problem_id
stringlengths 32
32
| link
stringlengths 75
84
| problem
stringlengths 14
5.33k
| solution
stringlengths 15
6.63k
| letter
stringclasses 5
values | answer
stringclasses 957
values |
|---|---|---|---|---|---|
bf8152d490bb5752c3e75b85a0e6d7da | https://artofproblemsolving.com/wiki/index.php/2021_AMC_12B_Problems/Problem_24 | Let $ABCD$ be a parallelogram with area $15$ . Points $P$ and $Q$ are the projections of $A$ and $C,$ respectively, onto the line $BD;$ and points $R$ and $S$ are the projections of $B$ and $D,$ respectively, onto the line $AC.$ See the figure, which also shows the relative locations of these points.
[asy] size(350); defaultpen(linewidth(0.8)+fontsize(11)); real theta = aTan(1.25/2); pair A = 2.5*dir(180+theta), B = (3.35,0), C = -A, D = -B, P = foot(A,B,D), Q = -P, R = foot(B,A,C), S = -R; draw(A--B--C--D--A^^B--D^^R--S^^rightanglemark(A,P,D,6)^^rightanglemark(C,Q,D,6)); draw(B--R^^C--Q^^A--P^^D--S,linetype("4 4")); dot("$A$",A,dir(270)); dot("$B$",B,E); dot("$C$",C,N); dot("$D$",D,W); dot("$P$",P,SE); dot("$Q$",Q,NE); dot("$R$",R,N); dot("$S$",S,dir(270)); [/asy]
Suppose $PQ=6$ and $RS=8,$ and let $d$ denote the length of $\overline{BD},$ the longer diagonal of $ABCD.$ Then $d^2$ can be written in the form $m+n\sqrt p,$ where $m,n,$ and $p$ are positive integers and $p$ is not divisible by the square of any prime. What is $m+n+p?$
$\textbf{(A) }81 \qquad \textbf{(B) }89 \qquad \textbf{(C) }97\qquad \textbf{(D) }105 \qquad \textbf{(E) }113$ | Let $BQ = PD = x.$ We know that the area of the parallelogram is $15,$ so it follows that $[\triangle{BCD}] = [\triangle{BAD}] = \tfrac{15}{2}$ and the height of each triangle, which are also the lengths of $QC$ and $AP,$ is $\tfrac{15}{2(x+3)}.$ Suppose that $E = RS \cap BD.$ Because $\angle{BRE} = \angle{CQE}$ and $\angle{BER} = \angle{CQD},$ we have $\triangle{BRE} \sim \triangle{CQE}.$ The length of $CE,$ by the Pythagorean Theorem is $\sqrt{3^2+(\tfrac{15}{2(x+3)})^2}$ and the length of $BR,$ by the Pythagorean Theorem on $\triangle{BRE},$ is $\sqrt{(x+3)^2 - 4}.$ Note that \[\sin{\angle QEC} = \frac{CQ}{CE} = \frac{BR}{BE}\] Substituting in our values, \[\frac{\frac{15}{2(x+3)}}{\sqrt{9+(\frac{15}{2(x+3)})^2}} = \frac{\sqrt{(x+3)^2 - 4^2}}{x+3}\] To rid unnecessary computation, we let $(x+3)^2 = a.$ The equation simplifies, after cross multiplying, to \[\sqrt{9+\frac{15^2}{4a}} \sqrt{a-16 } = \frac{15}{2}\] \[36a^2 - 576a - 15^2\cdot 16 = 0\] \[a^2-16a-100 =0\] By the quadratic formula, $a \in \{\tfrac{16 - \sqrt{656}}{2}, \tfrac{16 + \sqrt{656}}{2}\},$ so we discard the negative solution. The value of $BD^2$ is \[BD^2 = (2x+6)^2 = 4(x+3)^2 = 4a = 4 \cdot \frac{16 + \sqrt{656}}{2} = 32+8\sqrt{41}\] and the desired answer is $32+8+41 = \boxed{81}$ ~skyscraper | A | 81 |
bf8152d490bb5752c3e75b85a0e6d7da | https://artofproblemsolving.com/wiki/index.php/2021_AMC_12B_Problems/Problem_24 | Let $ABCD$ be a parallelogram with area $15$ . Points $P$ and $Q$ are the projections of $A$ and $C,$ respectively, onto the line $BD;$ and points $R$ and $S$ are the projections of $B$ and $D,$ respectively, onto the line $AC.$ See the figure, which also shows the relative locations of these points.
[asy] size(350); defaultpen(linewidth(0.8)+fontsize(11)); real theta = aTan(1.25/2); pair A = 2.5*dir(180+theta), B = (3.35,0), C = -A, D = -B, P = foot(A,B,D), Q = -P, R = foot(B,A,C), S = -R; draw(A--B--C--D--A^^B--D^^R--S^^rightanglemark(A,P,D,6)^^rightanglemark(C,Q,D,6)); draw(B--R^^C--Q^^A--P^^D--S,linetype("4 4")); dot("$A$",A,dir(270)); dot("$B$",B,E); dot("$C$",C,N); dot("$D$",D,W); dot("$P$",P,SE); dot("$Q$",Q,NE); dot("$R$",R,N); dot("$S$",S,dir(270)); [/asy]
Suppose $PQ=6$ and $RS=8,$ and let $d$ denote the length of $\overline{BD},$ the longer diagonal of $ABCD.$ Then $d^2$ can be written in the form $m+n\sqrt p,$ where $m,n,$ and $p$ are positive integers and $p$ is not divisible by the square of any prime. What is $m+n+p?$
$\textbf{(A) }81 \qquad \textbf{(B) }89 \qquad \textbf{(C) }97\qquad \textbf{(D) }105 \qquad \textbf{(E) }113$ | Let the intersection of $RS$ and $BD$ be $X$
$\because$ $\angle APX = \angle DSX$ and $\angle AXP = \angle DXS$ $\triangle APX \sim \triangle DSX$ by $AA$
$\therefore$ $\frac{PA}{DS} = \frac68 = \frac34$ $DS = \frac43 \cdot PA$
By the Pythagorean theorem and the property of projection, $BD^2 = (DS+BR)^2 + RS^2 = 4DS^2 + 64 = 4(\frac43 \cdot PA)^2 + 64 = \frac{64}{9} \cdot PA^2 + 64$ $\frac{64}{9} \cdot PA^2 = BD^2 - 64$
$\because [ABCD] = PA \cdot BD = 15$ $\therefore PA = \frac{15}{BD}$
\[\frac{64}{9} (\frac{15}{BD})^2 = BD^2 - 64\]
\[\frac{1600}{BD^2} = BD^2 - 64\]
\[BD^4 - 64 BD^2 - 1600 = 0\]
\[BD^2 = \frac{64 + \sqrt{64^2 - 4 (-1600)}}{2} = 32 + 8 \sqrt{41}\]
Therefore, the answer is $32 + 8 + 41 = \boxed{81}$ | A | 81 |
bd08ac7601d0fff5f9c27283846b0912 | https://artofproblemsolving.com/wiki/index.php/2021_AMC_12B_Problems/Problem_25 | Let $S$ be the set of lattice points in the coordinate plane, both of whose coordinates are integers between $1$ and $30,$ inclusive. Exactly $300$ points in $S$ lie on or below a line with equation $y=mx.$ The possible values of $m$ lie in an interval of length $\frac ab,$ where $a$ and $b$ are relatively prime positive integers. What is $a+b?$
$\textbf{(A)} ~31 \qquad \textbf{(B)} ~47 \qquad \textbf{(C)} ~62\qquad \textbf{(D)} ~72 \qquad \textbf{(E)} ~85$ | I know that I want about $\frac{2}{3}$ of the box of integer coordinates above my line. There are a total of 30 integer coordinates in the desired range for each axis which gives a total of 900 lattice points. I estimate that the slope, m, is $\frac{2}{3}$ . Now, although there is probably an easier solution, I would try to count the number of points above the line to see if there are 600 points above the line. The line $y=\frac{2}{3}x$ separates the area inside the box so that $\frac{2}{3}$ of the are is above the line.
I find that the number of coordinates with $x=1$ above the line is 30, and the number of coordinates with $x=2$ above the line is 29. Every time the line $y=\frac{2}{3}x$ hits a y-value with an integer coordinate, the number of points above the line decreases by one. I wrote out the sum of 30 terms in hopes of finding a pattern. I graphed the first couple positive integer x-coordinates, and found that the sum of the integers above the line is $30+29+28+28+27+26+26 \ldots+ 10$ . The even integer repeats itself every third term in the sum. I found that the average of each of the terms is 20, and there are 30 of them which means that exactly 600 above the line as desired. This give a lower bound because if the slope decreases a little bit, then the points that the line goes through will be above the line.
To find the upper bound, notice that each point with an integer-valued x-coordinate is either $\frac{1}{3}$ or $\frac{2}{3}$ above the line. Since the slope through a point is the y-coordinate divided by the x-coordinate, a shift in the slope will increase the y-value of the higher x-coordinates. We turn our attention to $x=28, 29, 30$ which the line $y=\frac{2}{3}x$ intersects at $y= \frac{56}{3}, \frac{58}{3}, 20$ . The point (30,20) is already counted below the line, and we can clearly see that if we slowly increase the slope of the line, we will hit the point (28,19) since (28, $\frac{56}{3}$ ) is closer to the lattice point. The slope of the line which goes through both the origin and (28,19) is $y=\frac{19}{28}x$ . This gives an upper bound of $m=\frac{19}{28}$
Taking the upper bound of m and subtracting the lower bound yields $\frac{19}{28}-\frac{2}{3}=\frac{1}{84}$ . This is answer $1+84=$ $\boxed{85}$ | E | 85 |
bd08ac7601d0fff5f9c27283846b0912 | https://artofproblemsolving.com/wiki/index.php/2021_AMC_12B_Problems/Problem_25 | Let $S$ be the set of lattice points in the coordinate plane, both of whose coordinates are integers between $1$ and $30,$ inclusive. Exactly $300$ points in $S$ lie on or below a line with equation $y=mx.$ The possible values of $m$ lie in an interval of length $\frac ab,$ where $a$ and $b$ are relatively prime positive integers. What is $a+b?$
$\textbf{(A)} ~31 \qquad \textbf{(B)} ~47 \qquad \textbf{(C)} ~62\qquad \textbf{(D)} ~72 \qquad \textbf{(E)} ~85$ | As the procedure shown in the Solution 1, the lower bound of $m$ is $\frac{2}{3}.$ Here I give a more logic way to show how to find the upper bound of $m.$ Denote $N=\sum_{x=1}^{30}(\lfloor mx \rfloor)$ as the number of lattice points in $S$
$N = \lfloor m \rfloor+\lfloor 2m \rfloor+\lfloor 3m \rfloor+\cdots+\lfloor 30m \rfloor = 300 .$
Let $m = \frac{2}{3}+k$ . for $\forall x_{i}\le 30, x\in N^{*}, \lfloor mx_{i} \rfloor = \lfloor \frac{2}{3}x+xk \rfloor.$
Our target is finding the minimum value of $k$ which can increase one unit of $\lfloor mx_{i} \rfloor .$
Notice that:
When $x_{i} = 3n, \lfloor mx_{i} \rfloor = \lfloor 2n+(3n)k \rfloor$ We don't have to discuss this case.
When $x_{i} = 3n+1, \lfloor mx_{i} \rfloor = \lfloor 2n+\frac{2}{3}+(3n+1)k \rfloor, k_{min1}=\frac{1}{3(3n+1)}.$
When $x_{i} = 3n+2, \lfloor mx_{i} \rfloor = \lfloor 2n+1+\frac{1}{3}+(3n+2)k \rfloor, k_{min2}=\frac{2}{3(3n+2)}.$
Here $n\in N^{*}, n \le 9.$
Denote $k_{min}=min\left \{k_{min1},k_{min2} \right \}.$
Obviously $k_{min1}$ and $k_{min2}$ are decreasing. Let's considering the situation when $n=9.$
$k_{min}=min\left\{\frac{1}{84},\frac{2}{87}\right\}=\frac{1}{84}.$
This means that the answer is just $\frac{1}{84}$ , so $a+b=85$ . Choose $\boxed{85}.$ | E | 85 |
bd08ac7601d0fff5f9c27283846b0912 | https://artofproblemsolving.com/wiki/index.php/2021_AMC_12B_Problems/Problem_25 | Let $S$ be the set of lattice points in the coordinate plane, both of whose coordinates are integers between $1$ and $30,$ inclusive. Exactly $300$ points in $S$ lie on or below a line with equation $y=mx.$ The possible values of $m$ lie in an interval of length $\frac ab,$ where $a$ and $b$ are relatively prime positive integers. What is $a+b?$
$\textbf{(A)} ~31 \qquad \textbf{(B)} ~47 \qquad \textbf{(C)} ~62\qquad \textbf{(D)} ~72 \qquad \textbf{(E)} ~85$ | It's easier to calculate the number of lattice points inside a rectangle with vertices $(0,0)$ $(p,0)$ $(p,q)$ $(0,q)$ . Those lattice points are divided by the diagonal $y = \frac{p}{q} \cdot x$ into $2$ halves. In this problem the number of lattice points on or below the diagonal and $x \ge 1$ is
$(p+1)(q+1)$ is the total number of lattice points inside the rectangle. Subtract the number of lattice points on the diagonal, divided by 2 is the number of lattice points below the diagonal, add the number of lattice points on the diagonal, and subtract the lattice points on the $x$ axis, then we get the total number of lattice points on or below the diagonal and $x \ge 1$
There are $900$ lattice points in total. $300$ is $\frac{1}{3}$ of $900$ . The $x$ coordinate of the top-right vertex of the rectangle is $30$ $\frac{1}{2} \cdot 30 \cdot 20 = 300$ . I guess the $y$ coordinate of the top-right vertex of the rectangle is $20$ . Now I am going to verify that. The slope of the diagonal is $\frac{20}{30} = \frac{2}{3}$ , there are $11$ lattice points on the diagonal. Substitute $(p,q)=(30, 20)$ $d=11$ to the above formula:
Because there are $11$ lattice points on line $y = \frac{2}{3}x$ , if $m < \frac{2}{3}$ , then the number of lattice points on or below the line is less than $300$ . So $m = \frac{2}{3}$ is the lower bound.
Now I am going to calculate the upper bound. From $\frac{b}{a} < \frac{b+1}{a+1}$
$\frac{2}{3} = \frac{18}{27} < \frac{19}{28}$
$\frac{2}{3} = \frac{20}{30} < \frac{21}{31} < \frac{19}{28}$
[asy] /* Created by Brendanb4321, modified by isabelchen */ import graph; size(18cm); defaultpen(fontsize(9pt)); xaxis(0,31,Ticks(1.0)); yaxis(0,22,Ticks(1.0)); draw((0,0)--(30,20)); draw((0,0)--(30,30*19/28), dotted); draw((0,0)--(31,31*21/31), dotted); for (int i = 1; i<=31; ++i) { for (int j = 1; j<=2/3*i+1; ++j) { dot((i,j)); } } dot((28,19), red); dot((31,21), blue); label("$m=2/3$", (33,20)); label("$m=21/31$", (33,21)); label("$m=19/28$", (33,22)); [/asy]
If $m = \frac{21}{31}$ , I will calculate by using the rectangle with blue vertex $(p,q) = (31, 21)$ , then subtract lattice points on line $x = 31$ , which is $21$ . There are 2 lattice points on the diagonal, $d=2$
If $m = \frac{19}{28}$ , I will calculate by using the rectangle with red vertex $(p,q) = (28, 19)$ , then add lattice points on line $x = 29$ and $x = 30$ , which is $19 + 20 = 39$ . There are 2 lattice points on the diagonal, $d=2$
When $m$ increases, more lattice points falls below the line $y = mx$ . Any value larger than $\frac{19}{28}$ has more than $301$ lattice points on or below $y = \frac{19}{28} x$ . So the upper bound is $\frac{19}{28}$
$\frac{19}{28}-\frac{2}{3}=\frac{1}{84}$ $\boxed{85}$ | E | 85 |
bd08ac7601d0fff5f9c27283846b0912 | https://artofproblemsolving.com/wiki/index.php/2021_AMC_12B_Problems/Problem_25 | Let $S$ be the set of lattice points in the coordinate plane, both of whose coordinates are integers between $1$ and $30,$ inclusive. Exactly $300$ points in $S$ lie on or below a line with equation $y=mx.$ The possible values of $m$ lie in an interval of length $\frac ab,$ where $a$ and $b$ are relatively prime positive integers. What is $a+b?$
$\textbf{(A)} ~31 \qquad \textbf{(B)} ~47 \qquad \textbf{(C)} ~62\qquad \textbf{(D)} ~72 \qquad \textbf{(E)} ~85$ | The lower bound of $m$ is $\frac23 = \frac{20}{30}$ . Inside the rectangle with vertices $(0,0)$ $(30,0)$ $(30,20)$ $(0, 20)$ and diagonal $y = \frac23 x$ , there are $(30-1)(20-1) = 551$ lattice points inside, not including the edges. There are $9$ lattice points on diagonal $y = \frac23 x$ inside the rectangle, $551 + 9 = 560$ . Half of the $560$ lattice points are below diagonal $y = \frac23 x$ $560 \cdot \frac12 = 280$ . There are $20$ lattice points on edge $x = 30$ $280 + 20 = 300$ . Once $m < \frac23$ , the $9$ lattice points on diagonal $y = \frac23 x$ will be above the new diagonal, making the number of lattice points on and below the diagonal less than $300$
Now we are going to calculate the upper bound by the following formula:
[asy] /* Created by isabelchen */ import graph; size(8cm); defaultpen(fontsize(9pt)); xaxis(0,8); yaxis(0,6); draw((0,0)--(7,5)); draw((7,0)--(7,5)); draw((0,5)--(7,5)); dot((0,0)); dot((1,0)); dot((2,0)); dot((3,0)); dot((4,0)); dot((5,0)); dot((6,0)); dot((7,0)); dot((8,0)); dot((0,1)); dot((1,1)); dot((2,1)); dot((3,1)); dot((4,1)); dot((5,1)); dot((6,1)); dot((7,1)); dot((8,1)); dot((0,2)); dot((1,2)); dot((2,2)); dot((3,2)); dot((4,2)); dot((5,2)); dot((6,2)); dot((7,2)); dot((8,2)); dot((0,3)); dot((1,3)); dot((2,3)); dot((3,3)); dot((4,3)); dot((5,3)); dot((6,3)); dot((7,3)); dot((8,3)); dot((0,4)); dot((1,4)); dot((2,4)); dot((3,4)); dot((4,4)); dot((5,4)); dot((6,4)); dot((7,4)); dot((8,4)); dot((0,5)); dot((1,5)); dot((2,5)); dot((3,5)); dot((4,5)); dot((5,5)); dot((6,5)); dot((7,5)); dot((8,5)); dot((0,6)); dot((1,6)); dot((2,6)); dot((3,6)); dot((4,6)); dot((5,6)); dot((6,6)); dot((7,6)); dot((8,6)); label("$(0,0)$", (0,0), SW); label("$(a, b)$", (7,5), NE); dot((7,0)); label("$a$", (7,0), S); dot((0,5)); label("$b$", (0,5), W); [/asy]
$\frac{2}{3} = \frac{20}{30} < \frac{21}{31} < \frac{19}{28}$
When $a = 31$ $b = 21$ $\frac{(31-1)(21-1)}{2} = 300$
When $a = 28$ $b = 19$ $\frac{(28-1)(19-1)}{2} = 243$ . Below the line $y = \frac{19}{28} x$ , there are $19$ lattice points on line $x = 28$ $19$ lattice points on line $x = 29$ $20$ lattice points on line $x = 30$ $243 + 19 + 19 + 20 = 301$
More lattice points fall below the line $y = mx$ as $m$ increases. There are more than $301$ lattice points on and below the line for any $m$ greater than $\frac{19}{28}$ . Therefore, the upper bound is $\frac{19}{28}$
$\frac{19}{28}-\frac{2}{3}=\frac{1}{84}$ , so $1+84=\boxed{85}$ | E | 85 |
e762e2028535b415ee6a806373e2837b | https://artofproblemsolving.com/wiki/index.php/2020_AMC_12A_Problems/Problem_1 | Carlos took $70\%$ of a whole pie. Maria took one third of the remainder. What portion of the whole pie was left?
$\textbf{(A)}\ 10\%\qquad\textbf{(B)}\ 15\%\qquad\textbf{(C)}\ 20\%\qquad\textbf{(D)}\ 30\%\qquad\textbf{(E)}\ 35\%$ | If Carlos took $70\%$ of the pie, there must be $(100 - 70)\% = 30\%$ left. After Maria takes $\frac{1}{3}$ of the remaining $30\%, \ 1 - \frac{1}{3} = \frac{2}{3}$ of the remaining $30\%$ is left.
Therefore, the answer is $30\% \cdot \frac{2}{3} = \boxed{20}.$ | C | 20 |
e762e2028535b415ee6a806373e2837b | https://artofproblemsolving.com/wiki/index.php/2020_AMC_12A_Problems/Problem_1 | Carlos took $70\%$ of a whole pie. Maria took one third of the remainder. What portion of the whole pie was left?
$\textbf{(A)}\ 10\%\qquad\textbf{(B)}\ 15\%\qquad\textbf{(C)}\ 20\%\qquad\textbf{(D)}\ 30\%\qquad\textbf{(E)}\ 35\%$ | Like solution 1, it is clear that there is $30\%$ of the pie remaining. Since Maria takes $\frac{1}{3}$ of the remainder, she takes $\frac{1}{3} \cdot 30\% = 10\%,$ meaning that there is $30\% - 10\% = \boxed{20}$ left. | C | 20 |
e762e2028535b415ee6a806373e2837b | https://artofproblemsolving.com/wiki/index.php/2020_AMC_12A_Problems/Problem_1 | Carlos took $70\%$ of a whole pie. Maria took one third of the remainder. What portion of the whole pie was left?
$\textbf{(A)}\ 10\%\qquad\textbf{(B)}\ 15\%\qquad\textbf{(C)}\ 20\%\qquad\textbf{(D)}\ 30\%\qquad\textbf{(E)}\ 35\%$ | We have \[\left(100\%-70\%\right)\cdot\left(1-\frac13\right)=30\%\cdot\frac23=\boxed{20}\] of the whole pie left. | C | 20 |
c0b72835a4e2b4946bf6e4a4ac6eb1e8 | https://artofproblemsolving.com/wiki/index.php/2020_AMC_12A_Problems/Problem_3 | A driver travels for $2$ hours at $60$ miles per hour, during which her car gets $30$ miles per gallon of gasoline. She is paid $$0.50$ per mile, and her only expense is gasoline at $$2.00$ per gallon. What is her net rate of pay, in dollars per hour, after this expense?
