, the maximum possible area of $AQRS$ is $620(1+\tfrac15)=\boxed{744}$ | null | 744 |
10b89ea3f13aa2c628bafff65bf48904 | https://artofproblemsolving.com/wiki/index.php/2016_AIME_I_Problems/Problem_9 | problem_id
10b89ea3f13aa2c628bafff65bf48904 Triangle $ABC$ has $AB=40,AC=31,$ and $\sin{A}...
10b89ea3f13aa2c628bafff65bf48904 It has been noted that this answer won't actua...
Name: Text, dtype: object | Let $\theta$ be the angle $\angle BAQ$ . The height of the rectangle then can be expressed as $h = 31 \sin (A+\theta)$ , and the length of the rectangle can be expressed as $l = 40\cos \theta$ . The area of the rectangle can then be written as a function of $\theta$ $[AQRS] = a(\theta) = 31\sin (A+\theta)\cdot 40 \cos \theta = 1240 \sin (A+\theta) \cos \theta$ . For now, we will ignore the $1240$ and focus on the function $f(\theta) = \sin (A+\theta) \cos \theta = (\sin A \cos \theta + \cos A \sin \theta)(\cos \theta) = \sin A \cos^2 \theta + \cos A \sin \theta \cos \theta = \sin A \cos^2 \theta + \frac{1}{2} \cos A \sin 2\theta$
Taking the derivative, $f'(\theta) = \sin A \cdot -2\cos \theta \sin \theta + \cos A \cos 2\theta = \cos A \cos 2\theta - \sin A \sin 2\theta = \cos(2\theta + A)$ . Setting this equal to $0$ , we get $\cos(2 \theta + A) = 0 \Rightarrow 2\theta +A = 90, 270 ^\circ$ . Since we know that $A+ \theta < 90$ , the $270^\circ$ solution is extraneous. Thus, we get that $\theta = \frac{90 - A}{2} = 45 - \frac{A}{2}$
Plugging this value into the original area equation, $a(45 - \frac{A}{2}) = 1240 \sin (45 - \frac{A}{2} + A) \cos (45 - \frac{A}{2}) = 1240\sin( 45+ \frac{A}{2})\cos(45 - \frac{A}{2})$ . Using a product-to-sum formula, we get that: \[1240\sin( 45+ \frac{A}{2})\cos(45 - \frac{A}{2}) =\] \[1240\cdot \frac{1}{2}\cdot(\sin((45 + \frac{A}{2}) + (45 -\frac{A}{2}))+\sin((45 +\frac{A}{2})-(45 - \frac{A}{2})))=\] \[620 (\sin 90^\circ + \sin A) = 620 \cdot \frac{6}{5} = \boxed{744}\] | null | 744 |
10b89ea3f13aa2c628bafff65bf48904 | https://artofproblemsolving.com/wiki/index.php/2016_AIME_I_Problems/Problem_9 | problem_id
10b89ea3f13aa2c628bafff65bf48904 Triangle $ABC$ has $AB=40,AC=31,$ and $\sin{A}...
10b89ea3f13aa2c628bafff65bf48904 It has been noted that this answer won't actua...
Name: Text, dtype: object | Let $\alpha$ be the angle $\angle CAS$ and $\beta$ be the angle $\angle BAQ$ . Then \[\alpha + \beta + \angle A = 90^\circ \Rightarrow \alpha + \beta = 90^\circ - \angle A\] \[\cos(\alpha + \beta) = \cos(90^\circ - \angle A)\] \[\cos(\alpha + \beta) = \sin(\angle A) = \frac{1}{5}\] \[\cos\alpha\cos\beta - \sin\alpha\sin\beta = \frac{1}{5}\] \[\cos\alpha\cos\beta - \sqrt{(1-\cos^2\alpha)(1-\cos^2\beta)} = \frac{1}{5}\] \[\cos\alpha\cos\beta - \sqrt{1-\cos^2\alpha-\cos^2\beta+\cos^2\alpha\cos^2\beta} = \frac{1}{5}\] However, by AM-GM: \[\cos^2\alpha+\cos^2\beta \ge 2\cos\alpha\beta\] Therefore, \[1-\cos^2\alpha-\cos^2\beta+\cos^2\alpha\cos^2\beta \le 1-2\cos\alpha\beta+\cos^2\alpha\cos^2\beta = (1-\cos\alpha\cos\beta)^2\] \[\sqrt{1-\cos^2\alpha-\cos^2\beta+\cos^2\alpha\cos^2\beta} \le 1-\cos\alpha\cos\beta\] So, \[\frac{1}{5} \ge \cos\alpha\cos\beta - (1-\cos\alpha\cos\beta) = 2\cos\alpha\cos\beta-1\] \[\frac{3}{5} \ge \cos\alpha\cos\beta\] .
However, the area of the rectangle is just $AS \cdot AQ = 31\cos\alpha \cdot 40\cos\beta \le 31 \cdot 40 \cdot \frac{3}{5} = \boxed{744}$ | null | 744 |
d86cb8912bd82f4c9715a15dd33ed851 | https://artofproblemsolving.com/wiki/index.php/2016_AIME_I_Problems/Problem_10 | A strictly increasing sequence of positive integers $a_1$ $a_2$ $a_3$ $\cdots$ has the property that for every positive integer $k$ , the subsequence $a_{2k-1}$ $a_{2k}$ $a_{2k+1}$ is geometric and the subsequence $a_{2k}$ $a_{2k+1}$ $a_{2k+2}$ is arithmetic. Suppose that $a_{13} = 2016$ . Find $a_1$ | Instead of setting $a_1$ equal to something and $a_2$ equal to something, note that it is rather easier to set $a_1=x^2$ and $a_3=y^2$ so that $a_2=xy,a_4=y(2y-x),a_5=(2y-x)^2$ and so on until you reached $a_{13}=(6y-5x)^2$ (Or simply notice the pattern), so $6y-5x=\sqrt{2016}=12\sqrt{14}$ . Note that since each of the terms has degree 2 so if you multiply $x$ and $y$ by $\sqrt{14}$ you multiply each term by $14$ so each term is still a integer if the terms are already integers before you multiply $x$ and $y$ by $\sqrt{14}$ , so let $w=\frac{x}{\sqrt{14}}$ and $z=\frac{y}{\sqrt{14}}$ so $6z-5w=12$ . Then, for the sequence to be strictly increasing positive integers we have $(w,z)=(6,7)$ so $x=6\sqrt{14}$ and $a_1=x^2=6^2 \cdot 14=\boxed{504}$ Ddk001 | null | 504 |
d86cb8912bd82f4c9715a15dd33ed851 | https://artofproblemsolving.com/wiki/index.php/2016_AIME_I_Problems/Problem_10 | A strictly increasing sequence of positive integers $a_1$ $a_2$ $a_3$ $\cdots$ has the property that for every positive integer $k$ , the subsequence $a_{2k-1}$ $a_{2k}$ $a_{2k+1}$ is geometric and the subsequence $a_{2k}$ $a_{2k+1}$ $a_{2k+2}$ is arithmetic. Suppose that $a_{13} = 2016$ . Find $a_1$ | The thirteenth term of the sequence is $2016$ , which makes that fourteenth term of the sequence $2016+r$ and the $15^{\text{th}}$ term $\frac{(2016+r)^2}{2016}$ . We note that $r$ is an integer so that means $\frac{r^2}{2016}$ is an integer. Thus, we assume the smallest value of $r$ , which is $168$ . We bash all the way back to the first term and get our answer of $\boxed{504}$ | null | 504 |
d86cb8912bd82f4c9715a15dd33ed851 | https://artofproblemsolving.com/wiki/index.php/2016_AIME_I_Problems/Problem_10 | A strictly increasing sequence of positive integers $a_1$ $a_2$ $a_3$ $\cdots$ has the property that for every positive integer $k$ , the subsequence $a_{2k-1}$ $a_{2k}$ $a_{2k+1}$ is geometric and the subsequence $a_{2k}$ $a_{2k+1}$ $a_{2k+2}$ is arithmetic. Suppose that $a_{13} = 2016$ . Find $a_1$ | Let $a_{2k-1}=s$ where $k=1$ . Then, $a_{2k}=sr \Longrightarrow a_{2k+1}=sr^2$ . Continuing on, we get $a_{2k+2}=sr(r-1)+sr^2=sr(2r-1) \Longrightarrow a_{2k+3}=sr^2(\frac{2r-1}{r})^2=s(2r-1)^2$ . Moreover, $a_{2k+4}=s(2r-1)(r-1)+s(2r-1)^2=s(2r-1)(3r-2) \Longrightarrow a_{2k+5}=s(2r-1)^2(\frac{3r-2}{2r-1})^2=s(3r-2)^2$
It is clear now that $a_{2k+2c}=s(cr-(c-1))((c+1)r-c)$ and $a_{2k+2c-1}=s(cr-(c-1))^2$ . Plugging in $c=6$ $a_{13}=s(6r-5)^2=2016$ . The prime factorization of $2016=2^5\cdot3^2\cdot7$ so we look for perfect squares.
$6r-5\equiv (6r-5)^2\equiv 1\pmod{6}$ if $r$ is an integer, and $\frac{\omega+5}{6}=r \Longrightarrow 6\mid{s}$ if $r$ is not an integer and $\omega$ is rational, so $6\mid{s}$ . This forces $s=2\cdot3^2\cdot7\cdot{N}$ . Assuming $(6r-5)$ is an integer, it can only be $2^x$ $x\in{1,2}$
If $6r-5=2^1$ $r=\frac{7}{6}$ . If $6r-5=2^2$ $r=\frac{3}{2}$ . Note that the latter cannot work since $a_{2k+1}=s(\frac{9}{4}) \Longrightarrow 4\mid{s}$ but $N=1 \Longrightarrow s=2\cdot3^2\cdot7$ in this scenario. Therefore, $r=\frac{7}{6} \Longrightarrow s=\frac{2016}{2^2}=504$ . Plugging back $k=1$ $a_1=s=\boxed{504}$ | null | 504 |
009f851bfba0dff4f3119f4ac00ccb6f | https://artofproblemsolving.com/wiki/index.php/2016_AIME_I_Problems/Problem_11 | Let $P(x)$ be a nonzero polynomial such that $(x-1)P(x+1)=(x+2)P(x)$ for every real $x$ , and $\left(P(2)\right)^2 = P(3)$ . Then $P(\tfrac72)=\tfrac{m}{n}$ , where $m$ and $n$ are relatively prime positive integers. Find $m + n$ | Plug in $x=1$ to get $(1-1)P(1+1) = 0 = (1+2)P(1) \Rightarrow P(1) = 0$ . Plug in $x=0$ to get $(0-1)P(0+1) = (0+2)P(0)\Rightarrow P(0) = -\frac{1}{2}P(1) = 0$ . Plug in $x=-1$ to get $(-1-1)P(-1+1) = (-1+2)P(-1)\Rightarrow (-2)P(0)=P(-1)\Rightarrow P(-1) = 0$ . So $P(x) = x(x-1)(x+1)Q(x)$ for some polynomial $Q(x)$ . Using the initial equation, once again, \[(x-1)P(x+1) = (x+2)P(x)\] \[(x-1)((x+1)(x+1-1)(x+1+1)Q(x+1)) = (x+2)((x)(x-1)(x+1)Q(x))\] \[(x-1)(x+1)(x)(x+2)Q(x+1) = (x+2)(x)(x-1)(x+1)Q(x)\] \[Q(x+1) = Q(x)\] From here, we know that $Q(x) = C$ for a constant $C$ $Q(x)$ cannot be periodic since it is a polynomial), so $P(x) = Cx(x-1)(x+1)$ . We know that $\left(P(2)\right)^2 = P(3)$ . Plugging those into our definition of $P(x)$ $(C \cdot 2 \cdot (2-1) \cdot (2+1))^2 = C \cdot 3 \cdot (3-1) \cdot (3+1) \Rightarrow (6C)^2 = 24C \Rightarrow 36C^2 - 24C = 0 \Rightarrow C = 0$ or $\frac{2}{3}$ . So we know that $P(x) = \frac{2}{3}x(x-1)(x+1)$ . So $P(\frac{7}{2}) = \frac{2}{3} \cdot \frac{7}{2} \cdot (\frac{7}{2} - 1) \cdot (\frac{7}{2} + 1) = \frac{105}{4}$ . Thus, the answer is $105 + 4 = \boxed{109}$ | null | 109 |
009f851bfba0dff4f3119f4ac00ccb6f | https://artofproblemsolving.com/wiki/index.php/2016_AIME_I_Problems/Problem_11 | Let $P(x)$ be a nonzero polynomial such that $(x-1)P(x+1)=(x+2)P(x)$ for every real $x$ , and $\left(P(2)\right)^2 = P(3)$ . Then $P(\tfrac72)=\tfrac{m}{n}$ , where $m$ and $n$ are relatively prime positive integers. Find $m + n$ | From the equation we see that $x-1$ divides $P(x)$ and $(x+2)$ divides $P(x+1)$ so we can conclude that $x-1$ and $x+1$ divide $P(x)$ (if we shift the function right by 1, we get $(x-2)P(x) = (x+1)P(x-1)$ , and from here we can see that $x+1$ divides $P(x)$ ). This means that $1$ and $-1$ are roots of $P(x)$ . Plug in $x = 0$ and we see that $P(0) = 0$ so $0$ is also a root.
Suppose we had another root that is not one of those $3$ . Notice that the equation above indicates that if $r$ is a root then $r+1$ and $r-1$ is also a root. Then we'd get an infinite amount of roots! So that is bad. So we cannot have any other roots besides those three.
That means $P(x) = cx(x-1)(x+1)$ . We can use $P(2)^2 = P(3)$ to get $c = \frac{2}{3}$ . Plugging in $\frac{7}{2}$ is now trivial and we see that it is $\frac{105}{4}$ so our answer is $\boxed{109}$ | null | 109 |
009f851bfba0dff4f3119f4ac00ccb6f | https://artofproblemsolving.com/wiki/index.php/2016_AIME_I_Problems/Problem_11 | Let $P(x)$ be a nonzero polynomial such that $(x-1)P(x+1)=(x+2)P(x)$ for every real $x$ , and $\left(P(2)\right)^2 = P(3)$ . Then $P(\tfrac72)=\tfrac{m}{n}$ , where $m$ and $n$ are relatively prime positive integers. Find $m + n$ | Although this may not be the most mathematically rigorous answer, we see that $\frac{P(x+1)}{P(x)}=\frac{x+2}{x-1}$ . Using a bit of logic, we can make a guess that $P(x+1)$ has a factor of $x+2$ , telling us $P(x)$ has a factor of $x+1$ . Similarly, we guess that $P(x)$ has a factor of $x-1$ , which means $P(x+1)$ has a factor of $x$ . Now, since $P(x)$ and $P(x+1)$ have so many factors that are off by one, we may surmise that when you plug $x+1$ into $P(x)$ , the factors "shift over," i.e. $P(x)=(A)(A+1)(A+2)...(A+n)$ , which goes to $P(x+1)=(A+1)(A+2)(A+3)...(A+n+1)$ . This is useful because these, when divided, result in $\frac{P(x+1)}{P(x)}=\frac{A+n+1}{A}$ . If $\frac{A+n+1}{A}=\frac{x+2}{x-1}$ , then we get $A=x-1$ and $A+n+1=x+2$ $n=2$ . This gives us $P(x)=(x-1)x(x+1)$ and $P(x+1)=x(x+1)(x+2)$ , and at this point we realize that there has to be some constant $a$ multiplied in front of the factors, which won't affect our fraction $\frac{P(x+1)}{P(x)}=\frac{x+2}{x-1}$ but will give us the correct values of $P(2)$ and $P(3)$ . Thus $P(x)=a(x-1)x(x+1)$ , and we utilize $P(2)^2=P(3)$ to find $a=\frac{2}{3}$ . Evaluating $P \left ( \frac{7}{2} \right )$ is then easy, and we see it equals $\frac{105}{4}$ , so the answer is $\boxed{109}$ | null | 109 |
009f851bfba0dff4f3119f4ac00ccb6f | https://artofproblemsolving.com/wiki/index.php/2016_AIME_I_Problems/Problem_11 | Let $P(x)$ be a nonzero polynomial such that $(x-1)P(x+1)=(x+2)P(x)$ for every real $x$ , and $\left(P(2)\right)^2 = P(3)$ . Then $P(\tfrac72)=\tfrac{m}{n}$ , where $m$ and $n$ are relatively prime positive integers. Find $m + n$ | Substituting $x=2$ into the given equation, we find that $P(3)=4P(2)=P(2)^2$ . Therefore, either $P(2)=0$ or $P(2)=4$ . Now for integers $n\ge 2$ , we know that \[P(n+1)=\frac{n+2}{n-1}P(n).\] Applying this repeatedly, we find that \[P(n+1)=\frac{(n+2)!/3!}{(n-1)!}P(2).\] If $P(2)=0$ , this shows that $P(x)$ has infinitely many roots, meaning that $P(x)$ is identically equal to zero. But this contradicts the problem statement. Therefore, $P(2)=4$ , and we find $P(n+1)=\frac{2}{3}(n+2)(n+1)n$ for all positive integers $n\ge2$ . This cubic polynomial matches the values $P(n+1)$ for infinitely many numbers, hence the two polynomials are identically equal. In particular, $P\left(\frac72\right)=\frac23\cdot\frac92\cdot\frac72\cdot\frac52=\frac{105}{4}$ , and the answer is $\boxed{109}$ | null | 109 |
009f851bfba0dff4f3119f4ac00ccb6f | https://artofproblemsolving.com/wiki/index.php/2016_AIME_I_Problems/Problem_11 | Let $P(x)$ be a nonzero polynomial such that $(x-1)P(x+1)=(x+2)P(x)$ for every real $x$ , and $\left(P(2)\right)^2 = P(3)$ . Then $P(\tfrac72)=\tfrac{m}{n}$ , where $m$ and $n$ are relatively prime positive integers. Find $m + n$ | We can find zeroes of the polynomial by making the first given equation $0 = 0$ . Plugging in $x = 1$ and $x = -2$ gives us the zeroes $1$ and $-1$ , respectively. Now we can plug in these zeros to get more zeroes. $x = -1$ gives us the zero $0$ (no pun intended). $x = 1$ makes the equation $0 \cdot P(2) = 0$ , which means $P(2)$ is not necessarily $0$ . If $P(2) = 0$ , then plugging in $2$ to the equation yields $P(3) = 0$ , plugging in $3$ to the equation yields $P(4) = 0$ , and so on, a contradiction of "nonzero polynomial". So $2$ is not a zero. Note that plugging in $x = 0$ to the equation does not yield any additional zeros. Thus, the only zeroes of $P(x)$ are $-1, 0,$ and $1$ , so $P(x) = a(x + 1)x(x - 1)$ for some nonzero constant $a$ . We can plug in $2$ and $3$ into the polynomial and use the second given equation to find an equation for $a$ $P(2) = 6a$ and $P(3) = 24a$ , so: \[(6a)^2 = 24a \implies 36a^2 = 24a \implies a = \frac23\] Plugging in $\frac72$ into the polynomial $\frac23(x + 1)x(x - 1)$ yields $\frac{105}{4}$ $105 + 4 = \boxed{109}$ | null | 109 |
009f851bfba0dff4f3119f4ac00ccb6f | https://artofproblemsolving.com/wiki/index.php/2016_AIME_I_Problems/Problem_11 | Let $P(x)$ be a nonzero polynomial such that $(x-1)P(x+1)=(x+2)P(x)$ for every real $x$ , and $\left(P(2)\right)^2 = P(3)$ . Then $P(\tfrac72)=\tfrac{m}{n}$ , where $m$ and $n$ are relatively prime positive integers. Find $m + n$ | Plug in $x=2$ yields $P(3)=4P(2)$ . Since also $(P(2))^2=P(3)$ , we have $P(2)=4$ and $P(3)=16$ . Plug in $x=3$ yields $2P(4)=5P(3)$ so $P(4)=40$
Repeat the action gives $P(2)=4$ $P(3)=16$ $P(4)=40$ $P(5)=80$ , and $P(6)=140$
Since $P(x)$ is a polynomial, the $k$ th difference is constant, where $k=\deg(P(x))$ . Thus we can list out the 0th, 1st, 2nd, 3rd, ... differences until we obtain a constant.
