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ee1bdfcb98da5de80d3a873baafff3e2 | https://artofproblemsolving.com/wiki/index.php/2009_AMC_12A_Problems/Problem_25 | The first two terms of a sequence are $a_1 = 1$ and $a_2 = \frac {1}{\sqrt3}$ . For $n\ge1$
What is $|a_{2009}|$
$\textbf{(A)}\ 0\qquad \textbf{(B)}\ 2 - \sqrt3\qquad \textbf{(C)}\ \frac {1}{\sqrt3}\qquad \textbf{(D)}\ 1\qquad \textbf{(E)}\ 2 + \sqrt3$ | First, some intuition. The given recurrence relation: $a_{n + 2} = \frac {a_n + a_{n + 1}}{1 - a_na_{n + 1}}$ looks much like the tangent addition formula. So, we let $a_n=\tan{\theta_n}$ . Then, we get: \[\tan{\theta_n}=\tan{(\theta_{n-1}+\theta_{n-2})}\] This gives us: \[\theta_n \equiv \theta_{n-1}+\theta_{n-2} \pmod{180}\] Now, observe that \[\theta_1=45\] \[\theta_2=30\] We know that this sequence of values for $\theta_n$ will repeat eventually because it is mod $180$ .
It is just a matter of when, so we start bashing:
\[\theta_3=75\] \[\theta_4=105\] \[\theta_5=0\] \[\theta_6=105\] \[\theta_7=105\] \[\theta_8=30\] \[\theta_9=135\] \[\theta_{10}=165\] \[\theta_{11}=120\] \[\theta_{12}=105\] \[\theta_{13}=45\] \[\theta_{14}=150\] \[\theta_{15}=15\] \[\theta_{16}=165\] \[\theta_{17}=0\] \[\theta_{18}=165\] \[\theta_{19}=165\] \[\theta_{20}=150\] \[\theta_{21}=135\] \[\theta_{22}=105\] \[\theta_{23}=60\] \[\theta_{24}=165\] \[\theta_{25}=45\] \[\theta_{26}=30\] And there is the repetition. So, this series has a period of 24. $2009 \equiv 17 \pmod{24}$ , so $|a_{2009}|=|\tan{\theta_{17}}|=|\tan{0}|=|0|= \boxed{0}$ ~Firebolt360 | A | 0 |
fe269dc41f6afbccf8122302c00b4c56 | https://artofproblemsolving.com/wiki/index.php/2009_AMC_12B_Problems/Problem_1 | Each morning of her five-day workweek, Jane bought either a $50$ -cent muffin or a $75$ -cent bagel. Her total cost for the week was a whole number of dollars. How many bagels did she buy?
$\textbf{(A) } 1\qquad\textbf{(B) } 2\qquad\textbf{(C) } 3\qquad\textbf{(D) } 4\qquad\textbf{(E) } 5$ | If Jane bought one more bagel but one fewer muffin, then her total cost for the week would increase by $25$ cents.
If Jane bought $1$ bagel, then she bought $4$ muffins. Her total cost for the week would be $75\cdot1+50\cdot4=275$ cents, or $2.75$ dollars. Clearly, she bought one more bagel but one fewer muffin at a total cost of $3.00$ dollars. Therefore, she bought $\boxed{2}$ bagels. | B | 2 |
66aaeac82ecfec08535112f980e62362 | https://artofproblemsolving.com/wiki/index.php/2009_AMC_12B_Problems/Problem_2 | Paula the painter had just enough paint for 30 identically sized rooms. Unfortunately, on the way to work, three cans of paint fell off her truck, so she had only enough paint for 25 rooms. How many cans of paint did she use for the 25 rooms?
$\mathrm{(A)}\ 10\qquad \mathrm{(B)}\ 12\qquad \mathrm{(C)}\ 15\qquad \mathrm{(D)}\ 18\qquad \mathrm{(E)}\ 25$ | Losing three cans of paint corresponds to being able to paint five fewer rooms. So $\frac 35 \cdot 25 = \boxed{15}$ | C | 15 |
f81f728a86002102c33e935c802936a1 | https://artofproblemsolving.com/wiki/index.php/2009_AMC_12B_Problems/Problem_3 | Twenty percent less than 60 is one-third more than what number?
$\mathrm{(A)}\ 16\qquad \mathrm{(B)}\ 30\qquad \mathrm{(C)}\ 32\qquad \mathrm{(D)}\ 36\qquad \mathrm{(E)}\ 48$ | Twenty percent less than 60 is $\frac 45 \cdot 60 = 48$ . One-third more than a number is $\frac 43n$ . Therefore $\frac 43n = 48$ and the number is $\boxed{36}$ | D | 36 |
e828b447da89623adbc406b6e327eb58 | https://artofproblemsolving.com/wiki/index.php/2009_AMC_12B_Problems/Problem_4 | A rectangular yard contains two flower beds in the shape of congruent isosceles right triangles. The remainder of the yard has a trapezoidal shape, as shown. The parallel sides of the trapezoid have lengths $15$ and $25$ meters. What fraction of the yard is occupied by the flower beds?
$\mathrm{(A)}\frac {1}{8}\qquad \mathrm{(B)}\frac {1}{6}\qquad \mathrm{(C)}\frac {1}{5}\qquad \mathrm{(D)}\frac {1}{4}\qquad \mathrm{(E)}\frac {1}{3}$ | Each triangle has leg length $\frac 12 \cdot (25 - 15) = 5$ meters and area $\frac 12 \cdot 5^2 = \frac {25}{2}$ square meters. Thus the flower beds have a total area of $25$ square meters. The entire yard has length $25$ m and width $5$ m, so its area is $125$ square meters. The fraction of the yard occupied by the flower beds is $\frac {25}{125} = \boxed{15}$ | C | 15 |
be5e38ab9682f34a3d391355863d0f22 | https://artofproblemsolving.com/wiki/index.php/2009_AMC_12B_Problems/Problem_5 | Kiana has two older twin brothers. The product of their three ages is 128. What is the sum of their three ages?
$\mathrm{(A)}\ 10\qquad \mathrm{(B)}\ 12\qquad \mathrm{(C)}\ 16\qquad \mathrm{(D)}\ 18\qquad \mathrm{(E)}\ 24$ | The age of each person is a factor of $128 = 2^7$ . So the twins could be $2^0 = 1, 2^1 = 2, 2^2 = 4, 2^3 = 8$ years of age and, consequently Kiana could be $128$ $32$ $8$ , or $2$ years old, respectively. Because Kiana is younger than her brothers, she must be $2$ years old. So the sum of their ages is $2 + 8 + 8 = \boxed{18}$ | D | 18 |
8cbf83bd8aa914782c54332e467f5ded | https://artofproblemsolving.com/wiki/index.php/2009_AMC_12B_Problems/Problem_6 | By inserting parentheses, it is possible to give the expression \[2\times3 + 4\times5\] several values. How many different values can be obtained?
$\text{(A) } 2 \qquad \text{(B) } 3 \qquad \text{(C) } 4 \qquad \text{(D) } 5 \qquad \text{(E) } 6$ | The three operations can be performed on any of $3! = 6$ orders. However, if the addition is performed either first or last, then multiplying in either order produces the same result. So at most four distinct values can be obtained. It is easy to check that the values of the four expressions \begin{align*} (2\times3) + (4\times5) &= 26,\\ (2\times3 + 4)\times5 &= 50,\\ 2\times(3 + 4\times5) &= 46,\\ 2\times(3 + 4)\times5 &= 70 \end{align*} are in fact all distinct. So the answer is $\boxed{4}$ | C | 4 |
901d16ede48414f811cd396e5d1231bc | https://artofproblemsolving.com/wiki/index.php/2009_AMC_12B_Problems/Problem_7 | In a certain year the price of gasoline rose by $20\%$ during January, fell by $20\%$ during February, rose by $25\%$ during March, and fell by $x\%$ during April. The price of gasoline at the end of April was the same as it had been at the beginning of January. To the nearest integer, what is $x$
$\mathrm{(A)}\ 12\qquad \mathrm{(B)}\ 17\qquad \mathrm{(C)}\ 20\qquad \mathrm{(D)}\ 25\qquad \mathrm{(E)}\ 35$ | Let $p$ be the price at the beginning of January. The price at the end of March was $(1.2)(0.8)(1.25)p = 1.2p.$ Because the price at the end of April was $p$ , the price decreased by $0.2p$ during April, and the percent decrease was \[x = 100 \cdot \frac{0.2p}{1.2p} = \frac {100}{6} \approx 16.7.\] So to the nearest integer $x$ is $\boxed{17}$ | B | 17 |
901d16ede48414f811cd396e5d1231bc | https://artofproblemsolving.com/wiki/index.php/2009_AMC_12B_Problems/Problem_7 | In a certain year the price of gasoline rose by $20\%$ during January, fell by $20\%$ during February, rose by $25\%$ during March, and fell by $x\%$ during April. The price of gasoline at the end of April was the same as it had been at the beginning of January. To the nearest integer, what is $x$
$\mathrm{(A)}\ 12\qquad \mathrm{(B)}\ 17\qquad \mathrm{(C)}\ 20\qquad \mathrm{(D)}\ 25\qquad \mathrm{(E)}\ 35$ | Without loss of generality, we can assume the price at the beginning of January was $$100$
When it rose by $20\%$ , it became $$120$ , when it fell by $20\%$ , it became $$96$ , and when it rose by $25\%$ , it became $$120$ again.
In order for the price at the end of April to be the same as it was at the beginning of January ( $$100$ ), the price must decrease by $$20$
20 is $\frac{1}{6}th$ of 120, and $\frac{1}{6} \approx 0.167 \approx 17\%$ So to the nearest integer, $x = 17$ and the answer is $\boxed{17}$ . ~azc1027 | B | 17 |
693057aa23e0da2dd625d017d8d55e5f | https://artofproblemsolving.com/wiki/index.php/2009_AMC_12B_Problems/Problem_9 | Triangle $ABC$ has vertices $A = (3,0)$ $B = (0,3)$ , and $C$ , where $C$ is on the line $x + y = 7$ . What is the area of $\triangle ABC$
$\mathrm{(A)}\ 6\qquad \mathrm{(B)}\ 8\qquad \mathrm{(C)}\ 10\qquad \mathrm{(D)}\ 12\qquad \mathrm{(E)}\ 14$ | Because the line $x + y = 7$ is parallel to $\overline {AB}$ , the area of $\triangle ABC$ is independent of the location of $C$ on that line. Therefore it may be assumed that $C$ is $(7,0)$ . In that case the triangle has base $AC = 4$ and altitude $3$ , so its area is $\frac 12 \cdot 4 \cdot 3 = \boxed{6}$ | null | 6 |
693057aa23e0da2dd625d017d8d55e5f | https://artofproblemsolving.com/wiki/index.php/2009_AMC_12B_Problems/Problem_9 | Triangle $ABC$ has vertices $A = (3,0)$ $B = (0,3)$ , and $C$ , where $C$ is on the line $x + y = 7$ . What is the area of $\triangle ABC$
$\mathrm{(A)}\ 6\qquad \mathrm{(B)}\ 8\qquad \mathrm{(C)}\ 10\qquad \mathrm{(D)}\ 12\qquad \mathrm{(E)}\ 14$ | The base of the triangle is $AB = \sqrt{3^2 + 3^2} = 3\sqrt 2$ . Its altitude is the distance between the point $A$ and the parallel line $x + y = 7$ , which is $\frac 4{\sqrt 2} = 2\sqrt 2$ . Therefore its area is $\frac 12 \cdot 3\sqrt 2 \cdot 2\sqrt 2 = \boxed{6}$ | A | 6 |
9b8aaeef58cf16d673c11473dcfdaeb3 | https://artofproblemsolving.com/wiki/index.php/2009_AMC_12B_Problems/Problem_10 | A particular $12$ -hour digital clock displays the hour and minute of a day. Unfortunately, whenever it is supposed to display a $1$ , it mistakenly displays a $9$ . For example, when it is 1:16 PM the clock incorrectly shows 9:96 PM. What fraction of the day will the clock show the correct time?