$\textbf{(A)}\ 20\qquad\textbf{(B)}\ 22\qquad\textbf{(C)}\ 24\qquad\textbf{(D)}\ 25\qquad\textbf{(E)}\ 26$ | Since the driver travels $60$ miles per hour and each hour she uses $2$ gallons of gasoline, she spends $$4$ per hour on gas. If she gets $$0.50$ per mile, then she gets $$30$ per hour of driving. Subtracting the gas cost, her net rate of money earned per hour is $\boxed{26}$ .
~mathsmiley | E | 26 |
c0b72835a4e2b4946bf6e4a4ac6eb1e8 | https://artofproblemsolving.com/wiki/index.php/2020_AMC_12A_Problems/Problem_3 | A driver travels for $2$ hours at $60$ miles per hour, during which her car gets $30$ miles per gallon of gasoline. She is paid $$0.50$ per mile, and her only expense is gasoline at $$2.00$ per gallon. What is her net rate of pay, in dollars per hour, after this expense?
$\textbf{(A)}\ 20\qquad\textbf{(B)}\ 22\qquad\textbf{(C)}\ 24\qquad\textbf{(D)}\ 25\qquad\textbf{(E)}\ 26$ | The driver is driving for $2$ hours at $60$ miles per hour, she drives $120$ miles. Therefore, she uses $\frac{120}{30}=4$ gallons of gasoline. So, total she has $$0.50\cdot120-$2.00\cdot4=$60-$8=$52$ . So, her rate is $\frac{52}{2}=\boxed{26}$ .
~sosiaops | E | 26 |
a2bc61907813361e77c1a7009a3fdc72 | https://artofproblemsolving.com/wiki/index.php/2020_AMC_12A_Problems/Problem_4 | How many $4$ -digit positive integers (that is, integers between $1000$ and $9999$ , inclusive) having only even digits are divisible by $5?$
$\textbf{(A) } 80 \qquad \textbf{(B) } 100 \qquad \textbf{(C) } 125 \qquad \textbf{(D) } 200 \qquad \textbf{(E) } 500$ | The units digit, for all numbers divisible by 5, must be either $0$ or $5$ . However, since all digits are even, the units digit must be $0$ . The middle two digits can be 0, 2, 4, 6, or 8, but the thousands digit can only be 2, 4, 6, or 8 since it cannot be zero. There is one choice for the units digit, 5 choices for each of the middle 2 digits, and 4 choices for the thousands digit, we have a total of $4\cdot5\cdot5\cdot1 = \boxed{100} \qquad$ numbers. | B | 100 |
5da039576715cc80d1f02f5e2cda05dc | https://artofproblemsolving.com/wiki/index.php/2020_AMC_12A_Problems/Problem_5 | The $25$ integers from $-10$ to $14,$ inclusive, can be arranged to form a $5$ -by- $5$ square in which the sum of the numbers in each row, the sum of the numbers in each column, and the sum of the numbers along each of the main diagonals are all the same. What is the value of this common sum?
$\textbf{(A) }2 \qquad\textbf{(B) } 5\qquad\textbf{(C) } 10\qquad\textbf{(D) } 25\qquad\textbf{(E) } 50$ | Without loss of generality, consider the five rows in the square. Each row must have the same sum of numbers, meaning that the sum of all the numbers in the square divided by $5$ is the total value per row. The sum of the $25$ integers is $-10+-9+...+14=11+12+13+14=50$ , and the common sum is $\frac{50}{5}=\boxed{10}$ | C | 10 |
5da039576715cc80d1f02f5e2cda05dc | https://artofproblemsolving.com/wiki/index.php/2020_AMC_12A_Problems/Problem_5 | The $25$ integers from $-10$ to $14,$ inclusive, can be arranged to form a $5$ -by- $5$ square in which the sum of the numbers in each row, the sum of the numbers in each column, and the sum of the numbers along each of the main diagonals are all the same. What is the value of this common sum?
$\textbf{(A) }2 \qquad\textbf{(B) } 5\qquad\textbf{(C) } 10\qquad\textbf{(D) } 25\qquad\textbf{(E) } 50$ | Take the sum of the middle 5 values of the set (they will turn out to be the mean of each row). We get $0 + 1 + 2 + 3 + 4 = \boxed{10}$ as our answer.
~Baolan | C | 10 |
5da039576715cc80d1f02f5e2cda05dc | https://artofproblemsolving.com/wiki/index.php/2020_AMC_12A_Problems/Problem_5 | The $25$ integers from $-10$ to $14,$ inclusive, can be arranged to form a $5$ -by- $5$ square in which the sum of the numbers in each row, the sum of the numbers in each column, and the sum of the numbers along each of the main diagonals are all the same. What is the value of this common sum?
$\textbf{(A) }2 \qquad\textbf{(B) } 5\qquad\textbf{(C) } 10\qquad\textbf{(D) } 25\qquad\textbf{(E) } 50$ | Taking the average of the first and last terms, $-10$ and $14$ , we have that the mean of the set is $2$ . There are 5 values in each row, column or diagonal, so the value of the common sum is $5\cdot2$ , or $\boxed{10}$ .
~Arctic_Bunny, edited by KINGLOGIC | C | 10 |
5da039576715cc80d1f02f5e2cda05dc | https://artofproblemsolving.com/wiki/index.php/2020_AMC_12A_Problems/Problem_5 | The $25$ integers from $-10$ to $14,$ inclusive, can be arranged to form a $5$ -by- $5$ square in which the sum of the numbers in each row, the sum of the numbers in each column, and the sum of the numbers along each of the main diagonals are all the same. What is the value of this common sum?
$\textbf{(A) }2 \qquad\textbf{(B) } 5\qquad\textbf{(C) } 10\qquad\textbf{(D) } 25\qquad\textbf{(E) } 50$ | Let us consider the horizontal rows. Since there are five of them, each with constant sum $x$ , we can add up the 25 numbers in 5 rows for a sum of $5x$ . Since the sum of the 25 numbers used is $-10-9-8-\cdots{}+12+13+14+15=11+12+13+14+15=50$ $5x=50$ and $x=\boxed{10}$ .
~cw357 | C | 10 |
5da039576715cc80d1f02f5e2cda05dc | https://artofproblemsolving.com/wiki/index.php/2020_AMC_12A_Problems/Problem_5 | The $25$ integers from $-10$ to $14,$ inclusive, can be arranged to form a $5$ -by- $5$ square in which the sum of the numbers in each row, the sum of the numbers in each column, and the sum of the numbers along each of the main diagonals are all the same. What is the value of this common sum?
$\textbf{(A) }2 \qquad\textbf{(B) } 5\qquad\textbf{(C) } 10\qquad\textbf{(D) } 25\qquad\textbf{(E) } 50$ | The mean of the set of numbers is $(14-10) \div 2 = 2$ . The numbers around it must be equal (i.e. if the mean of $1$ $2$ $3$ $4$ , and $5$ is $3$ , then $2+4=1+5$ .)
One row of the square would be \[\square \square 2 \square \square\]
Adding the numbers would be
\[0, 1, 2, 3, 4\]
with a sum of $\boxed{10}$ | C | 10 |
5da039576715cc80d1f02f5e2cda05dc | https://artofproblemsolving.com/wiki/index.php/2020_AMC_12A_Problems/Problem_5 | The $25$ integers from $-10$ to $14,$ inclusive, can be arranged to form a $5$ -by- $5$ square in which the sum of the numbers in each row, the sum of the numbers in each column, and the sum of the numbers along each of the main diagonals are all the same. What is the value of this common sum?
$\textbf{(A) }2 \qquad\textbf{(B) } 5\qquad\textbf{(C) } 10\qquad\textbf{(D) } 25\qquad\textbf{(E) } 50$ | If the sum of each row, column, and diagonal is x, then we have a total of 12x for the sum. The sum of the rows and columns is the sum of all the numbers doubled, which is $50\cdot2=100$ . Therefore $100+2x=12x$ $100=10x$ , and $x=\boxed{10}$ .
~MC413551 | null | 10 |
99163f150d5628b44fc0129a1ac54ba4 | https://artofproblemsolving.com/wiki/index.php/2020_AMC_12A_Problems/Problem_6 | In the plane figure shown below, $3$ of the unit squares have been shaded. What is the least number of additional unit squares that must be shaded so that the resulting figure has two lines of symmetry?
[asy] import olympiad; unitsize(25); filldraw((1,3)--(1,4)--(2,4)--(2,3)--cycle, gray(0.7)); filldraw((2,1)--(2,2)--(3,2)--(3,1)--cycle, gray(0.7)); filldraw((4,0)--(5,0)--(5,1)--(4,1)--cycle, gray(0.7)); for (int i = 0; i < 5; ++i) { for (int j = 0; j < 6; ++j) { pair A = (j,i); } } for (int i = 0; i < 5; ++i) { for (int j = 0; j < 6; ++j) { if (j != 5) { draw((j,i)--(j+1,i)); } if (i != 4) { draw((j,i)--(j,i+1)); } } } [/asy]
$\textbf{(A) } 4 \qquad \textbf{(B) } 5 \qquad \textbf{(C) } 6 \qquad \textbf{(D) } 7 \qquad \textbf{(E) } 8$ | The two lines of symmetry must be horizontally and vertically through the middle. We can then fill the boxes in like so:
[asy] import olympiad; unitsize(25); filldraw((1,3)--(1,4)--(2,4)--(2,3)--cycle, gray(0.7)); filldraw((0,3)--(0,4)--(1,4)--(1,3)--cycle, gray(0.9)); filldraw((3,3)--(3,4)--(4,4)--(4,3)--cycle, gray(0.9)); filldraw((4,3)--(4,4)--(5,4)--(5,3)--cycle, gray(0.9)); filldraw((2,2)--(2,3)--(3,3)--(3,2)--cycle, gray(0.9)); filldraw((2,1)--(2,2)--(3,2)--(3,1)--cycle, gray(0.7)); filldraw((3,0)--(4,0)--(4,1)--(3,1)--cycle, gray(0.9)); filldraw((1,0)--(2,0)--(2,1)--(1,1)--cycle, gray(0.9)); filldraw((4,0)--(5,0)--(5,1)--(4,1)--cycle, gray(0.7)); filldraw((0,0)--(1,0)--(1,1)--(0,1)--cycle, gray(0.9)); for (int i = 0; i < 5; ++i) { for (int j = 0; j < 6; ++j) { pair A = (j,i); } } for (int i = 0; i < 5; ++i) { for (int j = 0; j < 6; ++j) { if (j != 5) { draw((j,i)--(j+1,i)); } if (i != 4) { draw((j,i)--(j,i+1)); } } } [/asy]
where the light gray boxes are the ones we have filled. Counting these, we get $\boxed{7}$ total boxes. | D | 7 |
99163f150d5628b44fc0129a1ac54ba4 | https://artofproblemsolving.com/wiki/index.php/2020_AMC_12A_Problems/Problem_6 | In the plane figure shown below, $3$ of the unit squares have been shaded. What is the least number of additional unit squares that must be shaded so that the resulting figure has two lines of symmetry?
[asy] import olympiad; unitsize(25); filldraw((1,3)--(1,4)--(2,4)--(2,3)--cycle, gray(0.7)); filldraw((2,1)--(2,2)--(3,2)--(3,1)--cycle, gray(0.7)); filldraw((4,0)--(5,0)--(5,1)--(4,1)--cycle, gray(0.7)); for (int i = 0; i < 5; ++i) { for (int j = 0; j < 6; ++j) { pair A = (j,i); } } for (int i = 0; i < 5; ++i) { for (int j = 0; j < 6; ++j) { if (j != 5) { draw((j,i)--(j+1,i)); } if (i != 4) { draw((j,i)--(j,i+1)); } } } [/asy]
$\textbf{(A) } 4 \qquad \textbf{(B) } 5 \qquad \textbf{(C) } 6 \qquad \textbf{(D) } 7 \qquad \textbf{(E) } 8$ | We label the three shaded unit squares $A,B,$ and $C,$ then construct the two lines of symmetry of the resulting figure, as shown below: [asy] import olympiad; unitsize(25); filldraw((1,3)--(1,4)--(2,4)--(2,3)--cycle, gray(0.7)); filldraw((2,1)--(2,2)--(3,2)--(3,1)--cycle, gray(0.7)); filldraw((4,0)--(5,0)--(5,1)--(4,1)--cycle, gray(0.7)); for (int i = 0; i < 5; ++i) { for (int j = 0; j < 6; ++j) { pair A = (j,i); } } for (int i = 0; i < 5; ++i) { for (int j = 0; j < 6; ++j) { if (j != 5) { draw((j,i)--(j+1,i)); } if (i != 4) { draw((j,i)--(j,i+1)); } } } draw((-1,2)--(6,2),linewidth(2)+red); draw((2.5,-1)--(2.5,5),linewidth(2)+red); label("$A$",(1.5,3.5)); label("$B$",(2.5,1.5)); label("$C$",(4.5,0.5)); [/asy] Note that:
The shaded unit squares contributed by $A,B,$ and $C$ are all distinct, so we need to shade at least $4+4+2-3=\boxed{7} draw((-1,2)--(6,2),linewidth(2)+red); draw((2.5,-1)--(2.5,5),linewidth(2)+red); label("$A$",(1.5,3.5)); label("$B$",(2.5,1.5)); label("$C$",(4.5,0.5)); label("$A'$",(3.5,3.5)); label("$A'$",(1.5,0.5)); label("$A'$",(3.5,0.5)); label("$B'$",(2.5,2.5)); label("$C'$",(0.5,0.5)); label("$C'$",(0.5,3.5)); label("$C'$",(4.5,3.5)); [/asy] ~MRENTHUSIASM | D | 7 |
3f8199fa5a9e3f3675d5cb4ac05cad1d | https://artofproblemsolving.com/wiki/index.php/2020_AMC_12A_Problems/Problem_7 | Seven cubes, whose volumes are $1$ $8$ $27$ $64$ $125$ $216$ , and $343$ cubic units, are stacked vertically to form a tower in which the volumes of the cubes decrease from bottom to top. Except for the bottom cube, the bottom face of each cube lies completely on top of the cube below it. What is the total surface area of the tower (including the bottom) in square units?
$\textbf{(A)}\ 644\qquad\textbf{(B)}\ 658\qquad\textbf{(C)}\ 664\qquad\textbf{(D)}\ 720\qquad\textbf{(E)}\ 749$ | The volume of each cube follows the pattern of $n^3$ , for $n$ is between $1$ and $7$
We see that the total surface area can be comprised of three parts: the sides of the cubes, the tops of the cubes, and the bottom of the $7\times 7\times 7$ cube (which is just $7 \times 7 = 49$ ). The sides areas can be measured as the sum $4\sum_{n=1}^{7} n^2$ , giving us $560$ . Structurally, if we examine the tower from the top, we see that it really just forms a $7\times 7$ square of area $49$ . Therefore, we can say that the total surface area is $560 + 49 + 49 = \boxed{658}$ .
Alternatively, for the area of the tops, we could have found the sum $\sum_{n=2}^{7}((n)^{2}-(n-1)^{2})$ , giving us $49$ as well. | B | 658 |
3f8199fa5a9e3f3675d5cb4ac05cad1d | https://artofproblemsolving.com/wiki/index.php/2020_AMC_12A_Problems/Problem_7 | Seven cubes, whose volumes are $1$ $8$ $27$ $64$ $125$ $216$ , and $343$ cubic units, are stacked vertically to form a tower in which the volumes of the cubes decrease from bottom to top. Except for the bottom cube, the bottom face of each cube lies completely on top of the cube below it. What is the total surface area of the tower (including the bottom) in square units?
$\textbf{(A)}\ 644\qquad\textbf{(B)}\ 658\qquad\textbf{(C)}\ 664\qquad\textbf{(D)}\ 720\qquad\textbf{(E)}\ 749$ | It can quickly be seen that the side lengths of the cubes are the integers from 1 to 7 inclusive.
First, we will calculate the total surface area of the cubes, ignoring overlap. This value is $6 ( 1^2 + 2^2 + \cdots + 7^2 ) = 6\sum_{n=1}^{7} n^2 = 6 \left( \frac{7(7 + 1)(2 \cdot 7 + 1)}{6} \right) = 7 \cdot 8 \cdot 15 = 840$ . Then, we need to subtract out the overlapped parts of the cubes. Between each consecutive pair of cubes, one of the smaller cube's faces is completely covered, along with an equal area of one of the larger cube's faces. The total area of the overlapped parts of the cubes is thus equal to $2\sum_{n=1}^{6} n^2 = 182$ . Subtracting the overlapped surface area from the total surface area, we get $840 - 182 = \boxed{658}$ . ~ emerald_block | B | 658 |
3f8199fa5a9e3f3675d5cb4ac05cad1d | https://artofproblemsolving.com/wiki/index.php/2020_AMC_12A_Problems/Problem_7 | Seven cubes, whose volumes are $1$ $8$ $27$ $64$ $125$ $216$ , and $343$ cubic units, are stacked vertically to form a tower in which the volumes of the cubes decrease from bottom to top. Except for the bottom cube, the bottom face of each cube lies completely on top of the cube below it. What is the total surface area of the tower (including the bottom) in square units?
$\textbf{(A)}\ 644\qquad\textbf{(B)}\ 658\qquad\textbf{(C)}\ 664\qquad\textbf{(D)}\ 720\qquad\textbf{(E)}\ 749$ | It can be seen that the side lengths of the cubes using cube roots are all integers from $1$ to $7$ , inclusive.
Only the cubes with side length $1$ and $7$ have $5$ faces in the surface area and the rest have $4$ . Also, since the cubes are stacked, we have to find the difference between each $n^2$ and $(n-1)^2$ side length as $n$ ranges from $7$ to $2$
We then come up with this: $5(49)+13+4(36)+11+4(25)+9+4(16)+7+4(9)+5+4(4)+3+5(1)$
We then add all of this and get $\boxed{658}$ | B | 658 |
3f8199fa5a9e3f3675d5cb4ac05cad1d | https://artofproblemsolving.com/wiki/index.php/2020_AMC_12A_Problems/Problem_7 | Seven cubes, whose volumes are $1$ $8$ $27$ $64$ $125$ $216$ , and $343$ cubic units, are stacked vertically to form a tower in which the volumes of the cubes decrease from bottom to top. Except for the bottom cube, the bottom face of each cube lies completely on top of the cube below it. What is the total surface area of the tower (including the bottom) in square units?
$\textbf{(A)}\ 644\qquad\textbf{(B)}\ 658\qquad\textbf{(C)}\ 664\qquad\textbf{(D)}\ 720\qquad\textbf{(E)}\ 749$ | Notice that the surface area of the top cube is $6s^2$ and the others are $4s^2$ . Then we can directly compute. The edge length for the first cube is $7$ and has a surface area of $294$ . The surface area of the next cube is $144$ . The surface area of the next cube $100$ . The surface area of the next cube is $64$ . The surface area of the next cube is $36$ . The surface area of the next cube is $16$ . The surface area of the next cube is $4$ . We then sum up $294+144+100+64+36+16+4$ to get $\boxed{658}$ .
~smartatmath | B | 658 |
3f8199fa5a9e3f3675d5cb4ac05cad1d | https://artofproblemsolving.com/wiki/index.php/2020_AMC_12A_Problems/Problem_7 | Seven cubes, whose volumes are $1$ $8$ $27$ $64$ $125$ $216$ , and $343$ cubic units, are stacked vertically to form a tower in which the volumes of the cubes decrease from bottom to top. Except for the bottom cube, the bottom face of each cube lies completely on top of the cube below it. What is the total surface area of the tower (including the bottom) in square units?
$\textbf{(A)}\ 644\qquad\textbf{(B)}\ 658\qquad\textbf{(C)}\ 664\qquad\textbf{(D)}\ 720\qquad\textbf{(E)}\ 749$ | First of all, compute the area of the sides, excluding the top and bottoms, of the cubes. The side lengths (cube root the volumes) are 1, 2, 3, 4, 5, 6, 7.
Each cube's area of the sides can be calculated with $4($ area of one side $)$ $4(l^2)$ so in total that is $4(1+4+16+...+49)$ so $4(140)=560$ the area of all the sides of the cubes is $560$ .
Then, calculate the bottom face of the largest cube, $7*7=49$ .
Now, notice that if you stack the cubes up on top of each other, and look directly down on them, the tops of the cubes showing add up to the area of the bottom cube, the 7x7. Therefore, the sum of the area of the tops of the cubes is $7*7=49$
Now add them all up: $49+49+560=658.$ Therefore, the answer is $\boxed{658}$ | B | 658 |
b11d57aae2e15ddfb222adcf2313740f | https://artofproblemsolving.com/wiki/index.php/2020_AMC_12A_Problems/Problem_8 | What is the median of the following list of $4040$ numbers $?$ \[1, 2, 3, \ldots, 2020, 1^2, 2^2, 3^2, \ldots, 2020^2\] $\textbf{(A)}\ 1974.5\qquad\textbf{(B)}\ 1975.5\qquad\textbf{(C)}\ 1976.5\qquad\textbf{(D)}\ 1977.5\qquad\textbf{(E)}\ 1978.5$ | We can see that $44^2=1936$ which is less than 2020. Therefore, there are $2020-44=1976$ of the $4040$ numbers greater than $2020$ . Also, there are $2020+44=2064$ numbers that are less than or equal to $2020$
Since there are $44$ duplicates/extras, it will shift up our median's placement down $44$ . Had the list of numbers been $1,2,3, \dots, 4040$ , the median of the whole set would be $\dfrac{1+4040}{2}=2020.5$
Thus, our answer is $2020.5-44=\boxed{1976.5}$ | C | 1976.5 |
b11d57aae2e15ddfb222adcf2313740f | https://artofproblemsolving.com/wiki/index.php/2020_AMC_12A_Problems/Problem_8 | What is the median of the following list of $4040$ numbers $?$ \[1, 2, 3, \ldots, 2020, 1^2, 2^2, 3^2, \ldots, 2020^2\] $\textbf{(A)}\ 1974.5\qquad\textbf{(B)}\ 1975.5\qquad\textbf{(C)}\ 1976.5\qquad\textbf{(D)}\ 1977.5\qquad\textbf{(E)}\ 1978.5$ | As we are trying to find the median of a $4040$ -term set, we must find the average of the $2020$ th and $2021$ st terms.