\[4,16,40,80,140\] \[12,24,40,60\] \[12,16,20\] \[4,4,4\]
Since the 3rd difference of $P(x)$ is constant, we can conclude that $\deg(P(x))=3$
Let $P(x)=a_3x^3+a_2x^2+a_1x+a_0$ . Plug in the values for $x$ and solve the system of 4 equations gives $(a_3,a_2,a_1,a_0)=(\frac{2}{3},0,-\frac{2}{3},0)$
Thus $P(x)=\frac{2}{3}x^3-\frac{2}{3}x$ and $P(\frac{7}{2})=\frac{105}{4}\Longrightarrow m+n=\boxed{109}$ | null | 109 |
28255358ece8b3f88aeda0fd49680ec0 | https://artofproblemsolving.com/wiki/index.php/2016_AIME_I_Problems/Problem_12 | Find the least positive integer $m$ such that $m^2 - m + 11$ is a product of at least four not necessarily distinct primes. | Suppose $p=11$ ; then $m^2-m+11=11qrs$ . Reducing modulo 11, we get $m\equiv 1,0 \pmod{11}$ so $k(11k\pm 1)+1 = qrs$
Suppose $q=11$ . Then we must have $11k^2\pm k + 1 = 11rs$ , which leads to $k\equiv \mp 1 \pmod{11}$ , i.e., $k\in \{1,10,12,21,23,\ldots\}$
$k=1$ leads to $rs=1$ (impossible)! Then $k=10$ leads to $rs=103$ , a prime (impossible). Finally, for $k=12$ we get $rs=143=11\cdot 13$
Thus our answer is $m=11k= \boxed{132}$ | null | 132 |
28255358ece8b3f88aeda0fd49680ec0 | https://artofproblemsolving.com/wiki/index.php/2016_AIME_I_Problems/Problem_12 | Find the least positive integer $m$ such that $m^2 - m + 11$ is a product of at least four not necessarily distinct primes. | Let $m^2 - m + 11 = pqrs$ for primes $p, q, r, s\ge11$ . If $p, q, r, s = 11$ , then $m^2-m+11=11^4$ . We can multiply this by $4$ and complete the square to find $(2m-1)^2=4\cdot 11^4-43$ . But \[(2\cdot 11^2-1)^2=4\cdot 11^4-4\cdot 11^2+1 <4\cdot 11^4-43<(2\cdot 11^2)^2,\] hence we have pinned a perfect square $(2m-1)^2=4\cdot 11^4-43$ strictly between two consecutive perfect squares, a contradiction. Hence $pqrs \ge 11^3 \cdot 13$ . Thus $m^2-m+11\ge 11^3\cdot 13$ , or $(m-132)(m+131)\ge0$ . From the inequality, we see that $m \ge 132$ $132^2 - 132 + 11 = 11^3 \cdot 13$ , so $m = \boxed{132}$ and we are done. | null | 132 |
fa0f2373bc604d16dc6c7752c14c5acb | https://artofproblemsolving.com/wiki/index.php/2016_AIME_I_Problems/Problem_13 | Freddy the frog is jumping around the coordinate plane searching for a river, which lies on the horizontal line $y = 24$ . A fence is located at the horizontal line $y = 0$ . On each jump Freddy randomly chooses a direction parallel to one of the coordinate axes and moves one unit in that direction. When he is at a point where $y=0$ , with equal likelihoods he chooses one of three directions where he either jumps parallel to the fence or jumps away from the fence, but he never chooses the direction that would have him cross over the fence to where $y < 0$ . Freddy starts his search at the point $(0, 21)$ and will stop once he reaches a point on the river. Find the expected number of jumps it will take Freddy to reach the river. | Clearly Freddy's $x$ -coordinate is irrelevant, so we let $E(y)$ be the expected value of the number of jumps it will take him to reach the river from a given $y$ -coordinate. Observe that $E(24)=0$ , and \[E(y)=1+\frac{E(y+1)+E(y-1)+2E(y)}{4}\] for all $y$ such that $1\le y\le 23$ . Also note that $E(0)=1+\frac{2E(0)+E(1)}{3}$ . This gives $E(0)=E(1)+3$ . Plugging this into the equation for $E(1)$ gives that \[E(1)=1+\frac{E(2)+3E(1)+3}{4},\] or $E(1)=E(2)+7$ . Iteratively plugging this in gives that $E(n)=E(n+1)+4n+3$ . Thus $E(23)=E(24)+95$ $E(22)=E(23)+91=E(24)+186$ , and $E(21)=E(22)+87=E(24)+273=\boxed{273}$ | null | 273 |
b1cd2b981fe8be7f38f7a7436a1d098e | https://artofproblemsolving.com/wiki/index.php/2016_AIME_I_Problems/Problem_14 | Centered at each lattice point in the coordinate plane are a circle radius $\frac{1}{10}$ and a square with sides of length $\frac{1}{5}$ whose sides are parallel to the coordinate axes. The line segment from $(0,0)$ to $(1001, 429)$ intersects $m$ of the squares and $n$ of the circles. Find $m + n$ | First note that $1001 = 143 \cdot 7$ and $429 = 143 \cdot 3$ so every point of the form $(7k, 3k)$ is on the line. Then consider the line $l$ from $(7k, 3k)$ to $(7(k + 1), 3(k + 1))$ . Translate the line $l$ so that $(7k, 3k)$ is now the origin. There is one square and one circle that intersect the line around $(0,0)$ . Then the points on $l$ with an integral $x$ -coordinate are, since $l$ has the equation $y = \frac{3x}{7}$
\[(0,0), \left(1, \frac{3}{7}\right), \left(2, \frac{6}{7}\right), \left(3, 1 + \frac{2}{7}\right), \left(4, 1 + \frac{5}{7}\right), \left(5, 2 + \frac{1}{7}\right), \left(6, 2 + \frac{4}{7}\right), (7,3).\]
We claim that the lower right vertex of the square centered at $(2,1)$ lies on $l$ . Since the square has side length $\frac{1}{5}$ , the lower right vertex of this square has coordinates $\left(2 + \frac{1}{10}, 1 - \frac{1}{10}\right) = \left(\frac{21}{10}, \frac{9}{10}\right)$ . Because $\frac{9}{10} = \frac{3}{7} \cdot \frac{21}{10}$ $\left(\frac{21}{10}, \frac{9}{10}\right)$ lies on $l$ . Since the circle centered at $(2,1)$ is contained inside the square, this circle does not intersect $l$ . Similarly the upper left vertex of the square centered at $(5,2)$ is on $l$ . Since every other point listed above is farther away from a lattice point (excluding (0,0) and (7,3)) and there are two squares with centers strictly between $(0,0)$ and $(7,3)$ that intersect $l$ . Since there are $\frac{1001}{7} = \frac{429}{3} = 143$ segments from $(7k, 3k)$ to $(7(k + 1), 3(k + 1))$ , the above count is yields $143 \cdot 2 = 286$ squares. Since every lattice point on $l$ is of the form $(3k, 7k)$ where $0 \le k \le 143$ , there are $144$ lattice points on $l$ . Centered at each lattice point, there is one square and one circle, hence this counts $288$ squares and circles. Thus $m + n = 286 + 288 = \boxed{574}$ | null | 574 |
b1cd2b981fe8be7f38f7a7436a1d098e | https://artofproblemsolving.com/wiki/index.php/2016_AIME_I_Problems/Problem_14 | Centered at each lattice point in the coordinate plane are a circle radius $\frac{1}{10}$ and a square with sides of length $\frac{1}{5}$ whose sides are parallel to the coordinate axes. The line segment from $(0,0)$ to $(1001, 429)$ intersects $m$ of the squares and $n$ of the circles. Find $m + n$ | This is mostly a clarification to Solution 1, but let's take the diagram for the origin to $(7,3)$ . We have the origin circle and square intersected, then two squares, then the circle and square at $(7,3)$ . If we take the circle and square at the origin out of the diagram, we will be able to repeat the resulting segment (with its circles and squares) end to end from $(0,0)$ to $(1001,429)$ , which forms the line we need without overlapping. Since $143$ of these segments are needed to do this, and $3$ squares and $1$ circle are intersected with each, there are $143 \cdot (3+1) = 572$ squares and circles intersected. Adding the circle and square that are intersected at the origin back into the picture, we get that there are $572+2=\boxed{574}$ squares and circles intersected in total. | null | 574 |
b1cd2b981fe8be7f38f7a7436a1d098e | https://artofproblemsolving.com/wiki/index.php/2016_AIME_I_Problems/Problem_14 | Centered at each lattice point in the coordinate plane are a circle radius $\frac{1}{10}$ and a square with sides of length $\frac{1}{5}$ whose sides are parallel to the coordinate axes. The line segment from $(0,0)$ to $(1001, 429)$ intersects $m$ of the squares and $n$ of the circles. Find $m + n$ | This solution is a more systematic approach for finding when the line intersects the squares and circles. Because $1001 = 7*11*13$ and $429=3*11*13$ , the slope of our line is $\frac{3}{7}$ , and we only need to consider the line in the rectangle from the origin to $(7,3)$ , and we can iterate the line $11*13=143$ times. First, we consider how to figure out if the line intersects a square. Given a lattice point $(x_1, y_1)$ , we can think of representing a square centered at that lattice point as all points equal to $(x_1 \pm a, y_1 \pm b)$ s.t. $0 \leq a,b \leq \frac{1}{10}$ . If the line $y = \frac{3}{7}x$ intersects the square, then we must have $\frac{y_1 + b}{x_1 + a} = \frac{3}{7}$ . The line with the least slope that intersects the square intersects at the bottom right corner and the line with the greatest slope that intersects the square intersects at the top left corner; thus we must have that $\frac{3}{7}$ lies in between these slopes, or that $\frac{y_1-\frac{1}{10}}{x_1+\frac{1}{10}} \leq \frac{3}{7} \leq \frac{y_1+\frac{1}{10}}{x_1-\frac{1}{10}}$ . Simplifying, $3x_1 - 1 \leq 7y_1 \leq 3x_1 + 1$ . Because $y$ can only equal $0, 1, 2, 3$ , we just do casework based on the values of $y$ and find that the points $(2, 1)$ and $(5, 2)$ are intersected just at the corner of the square and $(0, 0), (7, 3)$ are intersected through the center of the square. However, we disregard one of $(0, 0)$ and $(7, 3)$ , WLOG $(0, 0)$ , since we just use it in our count for the next of the 143 segments. Therefore, in one of our "segments", 3 squares are intersected and 1 circle is intersected giving 4 total. Thus our answer is $143*4 = 572$ . HOWEVER, we cannot forget that we ignored $(0, 0)$ , which contributes another square and circle to our count, making the final answer $572 + 2 = \boxed{574}$ | null | 574 |
3d7ba8c297d2dec98f3dcee5666254c0 | https://artofproblemsolving.com/wiki/index.php/2016_AIME_I_Problems/Problem_15 | Circles $\omega_1$ and $\omega_2$ intersect at points $X$ and $Y$ . Line $\ell$ is tangent to $\omega_1$ and $\omega_2$ at $A$ and $B$ , respectively, with line $AB$ closer to point $X$ than to $Y$ . Circle $\omega$ passes through $A$ and $B$ intersecting $\omega_1$ again at $D \neq A$ and intersecting $\omega_2$ again at $C \neq B$ . The three points $C$ $Y$ $D$ are collinear, $XC = 67$ $XY = 47$ , and $XD = 37$ . Find $AB^2$ | Let $Z = XY \cap AB$ . By the radical axis theorem $AD, XY, BC$ are concurrent, say at $P$ . Moreover, $\triangle DXP \sim \triangle PXC$ by simple angle chasing. Let $y = PX, x = XZ$ . Then \[\frac{y}{37} = \frac{67}{y} \qquad \implies \qquad y^2 = 37 \cdot 67.\] Now, $AZ^2 = \tfrac 14 AB^2$ , and by power of a point, \begin{align*} x(y-x) &= \tfrac 14 AB^2, \quad \text{and} \\ x(47+x) &= \tfrac 14 AB^2 \end{align*} Solving, we get \[\tfrac 14 AB^2 = \tfrac 12 (y-47)\cdot \tfrac 12 (y+47) \qquad \implies\] \[\qquad AB ^ 2 = 37\cdot 67 - 47^2 = \boxed{270}\] | null | 270 |
3d7ba8c297d2dec98f3dcee5666254c0 | https://artofproblemsolving.com/wiki/index.php/2016_AIME_I_Problems/Problem_15 | Circles $\omega_1$ and $\omega_2$ intersect at points $X$ and $Y$ . Line $\ell$ is tangent to $\omega_1$ and $\omega_2$ at $A$ and $B$ , respectively, with line $AB$ closer to point $X$ than to $Y$ . Circle $\omega$ passes through $A$ and $B$ intersecting $\omega_1$ again at $D \neq A$ and intersecting $\omega_2$ again at $C \neq B$ . The three points $C$ $Y$ $D$ are collinear, $XC = 67$ $XY = 47$ , and $XD = 37$ . Find $AB^2$ | By the Radical Axis Theorem $AD, XY, BC$ concur at point $E$
Let $AB$ and $EY$ intersect at $S$ . Note that because $AXDY$ and $CYXB$ are cyclic, by Miquel's Theorem $AXBE$ is cyclic as well. Thus \[\angle AEX = \angle ABX = \angle XCB = \angle XYB\] and \[\angle XEB = \angle XAB = \angle XDA = \angle XYA.\] Thus $AY \parallel EB$ and $YB \parallel EA$ , so $AEBY$ is a parallelogram. Hence $AS = SB$ and $SE = SY$ . But notice that $DXE$ and $EXC$ are similar by $AA$ Similarity, so $XE^2 = XD \cdot XC = 37 \cdot 67$ . But \[XE^2 - XY^2 = (XE + XY)(XE - XY) = EY \cdot 2XS = 2SY \cdot 2SX = 4SA^2 = AB^2.\] Hence $AB^2 = 37 \cdot 67 - 47^2 = \boxed{270}.$ | null | 270 |
3d7ba8c297d2dec98f3dcee5666254c0 | https://artofproblemsolving.com/wiki/index.php/2016_AIME_I_Problems/Problem_15 | Circles $\omega_1$ and $\omega_2$ intersect at points $X$ and $Y$ . Line $\ell$ is tangent to $\omega_1$ and $\omega_2$ at $A$ and $B$ , respectively, with line $AB$ closer to point $X$ than to $Y$ . Circle $\omega$ passes through $A$ and $B$ intersecting $\omega_1$ again at $D \neq A$ and intersecting $\omega_2$ again at $C \neq B$ . The three points $C$ $Y$ $D$ are collinear, $XC = 67$ $XY = 47$ , and $XD = 37$ . Find $AB^2$ | First, we note that as $\triangle XDY$ and $\triangle XYC$ have bases along the same line, $\frac{[\triangle XDY]}{[\triangle XYC]}=\frac{DY}{YC}$ . We can also find the ratio of their areas using the circumradius area formula. If $R_1$ is the radius of $\omega_1$ and if $R_2$ is the radius of $\omega_2$ , then \[\frac{[\triangle XDY]}{[\triangle XYC]}=\frac{(37\cdot 47\cdot DY)/(4R_1)}{(47\cdot 67\cdot YC)/(4R_2)}=\frac{37\cdot DY\cdot R_2}{67\cdot YC\cdot R_1}.\] Since we showed this to be $\frac{DY}{YC}$ , we see that $\frac{R_2}{R_1}=\frac{67}{37}$
We extend $AD$ and $BC$ to meet at point $P$ , and we extend $AB$ and $CD$ to meet at point $Q$ as shown below. [asy] size(200); import olympiad; real R1=45,R2=67*R1/37; real m1=sqrt(R1^2-23.5^2); real m2=sqrt(R2^2-23.5^2); pair o1=(0,0),o2=(m1+m2,0),x=(m1,23.5),y=(m1,-23.5); draw(circle(o1,R1)); draw(circle(o2,R2)); pair q=(-R1/(R2-R1)*o2.x,0); pair a=tangent(q,o1,R1,2); pair b=tangent(q,o2,R2,2); pair d=intersectionpoints(circle(o1,R1),q--y+15*(y-q))[0]; pair c=intersectionpoints(circle(o2,R2),q--y+15*(y-q))[1]; pair p=extension(a,d,b,c); dot(q^^a^^b^^x^^y^^c^^d^^p); draw(q--b^^q--c); draw(p--d^^p--c^^x--y); draw(a--y^^b--y); draw(d--x--c); label("$A$",a,NW,fontsize(8)); label("$B$",b,NE,fontsize(8)); label("$C$",c,SE,fontsize(8)); label("$D$",d,SW,fontsize(8)); label("$X$",x,2*WNW,fontsize(8)); label("$Y$",y,3*S,fontsize(8)); label("$P$",p,N,fontsize(8)); label("$Q$",q,W,fontsize(8)); [/asy] As $ABCD$ is cyclic, we know that $\angle BCD=180-\angle DAB=\angle BAP$ . But then as $AB$ is tangent to $\omega_2$ at $B$ , we see that $\angle BCD=\angle ABY$ . Therefore, $\angle ABY=\angle BAP$ , and $BY\parallel PD$ . A similar argument shows $AY\parallel PC$ . These parallel lines show $\triangle PDC\sim\triangle ADY\sim\triangle BYC$ . Also, we showed that $\frac{R_2}{R_1}=\frac{67}{37}$ , so the ratio of similarity between $\triangle ADY$ and $\triangle BYC$ is $\frac{37}{67}$ , or rather \[\frac{AD}{BY}=\frac{DY}{YC}=\frac{YA}{CB}=\frac{37}{67}.\] We can now use the parallel lines to find more similar triangles. As $\triangle AQD\sim \triangle BQY$ , we know that \[\frac{QA}{QB}=\frac{QD}{QY}=\frac{AD}{BY}=\frac{37}{67}.\] Setting $QA=37x$ , we see that $QB=67x$ , hence $AB=30x$ , and the problem simplifies to finding $30^2x^2$ . Setting $QD=37^2y$ , we also see that $QY=37\cdot 67y$ , hence $DY=37\cdot 30y$ . Also, as $\triangle AQY\sim \triangle BQC$ , we find that \[\frac{QY}{QC}=\frac{YA}{CB}=\frac{37}{67}.\] As $QY=37\cdot 67y$ , we see that $QC=67^2y$ , hence $YC=67\cdot30y$
Applying Power of a Point to point $Q$ with respect to $\omega_2$ , we find \[67^2x^2=37\cdot 67^3 y^2,\] or $x^2=37\cdot 67 y^2$ . We wish to find $AB^2=30^2x^2=30^2\cdot 37\cdot 67y^2$
Applying Stewart's Theorem to $\triangle XDC$ , we find \[37^2\cdot (67\cdot 30y)+67^2\cdot(37\cdot 30y)=(67\cdot 30y)\cdot (37\cdot 30y)\cdot (104\cdot 30y)+47^2\cdot (104\cdot 30y).\] We can cancel $30\cdot 104\cdot y$ from both sides, finding $37\cdot 67=30^2\cdot 67\cdot 37y^2+47^2$ . Therefore, \[AB^2=30^2\cdot 37\cdot 67y^2=37\cdot 67-47^2=\boxed{270}.\] | null | 270 |
3d7ba8c297d2dec98f3dcee5666254c0 | https://artofproblemsolving.com/wiki/index.php/2016_AIME_I_Problems/Problem_15 | Circles $\omega_1$ and $\omega_2$ intersect at points $X$ and $Y$ . Line $\ell$ is tangent to $\omega_1$ and $\omega_2$ at $A$ and $B$ , respectively, with line $AB$ closer to point $X$ than to $Y$ . Circle $\omega$ passes through $A$ and $B$ intersecting $\omega_1$ again at $D \neq A$ and intersecting $\omega_2$ again at $C \neq B$ . The three points $C$ $Y$ $D$ are collinear, $XC = 67$ $XY = 47$ , and $XD = 37$ . Find $AB^2$ | [asy] size(9cm); import olympiad; real R1=45,R2=67*R1/37; real m1=sqrt(R1^2-23.5^2); real m2=sqrt(R2^2-23.5^2); pair o1=(0,0),o2=(m1+m2,0),x=(m1,23.5),y=(m1,-23.5); draw(circle(o1,R1)); draw(circle(o2,R2)); pair q=(-R1/(R2-R1)*o2.x,0); pair a=tangent(q,o1,R1,2); pair b=tangent(q,o2,R2,2); pair d=intersectionpoints(circle(o1,R1),q--y+15*(y-q))[0]; pair c=intersectionpoints(circle(o2,R2),q--y+15*(y-q))[1]; dot(a^^b^^x^^y^^c^^d); draw(x--y); draw(a--y^^b--y); draw(d--x--c); draw(a--b--c--d--cycle); draw(x--a^^x--b); label("$A$",a,NW,fontsize(9)); label("$B$",b,NE,fontsize(9)); label("$C$",c,SE,fontsize(9)); label("$D$",d,SW,fontsize(9)); label("$X$",x,2*N,fontsize(9)); label("$Y$",y,3*S,fontsize(9)); [/asy] First of all, since quadrilaterals $ADYX$ and $XYCB$ are cyclic, we can let $\angle DAX = \angle XYC = \theta$ , and $\angle XYD = \angle CBX = 180 - \theta$ , due to the properties of cyclic quadrilaterals. In addition, let $\angle BAX = x$ and $\angle ABX = y$ . Thus, $\angle ADX = \angle AYX = x$ and $\angle XYB = \angle XCB = y$ . Then, since quadrilateral $ABCD$ is cyclic as well, we have the following sums: \[\theta + x +\angle XCY + y = 180^{\circ}\] \[180^{\circ} - \theta + y + \angle XDY + x = 180^{\circ}\] Cancelling out $180^{\circ}$ in the second equation and isolating $\theta$ yields $\theta = y + \angle XDY + x$ . Substituting $\theta$ back into the first equation, we obtain \[2x + 2y + \angle XCY + \angle XDY = 180^{\circ}\] Since \[x + y +\angle XAY + \angle XCY + \angle DAY = 180^{\circ}\] \[x + y + \angle XDY + \angle XCY + \angle DAY = 180^{\circ}\] we can then imply that $\angle DAY = x + y$ . Similarly, $\angle YBC = x + y$ . So then $\angle DXY = \angle YXC = x + y$ , so since we know that $XY$ bisects $\angle DXC$ , we can solve for $DY$ and $YC$ with Stewart’s Theorem. Let $DY = 37n$ and $YC = 67n$ . Then \[37n \cdot 67n \cdot 104n + 47^2 \cdot 104n = 37^2 \cdot 67n + 67^2 \cdot 37n\] \[37n \cdot 67n + 47^2 = 37 \cdot 67\] \[n^2 = \frac{270}{2479}\] Now, since $\angle AYX = x$ and $\angle BYX = y$ $\angle AYB = x + y$ . From there, let $\angle AYD = \alpha$ and $\angle BYC = \beta$ . From angle chasing we can derive that $\angle YDX = \angle YAX = \beta - x$ and $\angle YCX = \angle YBX = \alpha - y$ . From there, since $\angle ADX = x$ , it is quite clear that $\angle ADY = \beta$ , and $\angle YAB = \beta$ can be found similarly. From there, since $\angle ADY = \angle YAB = \angle BYC = \beta$ and $\angle DAY = \angle AYB = \angle YBC = x + y$ , we have $AA$ similarity between $\triangle DAY$ $\triangle AYB$ , and $\triangle YBC$ . Therefore the length of $AY$ is the geometric mean of the lengths of $DA$ and $YB$ (from $\triangle DAY \sim \triangle AYB$ ). However, $\triangle DAY \sim \triangle AYB \sim \triangle YBC$ yields the proportion $\frac{AD}{DY} = \frac{YA}{AB} = \frac{BY}{YC}$ ; hence, the length of $AB$ is the geometric mean of the lengths of $DY$ and $YC$ .