$\mathrm{(A)}\ \frac 12\qquad \mathrm{(B)}\ \frac 58\qquad \mathrm{(C)}\ \frac 34\qquad \mathrm{(D)}\ \frac 56\qquad \mathrm{(E)}\ \frac {9}{10}$ | The clock will display the incorrect time for the entire hours of $1, 10, 11$ and $12$ . So the correct hour is displayed $\frac 23$ of the time. The minutes will not display correctly whenever either the tens digit or the ones digit is a $1$ , so the minutes that will not display correctly are $10, 11, 12, \dots, 19$ and $01, 21, 31, 41,$ and $51$ . This amounts to fifteen of the sixty possible minutes for any given hour. Hence the fraction of the day that the clock shows the correct time is $\frac 23 \cdot \left(1 - \frac {15}{60}\right) = \frac 23 \cdot \frac 34 = \boxed{12}$ | A | 12 |
d1c6fed604d161f05d1771497f3653b6 | https://artofproblemsolving.com/wiki/index.php/2009_AMC_12B_Problems/Problem_12 | The fifth and eighth terms of a geometric sequence of real numbers are $7!$ and $8!$ respectively. What is the first term?
$\mathrm{(A)}\ 60\qquad \mathrm{(B)}\ 75\qquad \mathrm{(C)}\ 120\qquad \mathrm{(D)}\ 225\qquad \mathrm{(E)}\ 315$ | Let the $n$ th term of the series be $ar^{n-1}$ . Because \[\frac {8!}{7!} = \frac {ar^7}{ar^4} = r^3 = 8,\] it follows that $r = 2$ and the first term is $a = \frac {7!}{r^4} = \frac {7!}{16} = \boxed{315}$ | E | 315 |
2c9861a6554b6d7ae3817100d95d3516 | https://artofproblemsolving.com/wiki/index.php/2009_AMC_12B_Problems/Problem_13 | Triangle $ABC$ has $AB = 13$ and $AC = 15$ , and the altitude to $\overline{BC}$ has length $12$ . What is the sum of the two possible values of $BC$
$\mathrm{(A)}\ 15\qquad \mathrm{(B)}\ 16\qquad \mathrm{(C)}\ 17\qquad \mathrm{(D)}\ 18\qquad \mathrm{(E)}\ 19$ | Let $D$ be the foot of the altitude to $\overline BC$ . Then $BD = \sqrt {13^2 - 12^2} = 5$ and $DC = \sqrt {15^2 - 12^2} = 9$ . Thus $BC = BD + BC = 5 + 9 = 14$ or assume that the triangle is obtuse at angle $B$ then $BC = DC - BD = 9 -5 = 4$ . The sum of the two possible values is $14 + 4 = \boxed{18}$ | D | 18 |
d92205045da3b22da4b0bf925e2f4ff1 | https://artofproblemsolving.com/wiki/index.php/2009_AMC_12B_Problems/Problem_14 | Five unit squares are arranged in the coordinate plane as shown, with the lower left corner at the origin. The slanted line, extending from $(c,0)$ to $(3,3)$ , divides the entire region into two regions of equal area. What is $c$
[asy] unitsize(0.2cm); defaultpen(linewidth(.8pt)+fontsize(8pt)); fill((2/3,0)--(3,3)--(3,1)--(2,1)--(2,0)--cycle,gray); xaxis("$x$",-0.5,4,EndArrow(HookHead,4)); yaxis("$y$",-0.5,4,EndArrow(4)); draw((0,1)--(3,1)--(3,3)--(2,3)--(2,0)); draw((1,0)--(1,2)--(3,2)); draw((2/3,0)--(3,3)); label("$(c,0)$",(2/3,0),S); label("$(3,3)$",(3,3),NE); [/asy]
$\textbf{(A)}\ \frac12\qquad \textbf{(B)}\ \frac35\qquad \textbf{(C)}\ \frac23\qquad \textbf{(D)}\ \frac34\qquad \textbf{(E)}\ \frac45$ | For $c\geq 1.5$ the shaded area is at most $1.5$ , which is too little. Hence $c<1.5$ , and therefore the point $(2,1)$ is indeed inside the shaded part, as shown in the picture.
Then the area of the shaded part is one less than the area of the triangle with vertices $(c,0)$ $(3,0)$ , and $(3,3)$ . The area of the entire triangle is $\frac{3(3-c)}2$ , therefore the area of the shaded part is $\frac{7-3c}{2}$
The entire figure has area $5$ , hence we want the shaded part to have area $\frac 52$ . Solving for $c$ , we get $c=\boxed{23}$ | C | 23 |
d92205045da3b22da4b0bf925e2f4ff1 | https://artofproblemsolving.com/wiki/index.php/2009_AMC_12B_Problems/Problem_14 | Five unit squares are arranged in the coordinate plane as shown, with the lower left corner at the origin. The slanted line, extending from $(c,0)$ to $(3,3)$ , divides the entire region into two regions of equal area. What is $c$
[asy] unitsize(0.2cm); defaultpen(linewidth(.8pt)+fontsize(8pt)); fill((2/3,0)--(3,3)--(3,1)--(2,1)--(2,0)--cycle,gray); xaxis("$x$",-0.5,4,EndArrow(HookHead,4)); yaxis("$y$",-0.5,4,EndArrow(4)); draw((0,1)--(3,1)--(3,3)--(2,3)--(2,0)); draw((1,0)--(1,2)--(3,2)); draw((2/3,0)--(3,3)); label("$(c,0)$",(2/3,0),S); label("$(3,3)$",(3,3),NE); [/asy]
$\textbf{(A)}\ \frac12\qquad \textbf{(B)}\ \frac35\qquad \textbf{(C)}\ \frac23\qquad \textbf{(D)}\ \frac34\qquad \textbf{(E)}\ \frac45$ | The unit square is of area 1, so the five unit squares have area 5.
Therefore the shaded space must occupy 2.5. The missing unit square is of area 1, and if reconstituted the original triangle would be of area 3.5.
It can then be inferred: $(3-c) * 3 = 7$
$3-c=\frac{7}{3}$ , so $3-\frac{7}{3}=c$
$3-\frac{7}{3} = \frac{9-7}{3} = \frac{2}{3}$ $\boxed{23}$ | C | 23 |
d92205045da3b22da4b0bf925e2f4ff1 | https://artofproblemsolving.com/wiki/index.php/2009_AMC_12B_Problems/Problem_14 | Five unit squares are arranged in the coordinate plane as shown, with the lower left corner at the origin. The slanted line, extending from $(c,0)$ to $(3,3)$ , divides the entire region into two regions of equal area. What is $c$
[asy] unitsize(0.2cm); defaultpen(linewidth(.8pt)+fontsize(8pt)); fill((2/3,0)--(3,3)--(3,1)--(2,1)--(2,0)--cycle,gray); xaxis("$x$",-0.5,4,EndArrow(HookHead,4)); yaxis("$y$",-0.5,4,EndArrow(4)); draw((0,1)--(3,1)--(3,3)--(2,3)--(2,0)); draw((1,0)--(1,2)--(3,2)); draw((2/3,0)--(3,3)); label("$(c,0)$",(2/3,0),S); label("$(3,3)$",(3,3),NE); [/asy]
$\textbf{(A)}\ \frac12\qquad \textbf{(B)}\ \frac35\qquad \textbf{(C)}\ \frac23\qquad \textbf{(D)}\ \frac34\qquad \textbf{(E)}\ \frac45$ | The shaded space of the object can become a triangle by adding a unit square to its bottom. This triangle has a base length of $3-c$ and a height of $3$ . The area of this triangle region is now (using the formula $A=bh/2$ for a triangle) $(9-3c)/2$ . But, remember that we have to subtract the area of the extra unit square from this area to get the area of the shaded region.
If you add 3 unit squares to the top of the unshaded portion, the figure is now a right trapezoid. Using the formula for a trapezoid, you can derive that its area is $(9+3c)/2$ . After you subtract the area of the 3 extra unit squares, we have derived the area of the unshaded portion.
Since these areas are both equal, we set them equal to each other and solve for $c$ $(9-3c)/2 -1 = (9+3c)/2 -3$ $c$ is now solved to be $\frac{2}{3}$ $\boxed{23}$ | C | 23 |
d92205045da3b22da4b0bf925e2f4ff1 | https://artofproblemsolving.com/wiki/index.php/2009_AMC_12B_Problems/Problem_14 | Five unit squares are arranged in the coordinate plane as shown, with the lower left corner at the origin. The slanted line, extending from $(c,0)$ to $(3,3)$ , divides the entire region into two regions of equal area. What is $c$
[asy] unitsize(0.2cm); defaultpen(linewidth(.8pt)+fontsize(8pt)); fill((2/3,0)--(3,3)--(3,1)--(2,1)--(2,0)--cycle,gray); xaxis("$x$",-0.5,4,EndArrow(HookHead,4)); yaxis("$y$",-0.5,4,EndArrow(4)); draw((0,1)--(3,1)--(3,3)--(2,3)--(2,0)); draw((1,0)--(1,2)--(3,2)); draw((2/3,0)--(3,3)); label("$(c,0)$",(2/3,0),S); label("$(3,3)$",(3,3),NE); [/asy]
$\textbf{(A)}\ \frac12\qquad \textbf{(B)}\ \frac35\qquad \textbf{(C)}\ \frac23\qquad \textbf{(D)}\ \frac34\qquad \textbf{(E)}\ \frac45$ | We are looking for the area of the shaded region to be $\frac 52$ . We start by testing $(A) \frac 12$ . The area of the shaded region would be $\frac{(3-\frac 12)(3) }{2}-1=\frac {11}{4}$ when $c=\frac 12$ . This does not match our wanted answer.
We try $(B) \frac 35$ next. The area of the shaded region would be $\frac{(3-\frac 35)(3) }{2}-1=\frac {13}{5}$ when $c=\frac 35$ . This also does not match our desired answer.