Since $45^2 = 2025$ is slightly greater than $2020$ , we know that the $44$ perfect squares $1^2$ through $44^2$ are less than $2020$ , and the rest are greater. Thus, from the number $1$ to the number $2020$ , there are $2020 + 44 = 2064$ terms. Since $44^2$ is $44 + 45 = 89$ less than $45^2 = 2025$ and $84$ less than $2020$ , we will only need to consider the perfect square terms going down from the $2064$ th term, $2020$ , after going down $84$ terms. Since the $2020$ th and $2021$ st terms are only $44$ and $43$ terms away from the $2064$ th term, we can simply subtract $44$ from $2020$ and $43$ from $2020$ to get the two terms, which are $1976$ and $1977$ . Averaging the two, we get $\boxed{1976.5}.$ | C | 1976.5 |
b11d57aae2e15ddfb222adcf2313740f | https://artofproblemsolving.com/wiki/index.php/2020_AMC_12A_Problems/Problem_8 | What is the median of the following list of $4040$ numbers $?$ \[1, 2, 3, \ldots, 2020, 1^2, 2^2, 3^2, \ldots, 2020^2\] $\textbf{(A)}\ 1974.5\qquad\textbf{(B)}\ 1975.5\qquad\textbf{(C)}\ 1976.5\qquad\textbf{(D)}\ 1977.5\qquad\textbf{(E)}\ 1978.5$ | We want to know the $2020$ th term and the $2021$ st term to get the median.
We know that $44^2=1936$ . So, numbers $1^2, 2^2, \ldots,44^2$ are in between $1$ and $1936$
So, the sum of $44$ and $1936$ will result in $1980$ , which means that $1936$ is the $1980$ th number.
Also, notice that $45^2=2025$ , which is larger than $2021$
Then the $2020$ th term will be $1936+40 = 1976$ , and similarly the $2021$ th term will be $1977$
Solving for the median of the two numbers, we get $\boxed{1976.5}$ | C | 1976.5 |
b11d57aae2e15ddfb222adcf2313740f | https://artofproblemsolving.com/wiki/index.php/2020_AMC_12A_Problems/Problem_8 | What is the median of the following list of $4040$ numbers $?$ \[1, 2, 3, \ldots, 2020, 1^2, 2^2, 3^2, \ldots, 2020^2\] $\textbf{(A)}\ 1974.5\qquad\textbf{(B)}\ 1975.5\qquad\textbf{(C)}\ 1976.5\qquad\textbf{(D)}\ 1977.5\qquad\textbf{(E)}\ 1978.5$ | We note that $44^2 = 1936$ , which is the first square less than $2020$ , which means that there are $44$ additional terms before $2020$ . This makes $2020$ the $2064$ th term. To find the median, we need the $2020$ th and $2021$ st term. We note that every term before $2020$ is one less than the previous term (That is, we subtract $1$ to get the previous term.). If $2020$ is the $2064$ th term, than $2020 - 44$ is the $(2064 - 44)$ th term. So, the $2020$ th term is $1976$ , and the $2021$ st term is $1977$ , and the average of these two terms is the median, or $\boxed{1976.5}$ | C | 1976.5 |
b11d57aae2e15ddfb222adcf2313740f | https://artofproblemsolving.com/wiki/index.php/2020_AMC_12A_Problems/Problem_8 | What is the median of the following list of $4040$ numbers $?$ \[1, 2, 3, \ldots, 2020, 1^2, 2^2, 3^2, \ldots, 2020^2\] $\textbf{(A)}\ 1974.5\qquad\textbf{(B)}\ 1975.5\qquad\textbf{(C)}\ 1976.5\qquad\textbf{(D)}\ 1977.5\qquad\textbf{(E)}\ 1978.5$ | To find the median, we sort the $4040$ numbers in decreasing order, then average the $2020$ th and the $2021$ st numbers of the sorted list.
Since $45^2=2025$ and $44^2=1936,$ the first $2021$ numbers of the sorted list are \[\underbrace{2020^2,2019^2,2018^2,\ldots,46^2,45^2}_{1976\mathrm{ \ numbers}}\phantom{ },\phantom{ }\underbrace{2020,2019,2018,\ldots,1977,1976}_{45\mathrm{ \ numbers}}\phantom{ },\] from which the answer is $\frac{1977+1976}{2}=\boxed{1976.5}.$ | C | 1976.5 |
249babb1866b3889dc7f73643531d32b | https://artofproblemsolving.com/wiki/index.php/2020_AMC_12A_Problems/Problem_9 | How many solutions does the equation $\tan(2x)=\cos(\tfrac{x}{2})$ have on the interval $[0,2\pi]?$
$\textbf{(A)}\ 1\qquad\textbf{(B)}\ 2\qquad\textbf{(C)}\ 3\qquad\textbf{(D)}\ 4\qquad\textbf{(E)}\ 5$ | We count the intersections of the graphs of $y=\tan(2x)$ and $y=\cos\left(\frac x2\right):$
The graphs of $y=\tan(2x)$ and $y=\cos\left(\frac x2\right)$ intersect once on each of the five branches of $y=\tan(2x),$ as shown below: [asy] /* Made by MRENTHUSIASM */ size(800,200); real f(real x) { return tan(2*x); } real g(real x) { return cos(x/2); } draw(graph(f,0,atan(3)/2),red,"$y=\tan(2x)$"); draw(graph(f,-atan(3)/2+pi/2,atan(3)/2+pi/2),red); draw(graph(f,-atan(3)/2+2*pi/2,atan(3)/2+2*pi/2),red); draw(graph(f,-atan(3)/2+3*pi/2,atan(3)/2+3*pi/2),red); draw(graph(f,-atan(3)/2+4*pi/2,2*pi),red); draw(graph(g,0,2pi),blue,"$y=\cos\left(\frac x2\right)$"); real xMin = 0; real xMax = 9/4*pi; real yMin = -3; real yMax = 3; //Draws the horizontal gridlines void horizontalLines() { for (real i = yMin+1; i < yMax; ++i) { draw((xMin,i)--(xMax,i), mediumgray+linewidth(0.4)); } } //Draws the vertical gridlines void verticalLines() { for (real i = xMin+pi/2; i < xMax; i+=pi/2) { draw((i,yMin)--(i,yMax), mediumgray+linewidth(0.4)); } } //Draws the horizontal ticks void horizontalTicks() { for (real i = yMin+1; i < yMax; ++i) { draw((-1/8,i)--(1/8,i), black+linewidth(1)); } } //Draws the vertical ticks void verticalTicks() { for (real i = xMin+pi/2; i < xMax; i+=pi/2) { draw((i,-1/8)--(i,1/8), black+linewidth(1)); } } horizontalLines(); verticalLines(); horizontalTicks(); verticalTicks(); draw((xMin,0)--(xMax,0),black+linewidth(1.5),EndArrow(5)); draw((0,yMin)--(0,yMax),black+linewidth(1.5),EndArrow(5)); label("$x$",(xMax,0),(2,0)); label("$y$",(0,yMax),(0,2)); pair A[], B[]; A[0] = (2pi,0); A[1] = (0,2); A[2] = (0,0); A[3] = (0,-2); B[0] = intersectionpoints(graph(f,0,atan(3)/2),graph(g,0,2pi))[0]; B[1] = intersectionpoints(graph(f,-atan(3)/2+pi/2,atan(3)/2+pi/2),graph(g,0,2pi))[0]; B[2] = intersectionpoints(graph(f,-atan(3)/2+2*pi/2,atan(3)/2+2*pi/2),graph(g,0,2pi))[0]; B[3] = intersectionpoints(graph(f,-atan(3)/2+3*pi/2,atan(3)/2+3*pi/2),graph(g,0,2pi))[0]; B[4] = intersectionpoints(graph(f,-atan(3)/2+4*pi/2,atan(3)/2+4*pi/2),graph(g,0,2pi))[0]; label("$2\pi$",A[0],(0,-2.5)); label("$2$",A[1],(-2.5,0)); label("$0$",A[2],(-2.5,0)); label("$-2$",A[3],(-2.5,0)); for (int i = 0; i < 5; ++i) { dot(B[i],black+linewidth(5)); } add(legend(),point(E),60E,UnFill); [/asy] Therefore, the answer is $\boxed{5}.$ | E | 5 |
cbc5269a02159c993f317dae6cd82366 | https://artofproblemsolving.com/wiki/index.php/2020_AMC_12A_Problems/Problem_10 | There is a unique positive integer $n$ such that \[\log_2{(\log_{16}{n})} = \log_4{(\log_4{n})}.\] What is the sum of the digits of $n?$
$\textbf{(A) } 4 \qquad \textbf{(B) } 7 \qquad \textbf{(C) } 8 \qquad \textbf{(D) } 11 \qquad \textbf{(E) } 13$ | We can use the fact that $\log_{a^b} c = \frac{1}{b} \log_a c.$ This can be proved by using change of base formula to base $a.$
So, the original equation $\log_2{(\log_{2^4}{n})} = \log_{2^2}{(\log_{2^2}{n})}$ becomes \[\log_2\left({\frac{1}{4}\log_{2}{n}}\right) = \frac{1}{2}\log_2\left({\frac{1}{2}\log_2{n}}\right).\] Using log property of addition, we expand both sides and then simplify: \begin{align*} \log_2{\frac{1}{4}}+\log_2{(\log_{2}{n}}) &= \frac{1}{2}\left[\log_2{\frac{1}{2}} +\log_{2}{(\log_2{n})}\right] \\ \log_2{\frac{1}{4}}+\log_2{(\log_{2}{n}}) &= \frac{1}{2}\left[-1 +\log_{2}{(\log_2{n})}\right] \\ -2+\log_2{(\log_{2}{n}}) &= -\frac{1}{2}+ \frac{1}{2}(\log_{2}{(\log_2{n})}). \end{align*} Subtracting $\frac{1}{2}(\log_{2}{(\log_2{n})})$ from both sides and adding $2$ to both sides gives us \[\frac{1}{2}(\log_{2}{(\log_2{n})}) = \frac{3}{2}.\] Multiplying by $2,$ exponentiating, and simplifying gives us \begin{align*} \log_{2}{(\log_2{n})} &= 3 \\ \log_2{n}&=8 \\ n&=256. \end{align*} Adding the digits together, we have $2+5+6=\boxed{13}.$ | E | 13 |
cbc5269a02159c993f317dae6cd82366 | https://artofproblemsolving.com/wiki/index.php/2020_AMC_12A_Problems/Problem_10 | There is a unique positive integer $n$ such that \[\log_2{(\log_{16}{n})} = \log_4{(\log_4{n})}.\] What is the sum of the digits of $n?$
$\textbf{(A) } 4 \qquad \textbf{(B) } 7 \qquad \textbf{(C) } 8 \qquad \textbf{(D) } 11 \qquad \textbf{(E) } 13$ | We will apply the following logarithmic identity: \[\log_{p^k}{q^k}=\log_{p}{q},\] which can be proven by the Change of Base Formula \[\log_{p^k}{q^k}=\frac{\log_{p}{q^k}}{\log_{p}{p^k}}=\frac{k\log_{p}{q}}{k}=\log_{p}{q}.\] Note that $\log_{16}{n}\neq0,$ so we rewrite the original equation as follows: \begin{align*} \log_4{(\log_{16}{n})^2}&=\log_4{(\log_4{n})} \\ (\log_{16}{n})^2&=\log_4{n} \\ (\log_{16}{n})^2&=\log_{16}{n^2} \\ (\log_{16}{n})^2&=2\log_{16}{n} \\ \log_{16}{n}&=2, \end{align*} from which $n=16^2=256.$ The sum of its digits is $2+5+6=\boxed{13}.$ | E | 13 |
cbc5269a02159c993f317dae6cd82366 | https://artofproblemsolving.com/wiki/index.php/2020_AMC_12A_Problems/Problem_10 | There is a unique positive integer $n$ such that \[\log_2{(\log_{16}{n})} = \log_4{(\log_4{n})}.\] What is the sum of the digits of $n?$
$\textbf{(A) } 4 \qquad \textbf{(B) } 7 \qquad \textbf{(C) } 8 \qquad \textbf{(D) } 11 \qquad \textbf{(E) } 13$ | Using the change of base formula on the RHS of the initial equation yields \[\log_2{(\log_{16}{n})} = \frac{\log_2{(\log_4{n})}}{\log_2{4}}.\] This means we can multiply each side by $2$ for \[\log_2{(\log_{16}{n})^2} = \log_2{(\log_4{n})}.\] Canceling out the logs gives \[(\log_{16}{n})^2=\log_4{n}.\] We use change of base on the RHS to see that \begin{align*} (\log_{16}{n})^2&=\frac{ \log_{16}{n}}{\log_{16}{4}} \\ (\log_{16}{n})^2&=2 \log_{16}{n}. \end{align*} Substituting in $m = \log_{16}{n}$ gives $m^2=2m,$ so $m$ is either $0$ or $2.$ Since $m=0$ yields no solution for $n$ (since this would lead to use taking the log of $0$ ), we get $2 = \log_{16}{n},$ or $n=16^2=256,$ for the digit-sum of $2 + 5 + 6 = \boxed{13}.$ | E | 13 |
cbc5269a02159c993f317dae6cd82366 | https://artofproblemsolving.com/wiki/index.php/2020_AMC_12A_Problems/Problem_10 | There is a unique positive integer $n$ such that \[\log_2{(\log_{16}{n})} = \log_4{(\log_4{n})}.\] What is the sum of the digits of $n?$
$\textbf{(A) } 4 \qquad \textbf{(B) } 7 \qquad \textbf{(C) } 8 \qquad \textbf{(D) } 11 \qquad \textbf{(E) } 13$ | Suppose $\log_2(\log_{16}n)=k\implies\log_{16}n=2^k\implies n=16^{2^k}.$ Similarly, we have $\log_4(\log_4 n)=k\implies \log_4 n=4^k\implies n=4^{4^k}.$ Thus, we have \[16^{2^k}=(4^2)^{2^k}=4^{2^{k+1}}\] and \[4^{4^k}=4^{2^{2k}},\] so $k+1=2k\implies k=1.$ Plugging this in to either one of the expressions for $n$ gives $256$ , and the requested answer is $2+5+6=\boxed{13}.$ | E | 13 |
cbc5269a02159c993f317dae6cd82366 | https://artofproblemsolving.com/wiki/index.php/2020_AMC_12A_Problems/Problem_10 | There is a unique positive integer $n$ such that \[\log_2{(\log_{16}{n})} = \log_4{(\log_4{n})}.\] What is the sum of the digits of $n?$
$\textbf{(A) } 4 \qquad \textbf{(B) } 7 \qquad \textbf{(C) } 8 \qquad \textbf{(D) } 11 \qquad \textbf{(E) } 13$ | We know that, as the answer is an integer, $n$ must be some power of $16.$ Testing $16$ yields \begin{align*} \log_2{(\log_{16}{16})} &= \log_4{(\log_4{16})} \\ \log_2{1} &= \log_4{2} \\ 0 &= \frac{1}{2}, \end{align*} which does not work. We then try $256,$ giving us \begin{align*} \log_2{(\log_{16}{256})} &= \log_4{(\log_4{256})} \\ \log_2{2} &= \log_4{4} \\ 1 &= 1, \end{align*} which holds true. Thus, $n = 256,$ so the answer is $2 + 5 + 6 = \boxed{13}.$ | E | 13 |
16abafd4e82e6fff95db809cfdfa2481 | https://artofproblemsolving.com/wiki/index.php/2020_AMC_12A_Problems/Problem_11 | A frog sitting at the point $(1, 2)$ begins a sequence of jumps, where each jump is parallel to one of the coordinate axes and has length $1$ , and the direction of each jump (up, down, right, or left) is chosen independently at random. The sequence ends when the frog reaches a side of the square with vertices $(0,0), (0,4), (4,4),$ and $(4,0)$ . What is the probability that the sequence of jumps ends on a vertical side of the square?
$\textbf{(A)}\ \frac12\qquad\textbf{(B)}\ \frac 58\qquad\textbf{(C)}\ \frac 23\qquad\textbf{(D)}\ \frac34\qquad\textbf{(E)}\ \frac 78$ | Drawing out the square, it's easy to see that if the frog goes to the left, it will immediately hit a vertical end of the square. Therefore, the probability of this happening is $\frac{1}{4} \cdot 1 = \frac{1}{4}$ . If the frog goes to the right, it will be in the center of the square at $(2,2)$ , and by symmetry (since the frog is equidistant from all sides of the square), the chance it will hit a vertical side of a square is $\frac{1}{2}$ . The probability of this happening is $\frac{1}{4} \cdot \frac{1}{2} = \frac{1}{8}$
If the frog goes either up or down, it will hit a line of symmetry along the corner it is closest to and furthest to, and again, is equidistant relating to the two closer sides and also equidistant relating the two further sides. The probability for it to hit a vertical wall is $\frac{1}{2}$ . Because there's a $\frac{1}{2}$ chance of the frog going up or down, the total probability for this case is $\frac{1}{2} \cdot \frac{1}{2} = \frac{1}{4}$ and summing up all the cases, $\frac{1}{4} + \frac{1}{8} + \frac{1}{4} = \frac{5}{8} \implies \boxed{58}$ | B | 58 |
16abafd4e82e6fff95db809cfdfa2481 | https://artofproblemsolving.com/wiki/index.php/2020_AMC_12A_Problems/Problem_11 | A frog sitting at the point $(1, 2)$ begins a sequence of jumps, where each jump is parallel to one of the coordinate axes and has length $1$ , and the direction of each jump (up, down, right, or left) is chosen independently at random. The sequence ends when the frog reaches a side of the square with vertices $(0,0), (0,4), (4,4),$ and $(4,0)$ . What is the probability that the sequence of jumps ends on a vertical side of the square?
$\textbf{(A)}\ \frac12\qquad\textbf{(B)}\ \frac 58\qquad\textbf{(C)}\ \frac 23\qquad\textbf{(D)}\ \frac34\qquad\textbf{(E)}\ \frac 78$ | If the frog is on one of the 2 diagonals, the chance of landing on vertical or horizontal each becomes $\frac{1}{2}$ . Since it starts on $(1,2)$ , there is a $\frac{3}{4}$ chance (up, down, or right) it will reach a diagonal on the first jump and $\frac{1}{4}$ chance (left) it will reach the vertical side. The probablity of landing on a vertical is $\frac{1}{4}+\frac{3}{4} \cdot \frac{1}{2}=\boxed{58}$ .
- Lingjun | B | 58 |
16abafd4e82e6fff95db809cfdfa2481 | https://artofproblemsolving.com/wiki/index.php/2020_AMC_12A_Problems/Problem_11 | A frog sitting at the point $(1, 2)$ begins a sequence of jumps, where each jump is parallel to one of the coordinate axes and has length $1$ , and the direction of each jump (up, down, right, or left) is chosen independently at random. The sequence ends when the frog reaches a side of the square with vertices $(0,0), (0,4), (4,4),$ and $(4,0)$ . What is the probability that the sequence of jumps ends on a vertical side of the square?
$\textbf{(A)}\ \frac12\qquad\textbf{(B)}\ \frac 58\qquad\textbf{(C)}\ \frac 23\qquad\textbf{(D)}\ \frac34\qquad\textbf{(E)}\ \frac 78$ | Let $P_{(x,y)}$ denote the probability of the frog's sequence of jumps ends with it hitting a vertical edge when it is at $(x,y)$ . Note that $P_{(1,2)}=P_{(3,2)}$ by reflective symmetry over the line $x=2$ . Similarly, $P_{(1,1)}=P_{(1,3)}=P_{(3,1)}=P_{(3,3)}$ , and $P_{(2,1)}=P_{(2,3)}$ .
Now we create equations for the probabilities at each of these points/states by considering the probability of going either up, down, left, or right from that point: \[P_{(1,2)}=\frac{1}{4}+\frac{1}{2}P_{(1,1)}+\frac{1}{4}P_{(2,2)}\] \[P_{(2,2)}=\frac{1}{2}P_{(1,2)}+\frac{1}{2}P_{(2,1)}\] \[P_{(1,1)}=\frac{1}{4}+\frac{1}{4}P_{(1,2)}+\frac{1}{4}P_{(2,1)}\] \[P_{(2,1)}=\frac{1}{2}P_{(1,1)}+\frac{1}{4}P_{(2,2)}\] We have a system of $4$ equations in $4$ variables, so we can solve for each of these probabilities. Plugging the second equation into the fourth equation gives \[P_{(2,1)}=\frac{1}{2}P_{(1,1)}+\frac{1}{4}\left(\frac{1}{2}P_{(1,2)}+\frac{1}{2}P_{(2,1)}\right)\] \[P_{(2,1)}=\frac{8}{7}\left(\frac{1}{2}P_{(1,1)}+\frac{1}{8}P_{(1,2)}\right)=\frac{4}{7}P_{(1,1)}+\frac{1}{7}P_{(1,2)}\] Plugging in the third equation into this gives \[P_{(2,1)}=\frac{4}{7}\left(\frac{1}{4}+\frac{1}{4}P_{(1,2)}+\frac{1}{4}P_{(2,1)}\right)+\frac{1}{7}P_{(1,2)}\] \[P_{(2,1)}=\frac{7}{6}\left(\frac{1}{7}+\frac{2}{7}P_{(1,2)}\right)=\frac{1}{6}+\frac{1}{3}P_{(1,2)}\text{ (*)}\] Next, plugging in the second and third equation into the first equation yields \[P_{(1,2)}=\frac{1}{4}+\frac{1}{2}\left(\frac{1}{4}+\frac{1}{4}P_{(1,2)}+\frac{1}{4}P_{(2,1)}\right)+\frac{1}{4}\left(\frac{1}{2}P_{(1,2)}+\frac{1}{2}P_{(2,1)}\right)\] \[P_{(1,2)}=\frac{3}{8}+\frac{1}{4}P_{(1,2)}+\frac{1}{4}P_{(2,1)}\] Now plugging in (*) into this, we get \[P_{(1,2)}=\frac{3}{8}+\frac{1}{4}P_{(1,2)}+\frac{1}{4}\left(\frac{1}{6}+\frac{1}{3}P_{(1,2)}\right)\] \[P_{(1,2)}=\frac{3}{2}\cdot\frac{5}{12}=\boxed{58}\] -mathisawesome2169 | B | 58 |
16abafd4e82e6fff95db809cfdfa2481 | https://artofproblemsolving.com/wiki/index.php/2020_AMC_12A_Problems/Problem_11 | A frog sitting at the point $(1, 2)$ begins a sequence of jumps, where each jump is parallel to one of the coordinate axes and has length $1$ , and the direction of each jump (up, down, right, or left) is chosen independently at random. The sequence ends when the frog reaches a side of the square with vertices $(0,0), (0,4), (4,4),$ and $(4,0)$ . What is the probability that the sequence of jumps ends on a vertical side of the square?