We can now simply use arithmetic to calculate $AB^2$ \[AB^2 = DY \cdot YC\] \[AB^2 = 37 \cdot 67 \cdot \frac{270}{2479}\] \[AB^2 = \boxed{270}\] | null | 270 |
3d7ba8c297d2dec98f3dcee5666254c0 | https://artofproblemsolving.com/wiki/index.php/2016_AIME_I_Problems/Problem_15 | Circles $\omega_1$ and $\omega_2$ intersect at points $X$ and $Y$ . Line $\ell$ is tangent to $\omega_1$ and $\omega_2$ at $A$ and $B$ , respectively, with line $AB$ closer to point $X$ than to $Y$ . Circle $\omega$ passes through $A$ and $B$ intersecting $\omega_1$ again at $D \neq A$ and intersecting $\omega_2$ again at $C \neq B$ . The three points $C$ $Y$ $D$ are collinear, $XC = 67$ $XY = 47$ , and $XD = 37$ . Find $AB^2$ | Let $E = DA \cap CB$ . By Radical Axes, $E$ lies on $XY$ . Note that $EAXB$ is cyclic as $X$ is the Miquel point of $\triangle EDC$ in this configuration.
Claim. $\triangle DXE \sim \triangle EXC$ Proof. We angle chase. \[\measuredangle XEC = \measuredangle XEB = \measuredangle XAB = \measuredangle XDA = \measuredangle XDE\] and \[\measuredangle XCE = \measuredangle XCB = \measuredangle XBA = \measuredangle XEA = \measuredangle XED. \square\]
Let $F = EX \cap AB$ . Note \[FA^2 = FX \cdot FY = FB^2\] and \[EF \cdot FX = AF \cdot FB = FA^2 = FX \cdot FY \implies EF = FY\] By our claim, \[\frac{DX}{XE} = \frac{EX}{XC} \implies EX^2 = DX \cdot XC = 67 \cdot 37 \implies FY = \frac{EY}{2} = \frac{EX+XY}{2} = \frac{\sqrt{67 \cdot 37}+47}{2}\] and \[FX=FY-47=\frac{\sqrt{67 \cdot 37}-47}{2}\] Finally, \[AB^2 = (2 \cdot FA)^2 = 4 \cdot FX \cdot FY = 4 \cdot \frac{(67 \cdot 37) - 47^2}{4} = \boxed{270}. \blacksquare\] ~Mathscienceclass | null | 270 |
3d7ba8c297d2dec98f3dcee5666254c0 | https://artofproblemsolving.com/wiki/index.php/2016_AIME_I_Problems/Problem_15 | Circles $\omega_1$ and $\omega_2$ intersect at points $X$ and $Y$ . Line $\ell$ is tangent to $\omega_1$ and $\omega_2$ at $A$ and $B$ , respectively, with line $AB$ closer to point $X$ than to $Y$ . Circle $\omega$ passes through $A$ and $B$ intersecting $\omega_1$ again at $D \neq A$ and intersecting $\omega_2$ again at $C \neq B$ . The three points $C$ $Y$ $D$ are collinear, $XC = 67$ $XY = 47$ , and $XD = 37$ . Find $AB^2$ | $AB^2 = 4 AM^2 =2x(2x+ 2 XY) =(XP - XY) (XP + XY) = XP^2 - XY^2 = XC \cdot XD - XY^2 = 67 \cdot 37 - 47^2 = \boxed{270}.$ | null | 270 |
3d7ba8c297d2dec98f3dcee5666254c0 | https://artofproblemsolving.com/wiki/index.php/2016_AIME_I_Problems/Problem_15 | Circles $\omega_1$ and $\omega_2$ intersect at points $X$ and $Y$ . Line $\ell$ is tangent to $\omega_1$ and $\omega_2$ at $A$ and $B$ , respectively, with line $AB$ closer to point $X$ than to $Y$ . Circle $\omega$ passes through $A$ and $B$ intersecting $\omega_1$ again at $D \neq A$ and intersecting $\omega_2$ again at $C \neq B$ . The three points $C$ $Y$ $D$ are collinear, $XC = 67$ $XY = 47$ , and $XD = 37$ . Find $AB^2$ | Extend $\overline{AD}$ and $\overline{BC}$ to meet at point $P$ . Let $M$ be the midpoint of segment $AB$ . Then by radical axis on $(ADY)$ $(BCY)$ and $(ABCD)$ $P$ lies on $XY$ . By the bisector lemma, $M$ lies on $XY$ . It is well-known that $P$ $A$ $X$ , and $B$ are concyclic. By Power of a point on $M$ with respect to $(PAXB)$ and $(ADY)$ \[|\text{Pow}(M, (PAXB))| = MX \cdot MP = MA^2 = |\text{Pow}(M, (ADY))| = MX \cdot MY,\] so $MP=MY$ . Thus $AB$ and $PY$ bisect each other, so $PAYB$ is a parallelogram. This implies that \[\angle DAY = \angle YBC,\] so by the inscribed angle theorem $\overline{XY}$ bisects $\angle DXC$
Claim: $AB^2 = DY \cdot YC$
Proof. Define the linear function $f(\bullet) := \text{Pow}(\bullet, (ADY)) - \text{Pow}(\bullet, (ABCD))$ . Since $\overline{BY}$ is parallel to the radical axis $\overline{AD}$ of $(ADY)$ and $(ABCD)$ by our previous parallelism, $f(B)=f(Y)$ . Note that $f(B)=AB^2$ while $f(Y)=DY \cdot YC$ , so we conclude. $\square$
By Stewart's theorem on $\triangle DXC$ $DY \cdot YC=37 \cdot 67 - 47^2 = 270$ , so $AB^2=\boxed{270}$ | null | 270 |
5f66b7202ec3c1935d023767ff4219f5 | https://artofproblemsolving.com/wiki/index.php/2016_AIME_II_Problems/Problem_1 | Initially Alex, Betty, and Charlie had a total of $444$ peanuts. Charlie had the most peanuts, and Alex had the least. The three numbers of peanuts that each person had formed a geometric progression. Alex eats $5$ of his peanuts, Betty eats $9$ of her peanuts, and Charlie eats $25$ of his peanuts. Now the three numbers of peanuts each person has forms an arithmetic progression. Find the number of peanuts Alex had initially. | Let $r$ be the common ratio, where $r>1$ . We then have $ar-9-(a-5)=a(r-1)-4=ar^{2}-25-(ar-9)=ar(r-1)-16$ . We now have, letting, subtracting the 2 equations, $ar^{2}+-2ar+a=12$ , so we have $3ar=432,$ or $ar=144$ , which is how much Betty had. Now we have $144+\dfrac{144}{r}+144r=444$ , or $144(r+\dfrac{1}{r})=300$ , or $r+\dfrac{1}{r}=\dfrac{25}{12}$ , which solving for $r$ gives $r=\dfrac{4}{3}$ , since $r>1$ , so Alex had $\dfrac{3}{4} \cdot 144=\boxed{108}$ peanuts. | null | 108 |
5f66b7202ec3c1935d023767ff4219f5 | https://artofproblemsolving.com/wiki/index.php/2016_AIME_II_Problems/Problem_1 | Initially Alex, Betty, and Charlie had a total of $444$ peanuts. Charlie had the most peanuts, and Alex had the least. The three numbers of peanuts that each person had formed a geometric progression. Alex eats $5$ of his peanuts, Betty eats $9$ of her peanuts, and Charlie eats $25$ of his peanuts. Now the three numbers of peanuts each person has forms an arithmetic progression. Find the number of peanuts Alex had initially. | Let $a$ be Alex's peanuts and $k$ the common ratio. Then we have $a(k^2+k+1)=444$ . Adding $k$ to both sides and factoring,
\[\frac{444}{a}+k=(k+1)^2\]
For the common difference, $ak=5-(a-5)=ak^2-25-(ak-9)$ . Simplifying, $k^2-2k+1=\frac{12}{a}$ . Factoring, \[(k-1)^2=\frac{12}{a}\]
\[(k+1)^2-(k-1)^2=4k \implies 4k=\frac{444-12}{a}+k \implies k=\frac{144}{a}\]
Substitute $k$ in the second equation to get $(\frac{144-a}{a})^2=\frac{12}{a}$ . Expanding and applying the quadratic formula, \[a=150\pm\frac{\sqrt{300^2-4(144^2)}}{2}\] Taking out $4^2\cdot3^2$ from under the radical leaves \[a=150\pm6\sqrt{625-576}=108, 192\] Since Alex's peanut number was the lowest of the trio, and $3*192>444$ , Alex initially had $\boxed{108}$ peanuts. | null | 108 |
5f66b7202ec3c1935d023767ff4219f5 | https://artofproblemsolving.com/wiki/index.php/2016_AIME_II_Problems/Problem_1 | Initially Alex, Betty, and Charlie had a total of $444$ peanuts. Charlie had the most peanuts, and Alex had the least. The three numbers of peanuts that each person had formed a geometric progression. Alex eats $5$ of his peanuts, Betty eats $9$ of her peanuts, and Charlie eats $25$ of his peanuts. Now the three numbers of peanuts each person has forms an arithmetic progression. Find the number of peanuts Alex had initially. | Let the initial numbers of peanuts Alex, Betty and Charlie had be $a$ $b$ , and $c$ respectively.
Let the final numbers of peanuts, after eating, be $a'$ $b'$ , and $c'$
We are given that $a + b + c = 444$ . Since a total of $5 + 9 + 25 = 39$ peanuts are eaten, we must have $a' + b' + c' = 444 - 39 = 405$ .
Since $a'$ $b'$ , and $c'$ form an arithmetic progression, we have that $a' = b' - x$ and $c' = b' + x$ for some integer $x$ .
Substituting yields $3b' = 405$ and so $b' = 135$ . Since Betty ate $9$ peanuts, it follows that $b = b' + 9 = 144$
Since $a$ $b$ , and $c$ form a geometric progression, we have that $a = \frac{b}{r}$ and $c = br$ .
Multiplying yields $ac = b^2 = 144^2$ .
Since $a + c = 444 - b = 300$ , it follows that $a = 150 - \lambda$ and $c = 150 + \lambda$ for some integer $\lambda$ .
Substituting yields $(150-\lambda)(150+\lambda) = 144^2$ , which expands and rearranges to $\lambda^2 = 150^2-144^2 = 42^2$ .
Since $\lambda > 0$ , we must have $\lambda = 42$ , and so $a = 150 - \lambda = \boxed{108}$ | null | 108 |
5f66b7202ec3c1935d023767ff4219f5 | https://artofproblemsolving.com/wiki/index.php/2016_AIME_II_Problems/Problem_1 | Initially Alex, Betty, and Charlie had a total of $444$ peanuts. Charlie had the most peanuts, and Alex had the least. The three numbers of peanuts that each person had formed a geometric progression. Alex eats $5$ of his peanuts, Betty eats $9$ of her peanuts, and Charlie eats $25$ of his peanuts. Now the three numbers of peanuts each person has forms an arithmetic progression. Find the number of peanuts Alex had initially. | Bashing is not difficult. All we have to consider is the first equation. We can write it as $x*(1+r+r^2) = 444$ . The variable $x$ must be an integer, and after trying all the factors of $444$ , it's clear that $r$ is a fraction smaller than $10$ . When calculating the coefficient of $x$ , we must consider that the fraction produced will very likely have a numerator that divides $444$ . Trying a couple will make it easier to find the fraction, and soon you will find that $\frac{4}{3}$ gives a numerator of $37$ , a rather specific factor of $444$ . Solving for the rest will give you an integer value of $\boxed{108}$ . This is by no means a good solution, but it may be faster in a competition if you don't want to mess with several other equations. This is purely up to different individuals. | null | 108 |
5f66b7202ec3c1935d023767ff4219f5 | https://artofproblemsolving.com/wiki/index.php/2016_AIME_II_Problems/Problem_1 | Initially Alex, Betty, and Charlie had a total of $444$ peanuts. Charlie had the most peanuts, and Alex had the least. The three numbers of peanuts that each person had formed a geometric progression. Alex eats $5$ of his peanuts, Betty eats $9$ of her peanuts, and Charlie eats $25$ of his peanuts. Now the three numbers of peanuts each person has forms an arithmetic progression. Find the number of peanuts Alex had initially. | Let $b$ be the finish number of Betty's peanuts. Then \[3b = 444-(5 + 9 + 25) = 405 = 3 \cdot 135 \implies b = 135, b+ 9 = 144.\]
Let $k > 1$ be the common ratio. Then \[\frac{144}{k} + 144 + k \cdot 144 = 444 \implies \frac{144}{k} + k \cdot 144 = 300\implies \frac{12}{k} + k \cdot 12 = 25\implies k = \frac{4}{3} \implies \frac {144\cdot 3}{4} = \boxed{108}.\] | null | 108 |
ad2b3e4b01bce79fb1f1944c7a7493b2 | https://artofproblemsolving.com/wiki/index.php/2016_AIME_II_Problems/Problem_2 | There is a $40\%$ chance of rain on Saturday and a $30\%$ chance of rain on Sunday. However, it is twice as likely to rain on Sunday if it rains on Saturday than if it does not rain on Saturday. The probability that it rains at least one day this weekend is $\frac{a}{b}$ , where $a$ and $b$ are relatively prime positive integers. Find $a+b$ | Let $x$ be the probability that it rains on Sunday given that it doesn't rain on Saturday. We then have $\dfrac{4}{5}x+\dfrac{2}{5}2x = \dfrac{3}{10} \implies \dfrac{7}{5}x=\dfrac{3}{10}$ $\implies x=\dfrac{3}{14}$ . Therefore, the probability that it doesn't rain on either day is $\left(1-\dfrac{3}{14}\right)\left(\dfrac{3}{5}\right)=\dfrac{33}{70}$ . Therefore, the probability that rains on at least one of the days is $1-\dfrac{33}{70}=\dfrac{37}{70}$ , so adding up the $2$ numbers, we have $37+70=\boxed{107}$ | null | 107 |
ca8b836f8bf029f66cee559bc9bd81d4 | https://artofproblemsolving.com/wiki/index.php/2016_AIME_II_Problems/Problem_3 | Let $x,y,$ and $z$ be real numbers satisfying the system \begin{align*} \log_2(xyz-3+\log_5 x)&=5,\\ \log_3(xyz-3+\log_5 y)&=4,\\ \log_4(xyz-3+\log_5 z)&=4.\\ \end{align*} Find the value of $|\log_5 x|+|\log_5 y|+|\log_5 z|$ | First, we get rid of logs by taking powers: $xyz-3+\log_5 x=2^{5}=32$ $xyz-3+\log_5 y=3^{4}=81$ , and $(xyz-3+\log_5 z)=4^{4}=256$ . Adding all the equations up and using the $\log {xy}=\log {x}+\log{y}$ property, we have $3xyz+\log_5{xyz} = 378$ , so we have $xyz=125$ . Solving for $x,y,z$ by substituting $125$ for $xyz$ in each equation, we get $\log_5 x=-90, \log_5 y=-41, \log_5 z=134$ , so adding all the absolute values we have $90+41+134=\boxed{265}$ | null | 265 |
1b523e1ca38c818ca4b3d924a4421c42 | https://artofproblemsolving.com/wiki/index.php/2016_AIME_II_Problems/Problem_4 | An $a \times b \times c$ rectangular box is built from $a \cdot b \cdot c$ unit cubes. Each unit cube is colored red, green, or yellow. Each of the $a$ layers of size $1 \times b \times c$ parallel to the $(b \times c)$ faces of the box contains exactly $9$ red cubes, exactly $12$ green cubes, and some yellow cubes. Each of the $b$ layers of size $a \times 1 \times c$ parallel to the $(a \times c)$ faces of the box contains exactly $20$ green cubes, exactly $25$ yellow cubes, and some red cubes. Find the smallest possible volume of the box. | By counting the number of green cubes $2$ different ways, we have $12a=20b$ , or $a=\dfrac{5}{3} b$ . Notice that there are only $3$ possible colors for unit cubes, so for each of the $1 \times b \times c$ layers, there are $bc-21$ yellow cubes, and similarly there are $ac-45$ red cubes in each of the $1 \times a \times c$ layers. Therefore, we have $a(bc-21)=25b$ and $b(ac-45)=9a$ . We check a few small values of $a,b$ and solve for $c$ , checking $(a,b)=(5,3)$ gives $c=12$ with a volume of $180$ $(a,b)=(10,6)$ gives $c=6$ with a volume of $360$ , and $(a,b)=(15,9)$ gives $c=4$ , with a volume of $540$ . Any higher $(a,b)$ will $ab>180$ , so therefore, the minimum volume is $\boxed{180}$ | null | 180 |
1b523e1ca38c818ca4b3d924a4421c42 | https://artofproblemsolving.com/wiki/index.php/2016_AIME_II_Problems/Problem_4 | An $a \times b \times c$ rectangular box is built from $a \cdot b \cdot c$ unit cubes. Each unit cube is colored red, green, or yellow. Each of the $a$ layers of size $1 \times b \times c$ parallel to the $(b \times c)$ faces of the box contains exactly $9$ red cubes, exactly $12$ green cubes, and some yellow cubes. Each of the $b$ layers of size $a \times 1 \times c$ parallel to the $(a \times c)$ faces of the box contains exactly $20$ green cubes, exactly $25$ yellow cubes, and some red cubes. Find the smallest possible volume of the box. | The total number of green cubes is given by $12a=20b\Longrightarrow a=\frac{5}{3}b$
Let $r$ be the number of red cubes on each one of the $b$ layers then the total number of red cubes is $9a=br$ . Substitute $a=\frac{5}{3}b$ gives $r=15$
Repeating the procedure on the number of yellow cubes $y$ on each of the $a$ layers gives $y=15$
Therefore $bc=9+12+15=36$ and $ac=15+20+25=60$ . Multiplying yields $abc^2=2160$
Since $abc^2$ is fixed, $abc$ is minimized when $c$ is maximized, which occurs when $a$ $b$ are minimized (since each of $ac$ $bc$ is fixed). Thus $(a,b,c)=(3,5,12)\Longrightarrow abc=\boxed{180}$ | null | 180 |
b1fa1eeee806ff8b30a5cd4e3707b129 | https://artofproblemsolving.com/wiki/index.php/2016_AIME_II_Problems/Problem_5 | Triangle $ABC_0$ has a right angle at $C_0$ . Its side lengths are pairwise relatively prime positive integers, and its perimeter is $p$ . Let $C_1$ be the foot of the altitude to $\overline{AB}$ , and for $n \geq 2$ , let $C_n$ be the foot of the altitude to $\overline{C_{n-2}B}$ in $\triangle C_{n-2}C_{n-1}B$ . The sum $\sum_{n=2}^\infty C_{n-2}C_{n-1} = 6p$ . Find $p$ | Do note that by counting the area in 2 ways, the first altitude is $x = \frac{ab}{c}$ . By similar triangles, the common ratio is $\rho = \frac{a}{c}$ for each height, so by the geometric series formula, we have \begin{align} 6p=\frac{x}{1-\rho} = \frac{ab}{c-a}. \end{align} Writing $p=a+b+c$ and clearing denominators, we get \[13a=6p .\] Thus $p=13q$ $a=6q$ , and $b+c=7q$ , i.e. $c=7q-b$ . Plugging these into $(1)$ , we get $78q(q-b)=6bq$ , i.e., $14b=13q$ . Thus $q=14r$ and $p=182r$ $b=13r$ $a=84r$ $c=85r$ . Taking $r=1$ (since $a,b,c$ are relatively prime) we get $p=\boxed{182}$ | null | 182 |
b1fa1eeee806ff8b30a5cd4e3707b129 | https://artofproblemsolving.com/wiki/index.php/2016_AIME_II_Problems/Problem_5 | Triangle $ABC_0$ has a right angle at $C_0$ . Its side lengths are pairwise relatively prime positive integers, and its perimeter is $p$ . Let $C_1$ be the foot of the altitude to $\overline{AB}$ , and for $n \geq 2$ , let $C_n$ be the foot of the altitude to $\overline{C_{n-2}B}$ in $\triangle C_{n-2}C_{n-1}B$ . The sum $\sum_{n=2}^\infty C_{n-2}C_{n-1} = 6p$ . Find $p$ | Note that by counting the area in 2 ways, the first altitude is $\dfrac{ab}{c}$ . By similar triangles, the common ratio is $\dfrac{a}{c}$ for reach height, so by the geometric series formula, we have $6p=\dfrac{\dfrac{ab}{c}}{1-\dfrac{a}{c}}$ . Multiplying by the denominator and expanding, the equation becomes $\dfrac{ab}{c}=6a+6b+6c-\dfrac{6a^2}{c}-\dfrac{6ab}{c}-6a$ . Cancelling $6a$ and multiplying by $c$ yields $ab=6bc+6c^2-6a^2-6ab$ , so $7ab = 6bc+6b^2$ and $7a=6b+6c$ . Checking for Pythagorean triples gives $13,84,$ and $85$ , so $p=13+84+85=\boxed{182}$ | null | 182 |
b1fa1eeee806ff8b30a5cd4e3707b129 | https://artofproblemsolving.com/wiki/index.php/2016_AIME_II_Problems/Problem_5 | Triangle $ABC_0$ has a right angle at $C_0$ . Its side lengths are pairwise relatively prime positive integers, and its perimeter is $p$ . Let $C_1$ be the foot of the altitude to $\overline{AB}$ , and for $n \geq 2$ , let $C_n$ be the foot of the altitude to $\overline{C_{n-2}B}$ in $\triangle C_{n-2}C_{n-1}B$ . The sum $\sum_{n=2}^\infty C_{n-2}C_{n-1} = 6p$ . Find $p$ | We start by splitting the sum of all $C_{n-2}C_{n-1}$ into two parts: those where $n-2$ is odd and those where $n-2$ is even.