We then try $(C) \frac 23$ . The area of the shaded region would be $\frac{(3-\frac 23)(3) }{2}-1=\frac {5}{2}$ when $c=\frac 23$ . Therefore, our answer is $\boxed{23}$ | C | 23 |
5510db61c37f538a0a02f99a1d891d09 | https://artofproblemsolving.com/wiki/index.php/2009_AMC_12B_Problems/Problem_16 | Trapezoid $ABCD$ has $AD||BC$ $BD = 1$ $\angle DBA = 23^{\circ}$ , and $\angle BDC = 46^{\circ}$ . The ratio $BC: AD$ is $9: 5$ . What is $CD$
$\mathrm{(A)}\ \frac 79\qquad \mathrm{(B)}\ \frac 45\qquad \mathrm{(C)}\ \frac {13}{15}\qquad \mathrm{(D)}\ \frac 89\qquad \mathrm{(E)}\ \frac {14}{15}$ | Extend $\overline {AB}$ and $\overline {DC}$ to meet at $E$ . Then
\begin{align*} \angle BED &= 180^{\circ} - \angle EDB - \angle DBE\\ &= 180^{\circ} - 134^{\circ} -23^{\circ} = 23^{\circ}. \end{align*}
Thus $\triangle BDE$ is isosceles with $DE = BD$ . Because $\overline {AD} \parallel \overline {BC}$ , it follows that the triangles $BCE$ and $ADE$ are similar. Therefore \[\frac 95 = \frac {BC}{AD} = \frac {CD + DE}{DE} = \frac {CD}{BD} + 1 = CD + 1,\] so $CD = \boxed{45}.$ | null | 45 |
b8d78cc1cb3aa0936d1e06877bdf153c | https://artofproblemsolving.com/wiki/index.php/2009_AMC_12B_Problems/Problem_19 | For each positive integer $n$ , let $f(n) = n^4 - 360n^2 + 400$ . What is the sum of all values of $f(n)$ that are prime numbers?
$\textbf{(A)}\ 794\qquad \textbf{(B)}\ 796\qquad \textbf{(C)}\ 798\qquad \textbf{(D)}\ 800\qquad \textbf{(E)}\ 802$ | To find the answer it was enough to play around with $f$ . One can easily find that $f(1)=41$ is a prime, then $f$ becomes negative for $n$ between $2$ and $18$ , and then $f(19)=761$ is again a prime number. And as $41 + 761 = 802$ is already the largest option, the answer must be $\boxed{802}$ | null | 802 |
b8d78cc1cb3aa0936d1e06877bdf153c | https://artofproblemsolving.com/wiki/index.php/2009_AMC_12B_Problems/Problem_19 | For each positive integer $n$ , let $f(n) = n^4 - 360n^2 + 400$ . What is the sum of all values of $f(n)$ that are prime numbers?
$\textbf{(A)}\ 794\qquad \textbf{(B)}\ 796\qquad \textbf{(C)}\ 798\qquad \textbf{(D)}\ 800\qquad \textbf{(E)}\ 802$ | We will now show a complete solution, with a proof that no other values are prime.
Consider the function $g(x) = x^2 - 360x + 400$ , then obviously $f(x) = g(x^2)$
The roots of $g$ are: \[x_{1,2} = \frac{ 360 \pm \sqrt{ 360^2 - 4\cdot 400 } }2 = 180 \pm 80 \sqrt 5\]
We can then write $g(x) = (x - 180 - 80\sqrt 5)(x - 180 + 80\sqrt 5)$ , and thus $f(x) = (x^2 - 180 - 80\sqrt 5)(x^2 - 180 + 80\sqrt 5)$
We would now like to factor the right hand side further, using the formula $(x^2 - y^2) = (x-y)(x+y)$ . To do this, we need to express both constants as squares of some other constants. Luckily, we have a pretty good idea how they look like.
We are looking for rational $a$ and $b$ such that $(a+b\sqrt 5)^2 = 180 + 80\sqrt 5$ . Expanding the left hand side and comparing coefficients, we get $ab=40$ and $a^2 + 5b^2 = 180$ . We can easily guess (or compute) the solution $a=10$ $b=4$
Hence $180 + 80\sqrt 5 = (10 + 4\sqrt 5)^2$ , and we can easily verify that also $180 - 80\sqrt 5 = (10 - 4\sqrt 5)^2$
We now know the complete factorization of $f(x)$
\[f(x) = (x - 10 - 4\sqrt 5)(x + 10 + 4\sqrt 5)(x - 10 + 4\sqrt 5)(x + 10 - 4\sqrt 5)\]
As the final step, we can now combine the factors in a different way, in order to get rid of the square roots.
We have $(x - 10 - 4\sqrt 5)(x - 10 + 4\sqrt 5) = (x-10)^2 - (4\sqrt 5)^2 = x^2 - 20x + 20$ ,
and $(x + 10 - 4\sqrt 5)(x + 10 + 4\sqrt 5) = x^2 + 20x + 20$
Hence we obtain the factorization $f(x) = (x^2 - 20x + 20)(x^2 + 20x + 20)$
For $x\geq 20$ both terms are positive and larger than one, hence $f(x)$ is not prime. For $1<x<19$ the second factor is positive and the first one is negative, hence $f(x)$ is not a prime. The remaining cases are $x=1$ and $x=19$ . In both cases, $f(x)$ is indeed a prime, and their sum is $f(1) + f(19) = 41 + 761 = \boxed{802}$ | null | 802 |
8a30996d98a8199aaa54356a55e0970e | https://artofproblemsolving.com/wiki/index.php/2009_AMC_12B_Problems/Problem_20 | A convex polyhedron $Q$ has vertices $V_1,V_2,\ldots,V_n$ , and $100$ edges. The polyhedron is cut by planes $P_1,P_2,\ldots,P_n$ in such a way that plane $P_k$ cuts only those edges that meet at vertex $V_k$ . In addition, no two planes intersect inside or on $Q$ . The cuts produce $n$ pyramids and a new polyhedron $R$ . How many edges does $R$ have?
$\mathrm{(A)}\ 200\qquad \mathrm{(B)}\ 2n\qquad \mathrm{(C)}\ 300\qquad \mathrm{(D)}\ 400\qquad \mathrm{(E)}\ 4n$ | Each edge of $Q$ is cut by two planes, so $R$ has $200$ vertices. Three edges of $R$ meet at each vertex, so $R$ has $\frac 12 \cdot 3 \cdot 200 = \boxed{300}$ edges. | null | 300 |
8a30996d98a8199aaa54356a55e0970e | https://artofproblemsolving.com/wiki/index.php/2009_AMC_12B_Problems/Problem_20 | A convex polyhedron $Q$ has vertices $V_1,V_2,\ldots,V_n$ , and $100$ edges. The polyhedron is cut by planes $P_1,P_2,\ldots,P_n$ in such a way that plane $P_k$ cuts only those edges that meet at vertex $V_k$ . In addition, no two planes intersect inside or on $Q$ . The cuts produce $n$ pyramids and a new polyhedron $R$ . How many edges does $R$ have?
$\mathrm{(A)}\ 200\qquad \mathrm{(B)}\ 2n\qquad \mathrm{(C)}\ 300\qquad \mathrm{(D)}\ 400\qquad \mathrm{(E)}\ 4n$ | At each vertex, as many new edges are created by this process as there are original edges meeting at that vertex. Thus the total number of new edges is the total number of endpoints of the original edges, which is $200$ . A middle portion of each original edge is also present in $R$ , so $R$ has $100 + 200 = \boxed{300}$ edges. | null | 300 |
8a30996d98a8199aaa54356a55e0970e | https://artofproblemsolving.com/wiki/index.php/2009_AMC_12B_Problems/Problem_20 | A convex polyhedron $Q$ has vertices $V_1,V_2,\ldots,V_n$ , and $100$ edges. The polyhedron is cut by planes $P_1,P_2,\ldots,P_n$ in such a way that plane $P_k$ cuts only those edges that meet at vertex $V_k$ . In addition, no two planes intersect inside or on $Q$ . The cuts produce $n$ pyramids and a new polyhedron $R$ . How many edges does $R$ have?
$\mathrm{(A)}\ 200\qquad \mathrm{(B)}\ 2n\qquad \mathrm{(C)}\ 300\qquad \mathrm{(D)}\ 400\qquad \mathrm{(E)}\ 4n$ | Euler's Polyhedron Formula applied to $Q$ gives $n - 100 + F = 2$ , where F is the number of faces of $Q$ . Each edge of $Q$ is cut by two planes, so $R$ has $200$ vertices. Each cut by a plane $P_k$ creates an additional face on $R$ , so Euler's Polyhedron Formula applied to $R$ gives $200 - E + (F+n) = 2$ , where $E$ is the number of edges of $R$ . Subtracting the first equation from the second gives $300 - E = 0$ , whence $E = \boxed{300}$ | C | 300 |
8a30996d98a8199aaa54356a55e0970e | https://artofproblemsolving.com/wiki/index.php/2009_AMC_12B_Problems/Problem_20 | A convex polyhedron $Q$ has vertices $V_1,V_2,\ldots,V_n$ , and $100$ edges. The polyhedron is cut by planes $P_1,P_2,\ldots,P_n$ in such a way that plane $P_k$ cuts only those edges that meet at vertex $V_k$ . In addition, no two planes intersect inside or on $Q$ . The cuts produce $n$ pyramids and a new polyhedron $R$ . How many edges does $R$ have?
$\mathrm{(A)}\ 200\qquad \mathrm{(B)}\ 2n\qquad \mathrm{(C)}\ 300\qquad \mathrm{(D)}\ 400\qquad \mathrm{(E)}\ 4n$ | Each edge connects two points. The plane cuts that edge so it splits into $2$ at each end (like two legs) for a total of $4$ new edges.
[asy] pair A,B,C,D,E,F; A=(0,50); B=(0,-50); C=(86.602540378443864676372317075294,0); D=(300,0); E=(370.7,70.7); F=(370.7,-70.7); draw(A--C--B); draw(C--D); draw(E--D--F); [/asy]
But because each new edge is shared by an adjacent original edge cut similarly, the additional edges are overcounted $\times 2$
[asy] pair A,B,C,D,E,F,G,H,I,J,K; A=(0,50); B=(0,-50); C=(86.602540378443864676372317075294,0); D=(300,0); E=(370.7,70.7); F=(370.7,-70.7); G=(-107.5,236.2); H=(-107.5,-236.2); I=(370.7,285); J=(370.7,-285); K=(441.4,0); draw(A--C--B); draw(C--D); draw(E--D--F); draw(G--A, red); draw(H--B, red); draw(I--E, red); draw(J--F, red); draw(A--B, dashed); draw(E--K, dashed); draw(F--K, dashed); [/asy]
Since there are $100$ edges to start with, $400/2=200$ new edges result. So there are $100+200=\boxed{300}$ edges in the figure. | C | 300 |
8a30996d98a8199aaa54356a55e0970e | https://artofproblemsolving.com/wiki/index.php/2009_AMC_12B_Problems/Problem_20 | A convex polyhedron $Q$ has vertices $V_1,V_2,\ldots,V_n$ , and $100$ edges. The polyhedron is cut by planes $P_1,P_2,\ldots,P_n$ in such a way that plane $P_k$ cuts only those edges that meet at vertex $V_k$ . In addition, no two planes intersect inside or on $Q$ . The cuts produce $n$ pyramids and a new polyhedron $R$ . How many edges does $R$ have?