$\textbf{(A)}\ \frac12\qquad\textbf{(B)}\ \frac 58\qquad\textbf{(C)}\ \frac 23\qquad\textbf{(D)}\ \frac34\qquad\textbf{(E)}\ \frac 78$ | this is basically another version of solution 4; shoutout to mathisawesome2169 :D
First, we note the different places the frog can go at certain locations in the square:
If the frog is at a border vertical point ( $(1,2),(3,2)$ ), it moves with probability $\frac{1}{4}$ to a vertical side of the square, probability $\frac{1}{4}$ to the center of the square, and probability $\frac{1}{2}$ to a corner square.
If the frog is at a border horizontal point ( $(2,1),(2,3)$ ), it moves with probability $\frac{1}{4}$ to a horizontal side of the square, probability $\frac{1}{4}$ to the center of the square, and probability $\frac{1}{2}$ to a corner square.
If the frog is at a center square ( $(2,2)$ ), it moves with probability $\frac{1}{2}$ to a border horizontal point and probability $\frac{1}{2}$ to a border vertical point.
If the frog is at a corner ( $(1,1),(1,3),(3,3),(3,1)$ ), it moves with probability $\frac{1}{4}$ to a vertical side of the square, probability $\frac{1}{4}$ to a horizontal side, probability $\frac{1}{4}$ to a border horizontal point, and probability $\frac{1}{4}$ to a border vertical point.
Now, let $x$ denote the probability of the frog reaching a vertical side when it is at a border vertical point. Similarly, let $y$ denote the probability of the frog reaching a vertical side when it is at a border horizontal point.
Now, the probability of the frog reaching a vertical side of the square at any location inside the square can be expressed in terms of $x$ and $y$
First, the two easier ones: $P_{center}=\frac{1}{2}x+\frac{1}{2}y$ , and $P_{corner}=\frac{1}{4}+\frac{1}{4}x+\frac{1}{4}y$ .
Now, we can write $x$ and $y$ in terms of $x$ and $y$ , allowing us to solve a system of two variables: \[x=\frac{1}{4}+\frac{1}{4}P_{center}+\frac{1}{2}P_{corner}=\frac{1}{4}+\frac{1}{4}\left(\frac{1}{2}x+\frac{1}{2}y\right)+\frac{1}{2}\left(\frac{1}{4}+\frac{1}{4}x+\frac{1}{4}y\right)\] and \[y=\frac{1}{4}P_{center}+\frac{1}{2}P_{corner}=\frac{1}{4}\left(\frac{1}{2}x+\frac{1}{2}y\right)+\frac{1}{2}\left(\frac{1}{4}+\frac{1}{4}x+\frac{1}{4}y\right).\] From these two equations, it is apparent that $y=x-\frac{1}{4}$ . We can then substitute this value for $y$ back into any of the two equations above to get \[x=\frac{1}{4}+\frac{1}{4}\left(\frac{1}{2}x+\frac{1}{2}\left(x-\frac{1}{4}\right)\right)+\frac{1}{2}\left(\frac{1}{4}+\frac{1}{4}x+\frac{1}{4}\left(x-\frac{1}{4}\right)\right).\] Although this certainly looks intimidating, we can expand the parentheses and multiply both sides by 16 to eliminate the fractions, which upon simplification yields the equation \[16x=5+8x,\] giving us the desired probability $x=\frac{5}{8}$ . The answer is then $\boxed{58}$ | B | 58 |
77f6170acd00cfd1eacfcdf4c8175d86 | https://artofproblemsolving.com/wiki/index.php/2020_AMC_12A_Problems/Problem_12 | Line $l$ in the coordinate plane has equation $3x-5y+40=0$ . This line is rotated $45^{\circ}$ counterclockwise about the point $(20,20)$ to obtain line $k$ . What is the $x$ -coordinate of the $x$ -intercept of line $k?$
$\textbf{(A) } 10 \qquad \textbf{(B) } 15 \qquad \textbf{(C) } 20 \qquad \textbf{(D) } 25 \qquad \textbf{(E) } 30$ | The slope of the line is $\frac{3}{5}$ . We must transform it by $45^{\circ}$
$45^{\circ}$ creates an isosceles right triangle, since the sum of the angles of the triangle must be $180^{\circ}$ and one angle is $90^{\circ}$ . This means the last leg angle must also be $45^{\circ}$
In the isosceles right triangle, the two legs are congruent. We can therefore construct an isosceles right triangle with a line of $\frac{3}{5}$ slope on graph paper. That line with $\frac{3}{5}$ slope starts at $(0,0)$ and will go to $(5,3)$ , the vector $<5, 3 >$
Construct another line from $(0,0)$ to $(3,-5)$ , the vector $<3,-5>$ . This is $\perp$ and equal to the original line segment. The difference between the two vectors is $<2,8>$ , which is the slope $4$ , and that is the slope of line $k$
Furthermore, the equation $3x-5y+40=0$ passes straight through $(20,20)$ since $3(20)-5(20)+40=60-100+40=0$ , which means that any rotations about $(20,20)$ would contain $(20,20)$ . We can create a line of slope $4$ through $(20,20)$ . The $x$ -intercept is therefore $20-\frac{20}{4} = \boxed{15.}$ ~lopkiloinm ~ShawnX (diagram) | B | 15. |
77f6170acd00cfd1eacfcdf4c8175d86 | https://artofproblemsolving.com/wiki/index.php/2020_AMC_12A_Problems/Problem_12 | Line $l$ in the coordinate plane has equation $3x-5y+40=0$ . This line is rotated $45^{\circ}$ counterclockwise about the point $(20,20)$ to obtain line $k$ . What is the $x$ -coordinate of the $x$ -intercept of line $k?$
$\textbf{(A) } 10 \qquad \textbf{(B) } 15 \qquad \textbf{(C) } 20 \qquad \textbf{(D) } 25 \qquad \textbf{(E) } 30$ | Since the slope of the line is $\frac{3}{5}$ , and the angle we are rotating around is x, then $\tan x = \frac{3}{5}$ $\tan(x+45^{\circ}) = \frac{\tan x + \tan(45^{\circ})}{1-\tan x*\tan(45^{\circ})} = \frac{0.6+1}{1-0.6} = \frac{1.6}{0.4} = 4$
Hence, the slope of the rotated line is $4$ . Since we know the line intersects the point $(20,20)$ , then we know the line is $y=4x-60$ . Set $y=0$ to find the x-intercept, and so $x=\boxed{15}$ | null | 15 |
77f6170acd00cfd1eacfcdf4c8175d86 | https://artofproblemsolving.com/wiki/index.php/2020_AMC_12A_Problems/Problem_12 | Line $l$ in the coordinate plane has equation $3x-5y+40=0$ . This line is rotated $45^{\circ}$ counterclockwise about the point $(20,20)$ to obtain line $k$ . What is the $x$ -coordinate of the $x$ -intercept of line $k?$
$\textbf{(A) } 10 \qquad \textbf{(B) } 15 \qquad \textbf{(C) } 20 \qquad \textbf{(D) } 25 \qquad \textbf{(E) } 30$ | [asy] draw((0,0)--(20, 0)--(20, 20)--(0, 20)--cycle); draw((20, 20)--(0, 8)); draw((15, 0)--(20, 20)); dot("$P$", (20, 20)); dot("$A$", (0, 8), dir(75)); dot("$B$", (15, 0), dir(45)); dot("$X$", (20, 0)); dot("$Y$", (0, 20), dir(50)); [/asy]
Let $P$ be $(20, 20)$ and $X, Y$ be $(20, 0)$ and $(0, 20)$ respectively. Since the slope of the line is $3/5$ we know that $\tan{\angle{YPA}} = 3/5.$ Segments $\overline{PA}$ and $\overline{PB}$ represent the before and after of rotating $l$ by 45 counterclockwise. Thus, $\angle{XPB} = 45 - \angle{YPA}$ and \[BX = 20 \tan{\angle{XPB}} = 20 \cdot \frac{1 - 3/5}{1 + 3/5} = 5\] by tangent addition formula. Since $BX$ is 5 and the sidelength of the square is 20 the answer is $20 - 5 \implies \boxed{15}.$ | B | 15 |
77f6170acd00cfd1eacfcdf4c8175d86 | https://artofproblemsolving.com/wiki/index.php/2020_AMC_12A_Problems/Problem_12 | Line $l$ in the coordinate plane has equation $3x-5y+40=0$ . This line is rotated $45^{\circ}$ counterclockwise about the point $(20,20)$ to obtain line $k$ . What is the $x$ -coordinate of the $x$ -intercept of line $k?$
$\textbf{(A) } 10 \qquad \textbf{(B) } 15 \qquad \textbf{(C) } 20 \qquad \textbf{(D) } 25 \qquad \textbf{(E) } 30$ | Using the protractor you brought, carefully graph the equation and rotate the given line $45^{\circ}$ counter-clockwise about the point $(20,20)$ . Scaling everything down by a factor of 5 makes this process easier.
It should then become fairly obvious that the x intercept is $x=\boxed{15}$ (only use this as a last resort). | null | 15 |
77f6170acd00cfd1eacfcdf4c8175d86 | https://artofproblemsolving.com/wiki/index.php/2020_AMC_12A_Problems/Problem_12 | Line $l$ in the coordinate plane has equation $3x-5y+40=0$ . This line is rotated $45^{\circ}$ counterclockwise about the point $(20,20)$ to obtain line $k$ . What is the $x$ -coordinate of the $x$ -intercept of line $k?$
$\textbf{(A) } 10 \qquad \textbf{(B) } 15 \qquad \textbf{(C) } 20 \qquad \textbf{(D) } 25 \qquad \textbf{(E) } 30$ | First note that the given line goes through $(20,20)$ with a slope of $\frac{3}{5}$ . This means that $(25,23)$ is on the line. Now consider translating the graph so that $(20,20)$ goes to the origin, then $(25,23)$ becomes $(5,3)$ . We now rotate the line $45^\circ$ about the origin using a rotation matrix. This maps $(5,3)$ to \[\begin{bmatrix} \frac{\sqrt{2}}{2} & -\frac{\sqrt{2}}{2} \\ \frac{\sqrt{2}}{2} & \frac{\sqrt{2}}{2}\end{bmatrix}\begin{bmatrix} 5 \\ 3\end{bmatrix}=\begin{bmatrix}\sqrt{2} \\ 4\sqrt{2}\end{bmatrix}\] The line through the origin and $(\sqrt{2}, 4\sqrt{2})$ has slope $4$ . Translating this line so that the origin is mapped to $(20,20)$ , we find that the equation for the new line is $4x-60$ , meaning that the $x$ -intercept is $x=\frac{60}{4}=\boxed{15}$ | B | 15 |
77f6170acd00cfd1eacfcdf4c8175d86 | https://artofproblemsolving.com/wiki/index.php/2020_AMC_12A_Problems/Problem_12 | Line $l$ in the coordinate plane has equation $3x-5y+40=0$ . This line is rotated $45^{\circ}$ counterclockwise about the point $(20,20)$ to obtain line $k$ . What is the $x$ -coordinate of the $x$ -intercept of line $k?$
$\textbf{(A) } 10 \qquad \textbf{(B) } 15 \qquad \textbf{(C) } 20 \qquad \textbf{(D) } 25 \qquad \textbf{(E) } 30$ | A quick check tells us that $(20,20)$ falls on the given line. Common sense tells us that if the slope of the original line is $1$ , or 45 degrees from the horizontal, a 45 counter clockwise rotation will result in a vertical line, and the x-intercept will be $(20,0)$ .
Thinking of a 45 degree counter clockwise rotation as a 90 degree counter clockwise rotation that is bisected will helps in visualizing this line of reasoning.
Therefore, it follows that if the original line is made steeper, then the x-intercept will move away from $(20,0)$ to the right. If the original line is made lower, then the opposite will happen. Our given line has slope $3/5$ , so the answer must be $A$ or $B$ $A$ can be eliminated because an x-intercept of $(10,0)$ can only occur when the original line is horizontal. In conclusion, the answer must be $\boxed{15}$ | B | 15 |
6b86a82b93d760ab07a001c66f7db4d7 | https://artofproblemsolving.com/wiki/index.php/2020_AMC_12A_Problems/Problem_13 | There are integers $a, b,$ and $c,$ each greater than $1,$ such that
\[\sqrt[a]{N\sqrt[b]{N\sqrt[c]{N}}} = \sqrt[36]{N^{25}}\]
for all $N \neq 1$ . What is $b$
$\textbf{(A) } 2 \qquad \textbf{(B) } 3 \qquad \textbf{(C) } 4 \qquad \textbf{(D) } 5 \qquad \textbf{(E) } 6$ | $\sqrt[a]{N\sqrt[b]{N\sqrt[c]{N}}}$ can be simplified to $N^{\frac{1}{a}+\frac{1}{ab}+\frac{1}{abc}}.$
The equation is then $N^{\frac{1}{a}+\frac{1}{ab}+\frac{1}{abc}}=N^{\frac{25}{36}}$ which implies that $\frac{1}{a}+\frac{1}{ab}+\frac{1}{abc}=\frac{25}{36}.$
$a$ has to be $2$ since $\frac{25}{36}>\frac{7}{12}$ $\frac{7}{12}$ is the result when $a, b,$ and $c$ are $3, 2,$ and $2$
$b$ being $3$ will make the fraction $\frac{2}{3}$ which is close to $\frac{25}{36}$
Finally, with $c$ being $6$ , the fraction becomes $\frac{25}{36}$ . In this case $a, b,$ and $c$ work, which means that $b$ must equal $\boxed{3.}$ ~lopkiloinm | B | 3. |
6b86a82b93d760ab07a001c66f7db4d7 | https://artofproblemsolving.com/wiki/index.php/2020_AMC_12A_Problems/Problem_13 | There are integers $a, b,$ and $c,$ each greater than $1,$ such that
\[\sqrt[a]{N\sqrt[b]{N\sqrt[c]{N}}} = \sqrt[36]{N^{25}}\]
for all $N \neq 1$ . What is $b$
$\textbf{(A) } 2 \qquad \textbf{(B) } 3 \qquad \textbf{(C) } 4 \qquad \textbf{(D) } 5 \qquad \textbf{(E) } 6$ | As above, notice that you get $\frac{1}{a}+\frac{1}{ab}+\frac{1}{abc}=\frac{25}{36}.$
Now, combine the fractions to get $\frac{bc+c+1}{abc}=\frac{25}{36}$
Let us assume $bc+c+1=25$ and $abc=36$ as this is the most convenient. (EDIT: This used to say WLOG but that is inaccurate)
From the first equation, we get $c(b+1)=24$ . Note also that from the second equation, $b$ and $c$ must be factors of 36.
After listing out the factors of 36 and utilising trial and error, we find that $c=6$ and $b=3$ works, with $a=2$ . So our answer is $\boxed{3.}$ | B | 3. |
6b86a82b93d760ab07a001c66f7db4d7 | https://artofproblemsolving.com/wiki/index.php/2020_AMC_12A_Problems/Problem_13 | There are integers $a, b,$ and $c,$ each greater than $1,$ such that
\[\sqrt[a]{N\sqrt[b]{N\sqrt[c]{N}}} = \sqrt[36]{N^{25}}\]
for all $N \neq 1$ . What is $b$
$\textbf{(A) } 2 \qquad \textbf{(B) } 3 \qquad \textbf{(C) } 4 \qquad \textbf{(D) } 5 \qquad \textbf{(E) } 6$ | Collapsed, $\sqrt[a]{N\sqrt[b]{N\sqrt[c]{N}}} = \sqrt[abc]{N^{bc+c+1}}$ . Comparing this to $\sqrt[36]{N^{25}}$ , observe that $bc+c+1=25$ and $abc=36$ . The first can be rewritten as $c(b+1)=24$ . Then, $b+1$ has to factor into 24 while 1 less than that also must factor into 36. The prime factorizations are as follows $36=2^2 3^2$ and $24=2^33$ . Then, $b=\boxed{3}$ , as only 4 and 3 factor into 36 and 24 while being 1 apart. | B | 3 |
3c2163a091c997a9b7db17061c31c823 | https://artofproblemsolving.com/wiki/index.php/2020_AMC_12A_Problems/Problem_16 | A point is chosen at random within the square in the coordinate plane whose vertices are $(0, 0), (2020, 0), (2020, 2020),$ and $(0, 2020)$ . The probability that the point is within $d$ units of a lattice point is $\tfrac{1}{2}$ . (A point $(x, y)$ is a lattice point if $x$ and $y$ are both integers.) What is $d$ to the nearest tenth $?$
$\textbf{(A) } 0.3 \qquad \textbf{(B) } 0.4 \qquad \textbf{(C) } 0.5 \qquad \textbf{(D) } 0.6 \qquad \textbf{(E) } 0.7$ | [asy] size(10cm); draw((0,0)--(1,0)--(1,1)--(0,1)--cycle); filldraw((arc((0,0), 0.3989, 0, 90))--(0,0)--cycle, gray); draw(arc((1,0), 0.3989, 90, 180)); filldraw((arc((1,0), 0.3989, 90, 180))--(1,0)--cycle, gray); draw(arc((1,1), 0.3989, 180, 270)); filldraw((arc((1,1), 0.3989, 180, 270))--(1,1)--cycle, gray); draw(arc((0,1), 0.3989, 270, 360)); filldraw(arc((0,1), 0.3989, 270, 360)--(0,1)--cycle, gray); [/asy]
The diagram represents each unit square of the given $2020 \times 2020$ square.
We consider an individual one-by-one block.
If we draw a quarter of a circle from each corner (where the lattice points are located), each with radius $d$ , the area covered by the circles should be $0.5$ . Because of this, and the fact that there are four circles, we write
\[4 \cdot \frac{1}{4} \cdot \pi d^2 = \frac{1}{2}\]
Solving for $d$ , we obtain $d = \frac{1}{\sqrt{2\pi}}$ , where with $\pi \approx 3$ , we get $d \approx \frac{1}{\sqrt{6}} \approx \dfrac{1}{2.5} = \dfrac{10}{25} = \dfrac{2}{5}$ , and from here, we see that $d \approx 0.4 \implies \boxed{0.4}.$ | B | 0.4 |
3c2163a091c997a9b7db17061c31c823 | https://artofproblemsolving.com/wiki/index.php/2020_AMC_12A_Problems/Problem_16 | A point is chosen at random within the square in the coordinate plane whose vertices are $(0, 0), (2020, 0), (2020, 2020),$ and $(0, 2020)$ . The probability that the point is within $d$ units of a lattice point is $\tfrac{1}{2}$ . (A point $(x, y)$ is a lattice point if $x$ and $y$ are both integers.) What is $d$ to the nearest tenth $?$
$\textbf{(A) } 0.3 \qquad \textbf{(B) } 0.4 \qquad \textbf{(C) } 0.5 \qquad \textbf{(D) } 0.6 \qquad \textbf{(E) } 0.7$ | As in the previous solution, we obtain the equation $4 \cdot \frac{1}{4} \cdot \pi d^2 = \frac{1}{2}$ , which simplifies to $\pi d^2 = \frac{1}{2} = 0.5$ . Since $\pi$ is slightly more than $3$ $d^2$ is slightly less than $\frac{0.5}{3} = 0.1\bar{6}$ . We notice that $0.1\bar{6}$ is slightly more than $0.4^2 = 0.16$ , so $d$ is roughly $\boxed{0.4}.$ emerald_block | B | 0.4 |
3c2163a091c997a9b7db17061c31c823 | https://artofproblemsolving.com/wiki/index.php/2020_AMC_12A_Problems/Problem_16 | A point is chosen at random within the square in the coordinate plane whose vertices are $(0, 0), (2020, 0), (2020, 2020),$ and $(0, 2020)$ . The probability that the point is within $d$ units of a lattice point is $\tfrac{1}{2}$ . (A point $(x, y)$ is a lattice point if $x$ and $y$ are both integers.) What is $d$ to the nearest tenth $?$
$\textbf{(A) } 0.3 \qquad \textbf{(B) } 0.4 \qquad \textbf{(C) } 0.5 \qquad \textbf{(D) } 0.6 \qquad \textbf{(E) } 0.7$ | As above, we find that we need to estimate $d = \frac{1}{\sqrt{2\pi}}$
Note that we can approximate $2\pi \approx 6.28318 \approx 6.25$ and so $\frac{1}{\sqrt{2\pi}}$ $\approx \frac{1}{\sqrt{6.25}}=\frac{1}{2.5}=0.4$
And so our answer is $\boxed{0.4}$ | B | 0.4 |
3c2163a091c997a9b7db17061c31c823 | https://artofproblemsolving.com/wiki/index.php/2020_AMC_12A_Problems/Problem_16 | A point is chosen at random within the square in the coordinate plane whose vertices are $(0, 0), (2020, 0), (2020, 2020),$ and $(0, 2020)$ . The probability that the point is within $d$ units of a lattice point is $\tfrac{1}{2}$ . (A point $(x, y)$ is a lattice point if $x$ and $y$ are both integers.) What is $d$ to the nearest tenth $?$
$\textbf{(A) } 0.3 \qquad \textbf{(B) } 0.4 \qquad \textbf{(C) } 0.5 \qquad \textbf{(D) } 0.6 \qquad \textbf{(E) } 0.7$ | We only need to figure out the probability for a unit square, as it will scale up to the $2020\times 2020$ square. Since we want to find the probability that a point inside a unit square that is $d$ units away from a lattice point (a corner of the square) is $\frac{1}{2}$ , we can find which answer will come the closest to covering $\frac{1}{2}$ of the area.