First consider the sum of the lengths of the segments for which $n-2$ is odd for each $n\geq2$ . The perimeters of these triangles can be expressed using $p$ and ratios that result because of similar triangles. Considering triangles where $n-2$ is odd, we find that the perimeter for each such $n$ is $p\left(\frac{C_{n-1}C_{n}}{C_{0}B}\right)$ . Thus,
$p\sum_{n=1}^{\infty}\frac{C_{2n-1}C_{2n}}{C_{0}B}=6p+C_{0}B$
Simplifying,
$\sum_{n=1}^{\infty}C_{2n-1}C_{2n}=6C_{0}B + \frac{(C_{0}B)^2}{p}=C_{0}B\left(6+\frac{C_{0}B}{p}\right)$ . (1)
Continuing with a similar process for the sum of the lengths of the segments for which $n-2$ is even,
$p\sum_{n=1}^{\infty}\frac{C_{2n-2}C_{2n-1}}{C_{0}B}=6p+C_{0}A+AB=7p-C_{0}B$
Simplifying,
$\sum_{n=1}^{\infty}C_{2n-2}C_{2n-1}=C_{0}B\left(7-\frac{C_{0}B}{p}\right)$ . (2)
Adding (1) and (2) together, we find that
$6p=13C_{0}B \Rightarrow p=\frac{13C_{0}B}{6}=C_{0}B+C_{0}A+AB \Rightarrow \frac{7C_{0}B}{6}=C_{0}A+AB \Rightarrow 7C_{0}B=6C_{0}A + 6AB$
Setting $a=C_{0}B$ $b=C_{0}A$ , and $c=AB$ , we can now proceed as in Shaddoll's solution, and our answer is $p=13+84+85=\boxed{182}$ | null | 182 |
b1fa1eeee806ff8b30a5cd4e3707b129 | https://artofproblemsolving.com/wiki/index.php/2016_AIME_II_Problems/Problem_5 | Triangle $ABC_0$ has a right angle at $C_0$ . Its side lengths are pairwise relatively prime positive integers, and its perimeter is $p$ . Let $C_1$ be the foot of the altitude to $\overline{AB}$ , and for $n \geq 2$ , let $C_n$ be the foot of the altitude to $\overline{C_{n-2}B}$ in $\triangle C_{n-2}C_{n-1}B$ . The sum $\sum_{n=2}^\infty C_{n-2}C_{n-1} = 6p$ . Find $p$ | [asy] size(10cm); // Setup pair A, B; pair C0, C1, C2, C3, C4, C5, C6, C7, C8; A = (5, 0); B = (0, 3); C0 = (0, 0); C1 = foot(C0, A, B); C2 = foot(C1, C0, B); C3 = foot(C2, C1, B); C4 = foot(C3, C2, B); C5 = foot(C4, C3, B); C6 = foot(C5, C4, B); C7 = foot(C6, C5, B); C8 = foot(C7, C6, B); // Labels label("$A$", A, SE); label("$B$", B, NW); label("$C_0$", C0, SW); label("$C_1$", C1, NE); label("$C_2$", C2, W); label("$C_3$", C3, NE); label("$C_4$", C4, W); label("$a$", (B+C0)/2, W); label("$b$", (A+C0)/2, S); label("$c$", (A+B)/2, NE); // Drawings draw(A--B--C0--cycle); draw(C0--C1--C2, red); draw(C2--C3--C4--C5--C6--C7--C8); draw(C0--C2, blue); [/asy]
Let $a = BC_0$ $b = AC_0$ , and $c = AB$ .
Note that the total length of the red segments in the figure above is equal to the length of the blue segment times $\frac{a+c}{b}$
The desired sum is equal to the total length of the infinite path $C_0 C_1 C_2 C_3 \cdots$ , shown in red in the figure below.
Since each of the triangles $\triangle C_0 C_1 C_2, \triangle C_2 C_3 C_4, \dots$ on the left are similar, it follows that the total length of the red segments in the figure below is equal to the length of the blue segment times $\frac{a+c}{b}$ .
In other words, we have that $a\left(\frac{a+c}{b}\right) = 6p$
Guessing and checking Pythagorean triples reveals that $a = 84$ $b=13$ $c = 85$ , and $p = a + b + c = \boxed{182}$ satisfies this equation. | null | 182 |
b1fa1eeee806ff8b30a5cd4e3707b129 | https://artofproblemsolving.com/wiki/index.php/2016_AIME_II_Problems/Problem_5 | Triangle $ABC_0$ has a right angle at $C_0$ . Its side lengths are pairwise relatively prime positive integers, and its perimeter is $p$ . Let $C_1$ be the foot of the altitude to $\overline{AB}$ , and for $n \geq 2$ , let $C_n$ be the foot of the altitude to $\overline{C_{n-2}B}$ in $\triangle C_{n-2}C_{n-1}B$ . The sum $\sum_{n=2}^\infty C_{n-2}C_{n-1} = 6p$ . Find $p$ | This solution proceeds from $\frac{\frac{ab}{c}}{1-\frac{a}{c}} = \frac{\frac{ab}{c}}{\frac{c-a}{c}} = \frac{ab}{c-a} = 6(a+b+c)$ . Note the general from for a primitive pythagorean triple, $m^2-n^2, 2mn, m^2+n^2$ and after substitution, letting $a = m^2-n^2, b = 2mn, c = m^2+n^2$ into the previous equation simplifies down very nicely into $m = 13n$ . Thus $a = 168n^2, b = 26n^2, c = 170n^2$ . Since we know all three side lengths are relatively prime, we must divide each by 2 and let n = 1 giving $a = 84, b = 13, c = 85$ yielding $p = a + b + c = \boxed{182}$ | null | 182 |
b1fa1eeee806ff8b30a5cd4e3707b129 | https://artofproblemsolving.com/wiki/index.php/2016_AIME_II_Problems/Problem_5 | Triangle $ABC_0$ has a right angle at $C_0$ . Its side lengths are pairwise relatively prime positive integers, and its perimeter is $p$ . Let $C_1$ be the foot of the altitude to $\overline{AB}$ , and for $n \geq 2$ , let $C_n$ be the foot of the altitude to $\overline{C_{n-2}B}$ in $\triangle C_{n-2}C_{n-1}B$ . The sum $\sum_{n=2}^\infty C_{n-2}C_{n-1} = 6p$ . Find $p$ | For this problem, first notice that its an infinite geometric series of $6(a+b+c)=\frac{ab}{c-b}$ if $c$ is the hypotenuse. WLOG $a<b$ , we can generalize a pythagorean triple of $x^2-y^2, 2xy, x^2+y^2$ . Let $b=2xy$ , then this generalization gives $6(a+b+c)(c-b)=ab$ \[(x^2-y^2)2xy=6(2x^2+2xy)(x-y)^2\] \[(x+y)xy=6(x^2+xy)(x-y)\] \[xy=6x(x-y)\] \[7xy=6x^2\] \[y=\frac{6}{7}x\]
Now this is just clear. Let $x=7m$ and $y=6m$ for $m$ to be a positive integer, the pythagorean triple is $13-84-85$ which yields $\boxed{182}$ | null | 182 |
ed28d7da8ff18e183b9afecf9a0a0ba9 | https://artofproblemsolving.com/wiki/index.php/2016_AIME_II_Problems/Problem_6 | For polynomial $P(x)=1-\dfrac{1}{3}x+\dfrac{1}{6}x^{2}$ , define $Q(x)=P(x)P(x^{3})P(x^{5})P(x^{7})P(x^{9})=\sum_{i=0}^{50} a_ix^{i}$ .
Then $\sum_{i=0}^{50} |a_i|=\dfrac{m}{n}$ , where $m$ and $n$ are relatively prime positive integers. Find $m+n$ | Note that all the coefficients of odd-powered terms is an odd number of odd degree terms multiplied together, and all coefficients of even-powered terms have an even number of odd degree terms multiplied together. Since every odd degree term is negative, and every even degree term is positive, the sum is just equal to $Q(-1)=P(-1)^{5}=\left( \dfrac{3}{2}\right)^{5}=\dfrac{243}{32}$ , so the desired answer is $243+32=\boxed{275}$ | null | 275 |
ed28d7da8ff18e183b9afecf9a0a0ba9 | https://artofproblemsolving.com/wiki/index.php/2016_AIME_II_Problems/Problem_6 | For polynomial $P(x)=1-\dfrac{1}{3}x+\dfrac{1}{6}x^{2}$ , define $Q(x)=P(x)P(x^{3})P(x^{5})P(x^{7})P(x^{9})=\sum_{i=0}^{50} a_ix^{i}$ .
Then $\sum_{i=0}^{50} |a_i|=\dfrac{m}{n}$ , where $m$ and $n$ are relatively prime positive integers. Find $m+n$ | We are looking for the sum of the absolute values of the coefficients of $Q(x)$ . By defining $P'(x) = 1 + \frac{1}{3}x+\frac{1}{6}x^2$ , and defining $Q'(x) = P'(x)P'(x^3)P'(x^5)P'(x^7)P'(x^9)$ , we have made it so that all coefficients in $Q'(x)$ are just the positive/absolute values of the coefficients of $Q(x)$ . .
To find the sum of the absolute values of the coefficients of $Q(x)$ , we can just take the sum of the coefficients of $Q'(x)$ . This sum is equal to \[Q'(1) = P'(1)P'(1)P'(1)P'(1)P'(1) = \left(1+\frac{1}{3}+\frac{1}{6}\right)^5 = \frac{243}{32},\]
so our answer is $243+32 = \boxed{275}$ | null | 275 |
ed28d7da8ff18e183b9afecf9a0a0ba9 | https://artofproblemsolving.com/wiki/index.php/2016_AIME_II_Problems/Problem_6 | For polynomial $P(x)=1-\dfrac{1}{3}x+\dfrac{1}{6}x^{2}$ , define $Q(x)=P(x)P(x^{3})P(x^{5})P(x^{7})P(x^{9})=\sum_{i=0}^{50} a_ix^{i}$ .
Then $\sum_{i=0}^{50} |a_i|=\dfrac{m}{n}$ , where $m$ and $n$ are relatively prime positive integers. Find $m+n$ | Multiply $P(x)P(x^3)$ and notice that the odd degree terms have a negative coefficient. Observing that this is probably true for all polynomials like this (including $P(x)P(x^3)P(x^5)P(x^7)P(x^9)$ ), we plug in $-1$ to get $\frac{243}{32} \implies \boxed{275}$ | null | 275 |
c13e7d679f067254a8ccc08b31dbb099 | https://artofproblemsolving.com/wiki/index.php/2016_AIME_II_Problems/Problem_7 | Squares $ABCD$ and $EFGH$ have a common center and $\overline{AB} || \overline{EF}$ . The area of $ABCD$ is 2016, and the area of $EFGH$ is a smaller positive integer. Square $IJKL$ is constructed so that each of its vertices lies on a side of $ABCD$ and each vertex of $EFGH$ lies on a side of $IJKL$ . Find the difference between the largest and smallest positive integer values for the area of $IJKL$ | Letting $AI=a$ and $IB=b$ , we have \[IJ^{2}=a^{2}+b^{2} \geq 1008\] by AM-GM inequality . Also, since $EFGH||ABCD$ , the angles that each square cuts another are equal, so all the triangles are formed by a vertex of a larger square and $2$ adjacent vertices of a smaller square are similar. Therefore, the areas form a geometric progression, so since \[2016=12^{2} \cdot 14\] we have the maximum area is \[2016 \cdot \dfrac{11}{12} = 1848\] (the areas of the squares from largest to smallest are $12^{2} \cdot 14, 11 \cdot 12 \cdot 14, 11^{2} \cdot 14$ forming a geometric progression).
The minimum area is $1008$ (every square is half the area of the square whose sides its vertices touch), so the desired answer is \[1848-1008=\boxed{840}\] | null | 840 |
b16ad281ca5df89b86a6beb9e2f674db | https://artofproblemsolving.com/wiki/index.php/2016_AIME_II_Problems/Problem_8 | Find the number of sets $\{a,b,c\}$ of three distinct positive integers with the property that the product of $a,b,$ and $c$ is equal to the product of $11,21,31,41,51,61$ | Note that the prime factorization of the product is $3^{2}\cdot 7 \cdot 11 \cdot 17 \cdot 31 \cdot 41 \cdot 61$ . Ignoring overcounting, by stars and bars there are $6$ ways to choose how to distribute the factors of $3$ , and $3$ ways to distribute the factors of the other primes, so we have $3^{6} \cdot 6$ ways. However, some sets have $2$ numbers that are the same, namely the ones in the form $1,1,x$ and $3,3,x$ , which are each counted $3$ times, and each other set is counted $6$ times, so the desired answer is $\dfrac{729 \cdot 6-6}{6} = \boxed{728}$ | null | 728 |
b16ad281ca5df89b86a6beb9e2f674db | https://artofproblemsolving.com/wiki/index.php/2016_AIME_II_Problems/Problem_8 | Find the number of sets $\{a,b,c\}$ of three distinct positive integers with the property that the product of $a,b,$ and $c$ is equal to the product of $11,21,31,41,51,61$ | Again, notice that the prime factors of the product are $3, 3, 7, 11, 17, 31, 41, 61$ . In this problem, we are asked to partition this set of distinct(ish) factors into three smaller indistinct sets. To do this, we can use Stirling numbers of the second kind, but Stirling numbers of the second kind would assume no empty parts, which isn't what we want. However, this is easy to fix. Denote Stirling numbers of the second kind with $S(n,k)$ . We may start at the situation when all three sets are nonempty. Then, the number of partitions is simply $S(8,3)$ . However, we are overcounting, since every time the two threes are in different sets, (unless they are both individually in their own sets), each one is double counted. To fix this, we can see that they were in different sets $S(8, 3) - S(7,3)$ times by complementary counting, but one of these times they were in their own individual sets, so the total overcount is $\dfrac{S(8,3) - S(7,3) - 1}{2}$ . Similarly, we can do the cases for if two of the sets are nonempty and one of the sets are nonempty, (but in those last two cases the threes cannot be individually in their own sets). We then find that the "answer" is given by: \[S(8,3) - \dfrac{S(8,3) - S(7,3) - 1}{2} + S(8,2) - \dfrac{S(8,2) - S(7,2)}{2}+ S(8,1) - \dfrac{S(8,1) - S(7,1)}{2}\] \[= \dfrac{S(8,3)+S(8,2) + S(8,1) + S(7,3) + S(7,2) + S(7,1) + 1}{2}\] Drawing out the Stirling number triangle and evaluating yields $730$ for the last expression. However, throughout the entire solution, we ignored the fact that the three numbers $a$ $b$ , and $c$ needed to be distinct. However, this is easy to fix, since there are only two sets in which they are not, namely $(1,1,x)$ and $(3,3,x)$ . Thus, the actual answer is $730 - 2 = \boxed{728}$ | null | 728 |
0bc911512b522dc06b6440fa374a0dbc | https://artofproblemsolving.com/wiki/index.php/2016_AIME_II_Problems/Problem_9 | The sequences of positive integers $1,a_2, a_3,...$ and $1,b_2, b_3,...$ are an increasing arithmetic sequence and an increasing geometric sequence, respectively. Let $c_n=a_n+b_n$ . There is an integer $k$ such that $c_{k-1}=100$ and $c_{k+1}=1000$ . Find $c_k$ | Since all the terms of the sequences are integers, and 100 isn't very big, we should just try out the possibilities for $b_2$ . When we get to $b_2=9$ and $a_2=91$ , we have $a_4=271$ and $b_4=729$ , which works, therefore, the answer is $b_3+a_3=81+181=\boxed{262}$ | null | 262 |
0bc911512b522dc06b6440fa374a0dbc | https://artofproblemsolving.com/wiki/index.php/2016_AIME_II_Problems/Problem_9 | The sequences of positive integers $1,a_2, a_3,...$ and $1,b_2, b_3,...$ are an increasing arithmetic sequence and an increasing geometric sequence, respectively. Let $c_n=a_n+b_n$ . There is an integer $k$ such that $c_{k-1}=100$ and $c_{k+1}=1000$ . Find $c_k$ | Using the same reasoning ( $100$ isn't very big), we can guess which terms will work. The first case is $k=3$ , so we assume the second and fourth terms of $c$ are $100$ and $1000$ . We let $r$ be the common ratio of the geometric sequence and write the arithmetic relationships in terms of $r$
The common difference is $100-r - 1$ , and so we can equate: $2(99-r)+100-r=1000-r^3$ . Moving all the terms to one side and the constants to the other yields
$r^3-3r = 702$ , or $r(r^2-3) = 702$ . Simply listing out the factors of $702$ shows that the only factor $3$ less than a square that works is $78$ . Thus $r=9$ and we solve from there to get $\boxed{262}$ | null | 262 |
0bc911512b522dc06b6440fa374a0dbc | https://artofproblemsolving.com/wiki/index.php/2016_AIME_II_Problems/Problem_9 | The sequences of positive integers $1,a_2, a_3,...$ and $1,b_2, b_3,...$ are an increasing arithmetic sequence and an increasing geometric sequence, respectively. Let $c_n=a_n+b_n$ . There is an integer $k$ such that $c_{k-1}=100$ and $c_{k+1}=1000$ . Find $c_k$ | The reason for bashing in this context can also be justified by the fact 100 isn't very big.