$\mathrm{(A)}\ 200\qquad \mathrm{(B)}\ 2n\qquad \mathrm{(C)}\ 300\qquad \mathrm{(D)}\ 400\qquad \mathrm{(E)}\ 4n$ | The question specifies the slices create as many pyramids as there are vertices, implying each vertex owns 4 edge ends. There are twice as many edge-ends as there are edges, and $2 \cdot 100 = 200$
$\frac{200}{4} = 50$ , so there are $50$ vertices.
The base of a pyramid has 4 edges, so each sliced vertex would add four edges to $R$
$100 + 4 \cdot 50$ $\boxed{300}$ | C | 300 |
8a30996d98a8199aaa54356a55e0970e | https://artofproblemsolving.com/wiki/index.php/2009_AMC_12B_Problems/Problem_20 | A convex polyhedron $Q$ has vertices $V_1,V_2,\ldots,V_n$ , and $100$ edges. The polyhedron is cut by planes $P_1,P_2,\ldots,P_n$ in such a way that plane $P_k$ cuts only those edges that meet at vertex $V_k$ . In addition, no two planes intersect inside or on $Q$ . The cuts produce $n$ pyramids and a new polyhedron $R$ . How many edges does $R$ have?
$\mathrm{(A)}\ 200\qquad \mathrm{(B)}\ 2n\qquad \mathrm{(C)}\ 300\qquad \mathrm{(D)}\ 400\qquad \mathrm{(E)}\ 4n$ | I doubt anyone has the mental capacity to imagine $Q$ in its full complexity in their mind. So, we begin by examining a simpler structure: a cube.
Let $E_C$ denote the number of edges of a polyhedron $C$ . Let $V_C$ denote the number of vertices of $C$ . Let $B_C$ denote the number of edges that meet at each vertex.
Suppose $C$ is a cube. Then
$E_C = 12$ $V_C = 8$ , and $B_C = 3$
(A cube has $12$ edges, $8$ vertices, and $3$ edges meet per vertex)
Notice that
$V_C*B_C = 2E_C$ $8*3 = 2*12$
Why? Well, imagine an ant walking along an edge of a cube. When it reaches the end of the edge, it will be standing on a vertex (by definition). But, what if it had walked the other way? Well, it would still be on the same edge, but it would reach the other endpoint of the edge, another vertex. Therefore, we can say that a polyhedron with $E$ edges has $2E$ total endpoints. In the case of $C$ , a cube, this means a cube has $2*12 = 24$ total endpoints. But a cube only has $8$ vertices, not $24$ ... This doesn't make any sense... Wait a minute, we're dealing with a polyhedron, which means multiple edges will be connected to the same vertex. In the case of a cube, $3$ edges meet at every vertex. So we have $24 \div 3$ , or $8$ vertices. The math works out. Let's apply this to $Q$ , our mystery polyhedron.
We are told that $E_Q = 100$ $V_Q = n$ and $B_Q$ isn't given explicitly. So we'll let $B_Q = b$ . Now our equation reads
$bn = 200$
What does this mean, exactly? Well $Q$ has $100$ edges, so it must has $200$ endpoints (endpoints, not vertices). So if we can figure out either $n$ , the number of vertices, or $b$ , the number of edges per vertex, we can figure out the other variable. Hmmmm... The problem doesn't give us any clues about either. How frustrating. Let's leave this alone and consider the other information given to us.
We are also told that "the polyhedron is cut by planes $P_1, P_2,...,P_n$ " (at this point I imagine $Q$ is a giant block of cheese, potato, etc.), and also that "plane $P_k$ cuts only those edges that meet at $V_k$ ". What does this mean?
Well, let's go back to our cube. Notice that there are exactly as many planes as vertices. Think of the planes as knives (or knife strokes) cutting into a big cube of cheese. We're cutting the cheese $8$ times, since there are $8$ vertices. Also, whenever we cut into the cheese, we have to direct the cuts near a vertex, cutting into only the edges connected to that vertex. How do we do that? Well, since we can't cut into any other edges, we're gonna have to settle for cutting corners (see what I did there? ...Ok, I admit it was cheesy... see what I did there, too?😂). We have to shave off just the corners of the cheese block, because this way it won't interfere with any of the other edges.
The problem also tells us that none of the planes overlap. In our cheese analogy, this means that every time we cut, we have to make sure the cuts are always into new cheese, and not into previously cut areas (all of the cuts are distinct, and don't overlap). Sure enough, this is in agreement with the information that exactly $n$ pyramids are cut off. Let's go back to $Q$
$Q$ has $n$ vertices, so we're gonna make $n$ cuts. The question we want to answer is: "After all of the cuts are made, how many total edges will $R$ (the figure that remains), have?". Well, we're gonna have to figure out either $n$ or $b$ ... or do we? We're told that $n$ pyramids were made, but it doesn't specify exactly how many sides the pyramid has. So... that implies that it doesn't matter, or else the problem wouldn't be solvable. Let's suppose $b = 4$ $4$ edges meet at every vertex). Then we would have $200 \div 4 = 50$ vertices, and we would make $50$ cuts. But how many more edges does it yield? (We are told that the planes don't overlap, so the original edges are all present.) Well, if we cut off the corner of a cube, it will create $3$ more edges... which is coincident with the fact that $3$ edges meet at every vertex. So for every vertex, we gain exactly $b$ new edges. Rewriting the equation to find the number of vertices, we have
$v = \frac{200}{b}$
But we just said that for every vertex/cut, we gain $b$ more edges. So the total number of new edges is $v*b \Rightarrow \frac{200}{b}*b \Rightarrow 200$ . It doesn't matter how many vertices $Q$ has, or how many edges meet at every vertex. Given the conditions in the problem, we will always gain $200$ more edges, regardless of the values of $n$ , or $b$ . We have $200$ new edges $+$ $100$ old edges to make $300$ edges $\Rightarrow \boxed{300}$ | C | 300 |
10e200fe6eed40bf88a238a10923fd53 | https://artofproblemsolving.com/wiki/index.php/2009_AMC_12B_Problems/Problem_21 | Ten women sit in $10$ seats in a line. All of the $10$ get up and then reseat themselves using all $10$ seats, each sitting in the seat she was in before or a seat next to the one she occupied before. In how many ways can the women be reseated?
$\textbf{(A)}\ 89\qquad \textbf{(B)}\ 90\qquad \textbf{(C)}\ 120\qquad \textbf{(D)}\ 2^{10}\qquad \textbf{(E)}\ 2^2 3^8$ | Notice that either a woman stays in her own seat after the rearrangement, or two adjacent women swap places. Thus, our answer is counting the number of ways to arrange 1x1 and 2x1 blocks to form a 1x10 rectangle. This can be done via casework depending on the number of 2x1 blocks. The cases of 0, 1, 2, 3, 4, 5 2x1 blocks correspond to 10, 8, 6, 4, 2, 0 1x1 blocks, and so the sum of the cases is \[\binom{10}{0} + \binom{9}{1} + \binom{8}{2} + \binom{7}{3} + \binom{6}{4} + \binom{5}{5} = 1 + 9 + 28 + 35 + 15 + 1 = \boxed{89}.\] | null | 89 |
10e200fe6eed40bf88a238a10923fd53 | https://artofproblemsolving.com/wiki/index.php/2009_AMC_12B_Problems/Problem_21 | Ten women sit in $10$ seats in a line. All of the $10$ get up and then reseat themselves using all $10$ seats, each sitting in the seat she was in before or a seat next to the one she occupied before. In how many ways can the women be reseated?
$\textbf{(A)}\ 89\qquad \textbf{(B)}\ 90\qquad \textbf{(C)}\ 120\qquad \textbf{(D)}\ 2^{10}\qquad \textbf{(E)}\ 2^2 3^8$ | Recall that the number of ways to arrange 1x1 and 2x1 blocks to form a 1xn rectangle results in Fibonacci numbers . Clearly, $\boxed{89}$ is the only fibonacci number, so no calculation is needed for this problem. | null | 89 |
10e200fe6eed40bf88a238a10923fd53 | https://artofproblemsolving.com/wiki/index.php/2009_AMC_12B_Problems/Problem_21 | Ten women sit in $10$ seats in a line. All of the $10$ get up and then reseat themselves using all $10$ seats, each sitting in the seat she was in before or a seat next to the one she occupied before. In how many ways can the women be reseated?
$\textbf{(A)}\ 89\qquad \textbf{(B)}\ 90\qquad \textbf{(C)}\ 120\qquad \textbf{(D)}\ 2^{10}\qquad \textbf{(E)}\ 2^2 3^8$ | Let $S_n$ be the number of possible seating arrangements with $n$ women. Consider $n \ge 3,$ and focus on the rightmost woman. If she returns back to her seat, then there are $S_{n-1}$ ways to seat the remaining $n-1$ women. If she sits in the second to last seat, then the woman who previously sat there must now sit at the rightmost seat. This gives us $S_{n-2}$ ways to seat the other $n-2$ women, so we obtain the recursion \[S_n = S_{n-1}+S_{n-2}.\]
Starting with $S_1=1$ and $S_2=2,$ we can calculate $S_{10}=\boxed{89}.$ | null | 89 |
581e5cd641338f04e2255c83c67ce0b0 | https://artofproblemsolving.com/wiki/index.php/2009_AMC_12B_Problems/Problem_22 | Parallelogram $ABCD$ has area $1,\!000,\!000$ . Vertex $A$ is at $(0,0)$ and all other vertices are in the first quadrant. Vertices $B$ and $D$ are lattice points on the lines $y = x$ and $y = kx$ for some integer $k > 1$ , respectively. How many such parallelograms are there? (A lattice point is any point whose coordinates are both integers.)
$\textbf{(A)}\ 49\qquad \textbf{(B)}\ 720\qquad \textbf{(C)}\ 784\qquad \textbf{(D)}\ 2009\qquad \textbf{(E)}\ 2048$ | The area of any parallelogram $ABCD$ can be computed as the size of the vector product of $\overrightarrow{AB}$ and $\overrightarrow{AD}$
In our setting where $A=(0,0)$ $B=(s,s)$ , and $D=(t,kt)$ this is simply $s\cdot kt - s\cdot t = (k-1)st$
In other words, we need to count the triples of integers $(k,s,t)$ where $k>1$ $s,t>0$ and $(k-1)st = 1,\!000,\!000 = 2^6 5^6$
These can be counted as follows: We have $6$ identical red balls (representing powers of $2$ ), $6$ blue balls (representing powers of $5$ ), and three labeled urns (representing the factors $k-1$ $s$ , and $t$ ). The red balls can be distributed in ${8\choose 2} = 28$ ways, and for each of these ways, the blue balls can then also be distributed in $28$ ways. (See Distinguishability for a more detailed explanation.)