Since the closest is $0.4$ which turns out to be $(0.4)^2\times \pi = 0.16 \times \pi$ which is about $0.502$ , we find that the answer rounded to the nearest tenth is $0.4$ or $\boxed{0.4}$ | B | 0.4 |
3c2163a091c997a9b7db17061c31c823 | https://artofproblemsolving.com/wiki/index.php/2020_AMC_12A_Problems/Problem_16 | A point is chosen at random within the square in the coordinate plane whose vertices are $(0, 0), (2020, 0), (2020, 2020),$ and $(0, 2020)$ . The probability that the point is within $d$ units of a lattice point is $\tfrac{1}{2}$ . (A point $(x, y)$ is a lattice point if $x$ and $y$ are both integers.) What is $d$ to the nearest tenth $?$
$\textbf{(A) } 0.3 \qquad \textbf{(B) } 0.4 \qquad \textbf{(C) } 0.5 \qquad \textbf{(D) } 0.6 \qquad \textbf{(E) } 0.7$ | As per the above diagram, realize that $\pi d^2 = \frac{1}{2}$ , so $d = \frac{1}{(\sqrt{2})(\sqrt{\pi})}$
$\sqrt{2} \approx 1.4 = \frac{7}{5}$
$\sqrt{\pi}$ is between $1.7$ and $1.8$ $((1.7)^2 = 2.89$ and $(1.8)^2 = 3.24)$ , so we can say $\sqrt{\pi} \approx 1.75 = \frac{7}{4}$
So $d \approx \frac{1}{(\frac{7}{5})(\frac{7}{4})} = \frac{1}{\frac{49}{20}} = \frac{20}{49}$ . This is slightly above $\boxed{0.4}$ | B | 0.4 |
3c2163a091c997a9b7db17061c31c823 | https://artofproblemsolving.com/wiki/index.php/2020_AMC_12A_Problems/Problem_16 | A point is chosen at random within the square in the coordinate plane whose vertices are $(0, 0), (2020, 0), (2020, 2020),$ and $(0, 2020)$ . The probability that the point is within $d$ units of a lattice point is $\tfrac{1}{2}$ . (A point $(x, y)$ is a lattice point if $x$ and $y$ are both integers.) What is $d$ to the nearest tenth $?$
$\textbf{(A) } 0.3 \qquad \textbf{(B) } 0.4 \qquad \textbf{(C) } 0.5 \qquad \textbf{(D) } 0.6 \qquad \textbf{(E) } 0.7$ | As above, we have the equation $\pi d^2 = \frac{1}{2}$ , and we want to find the most accurate value of $d$ . We resort to the answer choices and can plug those values of $d$ in and see which value of $d$ will lead to the most accurate value of $\pi$
Starting off in the middle, we try option C with $d=0.5$ . Plugging this in, we get $\pi \left(\frac{1}{2}\right)^2 = \frac{1}{2},$ and after simplifying we get $\pi = \frac{1}{2} \cdot 4 = 2.$ That's not very good. We know $\pi \approx 3.14.$
Let's see if we can do better. Trying option A with $d = 0.3,$ we get $\pi = \frac{1}{2} \cdot \frac{100}{9} = \frac{50}{9} = 5 \frac{5}{9}.$
Hm, let's try option B with $d = 0.4.$ We get $\pi = \frac{1}{2} \cdot \frac{25}{4} = \frac{25}{8} = 3 \frac{1}{8}$ . This is very close to $\pi$ and is the best estimate for $\pi$ of the 5 options.
Therefore, the answer is $\boxed{0.4}.$ ~ epiconan | B | 0.4 |
78f4782b5211c5e69ca3ec6288d11d22 | https://artofproblemsolving.com/wiki/index.php/2020_AMC_12A_Problems/Problem_17 | The vertices of a quadrilateral lie on the graph of $y=\ln{x}$ , and the $x$ -coordinates of these vertices are consecutive positive integers. The area of the quadrilateral is $\ln{\frac{91}{90}}$ . What is the $x$ -coordinate of the leftmost vertex?
$\textbf{(A) } 6 \qquad \textbf{(B) } 7 \qquad \textbf{(C) } 10 \qquad \textbf{(D) } 12 \qquad \textbf{(E) } 13$ | Let the coordinates of the quadrilateral be $(n,\ln(n)),(n+1,\ln(n+1)),(n+2,\ln(n+2)),(n+3,\ln(n+3))$ . We have by shoelace's theorem, that the area is \begin{align*} &\frac{\ln(n)(n+1) + \ln(n+1)(n+2) + \ln(n+2)(n+3)+n\ln(n+3)}{2} - \frac{\ln(n+1)(n) + \ln(n+2)(n+1) + \ln(n+3)(n+2)+\ln(n)(n+3)}{2} \\ &=\frac{\ln \left( \frac{n^{n+1}(n+1)^{n+2}(n+2)^{n+3}(n+3)^n}{(n+1)^n(n+2)^{n+1}(n+3)^{n+2}n^{n+3}}\right)}{2} \\ &= \ln \left( \sqrt{\frac{(n+1)^2(n+2)^2}{n^2(n+3)^2}} \right) \\ &= \ln \left(\frac{(n+1)(n+2)}{n(n+3)}\right) \\ &= \ln \left( \frac{91}{90} \right). \end{align*} We know that the numerator must have a factor of $13$ , so given the answer choices, $n$ is either $12$ or $11$ . If $n=11$ , the expression $\frac{(n+1)(n+2)}{n(n+3)}$ does not evaluate to $\frac{91}{90}$ , but if $n=12$ , the expression evaluates to $\frac{91}{90}$ . Hence, our answer is $\boxed{12}$ | null | 12 |
78f4782b5211c5e69ca3ec6288d11d22 | https://artofproblemsolving.com/wiki/index.php/2020_AMC_12A_Problems/Problem_17 | The vertices of a quadrilateral lie on the graph of $y=\ln{x}$ , and the $x$ -coordinates of these vertices are consecutive positive integers. The area of the quadrilateral is $\ln{\frac{91}{90}}$ . What is the $x$ -coordinate of the leftmost vertex?
$\textbf{(A) } 6 \qquad \textbf{(B) } 7 \qquad \textbf{(C) } 10 \qquad \textbf{(D) } 12 \qquad \textbf{(E) } 13$ | Like above, use the shoelace formula to find that the area of the quadrilateral is equal to $\ln\frac{(n+1)(n+2)}{n(n+3)}$ . Because the final area we are looking for is $\ln\frac{91}{90}$ , the numerator factors into $13$ and $7$ , which one of $n+1$ and $n+2$ has to be a multiple of $13$ and the other has to be a multiple of $7$ . Clearly, the only choice for that is $\boxed{12}$ | null | 12 |
83cbe96e1f28e8817a16e0501ed91b88 | https://artofproblemsolving.com/wiki/index.php/2020_AMC_12A_Problems/Problem_18 | Quadrilateral $ABCD$ satisfies $\angle ABC = \angle ACD = 90^{\circ}, AC=20,$ and $CD=30.$ Diagonals $\overline{AC}$ and $\overline{BD}$ intersect at point $E,$ and $AE=5.$ What is the area of quadrilateral $ABCD?$
$\textbf{(A) } 330 \qquad \textbf{(B) } 340 \qquad \textbf{(C) } 350 \qquad \textbf{(D) } 360 \qquad \textbf{(E) } 370$ | [asy] size(15cm,0); import olympiad; draw((0,0)--(0,2)--(6,4)--(4,0)--cycle); label("A", (0,2), NW); label("B", (0,0), SW); label("C", (4,0), SE); label("D", (6,4), NE); label("E", (1.714,1.143), N); label("F", (1,1.5), N); draw((0,2)--(4,0), dashed); draw((0,0)--(6,4), dashed); draw((0,0)--(1,1.5), dashed); label("20", (0,2)--(4,0), SW); label("30", (4,0)--(6,4), SE); label("$x$", (1,1.5)--(1.714,1.143), NE); label("5$-$$x$", (1,1.5)--(0,2), NE); draw(rightanglemark((0,2),(0,0),(4,0))); draw(rightanglemark((0,2),(4,0),(6,4))); draw(rightanglemark((0,0),(1,1.5),(0,2))); [/asy]
It's crucial to draw a good diagram for this one. Since $AC=20$ and $CD=30$ , we get $[ACD]=300$ . Now we need to find $[ABC]$ to get the area of the whole quadrilateral. Drop an altitude from $B$ to $AC$ and call the point of intersection $F$ . Let $FE=x$ . Since $AE=5$ , then $AF=5-x$
By dropping this altitude, we can also see two similar triangles, $\triangle BFE \sim \triangle DCE$ . Since $EC$ is $20-5=15$ , and $DC=30$ , we get that $BF=2x$
Now, if we redraw another diagram just of $ABC$ , we get that $(2x)^2=(5-x)(15+x)$ because of the altitude geometric mean theorem which states that in any right triangle, the altitude squared is equal to the product of the two lengths that it divides the base into.
Expanding, simplifying, and dividing by the GCF, we get $x^2+2x-15=0$ . This factors to $(x+5)(x-3)$ , which has roots of $x=-5, 3$ . Since lengths cannot be negative, $x=3$ . Since $x=3$ , that means the altitude $BF=2\cdot3=6$ , or $[ABC]=60$ . Thus $[ABCD]=[ACD]+[ABC]=300+60=\boxed{360}$ | D | 360 |
83cbe96e1f28e8817a16e0501ed91b88 | https://artofproblemsolving.com/wiki/index.php/2020_AMC_12A_Problems/Problem_18 | Quadrilateral $ABCD$ satisfies $\angle ABC = \angle ACD = 90^{\circ}, AC=20,$ and $CD=30.$ Diagonals $\overline{AC}$ and $\overline{BD}$ intersect at point $E,$ and $AE=5.$ What is the area of quadrilateral $ABCD?$
$\textbf{(A) } 330 \qquad \textbf{(B) } 340 \qquad \textbf{(C) } 350 \qquad \textbf{(D) } 360 \qquad \textbf{(E) } 370$ | [asy] size(10cm,0); draw((10,30)--(10,0)--(-8,-6)--(-10,0)--(10,30)); draw((-20,0)--(20,0)); draw((0,-15)--(0,35)); draw((10,30)--(-8,-6)); draw(circle((0,0),10)); label("E",(-4.05,-.25),S); label("D",(10,30),NE); label("C",(10,0),NE); label("B",(-8,-6),SW); label("A",(-10,0),NW); label("5",(-10,0)--(-5,0), NE); label("15",(-5,0)--(10,0), N); label("30",(10,0)--(10,30), E); dot((-5,0)); dot((-10,0)); dot((-8,-6)); dot((10,0)); dot((10,30)); [/asy] Let the points be $A(-10,0)$ $\:B(x,y)$ $\:C(10,0)$ $\:D(10,30)$ ,and $\:E(-5,0)$ , respectively. Since $B$ lies on line $DE$ , we know that $y=2x+10$ . Furthermore, since $\angle{ABC}=90^\circ$ $B$ lies on the circle with diameter $AC$ , so $x^2+y^2=100$ . Solving for $x$ and $y$ with these equations, we get the solutions $(0,10)$ and $(-8,-6)$ . We immediately discard the $(0,10)$ solution as $y$ should be negative. Thus, we conclude that $[ABCD]=[ACD]+[ABC]=\frac{20\cdot30}{2}+\frac{20\cdot6}{2}=\boxed{360}$ | D | 360 |
83cbe96e1f28e8817a16e0501ed91b88 | https://artofproblemsolving.com/wiki/index.php/2020_AMC_12A_Problems/Problem_18 | Quadrilateral $ABCD$ satisfies $\angle ABC = \angle ACD = 90^{\circ}, AC=20,$ and $CD=30.$ Diagonals $\overline{AC}$ and $\overline{BD}$ intersect at point $E,$ and $AE=5.$ What is the area of quadrilateral $ABCD?$
$\textbf{(A) } 330 \qquad \textbf{(B) } 340 \qquad \textbf{(C) } 350 \qquad \textbf{(D) } 360 \qquad \textbf{(E) } 370$ | Let $\angle C = \angle{ACB}$ and $\angle{B} = \angle{CBE}.$ Using Law of Sines on $\triangle{BCE}$ we get \[\dfrac{BE}{\sin{C}} = \dfrac{CE}{\sin{B}} = \dfrac{15}{\sin{B}}\] and LoS on $\triangle{ABE}$ yields \[\dfrac{BE}{\sin{(90 - C)}} = \dfrac{5}{\sin{(90 - B)}} = \dfrac{BE}{\cos{C}} = \dfrac{5}{\cos{B}}.\] Divide the two to get $\tan{B} = 3 \tan{C}.$ Now, \[\tan{\angle{CED}} = 2 = \tan{\angle{B} + \angle{C}} = \dfrac{4 \tan{C}}{1 - 3\tan^2{C}}\] and solve the quadratic, taking the positive solution (C is acute) to get $\tan{C} = \frac{1}{3}.$ So if $AB = a,$ then $BC = 3a$ and $[ABC] = \frac{3a^2}{2}.$ By Pythagorean Theorem, $10a^2 = 400 \iff \frac{3a^2}{2} = 60$ and the answer is $300 + 60 \iff \boxed{360}.$ | D | 360 |
83cbe96e1f28e8817a16e0501ed91b88 | https://artofproblemsolving.com/wiki/index.php/2020_AMC_12A_Problems/Problem_18 | Quadrilateral $ABCD$ satisfies $\angle ABC = \angle ACD = 90^{\circ}, AC=20,$ and $CD=30.$ Diagonals $\overline{AC}$ and $\overline{BD}$ intersect at point $E,$ and $AE=5.$ What is the area of quadrilateral $ABCD?$
$\textbf{(A) } 330 \qquad \textbf{(B) } 340 \qquad \textbf{(C) } 350 \qquad \textbf{(D) } 360 \qquad \textbf{(E) } 370$ | [asy] import olympiad; pair A = (0, 189), B = (0,0), C = (570,0), D = (798, 798); dot("$A$", A, W); dot("$B$", B, S); dot("$C$", C, E); dot("$D$", D, N);dot("$E$",(140, 140), N); draw(A--B--C--D--A); draw(A--C, dotted); draw(B--D, dotted); [/asy]
Denote $EB$ as $x$ . By the Law of Cosines: \[AB^2 = 25 + x^2 - 10x\cos(\angle DEC)\] \[BC^2 = 225 + x^2 + 30x\cos(\angle DEC)\]
Adding these up yields: \[400 = 250 + 2x^2 + 20x\cos(\angle DEC) \Longrightarrow x^2 + \frac{10x}{\sqrt{5}} - 75 = 0\] By the quadratic formula, $x = 3\sqrt5$
Observe: \[[AEB] + [BEC] = \frac{1}{2}(x)(5)\sin(\angle DEC) + \frac{1}{2}(x)(15)\sin(180-\angle DEC) = (3)(20) = 60\]
Thus the desired area is $\frac{1}{2}(30)(20) + 60 = \boxed{360}$ | D | 360 |
83cbe96e1f28e8817a16e0501ed91b88 | https://artofproblemsolving.com/wiki/index.php/2020_AMC_12A_Problems/Problem_18 | Quadrilateral $ABCD$ satisfies $\angle ABC = \angle ACD = 90^{\circ}, AC=20,$ and $CD=30.$ Diagonals $\overline{AC}$ and $\overline{BD}$ intersect at point $E,$ and $AE=5.$ What is the area of quadrilateral $ABCD?$
$\textbf{(A) } 330 \qquad \textbf{(B) } 340 \qquad \textbf{(C) } 350 \qquad \textbf{(D) } 360 \qquad \textbf{(E) } 370$ | Let $C = (0, 0)$ and $D = (0, 30)$ . Then $E = (-15, 0), A = (-20, 0),$ and $B$ lies on the line $y=2x+30.$ So the coordinates of $B$ are \[(x, 2x+30).\]
We can make this a vector problem. $\overrightarrow{\mathbf{B}} = \begin{pmatrix} x \\ 2x+30 \end{pmatrix}.$ We notice that point $B$ forms a right angle, meaning vectors $\overrightarrow{\mathbf{BC}}$ and $\overrightarrow{\mathbf{BA}}$ are orthogonal, and their dot-product is $0$
We determine $\overrightarrow{\mathbf{BC}}$ and $\overrightarrow{\mathbf{BA}}$ to be $\begin{pmatrix} -x \\ -2x-30 \end{pmatrix}$ and $\begin{pmatrix} -20-x \\ -2x-30 \end{pmatrix}$ , respectively. (To get this, we use the fact that $\overrightarrow{\mathbf{BC}} = \overrightarrow{\mathbf{C}}-\overrightarrow{\mathbf{B}}$ and similarly, $\overrightarrow{\mathbf{BA}} = \overrightarrow{\mathbf{A}} - \overrightarrow{\mathbf{B}}.$
Equating the cross-product to $0$ gets us the quadratic $-x(-20-x)+(-2x-30)(-2x-30)=0.$ The solutions are $x=-18, -10.$ Since $B$ clearly has a more negative x-coordinate than $E$ , we take $x=-18$ . So $B = (-18, -6).$
From here, there are multiple ways to get the area of $\Delta{ABC}$ to be $60$ , and since the area of $\Delta{ACD}$ is $300$ , we get our final answer to be \[60 + 300 = \boxed{360}.\] | D | 360 |
83cbe96e1f28e8817a16e0501ed91b88 | https://artofproblemsolving.com/wiki/index.php/2020_AMC_12A_Problems/Problem_18 | Quadrilateral $ABCD$ satisfies $\angle ABC = \angle ACD = 90^{\circ}, AC=20,$ and $CD=30.$ Diagonals $\overline{AC}$ and $\overline{BD}$ intersect at point $E,$ and $AE=5.$ What is the area of quadrilateral $ABCD?$
$\textbf{(A) } 330 \qquad \textbf{(B) } 340 \qquad \textbf{(C) } 350 \qquad \textbf{(D) } 360 \qquad \textbf{(E) } 370$ | [asy] import olympiad; pair A = (0, 189), B = (0,0), C = (570,0), D = (798, 798),F=(285,94.5),G=(361.2,361.2); dot("$A$", A, W); dot("$B$", B, S); dot("$C$", C, E); dot("$D$", D, N);dot("$E$",(140, 140), N);dot("$F$",F,N);dot("$G$",G,N); draw(A--B--C--D--A); draw(A--C, dotted); draw(B--D, dotted); draw(F--G, dotted); [/asy]
Let $F$ be the midpoint of $AC$ , and draw $FG // CD$ where $G$ is on $BD$ . We have $EF=5,FC=10$
$\Delta EFG \sim \Delta ECD \implies FG=10=FA=FC$ . Therefore $ABCG$ is a cyclic quadrilateral.
Notice that $\angle EFG=90^\circ, EG=\sqrt{5^2+10^2}=5\sqrt{5} \implies BE=\frac{AE\cdot EC}{EG}=\frac{5\cdot 15}{5\sqrt{5}}=3\sqrt{5}$ via Power of a Point.
The altitude from $B$ to $AC$ is then equal to $GF\cdot \frac{BE}{GE}=10\cdot \frac{3\sqrt 5}{5 \sqrt 5}=6$
Finally, the total area of $ABCD$ is equal to $\frac 12 \cdot 20 \left(30+6 \right) =\boxed{360}.$ | D | 360 |
83cbe96e1f28e8817a16e0501ed91b88 | https://artofproblemsolving.com/wiki/index.php/2020_AMC_12A_Problems/Problem_18 | Quadrilateral $ABCD$ satisfies $\angle ABC = \angle ACD = 90^{\circ}, AC=20,$ and $CD=30.$ Diagonals $\overline{AC}$ and $\overline{BD}$ intersect at point $E,$ and $AE=5.$ What is the area of quadrilateral $ABCD?$
$\textbf{(A) } 330 \qquad \textbf{(B) } 340 \qquad \textbf{(C) } 350 \qquad \textbf{(D) } 360 \qquad \textbf{(E) } 370$ | [asy] size(15cm,0); import olympiad; draw((0,0)--(0,2)--(6,4)--(4,0)--cycle); label("A", (0,2), NW); label("B", (0,0), SW); label("C", (4,0), SE); label("D", (6,4), NE); label("E", (1.714,1.143), N); label("F", (1.714,0), SE); draw((0,2)--(4,0), dashed); draw((0,0)--(6,4), dashed); draw((1.714,1.143)--(1.714,0), dashed); label("20", (0,2)--(4,0), SW); label("30", (4,0)--(6,4), SE); label("$x$", (-0.3,2)--(-0.3,0), N); label("$y$", (0,-0.3)--(4,-0.3), E); draw(rightanglemark((1.714,2),(1.714,0),(5.714,0))); draw(rightanglemark((0,2),(0,0),(4,0))); draw(rightanglemark((0,2),(4,0),(6,4))); [/asy]
Let $AB = x$ $BC = y$
Looking at the diagram we have $x^2 + y^2 = 20^2$ $DE = \sqrt{30^2+15^2} = 15\sqrt{5}$ $[ACD] = \frac{1}{2} \cdot 20 \cdot 30 = 300$
Because $\triangle CEF \sim \triangle CAB$ $EF = AB \cdot \frac{CE}{CA} = x \cdot \frac{15}{20} = \frac{3x}{4}$
$BF = BC - CF = BC - BC \cdot \frac{CE}{CA} = \frac{1}{4} \cdot BC = \frac{y}{4}$
$BE = \sqrt{ \left( \frac{3x}{4} \right) ^2 + \left( \frac{y}{4} \right) ^2 } = \frac{ \sqrt{9x^2 + y^2} }{4}$ , substituting $x^2 + y^2 = 400$ , we get $BE = \frac{ \sqrt{8x^2 + 400} }{4} = \frac{ \sqrt{2x^2 + 100} }{2}$
$[ABC] = \frac{1}{2} \cdot x \cdot y$
Because $\triangle ABC$ and $\triangle ACD$ share the same base, $\frac{[ABC]}{[ACD]} = \frac{BE}{DE}$
$[ABC] = [ACD] \cdot \frac{BE}{DE} = 300 \cdot \frac{ \frac{ \sqrt{2x^2 + 100} } {2} }{ 15 \sqrt{5} }$
$\frac{1}{2} \cdot x \cdot y = 20 \cdot \frac{ \frac{ \sqrt{2x^2 + 100} } {2} }{ \sqrt{5} }$
$xy = 4 \sqrt{10x^2 + 500}$
By $x^2 + y^2 = 400$ $y = \sqrt{400 - x^2}$ . So, $x \cdot \sqrt{400 - x^2} = 4 \sqrt{10x^2 + 500}$
$x^2 (400 - x^2) = 16 (10x^2 + 500)$
Let $x^2 = a$ $a (400 - a) = 16 (10a + 500)$ $400a - a^2 = 160a + 8000$ $a^2 - 240a + 8000 = 0$ $(a-200)(a-40) = 0$
Because $x < 20$ $a$ can only equal 40. $a = 40$ $x = 2 \sqrt{10}$ $y = 6 \sqrt{10}$
$[ABC] = \frac{1}{2} \cdot 2 \sqrt{10} \cdot 6 \sqrt{10} = 60$
$[ABCD] = [ABC] + [ACD] = 60 + 300 = \boxed{360}$ | D | 360 |
83cbe96e1f28e8817a16e0501ed91b88 | https://artofproblemsolving.com/wiki/index.php/2020_AMC_12A_Problems/Problem_18 | Quadrilateral $ABCD$ satisfies $\angle ABC = \angle ACD = 90^{\circ}, AC=20,$ and $CD=30.$ Diagonals $\overline{AC}$ and $\overline{BD}$ intersect at point $E,$ and $AE=5.$ What is the area of quadrilateral $ABCD?$
$\textbf{(A) } 330 \qquad \textbf{(B) } 340 \qquad \textbf{(C) } 350 \qquad \textbf{(D) } 360 \qquad \textbf{(E) } 370$ | Drop perpendiculars $\overline{AF}$ and $\overline{CG}$ to $\overline{BD}.$ Notice that since $\angle AEF=\angle CEG$ (since they are vertical angles) and $\angle AFE=\angle CGE=90^\circ,$ triangles $AEF$ and $CEG$ are similar. Therefore, we have
\[x/EF=CE/AE=15/5=3,\]
where $EG=x.$ Therefore, $EF=x/3.$
Additionally, angle chasing shows that triangles $CEG$ and $DCG$ are also similar. This gives $CG/x=DC/CE=30/15=2,$ so $CG=2x.$ Thus, applying the Pythagorean Theorem to triangle $CEG$ gives
\[x^2+(2x)^2=15^2,\]
so $EG=x=3\sqrt 5.$ Our pairs of similar triangles then allow us to fill in the following lengths (in this order):
\[EF=x/3=\sqrt 5, CG=2x=6\sqrt 5, AF=CG/3=2\sqrt 5, DG=2\cdot CG=12\sqrt 5.\]
Now, let $BF=y.$ Angle chasing shows that triangle $ABF$ and $BCG$ are similar, so $BG/AF=CG/BF.$ Plugging in known lengths gives
\[\dfrac{y+4\sqrt 5}{2\sqrt 5}=\dfrac{6\sqrt 5}{y}.\]
This gives $y=2\sqrt 5.$ Now we know all the lengths that make up $BD,$ which allows us to find
\[BD=2\sqrt 5+\sqrt 5+3\sqrt 5+12\sqrt 5=18\sqrt 5.\]
Therefore,
\begin{align*} [ABCD] &= [ABD]+[CBD] \\ &= (BD)(AF)/2+(BD)(CG)/2 \\ &= (18\sqrt 5)(2\sqrt 5)/2+(18\sqrt 5)(6\sqrt 5)/2 \\ &= \boxed{360} | D | 360 |
d7d0d9aafbb7532003e17d4c00ba6178 | https://artofproblemsolving.com/wiki/index.php/2020_AMC_12A_Problems/Problem_19 | There exists a unique strictly increasing sequence of nonnegative integers $a_1 < a_2 < … < a_k$ such that \[\frac{2^{289}+1}{2^{17}+1} = 2^{a_1} + 2^{a_2} + … + 2^{a_k}.\] What is $k?$
$\textbf{(A) } 117 \qquad \textbf{(B) } 136 \qquad \textbf{(C) } 137 \qquad \textbf{(D) } 273 \qquad \textbf{(E) } 306$ | First, substitute $2^{17}$ with $x$ .