Let the common difference for the arithmetic sequence be $a$ , and the common ratio for the geometric sequence be $b$ . The sequences are now $1, a+1, 2a+1, \ldots$ , and $1, b, b^2, \ldots$ . We can now write the given two equations as the following:
$1+(k-2)a+b^{k-2} = 100$
$1+ka+b^k = 1000$
Take the difference between the two equations to get $2a+(b^2-1)b^{k-2} = 900$ . Since 900 is divisible by 4, we can tell $a$ is even and $b$ is odd. Let $a=2m$ $b=2n+1$ , where $m$ and $n$ are positive integers. Substitute variables and divide by 4 to get:
$m+(n+1)(n)(2n+1)^{k-2} = 225$
Because very small integers for $n$ yield very big results, we can bash through all cases of $n$ . Here, we set an upper bound for $n$ by setting $k$ as 3. After trying values, we find that $n\leq 4$ , so $b=9, 7, 5, 3$ . Testing out $b=9$ yields the correct answer of $\boxed{262}$ . Note that even if this answer were associated with another b value like $b=3$ , the value of $k$ can still only be 3 for all of the cases. | null | 262 |
4b044161fc0e65e6335057a2820ffe3c | https://artofproblemsolving.com/wiki/index.php/2016_AIME_II_Problems/Problem_10 | Triangle $ABC$ is inscribed in circle $\omega$ . Points $P$ and $Q$ are on side $\overline{AB}$ with $AP<AQ$ . Rays $CP$ and $CQ$ meet $\omega$ again at $S$ and $T$ (other than $C$ ), respectively. If $AP=4,PQ=3,QB=6,BT=5,$ and $AS=7$ , then $ST=\frac{m}{n}$ , where $m$ and $n$ are relatively prime positive integers. Find $m+n$ | [asy] import cse5; pathpen = black; pointpen = black; pointfontsize = 9; size(8cm); pair A = origin, B = (13,0), P = (4,0), Q = (7,0), T = B + 5 dir(220), C = IP(circumcircle(A,B,T),Line(T,Q,-0.1,10)), S = IP(circumcircle(A,B,C),Line(C,P,-0.1,10)); Drawing(A--B--C--cycle); D(circumcircle(A,B,C),rgb(0,0.6,1)); DrawPathArray(C--S^^C--T,rgb(1,0.4,0.1)); DrawPathArray(A--S^^B--T,rgb(0,0.4,0)); D(S--T,rgb(1,0.2,0.4)); D("A",A,dir(215)); D("B",B,dir(330)); D("P",P,dir(240)); D("Q",Q,dir(240)); D("T",T,dir(290)); D("C",C,dir(120)); D("S",S,dir(250)); MP("4",(A+P)/2,dir(90)); MP("3",(P+Q)/2,dir(90)); MP("6",(Q+B)/2,dir(90)); MP("5",(B+T)/2,dir(140)); MP("7",(A+S)/2,dir(40)); [/asy] Let $\angle ACP=\alpha$ $\angle PCQ=\beta$ , and $\angle QCB=\gamma$ . Note that since $\triangle ACQ\sim\triangle TBQ$ we have $\tfrac{AC}{CQ}=\tfrac56$ , so by the Ratio Lemma \[\dfrac{AP}{PQ}=\dfrac{AC}{CQ}\cdot\dfrac{\sin\alpha}{\sin\beta}\quad\implies\quad \dfrac{\sin\alpha}{\sin\beta}=\dfrac{24}{15}.\] Similarly, we can deduce $\tfrac{PC}{CB}=\tfrac47$ and hence $\tfrac{\sin\beta}{\sin\gamma}=\tfrac{21}{24}$
Now Law of Sines on $\triangle ACS$ $\triangle SCT$ , and $\triangle TCB$ yields \[\dfrac{AS}{\sin\alpha}=\dfrac{ST}{\sin\beta}=\dfrac{TB}{\sin\gamma}.\] Hence \[\dfrac{ST^2}{\sin^2\beta}=\dfrac{TB\cdot AS}{\sin\alpha\sin\gamma},\] so \[TS^2=TB\cdot AS\left(\dfrac{\sin\beta}{\sin\alpha}\dfrac{\sin\beta}{\sin\gamma}\right)=\dfrac{15\cdot 21}{24^2}\cdot 5\cdot 7=\dfrac{35^2}{8^2}.\] Hence $ST=\tfrac{35}8$ and the requested answer is $35+8=\boxed{43}$ | null | 43 |
4b044161fc0e65e6335057a2820ffe3c | https://artofproblemsolving.com/wiki/index.php/2016_AIME_II_Problems/Problem_10 | Triangle $ABC$ is inscribed in circle $\omega$ . Points $P$ and $Q$ are on side $\overline{AB}$ with $AP<AQ$ . Rays $CP$ and $CQ$ meet $\omega$ again at $S$ and $T$ (other than $C$ ), respectively. If $AP=4,PQ=3,QB=6,BT=5,$ and $AS=7$ , then $ST=\frac{m}{n}$ , where $m$ and $n$ are relatively prime positive integers. Find $m+n$ | Projecting through $C$ we have \[\frac{3}{4}\times \frac{13}{6}=(A,Q;P,B)\stackrel{C}{=}(A,T;S,B)=\frac{ST}{7}\times \frac{13}{5}\] which easily gives $ST=\frac{35}{8}\Longrightarrow 35+8=\boxed{043}$ | null | 043 |
4b044161fc0e65e6335057a2820ffe3c | https://artofproblemsolving.com/wiki/index.php/2016_AIME_II_Problems/Problem_10 | Triangle $ABC$ is inscribed in circle $\omega$ . Points $P$ and $Q$ are on side $\overline{AB}$ with $AP<AQ$ . Rays $CP$ and $CQ$ meet $\omega$ again at $S$ and $T$ (other than $C$ ), respectively. If $AP=4,PQ=3,QB=6,BT=5,$ and $AS=7$ , then $ST=\frac{m}{n}$ , where $m$ and $n$ are relatively prime positive integers. Find $m+n$ | By Ptolemy's Theorem applied to quadrilateral $ASTB$ , we find \[5\cdot 7+13\cdot ST=AT\cdot BS.\] Therefore, in order to find $ST$ , it suffices to find $AT\cdot BS$ . We do this using similar triangles, which can be found by using Power of a Point theorem.
As $\triangle APS\sim \triangle CPB$ , we find \[\frac{4}{PC}=\frac{7}{BC}.\] Therefore, $\frac{BC}{PC}=\frac{7}{4}$
As $\triangle BQT\sim\triangle CQA$ , we find \[\frac{6}{CQ}=\frac{5}{AC}.\] Therefore, $\frac{AC}{CQ}=\frac{5}{6}$
As $\triangle ATQ\sim\triangle CBQ$ , we find \[\frac{AT}{BC}=\frac{7}{CQ}.\] Therefore, $AT=\frac{7\cdot BC}{CQ}$
As $\triangle BPS\sim \triangle CPA$ , we find \[\frac{9}{PC}=\frac{BS}{AC}.\] Therefore, $BS=\frac{9\cdot AC}{PC}$ . Thus we find \[AT\cdot BS=\left(\frac{7\cdot BC}{CQ}\right)\left(\frac{9\cdot AC}{PC}\right).\] But now we can substitute in our previously found values for $\frac{BC}{PC}$ and $\frac{AC}{CQ}$ , finding \[AT\cdot BS=63\cdot \frac{7}{4}\cdot \frac{5}{6}=\frac{21\cdot 35}{8}.\] Substituting this into our original expression from Ptolemy's Theorem, we find \begin{align*}35+13ST&=\frac{21\cdot 35}{8}\\13ST&=\frac{13\cdot 35}{8}\\ST&=\frac{35}{8}.\end{align*} Thus the answer is $\boxed{43}$ | null | 43 |
4b044161fc0e65e6335057a2820ffe3c | https://artofproblemsolving.com/wiki/index.php/2016_AIME_II_Problems/Problem_10 | Triangle $ABC$ is inscribed in circle $\omega$ . Points $P$ and $Q$ are on side $\overline{AB}$ with $AP<AQ$ . Rays $CP$ and $CQ$ meet $\omega$ again at $S$ and $T$ (other than $C$ ), respectively. If $AP=4,PQ=3,QB=6,BT=5,$ and $AS=7$ , then $ST=\frac{m}{n}$ , where $m$ and $n$ are relatively prime positive integers. Find $m+n$ | Extend $\overline{AB}$ past $B$ to point $X$ so that $CPTX$ is cyclic. Then, by Power of a Point on $CPTX$ $(CQ)(QT) = (PQ)(QX)$ . By Power of a Point on $CATB$ $(CQ)(QT) = (AQ)(QB) = 42$ . Thus, $(PQ)(QX) = 42$ , so $BX = 8$
By the Inscribed Angle Theorem on $CPTX$ $\angle SCT = \angle BXT$ . By the Inscribed Angle Theorem on $ASTC$ $\angle SCT = \angle SAT$ , so $\angle BXT = \angle SAT$ . Since $ASTB$ is cyclic, $\angle AST = \angle TBX$ . Thus, $\triangle AST \sim \triangle XBT$ , so $AS/XB = ST/BT$ . Solving for $ST$ yields $ST = \frac{35}{8}$ , for a final answer of $35+8 = \boxed{043}$ | null | 043 |
4b044161fc0e65e6335057a2820ffe3c | https://artofproblemsolving.com/wiki/index.php/2016_AIME_II_Problems/Problem_10 | Triangle $ABC$ is inscribed in circle $\omega$ . Points $P$ and $Q$ are on side $\overline{AB}$ with $AP<AQ$ . Rays $CP$ and $CQ$ meet $\omega$ again at $S$ and $T$ (other than $C$ ), respectively. If $AP=4,PQ=3,QB=6,BT=5,$ and $AS=7$ , then $ST=\frac{m}{n}$ , where $m$ and $n$ are relatively prime positive integers. Find $m+n$ | By Ptolemy's Theorem applied to quadrilateral $ASTB$ , we find \[AS\cdot BT+AB\cdot ST=AT\cdot BS.\] Projecting through $C$ we have \[\frac{AQ \cdot PB}{PQ \cdot AB} = (A,Q; P,B)\stackrel{C}{=}(A,T; S,B)=\frac{AT \cdot BS}{ST \cdot AB}.\] Therefore \[AT \cdot BS = \frac {AQ \cdot PB}{PQ} \times ST \implies\] \[\left(\frac {AQ \cdot PB}{PQ} - AB\right)\times ST = AS \cdot BT \implies\] \[ST = \frac {AS \cdot BT \cdot PQ}{AQ \cdot PB – AB \cdot PQ}\] \[ST = \frac {7\cdot 5 \cdot 3}{7\cdot 9 – 13 \cdot 3 } = \frac {35}{8} \implies 35 + 8 = \boxed{43}.\] | null | 43 |
4b044161fc0e65e6335057a2820ffe3c | https://artofproblemsolving.com/wiki/index.php/2016_AIME_II_Problems/Problem_10 | Triangle $ABC$ is inscribed in circle $\omega$ . Points $P$ and $Q$ are on side $\overline{AB}$ with $AP<AQ$ . Rays $CP$ and $CQ$ meet $\omega$ again at $S$ and $T$ (other than $C$ ), respectively. If $AP=4,PQ=3,QB=6,BT=5,$ and $AS=7$ , then $ST=\frac{m}{n}$ , where $m$ and $n$ are relatively prime positive integers. Find $m+n$ | Connect $AT$ and $\angle{SCT}=\angle{SAT}, \angle{ACS}=\angle{ATS}, \frac{ST}{\sin \angle{SAT}}=\frac{AS}{\sin \angle{ATS}}$
So we need to get the ratio of $\frac{\sin \angle{ACS}}{\sin \angle{SCT}}$
By clear observation $\triangle{CAQ}\sim \triangle{BTQ}$ , we have $\frac{CQ}{AC}=\frac{6}{5}$ , LOS tells $\frac{AC}{\sin \angle{CPA}}=\frac{4}{\sin \angle{ACS}}; \frac{CQ}{\sin \angle{CPQ}}=\frac{3}{\sin \angle{PCQ}}$ so we get $\frac{\sin \angle{PCQ}}{\sin \angle{ACS}}=\frac{5}{8}$ , the desired answer is $7\cdot \frac{\sin \angle{SAT}}{\sin \angle{ATS}}=\frac{35}{8}$ leads to $\boxed{043}$ | null | 043 |
1b8b184aeb492f68f8c561a60f276ace | https://artofproblemsolving.com/wiki/index.php/2016_AIME_II_Problems/Problem_11 | For positive integers $N$ and $k$ , define $N$ to be $k$ -nice if there exists a positive integer $a$ such that $a^{k}$ has exactly $N$ positive divisors. Find the number of positive integers less than $1000$ that are neither $7$ -nice nor $8$ -nice. | We claim that an integer $N$ is only $k$ -nice if and only if $N \equiv 1 \pmod k$ . By the number of divisors formula, the number of divisors of $\prod_{i=1}^n p_i^{a_i}$ is $\prod_{i=1}^n (a_i+1)$ . Since all the $a_i$ s are divisible by $k$ in a perfect $k$ power, the only if part of the claim follows. To show that all numbers $N \equiv 1 \pmod k$ are $k$ -nice, write $N=bk+1$ . Note that $2^{kb}$ has the desired number of factors and is a perfect kth power. By PIE, the number of positive integers less than $1000$ that are either $1 \pmod 7$ or $1\pmod 8$ is $142+125-17=250$ , so the desired answer is $999-250=\boxed{749}$ | null | 749 |
1b8b184aeb492f68f8c561a60f276ace | https://artofproblemsolving.com/wiki/index.php/2016_AIME_II_Problems/Problem_11 | For positive integers $N$ and $k$ , define $N$ to be $k$ -nice if there exists a positive integer $a$ such that $a^{k}$ has exactly $N$ positive divisors. Find the number of positive integers less than $1000$ that are neither $7$ -nice nor $8$ -nice. | All integers $a$ will have factorization $2^a3^b5^c7^d...$ . Therefore, the number of factors in $a^7$ is $(7a+1)(7b+1)...$ , and for $a^8$ is $(8a+1)(8b+1)...$ . The most salient step afterwards is to realize that all numbers $N$ not $1 \pmod{7}$ and also not $1 \pmod{8}$ satisfy the criterion. The cycle repeats every $56$ integers, and by PIE, $7+8-1=14$ of them are either $7$ -nice or $8$ -nice or both. Therefore, we can take $\frac{42}{56} * 1008 = 756$ numbers minus the $7$ that work between $1000-1008$ inclusive, to get $\boxed{749}$ positive integers less than $1000$ that are not nice for $k=7, 8$ | null | 749 |
6a40a96704aa68a4eddd48e7dbea9f48 | https://artofproblemsolving.com/wiki/index.php/2016_AIME_II_Problems/Problem_12 | The figure below shows a ring made of six small sections which you are to paint on a wall. You have four paint colors available and you will paint each of the six sections a solid color. Find the number of ways you can choose to paint the sections if no two adjacent sections can be painted with the same color.
[asy] draw(Circle((0,0), 4)); draw(Circle((0,0), 3)); draw((0,4)--(0,3)); draw((0,-4)--(0,-3)); draw((-2.598, 1.5)--(-3.4641, 2)); draw((-2.598, -1.5)--(-3.4641, -2)); draw((2.598, -1.5)--(3.4641, -2)); draw((2.598, 1.5)--(3.4641, 2)); [/asy] | Choose a section to start coloring. Assume, WLOG, that this section is color $1$ . We proceed coloring clockwise around the ring. Let $f(n,C)$ be the number of ways to color the first $n$ sections (proceeding clockwise) such that the last section has color $C$ . In general (except for when we complete the coloring), we see that \[f(n,C_i)=\sum_{j\ne i} f(n-1,C_j),\] i.e., $f(n,C_i)$ is equal to the number of colorings of $n-1$ sections that end in any color other than $C_i$ . Using this, we can compute the values of $f(n,C)$ in the following table.
$\begin{tabular}{c|c|c|c|c } \multicolumn{1}{c}{}&\multicolumn{4}{c}{\(C\)}\\ \(n\)&1 & 2 & 3& 4 \\ \hline 1& 1 & 0 & 0 & 0\\ 2 & 0 & 1 & 1 & 1 \\ 3& 3 & 2 & 2 & 2 \\ 4 & 6 & 7 & 7 & 7 \\ 5 & 21 & 20 & 20 & 20\\ 6& 0& 61 & 61 & 61\\ \end{tabular}$
Note that $f(6, 1)=0$ because then $2$ adjacent sections are both color $1$ . We multiply this by $4$ to account for the fact that the initial section can be any color. Thus the desired answer is $(61+61+61) \cdot 4 = \boxed{732}$ | null | 732 |
6a40a96704aa68a4eddd48e7dbea9f48 | https://artofproblemsolving.com/wiki/index.php/2016_AIME_II_Problems/Problem_12 | The figure below shows a ring made of six small sections which you are to paint on a wall. You have four paint colors available and you will paint each of the six sections a solid color. Find the number of ways you can choose to paint the sections if no two adjacent sections can be painted with the same color.
[asy] draw(Circle((0,0), 4)); draw(Circle((0,0), 3)); draw((0,4)--(0,3)); draw((0,-4)--(0,-3)); draw((-2.598, 1.5)--(-3.4641, 2)); draw((-2.598, -1.5)--(-3.4641, -2)); draw((2.598, -1.5)--(3.4641, -2)); draw((2.598, 1.5)--(3.4641, 2)); [/asy] | We use complementary counting. There are $4^6$ total colorings of the ring without restriction. To count the complement, we wish to count the number of colorings in which at least one set of adjacent sections are the same color. There are six possible sets of adjacent sections that could be the same color (think of these as borders). Call these $B_1,B_2,\dots,B_6$ . Let $\mathcal{A}_1, \mathcal{A}_2,\dots,\mathcal{A}_6$ be the sets of colorings of the ring where the sections on both sides of $B_1,B_2,\dots,B_6$ are the same color. We wish to determine $|\mathcal{A}_1\cup\mathcal{A}_2\cup\cdots\cup\mathcal{A}_6|$ . Note that all of these cases are symmetric, and in general, $|\mathcal{A}_i|=4^5$ . There are $6$ such sets $\mathcal{A}_i$ . Also, $|\mathcal{A}_i\cup\mathcal{A}_j|=4^4$ , because we can only change colors at borders, so if we have two borders along which we cannot change colors, then there are four borders along which we have a choice of color. There are $\binom{6}{2}$ such pairs $\mathcal{A}_i\cup\mathcal{A}_j$ . Similarly, $|\mathcal{A}_i\cup \mathcal{A}_j\cup \mathcal{A}_k|=4^3$ , with $\binom{6}{3}$ such triples, and we see that the pattern will continue for 4-tuples and 5-tuples. For 6-tuples, however, these cases occur when there are no changes of color along the borders, i.e., each section has the same color. Clearly, there are four such possibilities.
Therefore, by PIE, \[|\mathcal{A}_1\cup\mathcal{A}_2\cup\cdots\cup\mathcal{A}_6|=\binom{6}{1}\cdot 4^5-\binom{6}{2}\cdot 4^4+\binom{6}{3}\cdot 4^3-\binom{6}{4}\cdot 4^2+\binom{6}{5}\cdot 4^1-4.\] We wish to find the complement of this, or \[4^6-\left(\binom{6}{1}\cdot 4^5-\binom{6}{2}\cdot 4^4+\binom{6}{3}\cdot 4^3-\binom{6}{4}\cdot 4^2+\binom{6}{5}\cdot 4^1-4\right).\] By the Binomial Theorem, this is $(4-1)^6+3=\boxed{732}$ | null | 732 |
6a40a96704aa68a4eddd48e7dbea9f48 | https://artofproblemsolving.com/wiki/index.php/2016_AIME_II_Problems/Problem_12 | The figure below shows a ring made of six small sections which you are to paint on a wall. You have four paint colors available and you will paint each of the six sections a solid color. Find the number of ways you can choose to paint the sections if no two adjacent sections can be painted with the same color.
[asy] draw(Circle((0,0), 4)); draw(Circle((0,0), 3)); draw((0,4)--(0,3)); draw((0,-4)--(0,-3)); draw((-2.598, 1.5)--(-3.4641, 2)); draw((-2.598, -1.5)--(-3.4641, -2)); draw((2.598, -1.5)--(3.4641, -2)); draw((2.598, 1.5)--(3.4641, 2)); [/asy] | We use generating functions. Suppose that the colors are $0,1,2,3$ . Then as we proceed around a valid coloring of the ring in the clockwise direction, we know that between two adjacent sections with colors $s_i$ and $s_{i+1}$ , there exists a number $d_i\in\{1,2,3\}$ such that $s_{i+1}\equiv s_i+d_i\pmod{4}$ . Therefore, we can represent each border between sections by the generating function $(x+x^2+x^3)$ , where $x,x^2,x^3$ correspond to increasing the color number by $1,2,3\pmod4$ , respectively. Thus the generating function that represents going through all six borders is $A(x)=(x+x^2+x^3)^6$ , where the coefficient of $x^n$ represents the total number of colorings where the color numbers are increased by a total of $n$ as we proceed around the ring. But if we go through all six borders, we must return to the original section, which is already colored. Therefore, we wish to find the sum of the coefficients of $x^n$ in $A(x)$ with $n\equiv 0\pmod4$
Now we note that if $P(x)=x^n$ , then \[P(x)+P(ix)+P(-x)+P(-ix)=\begin{cases}4x^n&\text{if } n\equiv0\pmod{4}\\0&\text{otherwise}.\end{cases}\] Therefore, the sum of the coefficients of $A(x)$ with powers congruent to $0\pmod 4$ is \[\frac{A(1)+A(i)+A(-1)+A(-i)}{4}=\frac{3^6+(-1)^6+(-1)^6+(-1)^6}{4}=\frac{732}{4}.\] We multiply this by $4$ to account for the initial choice of color, so our answer is $\boxed{732}$ | null | 732 |
6a40a96704aa68a4eddd48e7dbea9f48 | https://artofproblemsolving.com/wiki/index.php/2016_AIME_II_Problems/Problem_12 | The figure below shows a ring made of six small sections which you are to paint on a wall. You have four paint colors available and you will paint each of the six sections a solid color. Find the number of ways you can choose to paint the sections if no two adjacent sections can be painted with the same color.
[asy] draw(Circle((0,0), 4)); draw(Circle((0,0), 3)); draw((0,4)--(0,3)); draw((0,-4)--(0,-3)); draw((-2.598, 1.5)--(-3.4641, 2)); draw((-2.598, -1.5)--(-3.4641, -2)); draw((2.598, -1.5)--(3.4641, -2)); draw((2.598, 1.5)--(3.4641, 2)); [/asy] | Let $f(n)$ be the number of valid ways to color a ring with $n$ sections (which we call an $n$ -ring), so the answer is given by $f(6)$ . For $n=2$ , we compute $f(n)=4\cdot3=12$ . For $n \ge 3$ , we can count the number of valid colorings as follows: choose one of the sections arbitrarily, which we may color in $4$ ways. Moving clockwise around the ring, there are $3$ ways to color each of the $n-1$ other sections. Therefore, we have $4 \cdot 3^{n-1}$ colorings of an $n$ -ring.
However, note that the first and last sections could be the same color under this method. To count these invalid colorings, we see that by "merging" the first and last sections into one, we get a valid coloring of an $(n-1)$ -ring. That is, there are $f(n-1)$ colorings of an $n$ -ring in which the first and last sections have the same color. Thus, $f(n) = 4 \cdot 3^{n-1} - f(n-1)$ for all $n \ge 3$
To compute the requested value $f(6)$ , we repeatedly apply this formula: \begin{align*} f(6)&=4\cdot3^5-f(5)\\&=4\cdot3^5-4\cdot3^4+f(4)\\&=4\cdot3^5-4\cdot3^4+4\cdot3^3-f(3)\\&=4\cdot3^5-4\cdot3^4+4\cdot3^3-4\cdot3^2+f(2)\\&=4(3^5-3^4+3^3-3^2+3)\\&=4\cdot3\cdot\frac{3^5+1}{3+1}\\&=\boxed{732} (Solution by MSTang.) | null | 732 |
6a40a96704aa68a4eddd48e7dbea9f48 | https://artofproblemsolving.com/wiki/index.php/2016_AIME_II_Problems/Problem_12 | The figure below shows a ring made of six small sections which you are to paint on a wall. You have four paint colors available and you will paint each of the six sections a solid color. Find the number of ways you can choose to paint the sections if no two adjacent sections can be painted with the same color.