Thus there are exactly $28^2 = 784$ ways how to break $1,\!000,\!000$ into three positive integer factors, and for each of them we get a single parallelogram. Hence the number of valid parallelograms is $784 \longrightarrow \boxed{784}$ | C | 784 |
581e5cd641338f04e2255c83c67ce0b0 | https://artofproblemsolving.com/wiki/index.php/2009_AMC_12B_Problems/Problem_22 | Parallelogram $ABCD$ has area $1,\!000,\!000$ . Vertex $A$ is at $(0,0)$ and all other vertices are in the first quadrant. Vertices $B$ and $D$ are lattice points on the lines $y = x$ and $y = kx$ for some integer $k > 1$ , respectively. How many such parallelograms are there? (A lattice point is any point whose coordinates are both integers.)
$\textbf{(A)}\ 49\qquad \textbf{(B)}\ 720\qquad \textbf{(C)}\ 784\qquad \textbf{(D)}\ 2009\qquad \textbf{(E)}\ 2048$ | We know that $A$ is $(0, 0)$ . Since $B$ is on the line $y=x$ , let it be represented by the point $(b, b)$ . Similarly, let $D$ be $(d, kd)$ . Since this is a parallelogram, sides $\overline{AD}$ and $\overline{BC}$ are parallel. Therefore, the distance and relative position of $D$ to $A$ is equivalent to that of $C$ to $B$ (if we take the translation of $A$ to $D$ and apply it to $B$ , we will get the coordinates of $C$ ). This yields $C (b+d, b+kd)$ . Using the Shoelace Theorem we get
$1,000,000 = \frac{1}{2}|\left(b(b+kd) + (b+d)(kd)\right) - \left(b(b+d) + (b+kd)d\right)|$
$\Rightarrow 2,000,000 = |2kbd - 2bd|$
$\Rightarrow 1,000,000 = |kbd - bd|$
Since $k > 1, kbd > bd$ . The equation becomes
$1,000,000 = (k-1)bd$
$\Rightarrow \frac{1,000,000}{k-1} = bd$
Since $k$ must be a positive integer greater than $1$ , we know $k-1$ will be a positive integer. We also know that $bd$ is an integer, so $k-1$ must be a factor of $1,000,000$ . Therefore $bd$ will also be a factor of $1,000,000$
Notice that $1,000,000 = 10^6 = 2^6*5^6$
Let $b$ be $2^x5^y$ such that $x, y$ are integers on the interval $[0, 6]$
Let $d$ be $2^w5^z$ such that $w, z$ are integers, $x+w \le 6$ , and $y+z \le 6$
For a pair $(x, y)$ , there are $7-x$ possibilities for $w$ and $7-y$ possibilites for $z$ $d$ doesn't have to be the co-factor of $1,000,000$ , it just can't be big enough such that $bd > 1,000,000$ ), for a total of $(7-x)(7-y)$ possibilities. So we want
$\sum_{k=0, i=0}^6 (7-k)(7-i)$
$\left(\text{the sum of the number of possible pairs}(w, z) \text{ for all pairs}(k , i)\text{ for } k[0, 6]\text{ and } i[0, 6]\right)$
Notice that if we "fix" the value of $k$ , at, say $6$ , then run through all of the values of $i$ , change the value of $k$ to $5$ , and run through all of the values of $i$ again, and so on until we exhaust all $49$ combinations of $(k, i)$ we get something like this:
$1*1 + 1*2 + ... + 1*6 + 1*7 + 2*1 + 2*2 + ... + 2*6 + 2*7 + ..... + 7*1 + 7*2 + ... + 7*6 + 7*7$
which can be rewritten
$1(1+2+...+7)+2(1+2+...+7)+.....+7(1+2+...+7)$
$\Rightarrow (1+2+...+7)(1+2+...+7)$
$\Rightarrow 28^2$
$\Rightarrow 784$
So there are $784$ possible sets of coordinates $B,$ $C$ , and $D \Rightarrow \boxed{784}$ | C | 784 |
a3ece1dbea30099755fb0412cd616c24 | https://artofproblemsolving.com/wiki/index.php/2009_AMC_12B_Problems/Problem_23 | A region $S$ in the complex plane is defined by \[S = \{x + iy: - 1\le x\le1, - 1\le y\le1\}.\] A complex number $z = x + iy$ is chosen uniformly at random from $S$ . What is the probability that $\left(\frac34 + \frac34i\right)z$ is also in $S$
$\textbf{(A)}\ \frac12\qquad \textbf{(B)}\ \frac23\qquad \textbf{(C)}\ \frac34\qquad \textbf{(D)}\ \frac79\qquad \textbf{(E)}\ \frac78$ | We multiply $z$ and $(\frac{3}{4}+\frac{3}{4}i)$ to get \[(\frac{3}{4}x-\frac{3}{4}y)+(\frac{3}{4}xi+\frac{3}{4}yi).\] Since we want to find the probability that this number is in $S$ , we need the real and complex coefficients of this number to be less than or equal to $1$ or greater than or equal to $-1.$ This gives us the equations \[-1\le \frac{3}{4}x-\frac{3}{4}y \le 1\] and \[-1\le \frac{3}{4}x+\frac{3}{4}y\le 1.\] Now, we see that we can solve this by graphing. We can graph our barriers $-1\le x\le 1$ and $-1\le y\le 1$ to form a $2$ by $2$ square centered at the origin. Graphing our two equations gives us the four lines \[x-y=\frac{4}{3},\] \[x-y=-\frac{4}{3},\] \[x+y=\frac{4}{3},\] \[x+y=-\frac{4}{3}.\] The square that is formed is the region that satisfies these four equations. Now, the barriers and this square gives us an octagon as the desired region. The area of this octagon is the total area of the square minus the 4 small triangles on each corner, each with $\frac{2}{9}$ area. Therefore, the octagon has area of $\frac{28}{9}.$ Finally, to find the probability of it working, we find the area of the octagon divided by the area of the entire square which is $\frac{\frac{28}{9}}{4}=\frac{7}{9}$ or $\boxed{79}.$ | D | 79 |
366bff86d307567641ee42ffd38d4c6a | https://artofproblemsolving.com/wiki/index.php/2009_AMC_12B_Problems/Problem_24 | For how many values of $x$ in $[0,\pi]$ is $\sin^{ - 1}(\sin 6x) = \cos^{ - 1}(\cos x)$ ?
Note: The functions $\sin^{ - 1} = \arcsin$ and $\cos^{ - 1} = \arccos$ denote inverse trigonometric functions.
$\textbf{(A)}\ 3\qquad \textbf{(B)}\ 4\qquad \textbf{(C)}\ 5\qquad \textbf{(D)}\ 6\qquad \textbf{(E)}\ 7$ | First of all, we have to agree on the range of $\sin^{-1}$ and $\cos^{-1}$ . This should have been a part of the problem statement -- but as it is missing, we will assume the most common definition: $\forall x: -\pi/2 \leq \sin^{-1}(x) \leq \pi/2$ and $\forall x: 0\leq \cos^{-1}(x) \leq \pi$
Hence we get that $\forall x\in[0,\pi]: \cos^{ - 1}(\cos x) = x$ , thus our equation simplifies to $\sin^{ - 1}(\sin 6x) = x$
Consider the function $f(x) = \sin^{ - 1}(\sin 6x) - x$ . We are looking for roots of $f$ on $[0,\pi]$
By analyzing properties of $\sin$ and $\sin^{-1}$ (or by computing the derivative of $f$ ) one can discover the following properties of $f$
For $x=\pi/6$ we have $f(x) = \sin^{-1} (\sin \pi) - \pi/6 = -\pi/6 < 0$ . Hence $f$ has exactly one root on $(0,\pi/6)$
For $x=2\pi/6$ we have $f(x) = \sin^{-1} (\sin 2\pi) - 2\pi/6 = -2\pi/6 < 0$ . Hence $f$ is negative on the entire interval $[\pi/6,2\pi/6]$
Now note that $\forall t: \sin^{-1}(t) \leq \pi/2$ . Hence for $x > 3\pi/6$ we have $f(x) < 0$ , and we can easily check that $f(3\pi/6)<0$ as well.
Thus the only unknown part of $f$ is the interval $(2\pi/6,3\pi/6)$ . On this interval, $f$ is negative in both endpoints, and we know that it is first increasing and then decreasing. Hence there can be zero, one, or two roots on this interval.
To prove that there are two roots, it is enough to find any $x$ from this interval such that $f(x)>0$
A good guess is its midpoint, $x=5\pi/12$ , where the function $\sin^{-1}(\sin 6x)$ has its local maximum. We can evaluate: $f(x) = \sin^{-1}(\sin 5\pi/2) - 5\pi/12 = \pi/2 - 5\pi/12 = \pi/12 > 0$
Summary: The function $f$ has $\boxed{4}$ roots on $[0,\pi]$ : the first one is $0$ , the second one is in $(0,\pi/6)$ , and the last two are in $(2\pi/6,3\pi/6)$ | B | 4 |
366bff86d307567641ee42ffd38d4c6a | https://artofproblemsolving.com/wiki/index.php/2009_AMC_12B_Problems/Problem_24 | For how many values of $x$ in $[0,\pi]$ is $\sin^{ - 1}(\sin 6x) = \cos^{ - 1}(\cos x)$ ?
Note: The functions $\sin^{ - 1} = \arcsin$ and $\cos^{ - 1} = \arccos$ denote inverse trigonometric functions.
$\textbf{(A)}\ 3\qquad \textbf{(B)}\ 4\qquad \textbf{(C)}\ 5\qquad \textbf{(D)}\ 6\qquad \textbf{(E)}\ 7$ | Since $0\leq \cos^{-1}(a) \leq \pi$ for all $a$ , the equation reduces to $\sin^{-1}(\sin(6x)) = x$ . Since $-\pi/2 \leq \sin^{-1}(a) \leq \pi/2$ for all $a$ $0 \leq x \leq \pi/2$ . To make the problem easier, we will measure angles in degrees. We will consider each sixth of the interval $[0, 90]$
For $0 \leq x \leq 15$ $6x$ is in the first quadrant. Thus, $\sin^{-1}(\sin(6x)) = 6x$ . Setting this equal to $x$ yields the solution $x = 0$
For $15 \leq x \leq 30$ $6x$ is in the second quadrant. Thus, $\sin^{-1}(\sin(6x)) = 180 - 6x$ . This yields the solution $x = \frac{180}7$
For $30 \leq x \leq 45$ $6x$ is in the third quadrant. Thus, $\sin^{-1}(\sin(6x)) = 180 - 6x$ . As $\frac{180}{7}$ is not on the interval $30 \leq x \leq 45$ , this yields no solution.