Then, the given equation becomes $\frac{x^{17}+1}{x+1}=x^{16}-x^{15}+x^{14}...-x^1+x^0$ by sum of powers factorization.
Now consider only $x^{16}-x^{15}$ . This equals $x^{15}(x-1)=x^{15} \cdot (2^{17}-1)$ .
Note that $2^{17}-1$ equals $2^{16}+2^{15}+...+1$ , by difference of powers factorization (or by considering the expansion of $2^{17}=2^{16}+2^{15}+...+2+2$ ).
Thus, we can see that $x^{16}-x^{15}$ forms the sum of 17 different powers of 2.
Applying the same method to each of $x^{14}-x^{13}$ $x^{12}-x^{11}$ , ... , $x^{2}-x^{1}$ , we can see that each of the pairs forms the sum of 17 different powers of 2. This gives us $17 \cdot 8=136$ .
But we must count also the $x^0$ term.
Thus, Our answer is $136+1=\boxed{137}$ | C | 137 |
d7d0d9aafbb7532003e17d4c00ba6178 | https://artofproblemsolving.com/wiki/index.php/2020_AMC_12A_Problems/Problem_19 | There exists a unique strictly increasing sequence of nonnegative integers $a_1 < a_2 < … < a_k$ such that \[\frac{2^{289}+1}{2^{17}+1} = 2^{a_1} + 2^{a_2} + … + 2^{a_k}.\] What is $k?$
$\textbf{(A) } 117 \qquad \textbf{(B) } 136 \qquad \textbf{(C) } 137 \qquad \textbf{(D) } 273 \qquad \textbf{(E) } 306$ | Multiply both sides by $2^{17}+1$ to get \[2^{289}+1=2^{a_1} + 2^{a_2} + … + 2^{a_k} + 2^{a_1+17} + 2^{a_2+17} + … + 2^{a_k+17}.\]
Notice that $a_1 = 0$ , since there is a $1$ on the LHS. However, now we have an extra term of $2^{18}$ on the right from $2^{a_1+17}$ . To cancel it, we let $a_2 = 18$ . The two $2^{18}$ 's now combine into a term of $2^{19}$ , so we let $a_3 = 19$ . And so on, until we get to $a_{18} = 34$ . Now everything we don't want telescopes into $2^{35}$ . We already have that term since we let $a_2 = 18 \implies a_2+17 = 35$ . Everything from now on will automatically telescope to $2^{52}$ . So we let $a_{19}$ be $52$
As you can see, we will have to add $17$ $a_n$ 's at a time, then "wait" for the sum to automatically telescope for the next $17$ numbers, etc, until we get to $2^{289}$ . We only need to add $a_n$ 's between odd multiples of $17$ and even multiples. The largest even multiple of $17$ below $289$ is $17\cdot16$ , so we will have to add a total of $17\cdot 8$ $a_n$ 's. However, we must not forget we let $a_1=0$ at the beginning, so our answer is $17\cdot8+1 = \boxed{137}$ | C | 137 |
d7d0d9aafbb7532003e17d4c00ba6178 | https://artofproblemsolving.com/wiki/index.php/2020_AMC_12A_Problems/Problem_19 | There exists a unique strictly increasing sequence of nonnegative integers $a_1 < a_2 < … < a_k$ such that \[\frac{2^{289}+1}{2^{17}+1} = 2^{a_1} + 2^{a_2} + … + 2^{a_k}.\] What is $k?$
$\textbf{(A) } 117 \qquad \textbf{(B) } 136 \qquad \textbf{(C) } 137 \qquad \textbf{(D) } 273 \qquad \textbf{(E) } 306$ | In order to shorten expressions, $\#$ will represent $16$ consecutive $0$ s when expressing numbers. Think of the problem in binary. We have $\frac{1\#0\#0\#0\#0\#0\#0\#0\#0\#0\#0\#0\#0\#0\#0\#0\#0\#1_2}{1\#1_2}$ Note that $(2^{17} + 1) (2^0 + 2^{34} + 2^{68} + \cdots + 2^{272}) = 2^0(2^{17} + 1) + 2^{34}(2^{17} + 1) + 2^{68}(2^{17} + 1) + \cdots 2^{272}(2^{17} + 1)$ $= 1\#1\#1\#1\#1\#1\#1\#1\#1\#1\#1\#1\#1\#1\#1\#1\#1\#1_2$ and $(2^{17} + 1) (2^{17} + 2^{51} + 2^{85} + \cdots + 2^{255}) = 2^{17}(2^{17} + 1) + 2^{51}(2^{17} + 1) + 2^{85}(2^{17} + 1) + \cdots 2^{255}(2^{17} + 1)$ $= 1\#1\#1\#1\#1\#1\#1\#1\#1\#1\#1\#1\#1\#1\#1\#1\#0_2$ Since $\phantom{=\ } 1\#1\#1\#1\#1\#1\#1\#1\#1\#1\#1\#1\#1\#1\#1\#1\#1\#1_2$ $-\ \phantom{1\#} 1\#1\#1\#1\#1\#1\#1\#1\#1\#1\#1\#1\#1\#1\#1\#1\#0_2$ $= 1\#0\#0\#0\#0\#0\#0\#0\#0\#0\#0\#0\#0\#0\#0\#0\#0\#1_2$ this means that $(2^{17} + 1) (2^0 + 2^{34} + 2^{68} + \cdots + 2^{272}) - (2^{17} + 1) (2^{17} + 2^{51} + 2^{85} + \cdots + 2^{255}) = 2^{289}$ so $\frac{2^{289}+1}{2^{17}+1} = (2^0 + 2^{34} + 2^{68} + \cdots + 2^{272}) - (2^{17} + 2^{51} + 2^{85} + \cdots + 2^{255})$ $= 2^0 + (2^{34} - 2^{17}) + (2^{68} - 2^{51}) + \cdots + (2^{272} - 2^{255})$ Expressing each of the pairs of the form $2^{n + 17} - 2^n$ in binary, we have $\phantom{=\ } 1000000000000000000 \cdots 0_2$ $-\ \phantom{10000000000000000} 10 \cdots 0_2$ $= \phantom{1} 111111111111111110 \cdots 0_2$ or $2^{n + 17} - 2^n = 2^{n + 16} + 2^{n + 15} + 2^{n + 14} + \cdots + 2^{n}$ This means that each pair has $17$ terms of the form $2^n$ Since there are $8$ of these pairs, there are a total of $8 \cdot 17 = 136$ terms. Accounting for the $2^0$ term, which was not in the pair, we have a total of $136 + 1 = \boxed{137}$ terms. ~ emerald_block | C | 137 |
d7d0d9aafbb7532003e17d4c00ba6178 | https://artofproblemsolving.com/wiki/index.php/2020_AMC_12A_Problems/Problem_19 | There exists a unique strictly increasing sequence of nonnegative integers $a_1 < a_2 < … < a_k$ such that \[\frac{2^{289}+1}{2^{17}+1} = 2^{a_1} + 2^{a_2} + … + 2^{a_k}.\] What is $k?$
$\textbf{(A) } 117 \qquad \textbf{(B) } 136 \qquad \textbf{(C) } 137 \qquad \textbf{(D) } 273 \qquad \textbf{(E) } 306$ | Notice that the only answer choices that are spaced one apart are $136$ and $137$ . It's likely that people will forget to include the final term so the answer is $\boxed{137}$ | null | 137 |
eabdfcd226357b0fa27430d8202a05be | https://artofproblemsolving.com/wiki/index.php/2020_AMC_12A_Problems/Problem_20 | Let $T$ be the triangle in the coordinate plane with vertices $(0,0), (4,0),$ and $(0,3).$ Consider the following five isometries (rigid transformations) of the plane: rotations of $90^{\circ}, 180^{\circ},$ and $270^{\circ}$ counterclockwise around the origin, reflection across the $x$ -axis, and reflection across the $y$ -axis. How many of the $125$ sequences of three of these transformations (not necessarily distinct) will return $T$ to its original position? (For example, a $180^{\circ}$ rotation, followed by a reflection across the $x$ -axis, followed by a reflection across the $y$ -axis will return $T$ to its original position, but a $90^{\circ}$ rotation, followed by a reflection across the $x$ -axis, followed by another reflection across the $x$ -axis will not return $T$ to its original position.)
$\textbf{(A) } 12 \qquad \textbf{(B) } 15 \qquad \textbf{(C) } 17 \qquad \textbf{(D) } 20 \qquad \textbf{(E) } 25$ | [asy] size(10cm); Label f; f.p=fontsize(6); xaxis(-6,6,Ticks(f, 2.0)); yaxis(-6,6,Ticks(f, 2.0)); filldraw(origin--(4,0)--(0,3)--cycle, gray, black+linewidth(1)); [/asy]
First, any combination of motions we can make must reflect $T$ an even number of times. This is because every time we reflect $T$ , it changes orientation. Once $T$ has been flipped once, no combination of rotations will put it back in place because it is the mirror image; however, flipping it again changes it back to the original orientation. Since we are only allowed $3$ transformations and an even number of them must be reflections, we either reflect $T$ $0$ times or $2$ times.
Case 1: $0$ reflections on $T$
In this case, we must use $3$ rotations to return $T$ to its original position. Notice that our set of rotations, $\{90^\circ,180^\circ,270^\circ\}$ , contains every multiple of $90^\circ$ except for $0^\circ$ . We can start with any two rotations $a,b$ in $\{90^\circ,180^\circ,270^\circ\}$ and there must be exactly one $c \equiv -a - b \pmod{360^\circ}$ such that we can use the three rotations $(a,b,c)$ which ensures that $a + b + c \equiv 0^\circ \pmod{360^\circ}$ . That way, the composition of rotations $a,b,c$ yields a full rotation. For example, if $a = b = 90^\circ$ , then $c \equiv -90^\circ - 90^\circ = -180^\circ \pmod{360^\circ}$ , so $c = 180^\circ$ and the rotations $(90^\circ,90^\circ,180^\circ)$ yields a full rotation.
The only case in which this fails is when $c$ would have to equal $0^\circ$ . This happens when $(a,b)$ is already a full rotation, namely, $(a,b) = (90^\circ,270^\circ),(180^\circ,180^\circ),$ or $(270^\circ,90^\circ)$ . However, we can simply subtract these three cases from the total. Selecting $(a,b)$ from $\{90^\circ,180^\circ,270^\circ\}$ yields $3 \cdot 3 = 9$ choices, and with $3$ that fail, we are left with $6$ combinations for case $1$
Case 2: $2$ reflections on $T$
In this case, we first eliminate the possibility of having two of the same reflection. Since two reflections across the x-axis maps $T$ back to itself, inserting a rotation before, between, or after these two reflections would change $T$ 's final location, meaning that any combination involving two reflections across the x-axis would not map $T$ back to itself. The same applies to two reflections across the y-axis.
Therefore, we must use one reflection about the x-axis, one reflection about the y-axis, and one rotation. Since a reflection about the x-axis changes the sign of the y component, a reflection about the y-axis changes the sign of the x component, and a $180^\circ$ rotation changes both signs, these three transformation composed (in any order) will suffice. It is therefore only a question of arranging the three, giving us $3! = 6$ combinations for case 2.
Combining both cases we get $6+6=\boxed{12}$ | A | 12 |
eabdfcd226357b0fa27430d8202a05be | https://artofproblemsolving.com/wiki/index.php/2020_AMC_12A_Problems/Problem_20 | Let $T$ be the triangle in the coordinate plane with vertices $(0,0), (4,0),$ and $(0,3).$ Consider the following five isometries (rigid transformations) of the plane: rotations of $90^{\circ}, 180^{\circ},$ and $270^{\circ}$ counterclockwise around the origin, reflection across the $x$ -axis, and reflection across the $y$ -axis. How many of the $125$ sequences of three of these transformations (not necessarily distinct) will return $T$ to its original position? (For example, a $180^{\circ}$ rotation, followed by a reflection across the $x$ -axis, followed by a reflection across the $y$ -axis will return $T$ to its original position, but a $90^{\circ}$ rotation, followed by a reflection across the $x$ -axis, followed by another reflection across the $x$ -axis will not return $T$ to its original position.)
$\textbf{(A) } 12 \qquad \textbf{(B) } 15 \qquad \textbf{(C) } 17 \qquad \textbf{(D) } 20 \qquad \textbf{(E) } 25$ | As in the previous solution, note that we must have either $0$ or $2$ reflections because of orientation since reflection changes orientation that is impossible to fix by rotation. We also know we can't have the same reflection twice, since that would give a net of no change and would require an identity rotation.
Suppose there are no reflections. Denote $90^{\circ}$ as $1$ $180^{\circ}$ as $2$ , and $270^{\circ}$ as $3$ , just for simplification purposes. We want a combination of $3$ of these that will sum to either $4$ or $8$ $0$ and $12$ are impossible since the minimum is $3$ and the max is $9$ ). $4$ can be achieved with any permutation of $(1-1-2)$ and $8$ can be achieved with any permutation of $(2-3-3)$ . This case can be done in $3+3=6$ ways.
Suppose there are two reflections. As noted already, they must be different, and as a result will take the triangle to the opposite side of the origin if we don't do any rotation. We have $1$ rotation left that we can do though, and the only one that will return to the original position is $2$ , which is $180^{\circ}$ AKA reflection across origin. Therefore, since all $3$ transformations are distinct. The three transformations can be applied anywhere since they are commutative (think quadrants). This gives $6$ ways.
$6+6=\boxed{12}$ | A | 12 |
eabdfcd226357b0fa27430d8202a05be | https://artofproblemsolving.com/wiki/index.php/2020_AMC_12A_Problems/Problem_20 | Let $T$ be the triangle in the coordinate plane with vertices $(0,0), (4,0),$ and $(0,3).$ Consider the following five isometries (rigid transformations) of the plane: rotations of $90^{\circ}, 180^{\circ},$ and $270^{\circ}$ counterclockwise around the origin, reflection across the $x$ -axis, and reflection across the $y$ -axis. How many of the $125$ sequences of three of these transformations (not necessarily distinct) will return $T$ to its original position? (For example, a $180^{\circ}$ rotation, followed by a reflection across the $x$ -axis, followed by a reflection across the $y$ -axis will return $T$ to its original position, but a $90^{\circ}$ rotation, followed by a reflection across the $x$ -axis, followed by another reflection across the $x$ -axis will not return $T$ to its original position.)
$\textbf{(A) } 12 \qquad \textbf{(B) } 15 \qquad \textbf{(C) } 17 \qquad \textbf{(D) } 20 \qquad \textbf{(E) } 25$ | Define $s$ as a reflection, and $r$ as a $90^{\circ}$ counterclockwise rotation. Thus, $r^4=s^2=e$ , and the five transformations can be represented as ${r, r^2, r^3, r^2s, s}$ , and $rs=sr^{-1}$
Now either $s$ doesn't appear at all or appears twice. For the former case, it's easy to see that only $r, r, r^2$ and $r^2, r^3, r^3$ will work. Both can be permuted in $3$ ways, giving $6$ ways in total.
For the latter case, note that $s$ can't appear twice, neither does $r^2s$ , else we need to get $e$ from ${r, r^2, r^3}$ , which is not possible. So $r^2s$ and $s$ must appear once each. The last transformation must be $r^2$ . A quick check shows that ${r^2, r^2s, s}$ is permutable, since $r^2s=sr^{-2}=sr^2$ (since $r^4=e$ ). This gives $6$ ways.
Thus the answer is $\boxed{12}$ | A | 12 |
c4df67fe04d32bea6ea998202b10fb02 | https://artofproblemsolving.com/wiki/index.php/2020_AMC_12A_Problems/Problem_21 | How many positive integers $n$ are there such that $n$ is a multiple of $5$ , and the least common multiple of $5!$ and $n$ equals $5$ times the greatest common divisor of $10!$ and $n?$
$\textbf{(A) } 12 \qquad \textbf{(B) } 24 \qquad \textbf{(C) } 36 \qquad \textbf{(D) } 48 \qquad \textbf{(E) } 72$ | We set up the following equation as the problem states:
\[\text{lcm}{(5!, n)} = 5\text{gcd}{(10!, n)}.\]
Breaking each number into its prime factorization, we see that the equation becomes
\[\text{lcm}{(2^3\cdot 3 \cdot 5, n)} = 5\text{gcd}{(2^8\cdot 3^4 \cdot 5^2 \cdot 7, n)}.\]
We can now determine the prime factorization of $n$ . We know that its prime factors belong to the set $\{2, 3, 5, 7\}$ , as no factor of $10!$ has $11$ in its prime factorization, nor anything greater. Next, we must find exactly how many different possibilities exist for each.
There can be anywhere between $3$ and $8$ $2$ 's and $1$ to $4$ $3$ 's. However, since $n$ is a multiple of $5$ , and we multiply the $\text{gcd}$ by $5$ , there can only be $3$ $5$ 's in $n$ 's prime factorization. Finally, there can either $0$ or $1$ $7$ 's.