[asy] draw(Circle((0,0), 4)); draw(Circle((0,0), 3)); draw((0,4)--(0,3)); draw((0,-4)--(0,-3)); draw((-2.598, 1.5)--(-3.4641, 2)); draw((-2.598, -1.5)--(-3.4641, -2)); draw((2.598, -1.5)--(3.4641, -2)); draw((2.598, 1.5)--(3.4641, 2)); [/asy] | Label the sections 1, 2, 3, 4, 5, 6 clockwise. We do casework on the colors of sections 1, 3, 5.
Case 1: the colors of the three sections are the same.
In this case, each of sections 2, 4, 6 can be one of 3 colors, so this case yields $4 \times 3^3 = 108$ ways.
Case 2: two of sections 1, 3, 5 are the same color.
Note that there are 3 ways for which two of the three sections have the same color, and $4 \times 3 = 12$ ways to determine their colors. After this, the section between the two with the same color can be one of 3 colors and the other two can be one of 2 colors. This case yields $3 \times (4 \times 3) \times (3 \times 2 \times 2) = 432$ ways.
Case 3: all three sections of 1, 3, 5 are of different colors.
Clearly, there are $4 \times 3 \times 2 = 24$ choices for which three colors to use, and there are 2 ways to choose the colors of each of sections 2, 4, 6. Thus, this case gives $4 \times 3 \times 2 \times 2^3 = 192$ ways.
In total, there are $108 + 432 + 192 = \boxed{732}$ valid colorings. | null | 732 |
6a40a96704aa68a4eddd48e7dbea9f48 | https://artofproblemsolving.com/wiki/index.php/2016_AIME_II_Problems/Problem_12 | The figure below shows a ring made of six small sections which you are to paint on a wall. You have four paint colors available and you will paint each of the six sections a solid color. Find the number of ways you can choose to paint the sections if no two adjacent sections can be painted with the same color.
[asy] draw(Circle((0,0), 4)); draw(Circle((0,0), 3)); draw((0,4)--(0,3)); draw((0,-4)--(0,-3)); draw((-2.598, 1.5)--(-3.4641, 2)); draw((-2.598, -1.5)--(-3.4641, -2)); draw((2.598, -1.5)--(3.4641, -2)); draw((2.598, 1.5)--(3.4641, 2)); [/asy] | We will take a recursive approach to this problem. We can start by writing out the number of colorings for a circle with $1, 2,$ and $3$ compartments, which are $4, 12,$ and $24.$ Now we will try to find a recursive formula, $C(n)$ , for a circle with an arbitrary number of compartments $n.$ We will do this by focusing on the $n-1$ section in the circle. This section can either be the same color as the first compartment, or it can be a different color as the first compartment. We will focus on each case separately.
Case 1:
If they are the same color, we can say there are $C(n-2)$ to fill the first $n-1$ compartments. The $nth$ compartment must be different from the first and second to last compartments, which are the same color. Hence this case adds $3*C(n-2)$ to our recursive formula.
Case 2:
If they are different colors, we can say that there are $C(n-1)$ to fill the first $n-1$ compartments, and for the the $nth$ compartment, there are $2$ ways to color it because the $n-1$ and $1$ compartments are different colors. Hence this case adds $2*C(n-1).$
So our recursive formula, $C(n)$ , is $3*C(n-2) + 2*C(n-1).$ Using the initial values we calculated, we can evaluate this recursive formula up to $n=6.$ When $n=6,$ we get $\boxed{732}$ valid colorings. | null | 732 |
6a40a96704aa68a4eddd48e7dbea9f48 | https://artofproblemsolving.com/wiki/index.php/2016_AIME_II_Problems/Problem_12 | The figure below shows a ring made of six small sections which you are to paint on a wall. You have four paint colors available and you will paint each of the six sections a solid color. Find the number of ways you can choose to paint the sections if no two adjacent sections can be painted with the same color.
[asy] draw(Circle((0,0), 4)); draw(Circle((0,0), 3)); draw((0,4)--(0,3)); draw((0,-4)--(0,-3)); draw((-2.598, 1.5)--(-3.4641, 2)); draw((-2.598, -1.5)--(-3.4641, -2)); draw((2.598, -1.5)--(3.4641, -2)); draw((2.598, 1.5)--(3.4641, 2)); [/asy] | WLOG, color the top left section $1$ and the top right section $2$ . Then the left section can be colored $2$ $3$ , or $4$ , and the right section can be colored $1$ $3$ , or $4$ . There are $3 \cdot 3 = 9$ ways to color the left and right sections. We split this up into two cases.
Case 1: The left and right sections are of the same color. There are $2$ ways this can happen: either they both are $3$ or they both are $4$ . We have $3$ colors to choose for the bottom left, and $2$ remaining colors to choose for the bottom right, for a total of $2 \cdot 3 \cdot 2$ cases.
Case 2: The left and right sections are of different colors. There are $9 - 2 = 7$ ways this can happen. Assume the left is $3$ and the right is $4$ . Then the bottom left can be $1$ $2$ , or $4$ , and the bottom right can be $1$ $2$ , or $3$ . However the bottom sections cannot both be $1$ or both be $2$ , so there are $3 \cdot 3 - 2 = 7$ ways to color the bottom sections, for a total for $7 \cdot 7 = 49$ colorings.
Since there were $4 \cdot 3 = 12$ ways to color the top sections, the answer is $12 \cdot (12 + 49) = \boxed{732}$ | null | 732 |
6a40a96704aa68a4eddd48e7dbea9f48 | https://artofproblemsolving.com/wiki/index.php/2016_AIME_II_Problems/Problem_12 | The figure below shows a ring made of six small sections which you are to paint on a wall. You have four paint colors available and you will paint each of the six sections a solid color. Find the number of ways you can choose to paint the sections if no two adjacent sections can be painted with the same color.
[asy] draw(Circle((0,0), 4)); draw(Circle((0,0), 3)); draw((0,4)--(0,3)); draw((0,-4)--(0,-3)); draw((-2.598, 1.5)--(-3.4641, 2)); draw((-2.598, -1.5)--(-3.4641, -2)); draw((2.598, -1.5)--(3.4641, -2)); draw((2.598, 1.5)--(3.4641, 2)); [/asy] | Label the four colors $1, 2, 3,$ and $4$ , respectively.
Now let's imagine a circle with the four numbers $1, 2, 3,$ and $4$ written clockwise. We'll say that a bug is standing on number $a$ . It is easy to see that for the bug to move to a different number, it must walk $1, 2,$ or $3$ steps clockwise. (This is since adjacent numbers can't be the same, as stated in the problem). Note that the sixth number in the bug's walking sequence must not equal the first number. Thus, our total number of ways, $N$ , given the bug's starting number $k$ , is simply the number of ordered quintuplets of positive integers $(a_1, a_2, a_3, a_4, a_5)$ that satisfy $a_i \in \{1, 2, 3\}$ for all $1 \leq i \leq 5$ and \[\sum_{i=1}^{5} a_i \neq 8, 12,\] since the bug cannot land on $k$ again on his fifth and last step.
We know that the number of ordered quintuplets of positive integers $(a_1, a_2, a_3, a_4, a_5)$ that satisfy $a_i \in \{1, 2, 3\}$ without the other restriction is just $3^5$ , so we aim to find the number of quintuplets such that \[a_1 + a_2 + a_3 + a_4 + a_5 = 8, 12.\] Note that the number of ordered quintuplets satisfying $\sum_{i=1}^{5} a_i = 8$ is the same as the number of them satisfying $\sum_{i=1}^{5} a_i = 12$ due to symmetry. By stars and bars, there are $\dbinom{7}{4} = 35$ ways to distribute the three "extra" units to the five variables $a_1, a_2, a_3, a_4, a_5$ (since $a_i \geq 1$ ), but ways of distribution such that one variable is equal to $4$ are illegal, so the actual number of ways is $35 - 5 = 30$ . Since there are four possible values of $k$ (or the starting position for the bug), we obtain \[N = 4 \times (3^5 - 2 \times 30) = \boxed{732}.\] -fidgetboss_4000 | null | 732 |
6a40a96704aa68a4eddd48e7dbea9f48 | https://artofproblemsolving.com/wiki/index.php/2016_AIME_II_Problems/Problem_12 | The figure below shows a ring made of six small sections which you are to paint on a wall. You have four paint colors available and you will paint each of the six sections a solid color. Find the number of ways you can choose to paint the sections if no two adjacent sections can be painted with the same color.
[asy] draw(Circle((0,0), 4)); draw(Circle((0,0), 3)); draw((0,4)--(0,3)); draw((0,-4)--(0,-3)); draw((-2.598, 1.5)--(-3.4641, 2)); draw((-2.598, -1.5)--(-3.4641, -2)); draw((2.598, -1.5)--(3.4641, -2)); draw((2.598, 1.5)--(3.4641, 2)); [/asy] | Let's number the regions $1,2,\dots 6$ . Suppose we color regions $1,2,3$ . Then, how many ways are there to color $4,5,6$
Note: the numbers are numbered as shown:
[asy] draw(Circle((0,0), 4)); draw(Circle((0,0), 3)); draw((0,4)--(0,3)); draw((0,-4)--(0,-3)); draw((-2.598, 1.5)--(-3.4641, 2)); draw((-2.598, -1.5)--(-3.4641, -2)); draw((2.598, -1.5)--(3.4641, -2)); draw((2.598, 1.5)--(3.4641, 2)); label("1",(-3.5,0)); label("2",(-1.6,3)); label("3",(1.6,3)); label("4",(3.5,0)); label("5",(1.6,-3)); label("6",(-1.6,-3)); [/asy]
$\textbf{Case 1:}$ The colors of $1,2,3$ are $BAB$ , in that order.
Then the colors of $6,5,4$ can be $AXA$ $AXC$ $CXA$ $CXC$ , or $CXD$ in that order, where $X$ is any color not equal to its surroundings. Then there are $4$ choices for $A$ $3$ choices for $B$ (it cannot be $A$ ), $2$ choices for $C$ , and $1$ for $D$ , the last color. So, summing up, we have \[4*3(1*3+2*2+2*2+2*3+2*1*2)=12*21=252\] colorings.
$\textbf{Case 2:}$ The colors of $1,2,3$ are $BAC$ , in that order.
Again, we list out the possible arrangements of $6,5,4$ $AXA$ $AXB$ $CXA$ $AXD$ $DXA$ $CXB$ $CXD$ $DXB$ , or $DXD$ . (Easily simplified; listed here for clarity.) Then there are $4$ choices for $A$ as usual, $3$ choices for $B$ , and so on. Hence we have \[4*3*2(1*3+1*2+1*2+1*2+1*2+1*2+1*2+1*2+1*3)=24*20=480\] colorings in this case.
Adding up, we have $252+480=\boxed{732}$ as our answer. | null | 732 |
6a40a96704aa68a4eddd48e7dbea9f48 | https://artofproblemsolving.com/wiki/index.php/2016_AIME_II_Problems/Problem_12 | The figure below shows a ring made of six small sections which you are to paint on a wall. You have four paint colors available and you will paint each of the six sections a solid color. Find the number of ways you can choose to paint the sections if no two adjacent sections can be painted with the same color.
[asy] draw(Circle((0,0), 4)); draw(Circle((0,0), 3)); draw((0,4)--(0,3)); draw((0,-4)--(0,-3)); draw((-2.598, 1.5)--(-3.4641, 2)); draw((-2.598, -1.5)--(-3.4641, -2)); draw((2.598, -1.5)--(3.4641, -2)); draw((2.598, 1.5)--(3.4641, 2)); [/asy] | We quickly notice that this is just the cycle graph with 6 vertices. The chromatic polynomial for a cycle is $(k-1)^n+(-1)^n (k-1)$ where we use $k$ colors on a cycle of $n$ vertices. Plugging in $k=4$ and $n=6$ we arrive at $\boxed{732}$ .
~chrisdiamond10 | null | 732 |
6a40a96704aa68a4eddd48e7dbea9f48 | https://artofproblemsolving.com/wiki/index.php/2016_AIME_II_Problems/Problem_12 | The figure below shows a ring made of six small sections which you are to paint on a wall. You have four paint colors available and you will paint each of the six sections a solid color. Find the number of ways you can choose to paint the sections if no two adjacent sections can be painted with the same color.
[asy] draw(Circle((0,0), 4)); draw(Circle((0,0), 3)); draw((0,4)--(0,3)); draw((0,-4)--(0,-3)); draw((-2.598, 1.5)--(-3.4641, 2)); draw((-2.598, -1.5)--(-3.4641, -2)); draw((2.598, -1.5)--(3.4641, -2)); draw((2.598, 1.5)--(3.4641, 2)); [/asy] | Let's label the regions as $1,2,3,4,5,6$ in that order. We start with region $1$ . There are no restrictions on the color of region $1$ so it can be any of the four colors. We know move on the region $2$ . It can be any color but color used for region $1$ , giving us $3$ choices. Section $3$ is where it gets a bit complicated; we will have to do casework based on whether the color of region $3$ is that of region $1$ or not.
If we have the color of region $3$ being different from that of region $1$ (in which we color do so in $2$ ways), then we have need for another casework at region $4$ . If the color of region $4$ is different from that of region $1$ (which can be achieved in $2$ ways), then we have yet another casework split.
If the color of region $5$ is different from that of region $1$ (which can be done in $2$ ways, then we would have a total of $2$ possible colorings for region $6$ (for it cannot be the color of regions $1$ nor $5$ ). Moving on to the case where the color of section $5$ is the same as that of section $1$ (which can be done in $1$ way), we will have $3$ ways (region $6$ cannot be that color of both region $1$ and $5$ ). Thus, if the color of region $4$ is different of that of region $1$ , then we have $2\cdot 2 + 3 = 7$ ways.
If the color of region $4$ is the same as that of region $1$ (which can be done in one way), then the color of region $5$ have to be different from that of sector $1$ $3$ ways). That means there will be $2$ choices for the color of region $6$ . So if the color of region $4$ is the same as region $1$ , then we will have $3\cdot 2 = 6$ . That means if the color of section $3$ is different from that of region $1$ , then there are $2\cdot 7 + 6 = 20$
Now, moving on to the case where section $3$ has the same color of region $1$ . That means section $4$ will have to be a different color of that of region $1$ $3$ ways). So, that means we have region $5$ to be either the same color or different color as region $1$ . If it is different (which can be done in $2$ ways), then there will be $2$ possibilities on the color of sector $6$ . If it is same (which can be done in $1$ way), then there are $3$ ways to color region $6$ . So, if section $3$ has the same color as section $1$ , then we have $3(2\cdot 2 + 3) = 3(7) = 21$
Now, in overall we will have $4\cdot 3 (2(20)+21) = 12(61) = \boxed{732}$ | null | 732 |
6a40a96704aa68a4eddd48e7dbea9f48 | https://artofproblemsolving.com/wiki/index.php/2016_AIME_II_Problems/Problem_12 | The figure below shows a ring made of six small sections which you are to paint on a wall. You have four paint colors available and you will paint each of the six sections a solid color. Find the number of ways you can choose to paint the sections if no two adjacent sections can be painted with the same color.
[asy] draw(Circle((0,0), 4)); draw(Circle((0,0), 3)); draw((0,4)--(0,3)); draw((0,-4)--(0,-3)); draw((-2.598, 1.5)--(-3.4641, 2)); draw((-2.598, -1.5)--(-3.4641, -2)); draw((2.598, -1.5)--(3.4641, -2)); draw((2.598, 1.5)--(3.4641, 2)); [/asy] | Let the top-right segment be segment $1,$ and remaining segments are numbered $2,3,4,5,6$ in clockwise order.
We have $4$ choices for segment $1,$ and $3$ choices for segments $2,3,4,5.$ For segment $6,$ we wish to find the expected value of the number of choices for segment $6$ 's color, which depends on whether segment $5$ is red. If we let $P(n)$ denote the probability of segment $P$ being the same color as segment $1$ (for simplicity, denote segment $1$ 's color as red), we get the following recurrence relation: \[P(n)=\dfrac{1-P(n-1)}{3}.\] This is because that you cannot have two reds in a row (hence the $1-P(n-1)$ ) and if segment $n-1$ is not red, there are three possible colors, one of which is red (hence the divide by $3$ ). Using the obvious fact that $P(1)=1$ by the definition of $P(n),$ we find that \[P(5)=\dfrac{7}{27}.\] If segment $5$ is red, then there are three possible colors for segment $6.$ If it is not, there are $2$ possible choices for segment $6.$ This means the expected number of color choices for segment $6$ is \[\dfrac{7}{27}\cdot3+\dfrac{20}{27}\cdot2,\] and the total number of colorings of the ring is \[4\cdot3\cdot3\cdot3\cdot3\cdot\left(\dfrac{7}{27}\cdot3+\dfrac{20}{27}\cdot2\right)=\boxed{732}.\] | null | 732 |
38c47f53906eecc3e6e802f1165da8ab | https://artofproblemsolving.com/wiki/index.php/2016_AIME_II_Problems/Problem_13 | Beatrix is going to place six rooks on a $6 \times 6$ chessboard where both the rows and columns are labeled $1$ to $6$ ; the rooks are placed so that no two rooks are in the same row or the same column. The $value$ of a square is the sum of its row number and column number. The $score$ of an arrangement of rooks is the least value of any occupied square.The average score over all valid configurations is $\frac{p}{q}$ , where $p$ and $q$ are relatively prime positive integers. Find $p+q$ | We casework to find the number of ways to get each possible score. Note that the lowest possible score is $2$ and the highest possible score is $7$ . Let the bijective function $f(x)=\{1,2,3,4,5,6\} \to \{1,2,3,4,5,6\}$ denote the row number of the rook for the corresponding column number.
Thus, the expected sum is $\dfrac{120 \cdot 2 + 216 \cdot 3 + 222 \cdot 4 + 130 \cdot 5 + 31 \cdot 6 + 1 \cdot 7}{720}= \dfrac{2619}{720}=\dfrac{291}{80}$ , so the desired answer is $291+80=\boxed{371}$ | null | 371 |
38c47f53906eecc3e6e802f1165da8ab | https://artofproblemsolving.com/wiki/index.php/2016_AIME_II_Problems/Problem_13 | Beatrix is going to place six rooks on a $6 \times 6$ chessboard where both the rows and columns are labeled $1$ to $6$ ; the rooks are placed so that no two rooks are in the same row or the same column. The $value$ of a square is the sum of its row number and column number. The $score$ of an arrangement of rooks is the least value of any occupied square.The average score over all valid configurations is $\frac{p}{q}$ , where $p$ and $q$ are relatively prime positive integers. Find $p+q$ | If the score is $n+1$ , then one of the rooks must appear in the $n$ th antidiagonal, and this is the first antidiagonal in which a rook can appear. To demonstrate this, we draw the following diagram when $n=4$
[asy] for (int i=0;i<7;++i) {draw((0,10*i)--(60,10*i));draw((10*i,0)--(10*i,60));} path x=(1,1)--(9,9),y=(1,9)--(9,1); for (int i=0;i<3;++i) {for (int j=3+i;j<6;++j) {draw(shift(10*i,10*j)*x,linewidth(1));draw(shift(10*i,10*j)*y,linewidth(1));}} for (int i=0;i<4;++i) {filldraw((10*i,20+10*i)--(10*i+10,20+10*i)--(10*i+10,20+10*i+10)--(10*i,20+10*i+10)--cycle,lightgray);} [/asy] We first count the number of arrangements that avoid the squares above the $n$ th diagonal, and then we subtract from these the number of arrangements that avoid all squares above the $(n+1)$ th diagonal. In the first column, there are $7-n$ rows in which to place the rook. In the second column, there is one more possible row, but one of the rows is used up by the rook in the first column, hence there are still $7-n$ places to place the rook. This pattern continues through the $n$ th column, so there are $(7-n)^n$ ways to place the first $n$ rooks while avoiding the crossed out squares. We can similarly compute that there are $(6-n)^n$ ways to place the rooks in the first $n$ columns that avoid both the crossed out and shaded squares. Therefore, there are $(7-n)^n-(6-n)^n$ ways to place the first $n$ rooks such that at least one of them appears in a shaded square.