For $45 \leq x \leq 60$ $6x$ is in the fourth quadrant. Thus, $\sin^{-1}(\sin(6x)) = 6x - 360$ . As $72$ is not on the interval $45 \leq x \leq 60$ , this yields no solution.
For $60 \leq x \leq 75$ $6x$ is in the first quadrant plus a full revolution. Thus, $\sin^{-1}(\sin(6x)) = 6x - 360$ . This yields the solution $x = 72$
For $75 \leq x \leq 90$ $6x$ is in the second quadrant plus a full revolution. Thus $\sin^{-1}(\sin(6x)) = 540 - 6x$ . This yields the solution $x = \frac{540}7$
There are $\boxed{4}$ solutions, $x=0$ $x=\pi/7$ $x=2\pi/5$ , and $x=3\pi/7$ | B | 4 |
366bff86d307567641ee42ffd38d4c6a | https://artofproblemsolving.com/wiki/index.php/2009_AMC_12B_Problems/Problem_24 | For how many values of $x$ in $[0,\pi]$ is $\sin^{ - 1}(\sin 6x) = \cos^{ - 1}(\cos x)$ ?
Note: The functions $\sin^{ - 1} = \arcsin$ and $\cos^{ - 1} = \arccos$ denote inverse trigonometric functions.
$\textbf{(A)}\ 3\qquad \textbf{(B)}\ 4\qquad \textbf{(C)}\ 5\qquad \textbf{(D)}\ 6\qquad \textbf{(E)}\ 7$ | Algebraically, the inverse function of a function should just cancel out, leaving $6x=x$ . However, upon inspection we find that the graphs of these "inverse function of the function" equations are also periodic, like their normal trig function counterparts, due to the fact that inverse trig functions will never return any angle value higher than $2\pi$ . But instead of a smooth wave, these graphs are made up of zigzags with slope $1$ and $-1$ . Trying a few values, we see that
$y = \arcsin{\sin{x}}$
peaks at $\frac{\pi}{2}$ and
$y = \arccos{\cos{x}}$
peaks at $\pi$
But we want the graph of $y = \arcsin{(\sin6{x})}$ , which has a period of $\frac{\pi}{3}$ instead of $2\pi$ . So this means the interval $[0, \pi]$ will show $3$ periods instead of $\frac{1}{2}$ of a period. Visually it would be lines with slopes $6$ and $-6$ . Using the graph paper given to us, we plot out the two equations according to the above and we see that they intersect $4$ times $\Rightarrow \boxed{4}$ | B | 4 |
366bff86d307567641ee42ffd38d4c6a | https://artofproblemsolving.com/wiki/index.php/2009_AMC_12B_Problems/Problem_24 | For how many values of $x$ in $[0,\pi]$ is $\sin^{ - 1}(\sin 6x) = \cos^{ - 1}(\cos x)$ ?
Note: The functions $\sin^{ - 1} = \arcsin$ and $\cos^{ - 1} = \arccos$ denote inverse trigonometric functions.
$\textbf{(A)}\ 3\qquad \textbf{(B)}\ 4\qquad \textbf{(C)}\ 5\qquad \textbf{(D)}\ 6\qquad \textbf{(E)}\ 7$ | The most conventional domain and range for $\cos^{-1} x$ is $x \in [-1, 1]$ and $\cos^{-1} x \in [0, \pi]$ . For $\cos x$ $x \in [0, \pi]$ and $\cos x \in [-1, 1]$ . Since the domain of $\cos x$ is equal to the range of $\cos^{-1} x$ , and the range of $\cos x$ is equal to the domain of $\cos^{-1} x$ $\cos^{-1} (\cos x) = x$
Meaning that $\sin^{ - 1}(\sin 6x) = x$ $\sin x = \sin (6x)$
By graphing $\sin x$ and $\sin 6x$ together, it can be seen that there are 7 intersections for $x \in [0, \pi]$
However, the range of $\sin^{-1} x$ is $[-\frac{\pi}{2}, \frac{\pi}{2}]$ . Therefore $x \in [0, \frac{\pi}{2}]$ for $x = \sin^{ - 1}(\sin 6x)$ . There are $\boxed{4}]$ | B | 4 |
d320ca53d9952b750fffb7c0851fc650 | https://artofproblemsolving.com/wiki/index.php/2009_AMC_12B_Problems/Problem_25 | The set $G$ is defined by the points $(x,y)$ with integer coordinates, $3\le|x|\le7$ $3\le|y|\le7$ . How many squares of side at least $6$ have their four vertices in $G$
[asy] defaultpen(black+0.75bp+fontsize(8pt)); size(5cm); path p = scale(.15)*unitcircle; draw((-8,0)--(8.5,0),Arrow(HookHead,1mm)); draw((0,-8)--(0,8.5),Arrow(HookHead,1mm)); int i,j; for (i=-7;i<8;++i) { for (j=-7;j<8;++j) { if (((-7 <= i) && (i <= -3)) || ((3 <= i) && (i<= 7))) { if (((-7 <= j) && (j <= -3)) || ((3 <= j) && (j<= 7))) { fill(shift(i,j)*p,black); }}}} draw((-7,-.2)--(-7,.2),black+0.5bp); draw((-3,-.2)--(-3,.2),black+0.5bp); draw((3,-.2)--(3,.2),black+0.5bp); draw((7,-.2)--(7,.2),black+0.5bp); draw((-.2,-7)--(.2,-7),black+0.5bp); draw((-.2,-3)--(.2,-3),black+0.5bp); draw((-.2,3)--(.2,3),black+0.5bp); draw((-.2,7)--(.2,7),black+0.5bp); label("$-7$",(-7,0),S); label("$-3$",(-3,0),S); label("$3$",(3,0),S); label("$7$",(7,0),S); label("$-7$",(0,-7),W); label("$-3$",(0,-3),W); label("$3$",(0,3),W); label("$7$",(0,7),W); [/asy]
$\textbf{(A)}\ 125\qquad \textbf{(B)}\ 150\qquad \textbf{(C)}\ 175\qquad \textbf{(D)}\ 200\qquad \textbf{(E)}\ 225$ | We need to find a reasonably easy way to count the squares.
First, obviously the maximum distance between two points in the same quadrant is $4\sqrt 2 < 6$ , hence each square has exactly one vertex in each quadrant.
Given any square, we can circumscribe another axes-parallel square around it. In the picture below, the original square is red and the circumscribed one is blue.
[asy] defaultpen(black+0.75bp+fontsize(8pt)); size(7.5cm); path p = scale(.15)*unitcircle; draw((-8,0)--(8.5,0),Arrow(HookHead,1mm)); draw((0,-8)--(0,8.5),Arrow(HookHead,1mm)); int i,j; for (i=-7;i<8;++i) { for (j=-7;j<8;++j) { if (((-7 <= i) && (i <= -3)) || ((3 <= i) && (i<= 7))) { if (((-7 <= j) && (j <= -3)) || ((3 <= j) && (j<= 7))) { fill(shift(i,j)*p,black); }}}} draw((-7,-.2)--(-7,.2),black+0.5bp); draw((-3,-.2)--(-3,.2),black+0.5bp); draw((3,-.2)--(3,.2),black+0.5bp); draw((7,-.2)--(7,.2),black+0.5bp); draw((-.2,-7)--(.2,-7),black+0.5bp); draw((-.2,-3)--(.2,-3),black+0.5bp); draw((-.2,3)--(.2,3),black+0.5bp); draw((-.2,7)--(.2,7),black+0.5bp); label("$-7$",(-7,0),S); label("$-3$",(-3,0),S); label("$3$",(3,0),S); label("$7$",(7,0),S); label("$-7$",(0,-7),W); label("$-3$",(0,-3),W); label("$3$",(0,3),W); label("$7$",(0,7),W); draw( (5,3) -- (-3,4) -- (-4,-4) -- (4,-5) -- cycle, red ); draw( (5,4) -- (-4,4) -- (-4,-5) -- (5,-5) -- cycle, dashed + blue ); [/asy]
Let's now consider the opposite direction. Assume that we picked the blue square, how many different red squares do share it?
Answering this question is not as simple as it may seem. Consider the picture below. It shows all three red squares that share the same blue square. In addition, the picture shows a green square that is not valid, as two of its vertices are in bad locations.
[asy] defaultpen(black+0.75bp+fontsize(8pt)); size(7.5cm); path p = scale(.15)*unitcircle; draw((-8,0)--(8.5,0),Arrow(HookHead,1mm)); draw((0,-8)--(0,8.5),Arrow(HookHead,1mm)); int i,j; for (i=-7;i<8;++i) { for (j=-7;j<8;++j) { if (((-7 <= i) && (i <= -3)) || ((3 <= i) && (i<= 7))) { if (((-7 <= j) && (j <= -3)) || ((3 <= j) && (j<= 7))) { fill(shift(i,j)*p,black); }}}} draw((-7,-.2)--(-7,.2),black+0.5bp); draw((-3,-.2)--(-3,.2),black+0.5bp); draw((3,-.2)--(3,.2),black+0.5bp); draw((7,-.2)--(7,.2),black+0.5bp); draw((-.2,-7)--(.2,-7),black+0.5bp); draw((-.2,-3)--(.2,-3),black+0.5bp); draw((-.2,3)--(.2,3),black+0.5bp); draw((-.2,7)--(.2,7),black+0.5bp); label("$-7$",(-7,0),S); label("$-3$",(-3,0),S); label("$3$",(3,0),S); label("$7$",(7,0),S); label("$-7$",(0,-7),W); label("$-3$",(0,-3),W); label("$3$",(0,3),W); label("$7$",(0,7),W); draw( (5,4) -- (-4,4) -- (-4,-5) -- (5,-5) -- cycle, red ); draw( (5,3) -- (-3,4) -- (-4,-4) -- (4,-5) -- cycle, red ); draw( (5,-3) -- (-2,-5) -- (-4,2) -- (3,4) -- cycle, dashed + green ); draw( (5,-4) -- (-3,-5) -- (-4,3) -- (4,4) -- cycle, red ); draw( scale(1.05)*((5,4) -- (-4,4) -- (-4,-5) -- (5,-5) -- cycle), dashed + blue ); [/asy]
The size of the blue square can range from $6\times 6$ to $14\times 14$ , and for the intermediate sizes, there is more than one valid placement. We will now examine the cases one after another. Also, we can use symmetry to reduce the number of cases.