Thus, we can multiply the total possibilities of $n$ 's factorization to determine the number of integers $n$ which satisfy the equation, giving us $6 \times 4 \times 1 \times 2 = \boxed{48}$ . ~ciceronii | D | 48 |
c4df67fe04d32bea6ea998202b10fb02 | https://artofproblemsolving.com/wiki/index.php/2020_AMC_12A_Problems/Problem_21 | How many positive integers $n$ are there such that $n$ is a multiple of $5$ , and the least common multiple of $5!$ and $n$ equals $5$ times the greatest common divisor of $10!$ and $n?$
$\textbf{(A) } 12 \qquad \textbf{(B) } 24 \qquad \textbf{(C) } 36 \qquad \textbf{(D) } 48 \qquad \textbf{(E) } 72$ | Like the Solution 1, we start from the equation:
\[\text{lcm}{(5!, n)} = 5\text{gcd}{(10!, n)}.\] Assume $\text{lcm}{(5!, n)}=k\cdot5!$ , with some integer $k$ . It follows that $k\cdot 4!=\text{gcd}{(10!, n)}$ . It means that $n$ has a divisor $4!$ . Since $n$ is a multiple of $5$ $n$ has a divisor $5!$ . Thus, $\text{lcm}{(5!, n)}=n=k\cdot5!$ . The equation can be changed as \[k\cdot5!=5\text{gcd}{(10!, k\cdot5!)}\] \[k=5\text{gcd}{(6\cdot7\cdot8\cdot9\cdot10, k)}\] We can see that $k$ is also a multiple of $5$ , with a form of $5\cdot m$ . Substituting it in the above equation, we have \[m=5\text{gcd}{(6\cdot7\cdot8\cdot9\cdot2, m)}\] Similarly, $m$ is a multiple of $5$ , with a form of $5\cdot s$ . We have \[s=\text{gcd}{(6\cdot7\cdot8\cdot9\cdot2, 5\cdot s)}=\text{gcd}{(2^5\cdot3^3\cdot7, s)}\] The equation holds, if $s$ is a divisor of $2^5\cdot3^3\cdot7$ , which has $(5+1)(3+1)(1+1)=\boxed{48}$ divisors. ~Linty Huang | D | 48 |
c4df67fe04d32bea6ea998202b10fb02 | https://artofproblemsolving.com/wiki/index.php/2020_AMC_12A_Problems/Problem_21 | How many positive integers $n$ are there such that $n$ is a multiple of $5$ , and the least common multiple of $5!$ and $n$ equals $5$ times the greatest common divisor of $10!$ and $n?$
$\textbf{(A) } 12 \qquad \textbf{(B) } 24 \qquad \textbf{(C) } 36 \qquad \textbf{(D) } 48 \qquad \textbf{(E) } 72$ | As in the previous solutions, we start with
\[\text{lcm}(5!,n) = 5\text{gcd}(10!,n)\]
From this we have that $\text{lcm}(5!,n) \,|\, 5\text{gcd}(10!,n)$ , and in particular, $n \,|\, 5\text{gcd}(10!,n)$ . However, $\text{gcd}(10!,n)\, |\, n$ , so we must have $\text{gcd}(10!,n) = n$ or $\text{gcd}(10!,n) = n/5$ . If $\text{gcd}(10!,n) = n$ , then we have $\text{lcm}(5!,n) = 5n$ ; because $5\, |\, 5!$ , this implies that 5 does not divide $n$ , so we must have $\text{gcd}(10!,n) = n/5$
Now we have $\text{lcm}(5!,n) = n$ , implying that $5!\, |\, n$ , and $n/5\, |\, 10!$ . Writing out prime factorizations, this gives us
\[2^3 \cdot 3 \cdot 5 \,|\, n\] \[n \,|\, 2^8 \cdot 3^4 \cdot 5^2 \cdot 7\]
So $n$ can have 3, 4, 5, 6, 7, or 8 factors of 2; 1, 2, 3, or 4 factors of two; and 0 or 1 factors of 7. Note that $\text{gcd}(2^8 \cdot 3^4 \cdot 5^2 \cdot 7,n) = n/5$ implies that $n$ has 2 factors of 5. Thus, there are $6 \cdot 4 \cdot 2 = 48$ possible choices for $n$ , and our answer is $\boxed{48}$ | D | 48 |
15449518d325df015d8e440769c7ba8a | https://artofproblemsolving.com/wiki/index.php/2020_AMC_12A_Problems/Problem_24 | Suppose that $\triangle{ABC}$ is an equilateral triangle of side length $s$ , with the property that there is a unique point $P$ inside the triangle such that $AP=1$ $BP=\sqrt{3}$ , and $CP=2$ . What is $s$
$\textbf{(A) } 1+\sqrt{2} \qquad \textbf{(B) } \sqrt{7} \qquad \textbf{(C) } \frac{8}{3} \qquad \textbf{(D) } \sqrt{5+\sqrt{5}} \qquad \textbf{(E) } 2\sqrt{2}$ | We begin by rotating $\triangle{ APB}$ counterclockwise by $60^{\circ}$ about $A$ , such that $P\mapsto Q$ and $B\mapsto C$ . We see that $\triangle{ APQ}$ is equilateral with side length $1$ , meaning that $\angle APQ = 60^{\circ}$ . We also see that $\triangle{CPQ}$ is a $30$ $60$ $90$ right triangle, meaning that $\angle CPQ= 60^{\circ}$ . Thus, by adding the two together, we see that $\angle APC = 120^{\circ}$ [asy] size(200); pen p = fontsize(10pt)+gray+0.5; pen q = fontsize(13pt); pair A,B,C,D,P,Q; real s=sqrt(7); B=origin; A=s*dir(60); C=s*right; P=IP(CR(A,1),CR(C,2)); Q=rotate(60,A)*P; draw(A--B--C--A, black+0.8); draw(A--P--B^^P--C^^A--Q--C, p); draw(P--Q, p+dashed); //draw(A--A+C--C, p); label("$A$", A, up, q); label("$B$", B, 0.5*(B-P), q); label("$C$", C, 0.5*(C-P), q); label("$P$", P, dir(180), q); label("$Q$", Q, 0.25*(Q-B), q); label("$\sqrt{3}$",B--P, right, p); label("$2$",C--P, 2*left, p); label("$1$",A--P, 1.5**dir(-10), p); label("$1$", A--Q, dir(250), p); label("$1$",P--Q, down, p); label("$\sqrt{3}$",C--Q, right, p); [/asy] We can now use the law of cosines as following: \begin{align*} s^2 &= (AP)^2 + (CP)^2 - 2\cdot AP\cdot CP\cdot \cos{\angle{APC}} \\ &= 1 + 4 - 2\cdot 1\cdot 2\cdot \cos{120^{\circ}} \\ &= 5 - 4\left(-\frac{1}{2}\right) \\ &= 7, \end{align*} giving us that $s = \boxed{7}$ | B | 7 |
15449518d325df015d8e440769c7ba8a | https://artofproblemsolving.com/wiki/index.php/2020_AMC_12A_Problems/Problem_24 | Suppose that $\triangle{ABC}$ is an equilateral triangle of side length $s$ , with the property that there is a unique point $P$ inside the triangle such that $AP=1$ $BP=\sqrt{3}$ , and $CP=2$ . What is $s$
$\textbf{(A) } 1+\sqrt{2} \qquad \textbf{(B) } \sqrt{7} \qquad \textbf{(C) } \frac{8}{3} \qquad \textbf{(D) } \sqrt{5+\sqrt{5}} \qquad \textbf{(E) } 2\sqrt{2}$ | Rotate $\triangle CPA$ counterclockwise $60^\circ$ around point $C$ to $\triangle CQB$ . Then $CP=CQ, \angle PCQ=60^\circ$ , so $\triangle CPQ$ is an equilateral triangle. [asy] size(200); pen p = fontsize(10pt)+gray+0.4; pen q = fontsize(13pt); pair A,B,C,D,P,Q; real s=sqrt(7); A=origin; B=s*right; C=s*dir(60); P=IP(CR(A,1),CR(C,2)); Q=rotate(60,C)*P; draw(A--B--C--A, black+0.8); draw(A--P--B^^P--C, p); draw(B--Q--C^^P--Q, p+dashed); label("$A$", A, A-P, q); label("$B$", B, B-P, q); label("$C$", C, up, q); label("$P$", P, dir(140), q); label("$Q$", Q, 0.25*(Q-P), q); label("$\sqrt{3}$",P--B, dir(60), p); label("$2$",C--P, 2*left, p); label("$1$",A--P, up, p); label("$1$", B--Q, dir(-30), p); label("$2$",P--Q, down, p); label("$2$",C--Q, dir(45), p); [/asy] Note that $\triangle PQB$ is a $30^\circ$ $60^\circ$ $90^\circ$ triangle, hence $\angle BPQ=30^\circ$ , and $\angle BPC=90^\circ$ , so \[BC^2=PC^2+PB^2=2^2+3=7,\] and the answer is $\boxed{7}$ | B | 7 |
15449518d325df015d8e440769c7ba8a | https://artofproblemsolving.com/wiki/index.php/2020_AMC_12A_Problems/Problem_24 | Suppose that $\triangle{ABC}$ is an equilateral triangle of side length $s$ , with the property that there is a unique point $P$ inside the triangle such that $AP=1$ $BP=\sqrt{3}$ , and $CP=2$ . What is $s$
$\textbf{(A) } 1+\sqrt{2} \qquad \textbf{(B) } \sqrt{7} \qquad \textbf{(C) } \frac{8}{3} \qquad \textbf{(D) } \sqrt{5+\sqrt{5}} \qquad \textbf{(E) } 2\sqrt{2}$ | Rotate $\triangle BPC$ counterclockwise by $60^\circ$ around point $B$ to $\triangle BQA$ . Then $BP=BQ$ , and $\angle PBQ=60^\circ$ , so $\triangle BPQ$ is an equilateral triangle. [asy] size(200); pen p = fontsize(10pt)+gray+0.4; pen q = fontsize(13pt); pair A,B,C,D,P,Q,R,X,Y,Z; real s=sqrt(7); C=origin; A=s*right; B=s*dir(60); P=IP(CR(A,1),CR(C,2)); Q=rotate(60,B)*P; draw(A--B--C--A, black+0.8); draw(A--P--B^^P--C, p); draw(B--Q--A^^P--Q, p+dashed); label("$A$", A, A-P, q); label("$B$", B, B-P, q); label("$C$", C, 0.6*(C-P), q); label("$P$", P, dir(250), q); label("$Q$", Q, 0.25*(Q-P), q); label("$\sqrt{3}$",B--Q, dir(60), p); label("$\sqrt{3}$",P--B, dir(210), p); label("$\sqrt{3}$",P--Q, dir(135), p); label("$2$",P--C, dir(120), p); label("$2$",A--Q, dir(-20), p); label("$1$",A--P, dir(210), p); [/asy] Note that $\triangle QAP$ is a $30^\circ$ $60^\circ$ $90^\circ$ triangle, hence $\angle PQA=30^\circ$ , and $\angle BQA=90^\circ$ , so \[AB^2=PQ^2+AQ^2=2^2+3=7,\] and the answer is $\boxed{7}$ | B | 7 |
15449518d325df015d8e440769c7ba8a | https://artofproblemsolving.com/wiki/index.php/2020_AMC_12A_Problems/Problem_24 | Suppose that $\triangle{ABC}$ is an equilateral triangle of side length $s$ , with the property that there is a unique point $P$ inside the triangle such that $AP=1$ $BP=\sqrt{3}$ , and $CP=2$ . What is $s$
$\textbf{(A) } 1+\sqrt{2} \qquad \textbf{(B) } \sqrt{7} \qquad \textbf{(C) } \frac{8}{3} \qquad \textbf{(D) } \sqrt{5+\sqrt{5}} \qquad \textbf{(E) } 2\sqrt{2}$ | [asy] unitsize(1inch); pen p = fontsize(10pt); dot((0.756,0.655)); dot((1.512,1.309)); dot((1.701,0.327)); pair A = origin, B = (1.323,2.291), C = (2.646,0), P = (0.756,0.655), Q = (1.512,1.309), R = (1.701,0.327); draw((0,0)--(1.323,2.291)--(2.646,0)--cycle); label("$A$", (0,0), SW, p); label("$C$", (2.646,0), SE, p); label("$B$", (1.323,2.291), N, p); label("$P'$", (0.756,0.655), NW, p); label("$1$", (2.174,0.164), N, p); label("$1$", (1.228,0.491), N, p); D(A--P, red); D(B--Q, red); D(C--R, red); D(P--Q--R--cycle, blue); D(B--P, magenta); [/asy]
Suppose that triangle $ABC$ had three segments of length $2$ , emanating from each of its vertices, making equal angles with each of its sides, and going into its interior. Suppose each of these segments intersected the segment clockwise to it precisely at its other endpoint and inside $ABC$ (as pictured in the diagram above). Clearly $s > 2$ and the triangle defined by these intersection points will be equilateral (pictured by the blue segments).
Take this equilateral triangle to have side length $1$ . The portions of each segment outside this triangle (in red) have length $1$ . Take $P'$ to be the intersection of the segments emanating from $A$ and $C$ . By Law of Cosines, \[BP' = \sqrt{1 + 1 - 2\cos{120^\circ}} = \sqrt{3}.\] So, $P'$ actually satisfies the conditions of the problem, and we can obtain again by Law of Cosines \[s = \sqrt{4 + 1 - 4\cos{120^\circ}} = \boxed{7}.\] | B | 7 |
15449518d325df015d8e440769c7ba8a | https://artofproblemsolving.com/wiki/index.php/2020_AMC_12A_Problems/Problem_24 | Suppose that $\triangle{ABC}$ is an equilateral triangle of side length $s$ , with the property that there is a unique point $P$ inside the triangle such that $AP=1$ $BP=\sqrt{3}$ , and $CP=2$ . What is $s$
$\textbf{(A) } 1+\sqrt{2} \qquad \textbf{(B) } \sqrt{7} \qquad \textbf{(C) } \frac{8}{3} \qquad \textbf{(D) } \sqrt{5+\sqrt{5}} \qquad \textbf{(E) } 2\sqrt{2}$ | [asy] unitsize(0.4inch); pen p = fontsize(10pt); draw((0,0)--(4,5.65)--(8,0)--cycle); label("$A$", (4,5.65), N, p); label("$C$", (8,0), SE, p); label("$B$", (0,0), SW, p); label("$P$", (3.5,3.5), E, p); label("$E$", (2.8191,3.982), NW, p); label("$F$", (4.848,4.452), NE, p); label("$G$", (3.5,0), down, p); draw((0,0)--(3.5,3.5)); label("$\sqrt{3}$",(0,0)--(3.5,3.5), SE,p); draw((8,0)--(3.5,3.5)); label("$2$",(8,0)--(3.5,3.5), SW,p); draw((4,5.65)--(3.5,3.5)); label("$1$",(4,5.65)--(3.5,3.5), E,p); draw((3.5,3.5)--(2.8191,3.982)); draw((3.5,3.5)--(4.848,4.452)); draw((3.5,3.5)--(3.5,0)); [/asy]
We begin by dropping altitudes from point $P$ down to all three sides of the triangle as shown above. We can therefore make equations regarding the areas of triangles $\triangle{APC}$ $\triangle{APB}$ , and $\triangle{BPC}$ . Let $s$ be the side of the equilateral triangle, we use the Heron's formula:
\[\triangle{APC} = \frac{s\cdot PF}{2} = \sqrt{\frac{s+3}{2}\left(\frac{s+3}{2}-s\right)\left(\frac{s+3}{2}-1\right)\left(\frac{s+3}{2}-2\right)}\] \[\implies PF = \frac{\sqrt{10s^2-s^4-9}}{2s}\]
Similarly, we obtain:
\[PE = \frac{\sqrt{8s^2-s^4-4}}{2s}\] \[PG = \frac{\sqrt{14s^2-s^4-1}}{2s}\]
By Viviani's theorem, \[\frac{\sqrt{10s^2-s^4-9}}{2s}+\frac{\sqrt{8s^2-s^4-4}}{2s}+\frac{\sqrt{14s^2-s^4-1}}{2s} = \frac{\sqrt{3}}{2}s\] \[\sqrt{10s^2-s^4-9}+\sqrt{8s^2-s^4-4}+\sqrt{14s^2-s^4-1} = \sqrt{3}s^2\]
Note that from now on, the algebra will get extremely ugly and almost impossible to do by hand within the time frame. However, we do see that it's extremely easy to check the answer choices with the equation in this form. Testing $s = \sqrt{7}$ , We obtain $7\sqrt{3}$ on both sides, revealing that our answer is in fact $\boxed{7}$ | B | 7 |
15449518d325df015d8e440769c7ba8a | https://artofproblemsolving.com/wiki/index.php/2020_AMC_12A_Problems/Problem_24 | Suppose that $\triangle{ABC}$ is an equilateral triangle of side length $s$ , with the property that there is a unique point $P$ inside the triangle such that $AP=1$ $BP=\sqrt{3}$ , and $CP=2$ . What is $s$
$\textbf{(A) } 1+\sqrt{2} \qquad \textbf{(B) } \sqrt{7} \qquad \textbf{(C) } \frac{8}{3} \qquad \textbf{(D) } \sqrt{5+\sqrt{5}} \qquad \textbf{(E) } 2\sqrt{2}$ | Instead of directly finding the side length of the equilateral triangle, we instead find the area and use it to find the side length.
Begin by reflecting $P$ over each of the sides. Label these reflected points $P', P'', P'''$ . Connect these points to the vertices of the equilateral triangle, as well as to each other.
[asy] size(300); draw((0,3.5)--(4,9.15)--(8,3.5)--cycle); label("$A$", (4,9.15), N, p = fontsize(10pt)); label("$C$", (8,3.5), SE, p = fontsize(10pt)); label("$B$", (0,3.5), SW, p = fontsize(10pt)); label("$P$", (3.5,7), NW, p = fontsize(10pt)); draw((0,3.5)--(3.5,7)); draw((8,3.5)--(3.5,7)); draw((4,9.15)--(3.5,7)); label("$P'$", (3.5, 0), S, p = fontsize(10pt)); draw((8,3.5)--(3.5,0)); draw((0,3.5)--(3.5,0)); label("$P''$",(6,8.5), NE, p = fontsize(10pt)); draw((4,9.15)--(6,8.5)); draw((8,3.5)--(6,8.5)); label("$P'''$",(2.25,8), NW, p = fontsize(10pt)); draw((0,3.5)--(2.25,8)); draw((4,9.15)--(2.25,8)); draw((3.5,0)--(6,8.5)--(2.25,8)--cycle); [/asy]
Observe that the area of the equilateral triangle $ABC$ is half that of the hexagon $AP''CP'BP'''$
Note that $AP=AP''=AP'''$ . The same goes for the other vertices. This means that $AP''P'''$ is isosceles. Using either the Law of Cosines or simply observing that $AP''P'''$ is comprised of two 30-60-90 triangles, we find that $P''P'''= \sqrt{3}$ . Similarly (pun intended), $P'P'''=3$ and $P'P''=2\sqrt{3}$ . Using the previous observation that $AP''P'''$ is two 30-60-90 triangles (as are the others) we find the areas of $AP''P''$ to be $\frac{\sqrt{3}}{4}$ . Again, using similarity we find the area of $BP'P'''$ to be $\frac{3\sqrt{3}}{4}$ and the area of $CP'P''$ to be $\sqrt{3}$
Next, observe that $P'P''P'''$ is a 30-60-90 right triangle. This right triangle therefore has an area of $\frac{3\sqrt{3}}{2}$
Adding these areas together, we get the area of the hexagon as $\frac{7\sqrt{3}}{2}$ . This means that the area of $ABC$ is $\frac{7\sqrt{3}}{4}$
The formula for the area of an equilateral triangle with side length $s$ is $\frac{s^2\sqrt{3}}{4}$ (if you don't have this memorized it's not hard to derive). Comparing this formula to the area of $ABC$ , we can easily find that $s^2=7$ , which means that the side length of $ABC$ is $\boxed{7}$ | B | 7 |
15449518d325df015d8e440769c7ba8a | https://artofproblemsolving.com/wiki/index.php/2020_AMC_12A_Problems/Problem_24 | Suppose that $\triangle{ABC}$ is an equilateral triangle of side length $s$ , with the property that there is a unique point $P$ inside the triangle such that $AP=1$ $BP=\sqrt{3}$ , and $CP=2$ . What is $s$
$\textbf{(A) } 1+\sqrt{2} \qquad \textbf{(B) } \sqrt{7} \qquad \textbf{(C) } \frac{8}{3} \qquad \textbf{(D) } \sqrt{5+\sqrt{5}} \qquad \textbf{(E) } 2\sqrt{2}$ | Suppose $A(0,\sqrt{3}a)$ $B(-a,0)$ $C(a,0)$ and $P(x,y)$ . So $s=2a$ . Since $BP = \sqrt{3}$ and $CP = 2$ , we have \[(x+a)^2+y^2=3\] \[(x-a)^2+y^2=4\] Solving the equations, we have \[x=-\frac{1}{4a},~~y=\sqrt{\frac72-a^2-\frac{1}{16a^2}}\] From $AP=1$ (and a fair amount of algebra), we can have $a=\sqrt{7}/2$ . The answer is $\boxed{7}$ | B | 7 |
15449518d325df015d8e440769c7ba8a | https://artofproblemsolving.com/wiki/index.php/2020_AMC_12A_Problems/Problem_24 | Suppose that $\triangle{ABC}$ is an equilateral triangle of side length $s$ , with the property that there is a unique point $P$ inside the triangle such that $AP=1$ $BP=\sqrt{3}$ , and $CP=2$ . What is $s$
$\textbf{(A) } 1+\sqrt{2} \qquad \textbf{(B) } \sqrt{7} \qquad \textbf{(C) } \frac{8}{3} \qquad \textbf{(D) } \sqrt{5+\sqrt{5}} \qquad \textbf{(E) } 2\sqrt{2}$ | Drawing out a rough sketch, it appears that $\angle BPC = 90^{\circ}$ . By Pythagorean, our answer is $\sqrt{\sqrt{3}^2 + 2^2} = \boxed{7}$ | B | 7 |
087b7c1747585a6ebaf360bb5e7f7e8a | https://artofproblemsolving.com/wiki/index.php/2020_AMC_12A_Problems/Problem_25 | The number $a=\frac{p}{q}$ , where $p$ and $q$ are relatively prime positive integers, has the property that the sum of all real numbers $x$ satisfying \[\lfloor x \rfloor \cdot \{x\} = a \cdot x^2\] is $420$ , where $\lfloor x \rfloor$ denotes the greatest integer less than or equal to $x$ and $\{x\}=x- \lfloor x \rfloor$ denotes the fractional part of $x$ . What is $p+q$
$\textbf{(A) } 245 \qquad \textbf{(B) } 593 \qquad \textbf{(C) } 929 \qquad \textbf{(D) } 1331 \qquad \textbf{(E) } 1332$ | Let $w=\lfloor x \rfloor$ and $f=\{x\}$ denote the whole part and the fractional part of $x,$ respectively, for which $0\leq f<1$ and $x=w+f.$
We rewrite the given equation as \[w\cdot f=a\cdot(w+f)^2. \hspace{38.75mm}(1)\] Since $a\cdot(w+f)^2\geq0,$ it follows that $w\cdot f\geq0,$ from which $w\geq0.$
We expand and rearrange $(1)$ as \[af^2+(2a-1)wf+aw^2=0, \hspace{23mm}(2)\] which is a quadratic with either $f$ or $w.$
For simplicity purposes, we will treat $w$ as some fixed nonnegative integer so that $(2)$ is a quadratic with $f.$ By the Quadratic Formula, we have \[f=w\Biggl(\frac{1-2a\pm\sqrt{1-4a}}{2a}\Biggr). \hspace{25mm}(3)\] If $w=0,$ then $f=0.$ We get $x=w+f=0,$ which does not affect the sum of the solutions. Therefore, we consider the case for $w\geq1:$
Recall that $0\leq f<1,$ so $\frac{1-2a\pm\sqrt{1-4a}}{2a}\geq0.$ From the discriminant, we require that $0\leq1-4a<1,$ or \[0<a\leq\frac14. \hspace{54mm}(4)\]
We consider each part of $0\leq f<1$ separately:
Now, we express $x$ in terms of $w$ and $k:$ \[x=w+f=w+wk=w(1+k).\] The sum of all solutions to the original equation is \[\sum_{w=1}^{W}w(1+k)=(1+k)\cdot\sum_{w=1}^{W}w=(1+k)\cdot\frac{W(W+1)}{2}=420. \hspace{10mm}(\bigstar)\] As $1+k<1+\frac1W,$ we conclude that $1+k$ is slightly above $1$ so that $\frac{W(W+1)}{2}$ is slightly below $420,$ or $W(W+1)$ is slightly below $840.$ By observations, we get $W=28.$ Substituting this into $(\bigstar)$ produces $k=\frac{1}{29},$ which satisfies $\frac{1}{W+1}\leq k<\frac1W,$ as required.