After this, there are $(6-n)$ rows and $(6-n)$ columns in which to place the remaining rooks, and we can do this in $(6-n)!$ ways. Hence the number of arrangements with a score of $n$ is $((7-n)^n-(6-n)^n)\cdot(6-n)!$ . We also know that $n$ can range from from $1$ to $6$ , so the average score is given by
\[\frac{2\cdot(6^1-5^1)\cdot5!+3\cdot(5^2-4^2)\cdot4!+4\cdot(4^3-3^3)\cdot3!+5\cdot(3^4-2^4)\cdot 2!+6\cdot(2^5-1^5)\cdot 1!+7\cdot(1^6-0^6)\cdot 0!}{6!}=\frac{291}{80}.\] Thus the answer is $\boxed{371}$ | null | 371 |
38c47f53906eecc3e6e802f1165da8ab | https://artofproblemsolving.com/wiki/index.php/2016_AIME_II_Problems/Problem_13 | Beatrix is going to place six rooks on a $6 \times 6$ chessboard where both the rows and columns are labeled $1$ to $6$ ; the rooks are placed so that no two rooks are in the same row or the same column. The $value$ of a square is the sum of its row number and column number. The $score$ of an arrangement of rooks is the least value of any occupied square.The average score over all valid configurations is $\frac{p}{q}$ , where $p$ and $q$ are relatively prime positive integers. Find $p+q$ | So we first count the number of permutations with score $\ge 2$ . This is obviously $6!=720$ . Then, the number of permutations with score $\ge 3$ can also be computed: in the first column, there are five ways to place a rook- anywhere but the place with score $1$ . In the next column, there are $5$ ways to place a rook- anywhere but the one in the same row as the previous row. We can continue this to obtain that the number of permutations with score $\ge 3$ is $600$ . Doing the same for scores $\ge 4$ $\ge 5$ $\ge 6$ , and $\ge 7$ we obtain that these respective numbers are $384$ $162$ $32$ $1$
Now, note that if $a_k$ is the number of permutations with score $\ge k$ , then $a_k-a_{k-1}=b_{k}$ , where $b_k$ is the number of permutations with score exactly $k$ . Thus, we can compute the number of permutations with scores $2$ $3$ , etc as $120,216,222,130,31,1$ . We then compute \[\frac{120(2)+216(3)+222(4)+130(5)+31(6)+1(7)}{720}=\frac{291}{80}\] leading us to the answer of $291+80=\boxed{371}$ $\blacksquare$ | null | 371 |
38c47f53906eecc3e6e802f1165da8ab | https://artofproblemsolving.com/wiki/index.php/2016_AIME_II_Problems/Problem_13 | Beatrix is going to place six rooks on a $6 \times 6$ chessboard where both the rows and columns are labeled $1$ to $6$ ; the rooks are placed so that no two rooks are in the same row or the same column. The $value$ of a square is the sum of its row number and column number. The $score$ of an arrangement of rooks is the least value of any occupied square.The average score over all valid configurations is $\frac{p}{q}$ , where $p$ and $q$ are relatively prime positive integers. Find $p+q$ | The problem is asking us to compute $\mathbb{E}[S]$ , where $S$ is the random variable that takes an arrangement of rooks and outputs its score, which is a non-negative integer quantity. For any random variable $S$ with non-negative integer values, we have the tail sum formula \[\mathbb{E}[S] = \sum_{n = 1}^{\infty}\mathbb{P}(S\geq n).\] These probabilities can be computed as in Solution 3, giving us the following table.
Hence \begin{align*} \mathbb{E}[S] &= \frac{720 + 720 + 600 + 384 + 162 + 32 + 1}{720} \\ &= 2 + \frac{1179}{720} = 2 + \frac{131}{80} = \frac{291}{80}, \end{align*} and the final answer is $291 + 80 = \boxed{371}$ as above. | null | 371 |
5766c196eb06bcaaaad6f53b37bea93e | https://artofproblemsolving.com/wiki/index.php/2016_AIME_II_Problems/Problem_14 | Equilateral $\triangle ABC$ has side length $600$ . Points $P$ and $Q$ lie outside the plane of $\triangle ABC$ and are on opposite sides of the plane. Furthermore, $PA=PB=PC$ , and $QA=QB=QC$ , and the planes of $\triangle PAB$ and $\triangle QAB$ form a $120^{\circ}$ dihedral angle (the angle between the two planes). There is a point $O$ whose distance from each of $A,B,C,P,$ and $Q$ is $d$ . Find $d$ | The inradius of $\triangle ABC$ is $100\sqrt 3$ and the circumradius is $200 \sqrt 3$ . Now, consider the line perpendicular to plane $ABC$ through the circumcenter of $\triangle ABC$ . Note that $P,Q,O$ must lie on that line to be equidistant from each of the triangle's vertices. Also, note that since $P, Q, O$ are collinear, and $OP=OQ$ , we must have $O$ is the midpoint of $PQ$ . Now, Let $K$ be the circumcenter of $\triangle ABC$ , and $L$ be the foot of the altitude from $A$ to $BC$ . We must have $\tan(\angle KLP+ \angle QLK)= \tan(120^{\circ})$ . Setting $KP=x$ and $KQ=y$ , assuming WLOG $x>y$ , we must have $\tan(120^{\circ})=-\sqrt{3}=\dfrac{\dfrac{x+y}{100 \sqrt{3}}}{\dfrac{30000-xy}{30000}}$ . Therefore, we must have $100(x+y)=xy-30000$ . Also, we must have $\left(\dfrac{x+y}{2}\right)^{2}=\left(\dfrac{x-y}{2}\right)^{2}+120000$ by the Pythagorean theorem, so we have $xy=120000$ , so substituting into the other equation we have $90000=100(x+y)$ , or $x+y=900$ . Since we want $\dfrac{x+y}{2}$ , the desired answer is $\boxed{450}$ | null | 450 |
5766c196eb06bcaaaad6f53b37bea93e | https://artofproblemsolving.com/wiki/index.php/2016_AIME_II_Problems/Problem_14 | Equilateral $\triangle ABC$ has side length $600$ . Points $P$ and $Q$ lie outside the plane of $\triangle ABC$ and are on opposite sides of the plane. Furthermore, $PA=PB=PC$ , and $QA=QB=QC$ , and the planes of $\triangle PAB$ and $\triangle QAB$ form a $120^{\circ}$ dihedral angle (the angle between the two planes). There is a point $O$ whose distance from each of $A,B,C,P,$ and $Q$ is $d$ . Find $d$ | Draw a good diagram. Draw $CH$ as an altitude of the triangle. Scale everything down by a factor of $100\sqrt{3}$ , so that $AB=2\sqrt{3}$ . Finally, call the center of the triangle U. Draw a cross-section of the triangle via line $CH$ , which of course includes $P, Q$ . From there, we can call $OU=h$ . There are two crucial equations we can thus generate. WLOG set $PU<QU$ , then we call $PU=d-h, QU=d+h$ . First equation: using the Pythagorean Theorem on $\triangle UOB$ $h^2+2^2=d^2$ . Next, using the tangent addition formula on angles $\angle PHU, \angle UHQ$ we see that after simplifying $-d^2+h^2=-4, 2d=3\sqrt{3}$ in the numerator, so $d=\frac{3\sqrt{3}}{2}$ . Multiply back the scalar and you get $\boxed{450}$ . Not that hard, was it? | null | 450 |
5766c196eb06bcaaaad6f53b37bea93e | https://artofproblemsolving.com/wiki/index.php/2016_AIME_II_Problems/Problem_14 | Equilateral $\triangle ABC$ has side length $600$ . Points $P$ and $Q$ lie outside the plane of $\triangle ABC$ and are on opposite sides of the plane. Furthermore, $PA=PB=PC$ , and $QA=QB=QC$ , and the planes of $\triangle PAB$ and $\triangle QAB$ form a $120^{\circ}$ dihedral angle (the angle between the two planes). There is a point $O$ whose distance from each of $A,B,C,P,$ and $Q$ is $d$ . Find $d$ | To make numbers more feasible, we'll scale everything down by a factor of $100$ so that $\overline{AB}=\overline{BC}=\overline{AC}=6$ . We should also note that $P$ and $Q$ must lie on the line that is perpendicular to the plane of $ABC$ and also passes through the circumcenter of $ABC$ (due to $P$ and $Q$ being equidistant from $A$ $B$ $C$ ), let $D$ be the altitude from $C$ to $AB$ . We can draw a vertical cross-section of the figure then: [asy]pair C, D, I, P, Q, O; D=(0,0); C=(5.196152,0); P=(1.732051,7.37228); I=(1.732051,0); Q=(1.732051,-1.62772); O=(1.732051,2.87228); draw(C--Q--D--P--cycle); draw(C--D, dashed); draw(P--Q, dotted); draw(O--C, dotted); label("$C$", C, E); label("$D$", D, W); label("$I$", I, NW); label("$P$", P, N); label("$Q$", Q, S); label("$O$", O, SW); dot(O); dot(I);[/asy] We let $\angle PDI=\alpha$ so $\angle QDI=120^{\circ}-\alpha$ , also note that $\overline{PO}=\overline{QO}=\overline{CO}=d$ . Because $I$ is the centroid of $ABC$ , we know that ratio of $\overline{CI}$ to $\overline{DI}$ is $2:1$ . Since we've scaled the figure down, the length of $CD$ is $3\sqrt{3}$ , from this it's easy to know that $\overline{CI}=2\sqrt{3}$ and $\overline{DI}=\sqrt{3}$ . The following two equations arise: \begin{align} \sqrt{3}\tan{\left(\alpha\right)}+\sqrt{3}\tan{\left(120^{\circ}-\alpha\right)}&=2d \\ \sqrt{3}\tan{\left(\alpha\right)} - d &= \sqrt{d^{2}-12} \end{align} Using trig identities for the tangent, we find that \begin{align*} \sqrt{3}\tan{\left(120^{\circ}-\alpha\right)}&=\sqrt{3}\left(\frac{\tan{\left(120^{\circ}\right)}+\tan{\left(\text{-}\alpha\right)}}{1-\tan{\left(120^{\circ}\right)}\tan{\left(\text{-}\alpha\right)}}\right) \\ &= \sqrt{3}\left(\frac{\text{-}\sqrt{3}+\tan{\left(\text{-}\alpha\right)}}{1+\sqrt{3}\tan{\left(\text{-}\alpha\right)}}\right) \\ &= \sqrt{3}\left(\frac{\text{-}\sqrt{3}-\tan{\left(\alpha\right)}}{1-\sqrt{3}\tan{\left(\alpha\right)}}\right) \\ &= \frac{\sqrt{3}\tan{\left(\alpha\right)}+3}{\sqrt{3}\tan{\left(\alpha\right)}-1}.\end{align*} Okay, now we can plug this into $\left(1\right)$ to get: \begin{align}\sqrt{3}\tan{\left(\alpha\right)}+\frac{\sqrt{3}\tan{\left(\alpha\right)}+3}{\sqrt{3}\tan{\left(\alpha\right)}-1}&=2d \\ \sqrt{3}\tan{\left(\alpha\right)} - d &= \sqrt{d^{2}-12} \end{align} Notice that $\alpha$ only appears in the above system of equations in the form of $\sqrt{3}\tan{\left(\alpha\right)}$ , we can set $\sqrt{3}\tan{\left(\alpha\right)}=a$ for convenience since we really only care about $d$ . Now we have \begin{align}a+\frac{a+3}{a-1}&=2d \\ a - d &= \sqrt{d^{2}-12} \end{align} Looking at $\left(2\right)$ , it's tempting to square it to get rid of the square-root so now we have: \begin{align*}a^{2}-2ad+d^{2}&=d^{2}-12 \\ a - 2ad &= \text{-}12 \end{align*} See the sneaky $2d$ in the above equation? That we means we can substitute it for $a+\frac{a+3}{a-1}$ \begin{align*}a^{2}-2ad+d^{2}&=d^{2}-12 \\ a^{2} - a\left(a+\frac{a+3}{a-1}\right) &= \text{-}12 \\ a^{2}-a^{2}-\frac{a^{2}+3a}{a-1} &=\text{-}12 \\ -\frac{a^{2}+3a}{a-1}&=\text{-}12 \\ \text{-}a^{2}-3a&=\text{-}12a+12 \\ 0 &= a^{2}-9a+12 \end{align*} Use the quadratic formula, we find that $a=\frac{9\pm\sqrt{9^{2}-4\left(1\right)\left(12\right)}}{2\left(1\right)}=\frac{9\pm\sqrt{33}}{2}$ - the two solutions were expected because $a$ can be $\angle PDI$ or $\angle QDI$ . We can plug this into $\left(1\right)$ \begin{align*}a+\frac{a+3}{a-1}&=2d \\ \frac{9\pm\sqrt{33}}{2}+\frac{\frac{9\pm\sqrt{33}}{2}+3}{\frac{9\pm\sqrt{33}}{2}-1}=2d \\ \frac{9\pm\sqrt{33}}{2}+\frac{15\pm\sqrt{33}}{7\pm\sqrt{33}} &= 2d\end{align*} I'll use $a=\frac{9+\sqrt{33}}{2}$ because both values should give the same answer for $d$ \begin{align*} \frac{9+\sqrt{33}}{2}+\frac{15+\sqrt{33}}{7+\sqrt{33}} &= 2d \\ \frac{\left(9+\sqrt{33}\right)\left(7+\sqrt{33}\right)+\left(2\right)\left(15+\sqrt{33}\right)}{\left(2\right)\left(7+\sqrt{33}\right)} &= 2d \\ \frac{63+33+16\sqrt{33}+30+2\sqrt{33}}{14+2\sqrt{33}} &= 2d \\ \frac{126+18\sqrt{33}}{14+2\sqrt{33}} &= 2d \\ 9 &= 2d \\ \frac{9}{2} &= d\end{align*} Wait! Before you get excited, remember that we scaled the entire figure by $100$ ?? That means that the answer is $d=100\times\frac{9}{2}=\boxed{450}$ .
-fatant | null | 450 |
5766c196eb06bcaaaad6f53b37bea93e | https://artofproblemsolving.com/wiki/index.php/2016_AIME_II_Problems/Problem_14 | Equilateral $\triangle ABC$ has side length $600$ . Points $P$ and $Q$ lie outside the plane of $\triangle ABC$ and are on opposite sides of the plane. Furthermore, $PA=PB=PC$ , and $QA=QB=QC$ , and the planes of $\triangle PAB$ and $\triangle QAB$ form a $120^{\circ}$ dihedral angle (the angle between the two planes). There is a point $O$ whose distance from each of $A,B,C,P,$ and $Q$ is $d$ . Find $d$ | We use the diagram from solution 3. From basic angle chasing, \[180=\angle{QOC}+\angle{CO}P=2\angle{OCP}+2\angle{OCQ}=2\angle{QCP}\] so triangle QCP is a right triangle. This means that triangles $CQI$ and $CPI$ are similar. If we let $\angle{IDQ}=x$ and $\angle{PDI}=y$ , then we know $x+y=120$ and \[\frac{PG}{GC}=\frac{GC}{GQ}\Rightarrow\frac{100\sqrt{3}\tan{y}}{200\sqrt{3}}=\frac{200\sqrt{3}}{100\sqrt{3}\tan{x}}\Rightarrow\tan{x}\tan{y}=4\] We also know that \[PQ=2d=100\sqrt{3}(\tan{x}+\tan{y})\] \[d=50\sqrt{3}(\tan{x}+\tan{y})\] \[\frac{d}{1-\tan{x}\tan{y}}=50\sqrt{3}\cdot\frac{\tan{x}+\tan{y}}{1-\tan{x}\tan{y}}\] \[\frac{d}{-3}=50\sqrt{3}\tan{(x+y)}\] \[d=-150\sqrt{3}\tan{120}=-150\sqrt{3}(-\sqrt{3})=\boxed{450}\] | null | 450 |
5766c196eb06bcaaaad6f53b37bea93e | https://artofproblemsolving.com/wiki/index.php/2016_AIME_II_Problems/Problem_14 | Equilateral $\triangle ABC$ has side length $600$ . Points $P$ and $Q$ lie outside the plane of $\triangle ABC$ and are on opposite sides of the plane. Furthermore, $PA=PB=PC$ , and $QA=QB=QC$ , and the planes of $\triangle PAB$ and $\triangle QAB$ form a $120^{\circ}$ dihedral angle (the angle between the two planes). There is a point $O$ whose distance from each of $A,B,C,P,$ and $Q$ is $d$ . Find $d$ | We use the diagram from solution 3.
Let $BP = a$ and $BQ = b$ . Then, by Stewart's on $BPQ$ , we find \[2x^3 + 2x^3 = a^2x + b^2x \implies a^2 + b^2 = 4x^2.\]
The altitude from $P$ to $ABC$ is $\sqrt{a^2 - (200\sqrt{3})^2}$ so \[PQ = 2x = \sqrt{a^2 - (200\sqrt{3})^2} + \sqrt{b^2 - (200\sqrt{3})^2}.\]
Furthermore, the altitude from $P$ to $AB$ is $\sqrt{a^2 - 300^2}$ , so, by LoC and the dihedral condition, \[a^2 - 300^2 + b^2 - 300^2 + \sqrt{a^2 - 300^2}\sqrt{b^2-300^2} = 4x^2.\]
Squaring the equation for $PQ$ and substituting $a^2 + b^2 = 4x^2$ yields \[2\sqrt{a^2 - (200\sqrt{3})^2}\sqrt{b^2 - (200\sqrt{3})^2} = 6\cdot 200^2.\]
Substituting $a^2 + b^2 = 4x^2$ into the other equation, \[\sqrt{a^2 - 300^2}\sqrt{b^2-300^2} = 2\cdot 300^2.\]
Squaring both of these gives \[a^2b^2-3\cdot 200^2(a^2 + b^2) + 9\cdot 200^4 = 9\cdot 200^4\] \[a^2b^2 - 300^2(a^2+b^2) + 300^4 = 4\cdot 300^4.\]
Substituting $a^2 + b^2 = 4x^2$ and solving for $x$ gives $\boxed{450}$ , as desired. | null | 450 |
5766c196eb06bcaaaad6f53b37bea93e | https://artofproblemsolving.com/wiki/index.php/2016_AIME_II_Problems/Problem_14 | Equilateral $\triangle ABC$ has side length $600$ . Points $P$ and $Q$ lie outside the plane of $\triangle ABC$ and are on opposite sides of the plane. Furthermore, $PA=PB=PC$ , and $QA=QB=QC$ , and the planes of $\triangle PAB$ and $\triangle QAB$ form a $120^{\circ}$ dihedral angle (the angle between the two planes). There is a point $O$ whose distance from each of $A,B,C,P,$ and $Q$ is $d$ . Find $d$ | Let $AB = a, M$ be midpoint $BC, I$ be the center of equilateral $\triangle ABC,$ $IM = b = \frac {a}{2\sqrt{3}}, O$ be the center of sphere $ABCPQ.$ Then \[AI = 2b, AO = BO = PO =QO = d.\] \[QA=QB=QC,PA=PB=PC \implies\] \[POIQ\perp ABC, \angle PMQ = 120^\circ.\] (See upper diagram).
We construct the circle PQMD, use the formulas for intersecting chords and get \[DI = 5b, FI = EO = \frac{3b}{2}\] \[\implies FM = \frac{5b}{2}.\] (See lower diagram).