Summing the last column, we get that the answer is $\boxed{225}$ | null | 225 |
d320ca53d9952b750fffb7c0851fc650 | https://artofproblemsolving.com/wiki/index.php/2009_AMC_12B_Problems/Problem_25 | The set $G$ is defined by the points $(x,y)$ with integer coordinates, $3\le|x|\le7$ $3\le|y|\le7$ . How many squares of side at least $6$ have their four vertices in $G$
[asy] defaultpen(black+0.75bp+fontsize(8pt)); size(5cm); path p = scale(.15)*unitcircle; draw((-8,0)--(8.5,0),Arrow(HookHead,1mm)); draw((0,-8)--(0,8.5),Arrow(HookHead,1mm)); int i,j; for (i=-7;i<8;++i) { for (j=-7;j<8;++j) { if (((-7 <= i) && (i <= -3)) || ((3 <= i) && (i<= 7))) { if (((-7 <= j) && (j <= -3)) || ((3 <= j) && (j<= 7))) { fill(shift(i,j)*p,black); }}}} draw((-7,-.2)--(-7,.2),black+0.5bp); draw((-3,-.2)--(-3,.2),black+0.5bp); draw((3,-.2)--(3,.2),black+0.5bp); draw((7,-.2)--(7,.2),black+0.5bp); draw((-.2,-7)--(.2,-7),black+0.5bp); draw((-.2,-3)--(.2,-3),black+0.5bp); draw((-.2,3)--(.2,3),black+0.5bp); draw((-.2,7)--(.2,7),black+0.5bp); label("$-7$",(-7,0),S); label("$-3$",(-3,0),S); label("$3$",(3,0),S); label("$7$",(7,0),S); label("$-7$",(0,-7),W); label("$-3$",(0,-3),W); label("$3$",(0,3),W); label("$7$",(0,7),W); [/asy]
$\textbf{(A)}\ 125\qquad \textbf{(B)}\ 150\qquad \textbf{(C)}\ 175\qquad \textbf{(D)}\ 200\qquad \textbf{(E)}\ 225$ | Consider any square that meets the requirements described in the problem. Then, take the vertices of the square and translate them to the first quadrant (This is the "mapping" described in Solution 2). For example, consider a square with vertices $(7, 7), (-7, 7), (-7, -7),$ and $(7, -7)$
[asy] defaultpen(black+0.75bp+fontsize(8pt)); size(5cm); path p = scale(.15)*unitcircle; draw((-8,0)--(8.5,0),Arrow(HookHead,1mm)); draw((0,-8)--(0,8.5),Arrow(HookHead,1mm)); int i,j; for (i=-7;i<8;++i) { for (j=-7;j<8;++j) { if (((-7 <= i) && (i <= -3)) || ((3 <= i) && (i<= 7))) { if (((-7 <= j) && (j <= -3)) || ((3 <= j) && (j<= 7))) { fill(shift(i,j)*p,black); }}}} draw((-7,-.2)--(-7,.2),black+0.5bp); draw((-3,-.2)--(-3,.2),black+0.5bp); draw((3,-.2)--(3,.2),black+0.5bp); draw((7,-.2)--(7,.2),black+0.5bp); draw((-.2,-7)--(.2,-7),black+0.5bp); draw((-.2,-3)--(.2,-3),black+0.5bp); draw((-.2,3)--(.2,3),black+0.5bp); draw((-.2,7)--(.2,7),black+0.5bp); label("$-7$",(-7,0),S); label("$-3$",(-3,0),S); label("$3$",(3,0),S); label("$7$",(7,0),S); label("$-7$",(0,-7),W); label("$-3$",(0,-3),W); label("$3$",(0,3),W); label("$7$",(0,7),W); draw( (7,7) -- (-7,7) -- (-7,-7) -- (7,-7) -- cycle, black ); [/asy]
After following the mapping described in Solution 2, the square looks like this:
[asy] defaultpen(black+0.75bp+fontsize(8pt)); size(5cm); path p = scale(.15)*unitcircle; draw((-8,0)--(8.5,0),Arrow(HookHead,1mm)); draw((0,-8)--(0,8.5),Arrow(HookHead,1mm)); int i,j; for (i=-7;i<8;++i) { for (j=-7;j<8;++j) { if (((-7 <= i) && (i <= -3)) || ((3 <= i) && (i<= 7))) { if (((-7 <= j) && (j <= -3)) || ((3 <= j) && (j<= 7))) { fill(shift(i,j)*p,black); }}}} draw((-7,-.2)--(-7,.2),black+0.5bp); draw((-3,-.2)--(-3,.2),black+0.5bp); draw((3,-.2)--(3,.2),black+0.5bp); draw((7,-.2)--(7,.2),black+0.5bp); draw((-.2,-7)--(.2,-7),black+0.5bp); draw((-.2,-3)--(.2,-3),black+0.5bp); draw((-.2,3)--(.2,3),black+0.5bp); draw((-.2,7)--(.2,7),black+0.5bp); label("$-7$",(-7,0),S); label("$-3$",(-3,0),S); label("$3$",(3,0),S); label("$7$",(7,0),S); label("$-7$",(0,-7),W); label("$-3$",(0,-3),W); label("$3$",(0,3),W); label("$7$",(0,7),W); draw( (3,3) -- (3,7) -- (7,7) -- (7,3) -- cycle, black ); [/asy]
The position of each vertex within their corresponding grid has not changed.For example, the point $(-7, 7)$ is still the top-left point in a grid, albeit a change in quadrant. Trying this out with a couple of other squares, we see that the following property holds:
\[\text{"For any and all squares with vertices in each of the four quadrants, the 'mapped' version is also a square."}\]
Therefore the logical inverse is true:
\[\text{"For any 'mapped' square with all four vertices in the first quadrant,}\] \[\text{there exists at least one corresponding 'unmapped' square with a vertex in each of the four quadrants."}\]
But how many "unmapped" squares, to be exact?
This might seem complicated at first, but with some intuitive thinking, we realize that there are exactly $4$ "unmapped" squares that correspond with a "mapped" square. This is because given a "mapped" square, there are $4$ choices for the vertex that will remain in the first quadrant; but once that point is chosen, there is only $1$ distribution of the other $3$ vertices that will result in a square. So, we want four times the number of squares we can make in the first quadrant grid.
We divide our counting method into two cases: squares with side length $0$ after mapping (which means all four vertices are in the same position relative to their own grids) and squares with side length $1-4$ after mapping.
Case 1: There are $25$ such squares of length $0$ (this is equivalent to counting the number of points on the grid). However, in this scenario, all of the vertices have been mapped onto the same point. So instead of $4$ choices for the first quadrant vertex, there is only one. Subsequently there are only $25$ such squares that correspond to them.
Case 2: Let a square with sides parallel to the axes be known as $A$ squares. These $A$ squares can have side length $1, 2, 3,$ or $4$ . However, the number of $A$ squares possible depends on the side length. For example, there is only $1$ possible $A$ square of side length $4$ , but $16$ squares of side length $1$ . To be exact, there are $(5-s)^2$ possible $A$ squares of side length $s$ . So, the total number of $A$ squares is
$\sum_{s=1}^4 (5-s)^2$
But what about "tilted" squares? Notice that "tilted squares" can always be inscribed (drawn within) another, bigger square. Let a square inscribed within an $A$ square be called a $B$ square. How many $B$ squares are there? Well, this also depends on the side length. We only want squares whose vertices are lattice points (integer value coordinates), so the number of $B$ squares should increase along with side length. We defined $B$ squares to be inscribed within $A$ squares, so we can say that all $B$ squares have their vertices on the side on an $A$ square. Consider an $A$ square with side length $4$ . There are $3$ other lattice points along the side of the $A$ square, not counting the vertices. Therefore, we can say that there are $s-1$ possible $B$ squares for every $A$ square with side length $s$ . We can multiply $(s-1)$ times the number of $A$ squares to get the number of $B$ squares. This is
$\sum_{s=1}^4 (s-1)(5-s)^2$
total $B$ squares. But we need to add these two quantities to get the number of squares for Case 2:
$\sum_{s=1}^4 (s-1)(5-s)^2 + \sum_{s=1}^4 (5-s)^2$
By distributive property, the expression becomes
$\sum_{s=1}^4 s(5-s)^2$
Solving, we get $50$ "mapped" squares, both $A$ and $B$ . Multiplying this by $4$ to get the corresponding number of "unmapped" squares, then adding to get the number of squares for Case 1, we get
$50*4 + 25 = 225 \Rightarrow \boxed{225}$ | E | 225 |
3a49d5542ac5aecf39bf63cf341ed90c | https://artofproblemsolving.com/wiki/index.php/2008_AMC_12A_Problems/Problem_11 | Three cubes are each formed from the pattern shown. They are then stacked on a table one on top of another so that the $13$ visible numbers have the greatest possible sum. What is that sum?
[asy] unitsize(.8cm); pen p = linewidth(1); draw(shift(-2,0)*unitsquare,p); label("1",(-1.5,0.5)); draw(shift(-1,0)*unitsquare,p); label("2",(-0.5,0.5)); draw(unitsquare,p); label("32",(0.5,0.5)); draw(shift(1,0)*unitsquare,p); label("16",(1.5,0.5)); draw(shift(0,1)*unitsquare,p); label("4",(0.5,1.5)); draw(shift(0,-1)*unitsquare,p); label("8",(0.5,-0.5)); [/asy]
$\mathrm{(A)}\ 154\qquad\mathrm{(B)}\ 159\qquad\mathrm{(C)}\ 164\qquad\mathrm{(D)}\ 167\qquad\mathrm{(E)}\ 189$ | Conversely, maximize the sum. Two cubes have 4 exposed faces. Since $32>16+8+4+2$ , 32 must be on the side. There are two distinct (asymmetrical) configurations with 32 on the side, but $(32,16,2,1)$ is the greatest at 51. There are 2 such cubes, so 51*2. The top cube has one unexposed face, so use 1 as the unexposed face. $2(51)+32+16+8+4+2=164 \implies \boxed{164}$ | C | 164 |
bc927c4a3cda140b72dab6758a55fa39 | https://artofproblemsolving.com/wiki/index.php/2008_AMC_12A_Problems/Problem_15 | Let $k={2008}^{2}+{2}^{2008}$ . What is the units digit of $k^2+2^k$
$\mathrm{(A)}\ 0\qquad\mathrm{(B)}\ 2\qquad\mathrm{(C)}\ 4\qquad\mathrm{(D)}\ 6\qquad\mathrm{(E)}\ 8$ | $k \equiv 2008^2 + 2^{2008} \equiv 8^2 + 2^4 \equiv 4+6 \equiv 0 \pmod{10}$
So, $k^2 \equiv 0 \pmod{10}$ . Since $k = 2008^2+2^{2008}$ is a multiple of four and the units digit of powers of two repeat in cycles of four, $2^k \equiv 2^4 \equiv 6 \pmod{10}$
Therefore, $k^2+2^k \equiv 0+6 \equiv 6 \pmod{10}$ . So the units digit is $6 \Rightarrow \boxed{6}$ | D | 6 |
bc927c4a3cda140b72dab6758a55fa39 | https://artofproblemsolving.com/wiki/index.php/2008_AMC_12A_Problems/Problem_15 | Let $k={2008}^{2}+{2}^{2008}$ . What is the units digit of $k^2+2^k$
$\mathrm{(A)}\ 0\qquad\mathrm{(B)}\ 2\qquad\mathrm{(C)}\ 4\qquad\mathrm{(D)}\ 6\qquad\mathrm{(E)}\ 8$ | I am going to share another approach to this problem.