Finally, we solve for $a$ in $k=\frac{1}{2a}-1-\frac{\sqrt{1-4a}}{2a}:$ \begin{align*} \frac{1}{29}&=\frac{1}{2a}-1-\frac{\sqrt{1-4a}}{2a} \\ \frac{2}{29}a&=1-2a-\sqrt{1-4a} \\ \frac{60}{29}a-1&=-\sqrt{1-4a} \\ \frac{60^2}{29^2}a^2-\frac{120}{29}a+1&=1-4a \\ \frac{60^2}{29^2}a^2-\frac{4}{29}a&=0 \\ a\left(\frac{60^2}{29^2}a-\frac{4}{29}\right)&=0. \end{align*} Since $a>0,$ we obtain $\frac{60^2}{29^2}a-\frac{4}{29}=0,$ from which \[a=\frac{4}{29}\cdot\frac{29^2}{60^2}=\frac{29}{900}.\] The answer is $29+900=\boxed{929}.$ | C | 929 |
087b7c1747585a6ebaf360bb5e7f7e8a | https://artofproblemsolving.com/wiki/index.php/2020_AMC_12A_Problems/Problem_25 | The number $a=\frac{p}{q}$ , where $p$ and $q$ are relatively prime positive integers, has the property that the sum of all real numbers $x$ satisfying \[\lfloor x \rfloor \cdot \{x\} = a \cdot x^2\] is $420$ , where $\lfloor x \rfloor$ denotes the greatest integer less than or equal to $x$ and $\{x\}=x- \lfloor x \rfloor$ denotes the fractional part of $x$ . What is $p+q$
$\textbf{(A) } 245 \qquad \textbf{(B) } 593 \qquad \textbf{(C) } 929 \qquad \textbf{(D) } 1331 \qquad \textbf{(E) } 1332$ | Let $x_n$ be a root in the interval $(n,n+1)$ . In this interval, $\lfloor x_n \rfloor = n$ and $\{x_n\}=x_n-n$ , so we must have $ax_n^2 = nx_n-n^2$ , i.e., $ax_n^2-nx_n+n^2=0$ . We can homogenize this equation by setting $x_n=n\zeta$ ; then $x_1=\zeta$ , and $\zeta$ is a root of $a\zeta^2-\zeta+1=0$
Suppose $N$ is the largest integer for which there is such a root; we have, for $n=1,2,\ldots , N$ \[n < x_n = n\zeta < n+1\] Summing over $n\in \{1,2,\ldots , N\}$ we get \[\tfrac 12 N(N+1) < 420 = \tfrac 12 N(N+1)\zeta < \tfrac 12 N(N+3)\] From the right inequality we get $27< N$ and from the left one we get $N<29$ . Thus $N=28$ . Using this in the middle equality we get $\zeta = \tfrac{30}{29}$ . Since $\zeta$ satisfies $a\zeta^2-\zeta+1=0$ , we get \[a = \zeta^{-2}(\zeta-1)= \tfrac{29^2}{30^2}\cdot \tfrac 1{29}= \tfrac{29}{900}.\] The answer is $29+900=\boxed{929}.$ | C | 929 |
087b7c1747585a6ebaf360bb5e7f7e8a | https://artofproblemsolving.com/wiki/index.php/2020_AMC_12A_Problems/Problem_25 | The number $a=\frac{p}{q}$ , where $p$ and $q$ are relatively prime positive integers, has the property that the sum of all real numbers $x$ satisfying \[\lfloor x \rfloor \cdot \{x\} = a \cdot x^2\] is $420$ , where $\lfloor x \rfloor$ denotes the greatest integer less than or equal to $x$ and $\{x\}=x- \lfloor x \rfloor$ denotes the fractional part of $x$ . What is $p+q$
$\textbf{(A) } 245 \qquad \textbf{(B) } 593 \qquad \textbf{(C) } 929 \qquad \textbf{(D) } 1331 \qquad \textbf{(E) } 1332$ | First note that $\lfloor x\rfloor \cdot \{x\}<0$ when $x<0$ while $ax^2\ge 0\forall x\in \mathbb{R}$ . Thus we only need to look at positive solutions ( $x=0$ doesn't affect the sum of the solutions).
Next, we break $\lfloor x\rfloor\cdot \{x\}$ down for each interval $[n,n+1)$ , where $n$ is a positive integer. Assume $\lfloor x\rfloor=n$ , then $\{x\}=x-n$ . This means that when $x\in [n,n+1)$ $\lfloor x\rfloor \cdot \{x\}=n(x-n)=nx-n^2$ . Setting this equal to $ax^2$ gives \[nx-n^2=ax^2\implies ax^2-nx+n^2=0 \implies x=\frac{n\pm \sqrt{n^2-4an^2}}{2a}\] We're looking at the solution with the positive $x$ , which is $x=\frac{n-n\sqrt{1-4a}}{2a}=\frac{n}{2a}\left(1-\sqrt{1-4a}\right)$ . Note that if $\lfloor x\rfloor=n$ is the greatest $n$ such that $\lfloor x\rfloor \cdot \{x\}=ax^2$ has a solution, the sum of all these solutions is slightly over $\sum_{k=1}^{n}k=\frac{n(n+1)}{2}$ , which is $406$ when $n=28$ , just under $420$ . Checking this gives \begin{align*} \sum_{k=1}^{28}\frac{k}{2a}\left(1-\sqrt{1-4a}\right)&=\frac{1-\sqrt{1-4a}}{2a}\cdot 406=420 \\ \frac{1-\sqrt{1-4a}}{2a}&=\frac{420}{406}=\frac{30}{29} \\ 29-29\sqrt{1-4a}&=60a \\ 29\sqrt{1-4a}&=29-60a \\ 29^2-4\cdot 29^2a&=29^2+3600a^2-120\cdot 29a \\ 3600a^2&=116a \\ a&=\frac{116}{3600}=\frac{29}{900} \implies \boxed{929} ~ktong | C | 929 |
7f335cf538d7ac1905a7b7d8f12fb468 | https://artofproblemsolving.com/wiki/index.php/2020_AMC_12B_Problems/Problem_1 | What is the value in simplest form of the following expression? \[\sqrt{1} + \sqrt{1+3} + \sqrt{1+3+5} + \sqrt{1+3+5+7}\]
$\textbf{(A) }5 \qquad \textbf{(B) }4 + \sqrt{7} + \sqrt{10} \qquad \textbf{(C) } 10 \qquad \textbf{(D) } 15 \qquad \textbf{(E) } 4 + 3\sqrt{3} + 2\sqrt{5} + \sqrt{7}$ | We have \[\sqrt{1} + \sqrt{1+3} + \sqrt{1+3+5} + \sqrt{1+3+5+7} = \sqrt{1} + \sqrt{4} + \sqrt{9} + \sqrt{16}\ = 1 + 2 + 3 + 4 = \boxed{10}.\] Note: This comes from the fact that the sum of the first $n$ odds is $n^2$ | C | 10 |
521ddd4624343366e4b42dbd7a615700 | https://artofproblemsolving.com/wiki/index.php/2020_AMC_12B_Problems/Problem_2 | What is the value of the following expression?
\[\frac{100^2-7^2}{70^2-11^2} \cdot \frac{(70-11)(70+11)}{(100-7)(100+7)}\]
$\textbf{(A) } 1 \qquad \textbf{(B) } \frac{9951}{9950} \qquad \textbf{(C) } \frac{4780}{4779} \qquad \textbf{(D) } \frac{108}{107} \qquad \textbf{(E) } \frac{81}{80}$ | Using difference of squares to factor the left term, we get \[\frac{100^2-7^2}{70^2-11^2} \cdot \frac{(70-11)(70+11)}{(100-7)(100+7)} = \frac{(100-7)(100+7)}{(70-11)(70+11)} \cdot \frac{(70-11)(70+11)}{(100-7)(100+7)}.\] Cancelling all the terms, we get $\boxed{1}$ as the answer. | A | 1 |
a45d852737dffc94a54a569e683ccb11 | https://artofproblemsolving.com/wiki/index.php/2020_AMC_12B_Problems/Problem_4 | The acute angles of a right triangle are $a^{\circ}$ and $b^{\circ}$ , where $a>b$ and both $a$ and $b$ are prime numbers. What is the least possible value of $b$
$\textbf{(A)}\ 2 \qquad\textbf{(B)}\ 3 \qquad\textbf{(C)}\ 5 \qquad\textbf{(D)}\ 7 \qquad\textbf{(E)}\ 11$ | Since the three angles of a triangle add up to $180^{\circ}$ and one of the angles is $90^{\circ}$ because it's a right triangle, $a^{\circ} + b^{\circ} = 90^{\circ}$
The greatest prime number less than $90$ is $89$ . If $a=89^{\circ}$ , then $b=90^{\circ}-89^{\circ}=1^{\circ}$ , which is not prime.
The next greatest prime number less than $90$ is $83$ . If $a=83^{\circ}$ , then $b=7^{\circ}$ , which IS prime, so we have our answer $\boxed{7}$ ~quacker88 | D | 7 |
a45d852737dffc94a54a569e683ccb11 | https://artofproblemsolving.com/wiki/index.php/2020_AMC_12B_Problems/Problem_4 | The acute angles of a right triangle are $a^{\circ}$ and $b^{\circ}$ , where $a>b$ and both $a$ and $b$ are prime numbers. What is the least possible value of $b$
$\textbf{(A)}\ 2 \qquad\textbf{(B)}\ 3 \qquad\textbf{(C)}\ 5 \qquad\textbf{(D)}\ 7 \qquad\textbf{(E)}\ 11$ | Looking at the answer choices, only $7$ and $11$ are coprime to $90$ . Testing $7$ , the smaller angle, makes the other angle $83$ which is prime, therefore our answer is $\boxed{7}$ | D | 7 |
a45d852737dffc94a54a569e683ccb11 | https://artofproblemsolving.com/wiki/index.php/2020_AMC_12B_Problems/Problem_4 | The acute angles of a right triangle are $a^{\circ}$ and $b^{\circ}$ , where $a>b$ and both $a$ and $b$ are prime numbers. What is the least possible value of $b$
$\textbf{(A)}\ 2 \qquad\textbf{(B)}\ 3 \qquad\textbf{(C)}\ 5 \qquad\textbf{(D)}\ 7 \qquad\textbf{(E)}\ 11$ | It is clear that $\gcd(a,b)=1.$ By the Euclidean Algorithm, we have \[\gcd(a,b)=\gcd(a+b,b)=\gcd(90,b)=1,\] so $90$ and $b$ are relatively prime.
The least such prime number $b$ is $7,$ from which $a=90-b=83$ is also a prime number. Therefore, the answer is $\boxed{7}.$ | D | 7 |
d072eb1b77b22ef845cf95aaf1388f06 | https://artofproblemsolving.com/wiki/index.php/2020_AMC_12B_Problems/Problem_5 | Teams $A$ and $B$ are playing in a basketball league where each game results in a win for one team and a loss for the other team. Team $A$ has won $\tfrac{2}{3}$ of its games and team $B$ has won $\tfrac{5}{8}$ of its games. Also, team $B$ has won $7$ more games and lost $7$ more games than team $A.$ How many games has team $A$ played?
$\textbf{(A) } 21 \qquad \textbf{(B) } 27 \qquad \textbf{(C) } 42 \qquad \textbf{(D) } 48 \qquad \textbf{(E) } 63$ | Suppose team $A$ has played $g$ games in total so that it has won $\frac23g$ games.
It follows that team $B$ has played $g+14$ games in total so that it has won $\frac23g+7$ games.
We set up and solve an equation for team $B$ 's win ratio: \begin{align*} \frac{\frac23g+7}{g+14}&=\frac58 \\ \frac{16}{3}g+56&=5g+70 \\ \frac13g&=14 \\ g&=\boxed{42} ~MRENTHUSIASM | C | 42 |
d072eb1b77b22ef845cf95aaf1388f06 | https://artofproblemsolving.com/wiki/index.php/2020_AMC_12B_Problems/Problem_5 | Teams $A$ and $B$ are playing in a basketball league where each game results in a win for one team and a loss for the other team. Team $A$ has won $\tfrac{2}{3}$ of its games and team $B$ has won $\tfrac{5}{8}$ of its games. Also, team $B$ has won $7$ more games and lost $7$ more games than team $A.$ How many games has team $A$ played?
$\textbf{(A) } 21 \qquad \textbf{(B) } 27 \qquad \textbf{(C) } 42 \qquad \textbf{(D) } 48 \qquad \textbf{(E) } 63$ | If we consider the number of games team $B$ has played as $x$ and the number of games that team $A$ has played as $y$ , then we can set up the following system of equations: \begin{align*} \frac{5}{8}x &= \frac{2}{3}y+7, \\ \frac{3}{8}x &= \frac{1}{3}y+7. \end{align*} The first system equated the number of wins of each team, while the second system equates the number of losses by each team. By multiplying the second equation by $2$ and solving the system, we get $y = 42$ or answer choice $\boxed{42}.$ | C | 42 |
d072eb1b77b22ef845cf95aaf1388f06 | https://artofproblemsolving.com/wiki/index.php/2020_AMC_12B_Problems/Problem_5 | Teams $A$ and $B$ are playing in a basketball league where each game results in a win for one team and a loss for the other team. Team $A$ has won $\tfrac{2}{3}$ of its games and team $B$ has won $\tfrac{5}{8}$ of its games. Also, team $B$ has won $7$ more games and lost $7$ more games than team $A.$ How many games has team $A$ played?
$\textbf{(A) } 21 \qquad \textbf{(B) } 27 \qquad \textbf{(C) } 42 \qquad \textbf{(D) } 48 \qquad \textbf{(E) } 63$ | First, let us assign some variables. Let \[A_w=2x, \ A_l=x, \ A_g=3x,\] \[B_w=5y, \ B_l=3y, \ B_g=8y,\] where $X_w$ denotes number of games won, $X_l$ denotes number of games lost, and $X_g$ denotes total games played for $X\in \{A, B\}$ . Using the given information, we can set up the following two equations: \begin{align*} B_w=A_w+7&\implies 5y=2x+7, \\ B_l=A_l+7&\implies 3y=x+7. \end{align*} We can solve through substitution, as the second equation can be written as $x=3y-7$ , and plugging this into the first equation gives $5y=6y-7\implies y=7$ , which means $x=3(7)-7=14$ . Finally, we want the total number of games team $A$ has played, which is $A_g=3(14)=\boxed{42}$ | C | 42 |
d072eb1b77b22ef845cf95aaf1388f06 | https://artofproblemsolving.com/wiki/index.php/2020_AMC_12B_Problems/Problem_5 | Teams $A$ and $B$ are playing in a basketball league where each game results in a win for one team and a loss for the other team. Team $A$ has won $\tfrac{2}{3}$ of its games and team $B$ has won $\tfrac{5}{8}$ of its games. Also, team $B$ has won $7$ more games and lost $7$ more games than team $A.$ How many games has team $A$ played?
$\textbf{(A) } 21 \qquad \textbf{(B) } 27 \qquad \textbf{(C) } 42 \qquad \textbf{(D) } 48 \qquad \textbf{(E) } 63$ | Using the information from the problem, we can note that team $A$ has lost $\frac{1}{3}$ of their matches. Using the answer choices, we can construct the following list of possible win-lose scenarios for $A,$ represented in the form $(w, l)$ for convenience: \begin{align*} \textbf{(A)} &\implies (14, 7) \\ \textbf{(B)} &\implies (18, 9) \\ \textbf{(C)} &\implies (28, 14) \\ \textbf{(D)} &\implies (32, 16) \\ \textbf{(E)} &\implies (42, 21) \end{align*} Thus, we have $5$ matching $B$ scenarios, simply adding $7$ to $w$ and $l.$ We can then test each of the five $B$ scenarios for $\frac{w}{w+l} = \frac{5}{8}$ and find that $(35, 21)$ fits this description. Then working backwards and subtracting $7$ from $w$ and $l$ gives us the point $(28, 14),$ making the answer $\boxed{42}.$ | C | 42 |
d072eb1b77b22ef845cf95aaf1388f06 | https://artofproblemsolving.com/wiki/index.php/2020_AMC_12B_Problems/Problem_5 | Teams $A$ and $B$ are playing in a basketball league where each game results in a win for one team and a loss for the other team. Team $A$ has won $\tfrac{2}{3}$ of its games and team $B$ has won $\tfrac{5}{8}$ of its games. Also, team $B$ has won $7$ more games and lost $7$ more games than team $A.$ How many games has team $A$ played?
$\textbf{(A) } 21 \qquad \textbf{(B) } 27 \qquad \textbf{(C) } 42 \qquad \textbf{(D) } 48 \qquad \textbf{(E) } 63$ | Let's say that team $A$ plays $n$ games in total. Therefore, team $B$ must play $n + 14$ games in total (7 wins, 7 losses) Since the ratio of $A$ is \[\frac{2}{3} \implies n \equiv 0 \pmod{3}\] Similarly, since the ratio of $B$ is \[\frac{5}{8} \implies n + 14 \equiv 0 \pmod{8}\] Now, we can go through the answer choices and see which ones work: \begin{align*} \textbf{(A) } 21 &\implies 21 + 14 = 35 \not \equiv 0\pmod{8} \\ \textbf{(B) } 27 &\implies 27 + 14 = 41 \not \equiv 0\pmod{8} \\ \textbf{(C) } 42 &\implies 42 + 14 = 56 \equiv 0\pmod{8} \\ \textbf{(D) } 48 &\implies 48 + 14 = 62 \not \equiv 0\pmod{8} \\ \textbf{(E) } 63 &\implies 63 + 14 = 77 \not \equiv 0\pmod{8} \\ \end{align*} So we can see $\boxed{42}$ is the only valid answer. | C | 42 |
a6357ed0d92ba2c50d3ade3b9501df5b | https://artofproblemsolving.com/wiki/index.php/2020_AMC_12B_Problems/Problem_7 | Two nonhorizontal, non vertical lines in the $xy$ -coordinate plane intersect to form a $45^{\circ}$ angle. One line has slope equal to $6$ times the slope of the other line. What is the greatest possible value of the product of the slopes of the two lines?
$\textbf{(A)}\ \frac16 \qquad\textbf{(B)}\ \frac23 \qquad\textbf{(C)}\ \frac32 \qquad\textbf{(D)}\ 3 \qquad\textbf{(E)}\ 6$ | Intersect at the origin and select a point on each line to define vectors $\mathbf{v}_{i}=(x_{i},y_{i})$ .
Note that $\theta=45^{\circ}$ gives equal magnitudes of the vector products \[\mathbf{v}_1\cdot\mathbf{v}_2 = v_{1}v_{2}\cos\theta \quad\mathrm{and}\quad |\mathbf{v}_1\times\mathbf{v}_2| = v_{1}v_{2}\sin\theta .\]
Substituting coordinate expressions for vector products, we find \[\mathbf{v}_1\cdot\mathbf{v}_2 = |\mathbf{v}_1\times\mathbf{v}_2| \ \implies\ x_{1}x_{2}+y_{1}y_{2} = x_{1}y_{2}-x_{2}y_{1} .\] Divide this equation by $x_{1}x_{2}$ to obtain \[1+m_{1}m_{2} = m_{2}-m_{1} ,\] where $m_{i}=y_{i}/x_{i}$ is the slope of line $i$ .
Taking $m_{2}=6m_{1}$ , we obtain \[6m_{1}^{2}-5m_{1}+1 = 0 \ \implies\ m_{1} \in \{\frac{1}{3},\frac{1}{2}\} .\] The latter solution gives the largest product of slopes $m_{1}m_{2} = 6m_{1}^2 = \frac{3}{2} . \quad \boxed{32}$ | C | 32 |
a6357ed0d92ba2c50d3ade3b9501df5b | https://artofproblemsolving.com/wiki/index.php/2020_AMC_12B_Problems/Problem_7 | Two nonhorizontal, non vertical lines in the $xy$ -coordinate plane intersect to form a $45^{\circ}$ angle. One line has slope equal to $6$ times the slope of the other line. What is the greatest possible value of the product of the slopes of the two lines?
$\textbf{(A)}\ \frac16 \qquad\textbf{(B)}\ \frac23 \qquad\textbf{(C)}\ \frac32 \qquad\textbf{(D)}\ 3 \qquad\textbf{(E)}\ 6$ | Place on coordinate plane.
Lines are $y=mx, y=6mx.$ The intersection point at the origin.
Goes through $(0,0),(1,m),(1,6m),(1,0).$ So by law of sines, $\frac{5m}{\sin{45^{\circ}}} = \frac{\sqrt{1+m^2}}{1/(\sqrt{1+36m^2})},$ lettin $a=m^2$ we want $6a.$ Simplifying gives $50a = (1+a)(1+36a),$ so $36a^2-13a+1=0 \implies 36(a-1/4)(a-1/9)=0,$ so max $a=1/4,$ and $6a=3/2 \quad \boxed{32}.$ | C | 32 |
a6357ed0d92ba2c50d3ade3b9501df5b | https://artofproblemsolving.com/wiki/index.php/2020_AMC_12B_Problems/Problem_7 | Two nonhorizontal, non vertical lines in the $xy$ -coordinate plane intersect to form a $45^{\circ}$ angle. One line has slope equal to $6$ times the slope of the other line. What is the greatest possible value of the product of the slopes of the two lines?
$\textbf{(A)}\ \frac16 \qquad\textbf{(B)}\ \frac23 \qquad\textbf{(C)}\ \frac32 \qquad\textbf{(D)}\ 3 \qquad\textbf{(E)}\ 6$ | Let one of the lines have equation $y=ax$ . Let $\theta$ be the angle that line makes with the x-axis, so $\tan(\theta)=a$ . The other line will have a slope of $\tan(45^{\circ}+\theta)=\frac{\tan(45^{\circ})+\tan(\theta)}{1-\tan(45^{\circ})\tan(\theta)} = \frac{1+a}{1-a}$ . Since the slope of one line is $6$ times the other, and $a$ is the smaller slope, $6a = \frac{1+a}{1-a} \implies 6a-6a^2=1+a \implies 6a^2-5a+1=0 \implies a=\frac{1}{2},\frac{1}{3}$ . If $a = \frac{1}{2}$ , the other line will have slope $\frac{1+\frac{1}{2}}{1-\frac{1}{2}} = 3$ . If $a = \frac{1}{3}$ , the other line will have slope $\frac{1+\frac{1}{3}}{1-\frac{1}{3}} = 2$ . The first case gives the bigger product of $\frac{3}{2}$ , so our answer is $\boxed{32}$ | C | 32 |
a6357ed0d92ba2c50d3ade3b9501df5b | https://artofproblemsolving.com/wiki/index.php/2020_AMC_12B_Problems/Problem_7 | Two nonhorizontal, non vertical lines in the $xy$ -coordinate plane intersect to form a $45^{\circ}$ angle. One line has slope equal to $6$ times the slope of the other line. What is the greatest possible value of the product of the slopes of the two lines?
$\textbf{(A)}\ \frac16 \qquad\textbf{(B)}\ \frac23 \qquad\textbf{(C)}\ \frac32 \qquad\textbf{(D)}\ 3 \qquad\textbf{(E)}\ 6$ | Let the smaller slope be $m$ , then the larger slope is $6m$ . Since we want the greatest product we begin checking each answer choice, starting with (E).
$6m^2=6$
$m^2=1$
This gives $m=1$ and $6m=6$ . Checking with a protractor we see that this does not form a 45 degree angle.
Using this same method for the other answer choices, we eventually find that the answer is $\boxed{32}$ since our slopes are $\frac12$ and $3$ which forms a perfect 45 degree angle. | C | 32 |
Subsets and Splits
No community queries yet
The top public SQL queries from the community will appear here once available.