We apply the Law of Sine to $\triangle PMQ$ and get \[2EM \sin 120^\circ =PQ\] \[\implies r \sqrt{3} = 2d\] \[\implies 3r^2 = 4d^2.\] We apply the Pythagorean Law on $\triangle AOI$ and $\triangle EFM$ and get \[d^2 = 4b^2 + OI^2, r^2 = \frac {25b^2}{4} + EF^2 \implies\] \[r = 3b\implies d = \frac {3a}{2} = \boxed{450}.\] [email protected], vvsss | null | 450 |
5766c196eb06bcaaaad6f53b37bea93e | https://artofproblemsolving.com/wiki/index.php/2016_AIME_II_Problems/Problem_14 | Equilateral $\triangle ABC$ has side length $600$ . Points $P$ and $Q$ lie outside the plane of $\triangle ABC$ and are on opposite sides of the plane. Furthermore, $PA=PB=PC$ , and $QA=QB=QC$ , and the planes of $\triangle PAB$ and $\triangle QAB$ form a $120^{\circ}$ dihedral angle (the angle between the two planes). There is a point $O$ whose distance from each of $A,B,C,P,$ and $Q$ is $d$ . Find $d$ | Let $M$ be the midpoint of $\overline{AB}$ and $X$ the center of $\triangle ABC$ . Then \[P, O, Q, M, X, C\] all lie in the same vertical plane. We can make the following observations:
To make calculations easier, we will denote $100\sqrt{3}=m$ , so that $MX=m$ and $XC=2m$
[asy] unitsize(20); pair P = (0, 12); pair Q = (0, -3); pair O = (P+Q)/2; pair M = (-3, 0); pair X = (0, 0); pair C = (6, 0); draw(P--O--Q); draw(M--X--C); draw(P--M--Q, blue); draw(Q--C--P); draw(circle((0, 4.5), 7.5)); label("$P$", P, N); label("$Q$", Q, S); label("$O$", O, E); dot(O); label("$M$", M, W); label("$X$", X, NE); label("$C$", C, E); label("$m$", (M+X)/2, N); label("$2m$", (X+C)/2, N); [/asy]
Denote $PX=p$ and $QX=q$ , where the tangent addition formula on $\triangle PMQ$ yields \[\frac{\tan\measuredangle PMX+\tan\measuredangle QMX}{1-\tan\measuredangle PMX\tan\measuredangle QMX}=\tan(120^{\circ})=-\sqrt{3}.\] Using $\tan\measuredangle PMX=\frac{p}{m}$ and $\tan\measuredangle QMX=\frac{q}{m}$ , we have \[\frac{\frac{p}{m}+\frac{q}{m}}{1-\frac{p}{m}\cdot\frac{q}{m}}=-\sqrt{3}.\] After multiplying both numerator and denominator by $m^{2}$ we have \[\frac{(p+q)m}{m^{2}-pq}=-\sqrt{3}.\] But note that $pq=(2m)(2m)=4m^{2}$ by power of a point at $X$ , where we deduce by symmetry that $MM^{\prime}=MX=m$ on the diagram below: [asy] unitsize(20); pair P = (0, 12); pair Q = (0, -3); pair O = (P+Q)/2; pair M = (-3, 0); pair Mprime = (-6, 0); pair X = (0, 0); pair C = (6, 0); draw(P--O--Q); draw(Mprime--M--X--C); draw(P--M--Q, blue); draw(Q--C--P); draw(circle((0, 4.5), 7.5)); label("$P$", P, N); label("$Q$", Q, S); label("$O$", O, E); dot(O); label("$M$", M, S); label("$M^{\prime}$", Mprime, W); label("$X$", X, SE); label("$C$", C, E); label("$m$", (Mprime+M)/2, N); label("$m$", (M+X)/2, N); label("$2m$", (X+C)/2, N); [/asy]
Thus \begin{align*} \frac{(p+q)m}{m^{2}-4m^{2}}=-\sqrt{3} \\ \frac{p+q}{-3m}=-\sqrt{3} \\ p+q=\left(-\sqrt{3}\right)\left(-3m\right) \\ p+q=3\sqrt{3}\cdot m.\end{align*} Earlier we assigned the variable $m$ to the length $100\sqrt{3}$ which implies $PQ=\left(3\sqrt{3}\right)\left(100\sqrt{3}\right)=900$ . Thus the distance $d$ is equal to $\frac{PQ}{2}=\boxed{450}$ | null | 450 |
5766c196eb06bcaaaad6f53b37bea93e | https://artofproblemsolving.com/wiki/index.php/2016_AIME_II_Problems/Problem_14 | Equilateral $\triangle ABC$ has side length $600$ . Points $P$ and $Q$ lie outside the plane of $\triangle ABC$ and are on opposite sides of the plane. Furthermore, $PA=PB=PC$ , and $QA=QB=QC$ , and the planes of $\triangle PAB$ and $\triangle QAB$ form a $120^{\circ}$ dihedral angle (the angle between the two planes). There is a point $O$ whose distance from each of $A,B,C,P,$ and $Q$ is $d$ . Find $d$ | Let $Z$ be the center of $\triangle ABC$ . Let $A’$ be the midpoint of $BC$ . Let $ZA’ = c = 100\sqrt{3}$ and $ZA = 2c = 200\sqrt{3}$ . Let $PZ = a$ and $QZ = b$ . We will be working in the plane that contains the points: $A$ $P$ $A’$ $Q$ $O$ , and $Z$
Since $P$ $O$ , and $Q$ are collinear and $PO = QO = AO$ $\triangle PAQ$ is a right triangle with $\angle PAQ = 90^{\circ}$ . Since $AZ \perp PQ$ $(PZ)(QZ) = (AZ)^2 = ab = (2c)^2 = 120000$
$PA’ = \sqrt{a^2 + c^2}$ $QA’ = \sqrt{b^2 + c^2}$ $PQ = a + b$ , and $\angle PAQ = 120^{\circ}$ . By Law of Cosines \[(a + b)^2 = a^2 + b^2 + 2c^2 + \sqrt{a^2b^2 + a^2c^2 + b^2c^2 + c^4}\] . Substituting $4c^2$ for $ab$ and simplifying, we get \[6c = \sqrt{17c^2 + a^2 + b^2}\] . Squaring and simplifying, we get \[a^2 + b^2 = 19c^2 = 570000\] . Adding $2ab = 8c^2$ to both sides we get $PQ = a + b = 900$ . Since $O$ is the midpoint of $PQ$ $d = PO = \boxed{450}$ | null | 450 |
c9126f36133a84da75c0dc5a7b649302 | https://artofproblemsolving.com/wiki/index.php/2016_AIME_II_Problems/Problem_15 | For $1 \leq i \leq 215$ let $a_i = \dfrac{1}{2^{i}}$ and $a_{216} = \dfrac{1}{2^{215}}$ . Let $x_1, x_2, ..., x_{216}$ be positive real numbers such that $\sum_{i=1}^{216} x_i=1$ and $\sum_{1 \leq i < j \leq 216} x_ix_j = \dfrac{107}{215} + \sum_{i=1}^{216} \dfrac{a_i x_i^{2}}{2(1-a_i)}$ . The maximum possible value of $x_2=\dfrac{m}{n}$ , where $m$ and $n$ are relatively prime positive integers. Find $m+n$ | Note that \begin{align*}\sum_{1 \leq i < j \leq 216} x_ix_j &= \frac12\left(\left(\sum_{i=1}^{216} x_i\right)^2-\sum_{i=1}^{216} x_i^2\right)\\&=\frac12\left(1-\sum x_i^2\right).\end{align*} Substituting this into the second equation and collecting $x_i^2$ terms, we find \[\sum_{i=1}^{216}\frac{x_i^2}{1-a_i}=\frac{1}{215}.\] Conveniently, $\sum_{i=1}^{216} 1-a_i=215$ so we find \[\left(\sum_{i=1}^{216} 1-a_i\right)\left(\sum_{i=1}^{216}\frac{x_i^2}{1-a_i}\right)=1=\left(\sum_{i=1}^{216} x_i\right)^2.\] This is the equality case of the Cauchy-Schwarz Inequality, so $x_i=c(1-a_i)$ for some constant $c$ . Summing these equations and using the facts that $\sum_{i=1}^{216} x_i=1$ and $\sum_{i=1}^{216} 1-a_i=215$ , we find $c=\frac{1}{215}$ and thus $x_2=c(1-a_2)=\frac{1}{215}\cdot \left(1-\frac{1}{4}\right)=\frac{3}{860}$ . Hence the desired answer is $860+3=\boxed{863}$ | null | 863 |
a50fab9b2bd9317789554370f93fa4f3 | https://artofproblemsolving.com/wiki/index.php/2015_AIME_I_Problems/Problem_4 | Point $B$ lies on line segment $\overline{AC}$ with $AB=16$ and $BC=4$ . Points $D$ and $E$ lie on the same side of line $AC$ forming equilateral triangles $\triangle ABD$ and $\triangle BCE$ . Let $M$ be the midpoint of $\overline{AE}$ , and $N$ be the midpoint of $\overline{CD}$ . The area of $\triangle BMN$ is $x$ . Find $x^2$ | Let $A$ be the origin, so $B=(16,0)$ and $C=(20,0).$ Using equilateral triangle properties tells us that $D=(8,8\sqrt3)$ and $E=(18,2\sqrt3)$ as well. Therefore, $M=(9,\sqrt3)$ and $N=(14,4\sqrt3).$ Applying the Shoelace Theorem to triangle $BMN$ gives
\[x=\dfrac 1 2 |16\sqrt3+36\sqrt3+0-(0+14\sqrt3+64\sqrt3)| =13\sqrt3,\]
so $x^2=\boxed{507}.$ | null | 507 |
a50fab9b2bd9317789554370f93fa4f3 | https://artofproblemsolving.com/wiki/index.php/2015_AIME_I_Problems/Problem_4 | Point $B$ lies on line segment $\overline{AC}$ with $AB=16$ and $BC=4$ . Points $D$ and $E$ lie on the same side of line $AC$ forming equilateral triangles $\triangle ABD$ and $\triangle BCE$ . Let $M$ be the midpoint of $\overline{AE}$ , and $N$ be the midpoint of $\overline{CD}$ . The area of $\triangle BMN$ is $x$ . Find $x^2$ | Note that $AB=DB=16$ and $BE=BC=4$ . Also, $\angle ABE = \angle DBC = 120^{\circ}$ . Thus, $\triangle ABE \cong \triangle DBC$ by SAS.
From this, it is clear that a $60^{\circ}$ rotation about $B$ will map $\triangle ABE$ to $\triangle DBC$ .
This rotation also maps $M$ to $N$ . Thus, $BM=BN$ and $\angle MBN=60^{\circ}$ . Thus, $\triangle BMN$ is equilateral.
Using the Law of Cosines on $\triangle ABE$ \[AE^2 = 16^2 + 4^2 - 2\cdot 16\cdot 4\cdot\left(-\frac{1}{2}\right)\] \[AE = 4\sqrt{21}\] Thus, $AM=ME=2\sqrt{21}$
Using Stewart's Theorem on $\triangle ABE$ \[AM\cdot ME\cdot AE + AE\cdot BM^2 = BE^2\cdot AM + BA^2\cdot ME\] \[BM = 2\sqrt{13}\]
Calculating the area of $\triangle BMN$ \[[BMN] = \frac{\sqrt{3}}{4} BM^2\] \[[BMN] = 13\sqrt{3}\] Thus, $x=13\sqrt{3}$ , so $x^2 = 507$ . Our final answer is $\boxed{507}$ | null | 507 |
a50fab9b2bd9317789554370f93fa4f3 | https://artofproblemsolving.com/wiki/index.php/2015_AIME_I_Problems/Problem_4 | Point $B$ lies on line segment $\overline{AC}$ with $AB=16$ and $BC=4$ . Points $D$ and $E$ lie on the same side of line $AC$ forming equilateral triangles $\triangle ABD$ and $\triangle BCE$ . Let $M$ be the midpoint of $\overline{AE}$ , and $N$ be the midpoint of $\overline{CD}$ . The area of $\triangle BMN$ is $x$ . Find $x^2$ | $AB = BD, BE = BC, \angle ABE = \angle CBD \implies \triangle ABE \cong \triangle DBC$
Medians are equal, so $MB = MN, \angle ABM = \angle DBN \implies$ $\angle MBN = \angle ABD - \angle ABM + \angle DBN = 60^\circ \implies$
$\triangle MNB$ is equilateral triangle.
The height of $\triangle BCE$ is $2 \sqrt{3},$ distance from $A$ to midpoint $BC$ is $16 + 2 = 18 \implies \frac {AE^2}{4} =\frac{ (16 + 2)^2 +2^2 \cdot 3}{4} = 81 + 3 = 84.$
$BM$ is the median of $\triangle ABE \implies$ $BM^2 = \frac {AB^2}{2} + \frac {BE^2}{2} - \frac {AE^2}{4}=16 \cdot 8 + 4 \cdot 2 - 84 = 52.$
The area of $\triangle BMN$
\[[BMN] = \frac{\sqrt{3}}{4} BM^2 =13 \sqrt{3} \implies \boxed{507}.\] | null | 507 |
85d0626540c792193c4ba56fdf9671fb | https://artofproblemsolving.com/wiki/index.php/2015_AIME_I_Problems/Problem_5 | In a drawer Sandy has $5$ pairs of socks, each pair a different color. On Monday Sandy selects two individual socks at random from the $10$ socks in the drawer. On Tuesday Sandy selects $2$ of the remaining $8$ socks at random and on Wednesday two of the remaining $6$ socks at random. The probability that Wednesday is the first day Sandy selects matching socks is $\frac{m}{n}$ , where $m$ and $n$ are relatively prime positive integers, Find $m+n$ | Let the fifth sock be arbitrary; the probability that the sixth sock matches in color is $\dfrac{1}{9}$
Assuming this, then let the first sock be arbitrary; the probability that the second sock does not match is $\dfrac{6}{7}.$
The only "hard" part is the third and fourth sock. But that is simple casework. If the third sock's color matches the color of one of the first two socks (which occurs with probability $\dfrac{2}{6} = \dfrac{1}{3}$ ), then the fourth sock can be arbitrary. Otherwise (with probability $\dfrac{2}{3}$ ), the fourth sock can be chosen with probability $\dfrac{4}{5}$ (5 socks left, 1 sock that can possibly match the third sock's color). The desired probability is thus \[\frac{1}{9} \cdot \frac{6}{7} \cdot \left(\dfrac{1}{3} + \dfrac{2}{3} \cdot \dfrac{4}{5}\right) = \frac{26}{315}.\] The sum is therefore $26+315=\boxed{341}.$ | null | 341 |
85d0626540c792193c4ba56fdf9671fb | https://artofproblemsolving.com/wiki/index.php/2015_AIME_I_Problems/Problem_5 | In a drawer Sandy has $5$ pairs of socks, each pair a different color. On Monday Sandy selects two individual socks at random from the $10$ socks in the drawer. On Tuesday Sandy selects $2$ of the remaining $8$ socks at random and on Wednesday two of the remaining $6$ socks at random. The probability that Wednesday is the first day Sandy selects matching socks is $\frac{m}{n}$ , where $m$ and $n$ are relatively prime positive integers, Find $m+n$ | The key is to count backwards. First, choose the pair which you pick on Wednesday in $5$ ways. Then there are four pairs of socks for you to pick a pair of on Tuesday, and you don't want to pick a pair. Since there are $4$ pairs, the number of ways to do this is $\dbinom{8}{2}-4$ . Then, there are two pairs and two nonmatching socks for you to pick from on Monday, a total of $6$ socks. Since you don't want to pick a pair, the number of ways to do this is $\dbinom{6}{2}-2$ . Thus the answer is \[\dfrac{\left(5\right)\left(\dbinom{8}{2}-4\right)\left(\dbinom{6}{2}-2\right)}{\dbinom{10}{2}\dbinom{8}{2}\dbinom{6}{2}}=\dfrac{26}{315}.\] $26 + 315 = \boxed{341}$ | null | 341 |
85d0626540c792193c4ba56fdf9671fb | https://artofproblemsolving.com/wiki/index.php/2015_AIME_I_Problems/Problem_5 | In a drawer Sandy has $5$ pairs of socks, each pair a different color. On Monday Sandy selects two individual socks at random from the $10$ socks in the drawer. On Tuesday Sandy selects $2$ of the remaining $8$ socks at random and on Wednesday two of the remaining $6$ socks at random. The probability that Wednesday is the first day Sandy selects matching socks is $\frac{m}{n}$ , where $m$ and $n$ are relatively prime positive integers, Find $m+n$ | For the first sock, note that to pick two different socks, can complementary count to get the total, $\binom{10}{2}$ minus the number of pairs (5) to get \[\frac{\binom{10}{2} - 5}{\binom{10}{2}}\] The next steps aren't quite as simple, though.
WLOG suppose the socks are (a, a, b, b, c, c, d, d, e, e) and that we chose ab on the first day.
This leaves abccddee as the remaining socks. We can think of our next choice as how many "pairs" we "break".
Note that right now, in abccddee there are 3 pairs, namely {cc, dd, ee}. Our next choice could either leave all of these pairs (if we choose ab), leave 2 of these pairs, (choose say ac to keep dd and ee as pairs), or leave only 1 as a pair, say choosing de to get abccde. \[\bold{\text{Case 1: Break zero pairs}}.\] In this case, we want to keep 3 pairs remaining, so our only choice is to choose AB, which would make the remaining ccddee. The probability that we choose AB is $\frac{1}{\binom{8}{2}}$ and the probability that we get a pair after this is $\frac{3}{\binom{6}{2}}$ since we just choose CC, DD, or EE. This case has a probability $\frac{1}{\binom{8}{2}}\frac{3}{\binom{6}{2}}.$ \[\bold{\text{Case 2: Break one pair}}.\] In this case, we want to choose A and one of {C,C,D,D,E,E} or B and one of {C,C,D,D,E,E}. This yields $\frac{6 + 6}{\binom{8}{2}}$ probability. After this, we end up with 2 pairs, so the total probability of choosing a pair on Wednesday in this case is $\frac{6 + 6}{\binom{8}{2}}\frac{2}{\binom{6}{2}}.$ \[\bold{\text{case 3: Break 2 pairs}}.\] In this case we'd choose two distinct characters from {C,C,D,D,E,E} which would be $\frac{\binom{6}{2} - 3}{\binom{6}{2}}.$ After this, only one pair remains so the total probability for this case is
$\frac{\binom{6}{2} - 3}{\binom{6}{2}} \frac{6}{\binom{8}{2}}\frac{1}{\binom{6}{2}}.$ Adding all the cases and multiplying by $\frac{\binom{10}{2} - 5}{\binom{10}{2}}$ yields \[\frac{\binom{10}{2} - 5}{\binom{10}{2}}\left(\frac{1}{\binom{8}{2}}\frac{3}{\binom{6}{2}}+\frac{6 + 6}{\binom{8}{2}}\frac{2}{\binom{6}{2}}+ \frac{\binom{6}{2} - 3}{\binom{6}{2}}\frac{6}{\binom{8}{2}}\frac{1}{\binom{6}{2}}\right) = \frac{26}{315} \rightarrow 26 + 315 = \boxed{341}\] | null | 341 |
a068f8d52bea29a125a622180ddb4e63 | https://artofproblemsolving.com/wiki/index.php/2015_AIME_I_Problems/Problem_6 | Point $A,B,C,D,$ and $E$ are equally spaced on a minor arc of a circle. Points $E,F,G,H,I$ and $A$ are equally spaced on a minor arc of a second circle with center $C$ as shown in the figure below. The angle $\angle ABD$ exceeds $\angle AHG$ by $12^\circ$ . Find the degree measure of $\angle BAG$
[asy] pair A,B,C,D,E,F,G,H,I,O; O=(0,0); C=dir(90); B=dir(70); A=dir(50); D=dir(110); E=dir(130); draw(arc(O,1,50,130)); real x=2*sin(20*pi/180); F=x*dir(228)+C; G=x*dir(256)+C; H=x*dir(284)+C; I=x*dir(312)+C; draw(arc(C,x,200,340)); label("$A$",A,dir(0)); label("$B$",B,dir(75)); label("$C$",C,dir(90)); label("$D$",D,dir(105)); label("$E$",E,dir(180)); label("$F$",F,dir(225)); label("$G$",G,dir(260)); label("$H$",H,dir(280)); label("$I$",I,dir(315)); [/asy] | Let $O$ be the center of the circle with $ABCDE$ on it.
Let $x$ be the degree measurement of $\overarc{ED}=\overarc{DC}=\overarc{CB}=\overarc{BA}$ in circle $O$
and $y$ be the degree measurement of $\overarc{EF}=\overarc{FG}=\overarc{GH}=\overarc{HI}=\overarc{IA}$ in circle $C$
$\angle ECA$ is, therefore, $5y$ by way of circle $C$ and \[\frac{360-4x}{2}=180-2x\] by way of circle $O$ $\angle ABD$ is $180 - \frac{3x}{2}$ by way of circle $O$ , and \[\angle AHG = 180 - \frac{3y}{2}\] by way of circle $C$
This means that:
\[180-\frac{3x}{2}=180-\frac{3y}{2}+12\]
which when simplified yields \[\frac{3x}{2}+12=\frac{3y}{2}\] or \[x+8=y\] Since: \[5y=180-2x\] and \[5x+40=180-2x\] So: \[7x=140\Longleftrightarrow x=20\] \[y=28\] $\angle BAG$ is equal to $\angle BAE$ $\angle EAG$ , which equates to $\frac{3x}{2} + y$ .
Plugging in yields $30+28$ , or $\boxed{058}$ | null | 058 |
a068f8d52bea29a125a622180ddb4e63 | https://artofproblemsolving.com/wiki/index.php/2015_AIME_I_Problems/Problem_6 | Point $A,B,C,D,$ and $E$ are equally spaced on a minor arc of a circle. Points $E,F,G,H,I$ and $A$ are equally spaced on a minor arc of a second circle with center $C$ as shown in the figure below. The angle $\angle ABD$ exceeds $\angle AHG$ by $12^\circ$ . Find the degree measure of $\angle BAG$
[asy] pair A,B,C,D,E,F,G,H,I,O; O=(0,0); C=dir(90); B=dir(70); A=dir(50); D=dir(110); E=dir(130); draw(arc(O,1,50,130)); real x=2*sin(20*pi/180); F=x*dir(228)+C; G=x*dir(256)+C; H=x*dir(284)+C; I=x*dir(312)+C; draw(arc(C,x,200,340)); label("$A$",A,dir(0)); label("$B$",B,dir(75)); label("$C$",C,dir(90)); label("$D$",D,dir(105)); label("$E$",E,dir(180)); label("$F$",F,dir(225)); label("$G$",G,dir(260)); label("$H$",H,dir(280)); label("$I$",I,dir(315)); [/asy] | Let $m$ be the degree measurement of $\angle GCH$ . Since $G,H$ lie on a circle with center $C$ $\angle GHC=\frac{180-m}{2}=90-\frac{m}{2}$
Since $\angle ACH=2 \angle GCH=2m$ $\angle AHC=\frac{180-2m}{2}=90-m$ . Adding $\angle GHC$ and $\angle AHC$ gives $\angle AHG=180-\frac{3m}{2}$ , and $\angle ABD=\angle AHG+12=192-\frac{3m}{2}$ . Since $AE$ is parallel to $BD$ $\angle DBA=180-\angle ABD=\frac{3m}{2}-12=$ $\overarc{BE}$
We are given that $A,B,C,D,E$ are evenly distributed on a circle. Hence,
$\overarc{ED}=\overarc{DC}=\overarc{CB}=\overarc{BA}$ $=\frac{\angle DBA}{3}=\frac{m}{2}-4$
Here comes the key: Draw a line through $C$ parallel to $AE$ , and select a point $X$ to the right of point $C$
$\angle ACX$ $\overarc{AB}$ $\overarc{BC}$ $m-8$
Let the midpoint of $\overline{HG}$ be $Y$ , then $\angle YCX=\angle ACX+\angle ACY=(m-8)+\frac{5m}{2}=90$ . Solving gives $m=28$
The rest of the solution proceeds as in solution 1, which gives $\boxed{058}$ | null | 058 |
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