A units digit $k$ for an integer $n$ implies $n \equiv k \pmod{10}$
Let us take this step by step. First, we consider $k^2.$
Note that $k^2 = \left(2008^2 + 2^{2008}\right)^2 = 2008^4 + 2 \cdot 2008^2 \cdot 2^{5016}.$ Now we calculate $k^2 \pmod{10}$
Before continuing, though, we must take note of the following:
\begin{align*} 2^1 &\equiv 2 \pmod {10} \\ 2^2 &\equiv 4 \pmod {10} \\ 2^3 &\equiv 8 \pmod {10} \\ 2^4 &\equiv 6 \pmod {10} \end{align*}
Now, we continue with the calculation.
\begin{align*} 2008^4 + 2 \cdot 2008^2 \cdot 2^{5016} &\equiv 8^4 + 2 \cdot 8^2 \cdot 2^{2008} + 2^{5016} \pmod{10} \\ &\equiv 8^4 + 2^1 \cdot 2^6 \cdot 2^{2008} + 2^{5016} \pmod{10} \\ &\equiv 8^4 + 2^{1+6+2008} + 2^{5016} \pmod{10} \\ &\equiv 6 + 2^{2015} + 6 \pmod{10} \\ &\equiv 6 + 2^{3} + 6 \pmod{10} \\ &\equiv 6 + 8 + 6 \pmod{10} \\ &\equiv 20 \pmod{10} \\ &\equiv 0 \pmod{10} \\ \end{align*}
We do the same with $2^k.$ However, we just need to find $k \pmod 4$ in order to do this calculation since we have the table of $2^k \pmod 10.$
\begin{align*} 2008^2 + 2^{2008} &\equiv 8^2 \pmod{4} \\ &\equiv 64 \pmod 4\\ &\equiv 0 \pmod 4 \end{align*}
This implies that \begin{align*} 2^k &\equiv 2^{4} \pmod{10} \\ &\equiv 6 \pmod{10} \end{align*}
Thus, \begin{align*} k^2 + 2^k &\equiv 6+0 \pmod{10}\\ &\equiv \boxed{6} | D | 6 |
d08df9586cb6544f1e5bfe23090766be | https://artofproblemsolving.com/wiki/index.php/2008_AMC_12A_Problems/Problem_16 | The numbers $\log(a^3b^7)$ $\log(a^5b^{12})$ , and $\log(a^8b^{15})$ are the first three terms of an arithmetic sequence , and the $12^\text{th}$ term of the sequence is $\log{b^n}$ . What is $n$
$\mathrm{(A)}\ 40\qquad\mathrm{(B)}\ 56\qquad\mathrm{(C)}\ 76\qquad\mathrm{(D)}\ 112\qquad\mathrm{(E)}\ 143$ | Let $A = \log(a)$ and $B = \log(b)$
The first three terms of the arithmetic sequence are $3A + 7B$ $5A + 12B$ , and $8A + 15B$ , and the $12^\text{th}$ term is $nB$
Thus, $2(5A + 12B) = (3A + 7B) + (8A + 15B) \Rightarrow A = 2B$
Since the first three terms in the sequence are $13B$ $22B$ , and $31B$ , the $k$ th term is $(9k + 4)B$
Thus the $12^\text{th}$ term is $(9\cdot12 + 4)B = 112B = nB \Rightarrow n = 112\Rightarrow \boxed{112}$ | D | 112 |
d08df9586cb6544f1e5bfe23090766be | https://artofproblemsolving.com/wiki/index.php/2008_AMC_12A_Problems/Problem_16 | The numbers $\log(a^3b^7)$ $\log(a^5b^{12})$ , and $\log(a^8b^{15})$ are the first three terms of an arithmetic sequence , and the $12^\text{th}$ term of the sequence is $\log{b^n}$ . What is $n$
$\mathrm{(A)}\ 40\qquad\mathrm{(B)}\ 56\qquad\mathrm{(C)}\ 76\qquad\mathrm{(D)}\ 112\qquad\mathrm{(E)}\ 143$ | If $\log(a^3b^7)$ $\log(a^5b^{12})$ , and $\log(a^8b^{15})$ are in arithmetic progression , then $a^3b^7$ $a^5b^{12}$ , and $a^8b^{15}$ are in geometric progression . Therefore,
\[a^2b^5=a^3b^3 \Rightarrow a=b^2\]
Therefore, $a^3b^7=b^{13}$ $a^5b^{12}=b^{22}$ , therefore the 12th term in the sequence is $b^{13+9*11}=b^{112} \Rightarrow \boxed{112}$ | D | 112 |
d08df9586cb6544f1e5bfe23090766be | https://artofproblemsolving.com/wiki/index.php/2008_AMC_12A_Problems/Problem_16 | The numbers $\log(a^3b^7)$ $\log(a^5b^{12})$ , and $\log(a^8b^{15})$ are the first three terms of an arithmetic sequence , and the $12^\text{th}$ term of the sequence is $\log{b^n}$ . What is $n$
$\mathrm{(A)}\ 40\qquad\mathrm{(B)}\ 56\qquad\mathrm{(C)}\ 76\qquad\mathrm{(D)}\ 112\qquad\mathrm{(E)}\ 143$ | Given the first three terms form an arithmetic progression, we have: \[a = \log(a^3b^7)\] \[a+d = \log(a^5b^{12})\] \[a+2d = \log(a^8b^{15}).\] Subtracting the first equation from the second and the third from the second, respectively, gives us these two expressions for $d$ \[d = \log \left( \frac{a^5b^{12}}{a^3b^7} \right) = \log a^2b^5\] \[d = \log \left( \frac{a^8b^{15}}{a^5b^{12}} \right) = \log a^3b^3.\] The desired $12$ th term in the sequence is $a+11d$ , so we can substitute our values for $a$ and $d$ (using either one of our two expressions for $d$ ): \[a+11d = \log a^3b^7 + 11\log(a^2b^5)\] \[= \log a^3b^7 + \log(a^{22}b^{55})\] \[= \log a^{25}b^{62}.\] The answer must be expressed as $\log(b^n)$ , however. We're in luck: the two different yet equal expressions for $d$ allow us to express $a$ and $b$ in terms of each other: \[\log a^2b^5 = \log a^3b^3\] \[a^2b^5 = a^3b^3\] \[a=b^2.\] Plugging in $a=b^2$ , we have: \[a+11d = \log b^{50}b^{62}\] \[= \log b^{112} \Rightarrow \boxed{112}.\] ~ Jingwei325 $\smiley$ | D | 112 |
5c02d22d052a3edf445adf6a73791ead | https://artofproblemsolving.com/wiki/index.php/2008_AMC_12A_Problems/Problem_18 | Triangle $ABC$ , with sides of length $5$ $6$ , and $7$ , has one vertex on the positive $x$ -axis, one on the positive $y$ -axis, and one on the positive $z$ -axis. Let $O$ be the origin . What is the volume of tetrahedron $OABC$
$\mathrm{(A)}\ \sqrt{85}\qquad\mathrm{(B)}\ \sqrt{90}\qquad\mathrm{(C)}\ \sqrt{95}\qquad\mathrm{(D)}\ 10\qquad\mathrm{(E)}\ \sqrt{105}$ | Without loss of generality, let $A$ be on the $x$ axis, $B$ be on the $y$ axis, and $C$ be on the $z$ axis, and let $AB, BC, CA$ have respective lengths of 5, 6, and 7. Let $a,b,c$ denote the lengths of segments $OA,OB,OC,$ respectively. Then by the Pythagorean Theorem \begin{align*} a^2+b^2 &=5^2 , \\ b^2+c^2&=6^2, \\ c^2+a^2 &=7^2 , \end{align*} so $a^2 = (5^2+7^2-6^2)/2 = 19$ ; similarly, $b^2 = 6$ and $c^2 = 30$ . Since $OA$ $OB$ , and $OC$ are mutually perpendicular, the tetrahedron's volume is \[abc/6\] because we can consider the tetrahedron to be a right triangular pyramid. \[abc/6 = \sqrt{a^2b^2c^2}/6 = \frac{\sqrt{19 \cdot 6 \cdot 30}}{6} = \sqrt{95},\] which is answer choice $\boxed{95}$ | C | 95 |
773e5f6a995c8c85da6847bcaf18d904 | https://artofproblemsolving.com/wiki/index.php/2008_AMC_12A_Problems/Problem_19 | In the expansion of \[\left(1 + x + x^2 + \cdots + x^{27}\right)\left(1 + x + x^2 + \cdots + x^{14}\right)^2,\] what is the coefficient of $x^{28}$
$\mathrm{(A)}\ 195\qquad\mathrm{(B)}\ 196\qquad\mathrm{(C)}\ 224\qquad\mathrm{(D)}\ 378\qquad\mathrm{(E)}\ 405$ | Let $A = \left(1 + x + x^2 + \cdots + x^{14}\right)$ and $B = \left(1 + x + x^2 + \cdots + x^{27}\right)$ . We are expanding $A \cdot A \cdot B$
Since there are $15$ terms in $A$ , there are $15^2 = 225$ ways to choose one term from each $A$ . The product of the selected terms is $x^n$ for some integer $n$ between $0$ and $28$ inclusive. For each $n \neq 0$ , there is one and only one $x^{28 - n}$ in $B$ . For example, if I choose $x^2$ from $A$ , then there is exactly one power of $x$ in $B$ that I can choose; in this case, it would be $x^{24}$ . Since there is only one way to choose one term from each $A$ to get a product of $x^0$ , there are $225 - 1 = 224$ ways to choose one term from each $A$ and one term from $B$ to get a product of $x^{28}$ . Thus the coefficient of the $x^{28}$ term is $224 \Rightarrow \boxed{224}$ | C | 224 |
773e5f6a995c8c85da6847bcaf18d904 | https://artofproblemsolving.com/wiki/index.php/2008_AMC_12A_Problems/Problem_19 | In the expansion of \[\left(1 + x + x^2 + \cdots + x^{27}\right)\left(1 + x + x^2 + \cdots + x^{14}\right)^2,\] what is the coefficient of $x^{28}$
$\mathrm{(A)}\ 195\qquad\mathrm{(B)}\ 196\qquad\mathrm{(C)}\ 224\qquad\mathrm{(D)}\ 378\qquad\mathrm{(E)}\ 405$ | Let $P(x) = \left(1 + x + x^2 + \cdots + x^{14}\right)^2 = a_0 + a_1x + a_2x^2 + \cdots + a_{28}x^{28}$ . Then the $x^{28}$ term from the product in question $\left(1 + x + x^2 + \cdots + x^{27}\right)(a_0 + a_1x + a_2x^2 + \cdots + a_{28}x^{28})$ is
$1a_{28}x^{28} + xa_{27}x^{27} + x^2a_{26}x^{26} + \cdots + x^{27}a_1x = \left(a_1 + a_2 + \cdots a_{28}\right)x^{28}$
So we are trying to find the sum of the coefficients of $P(x)$ minus $a_0$ . Since the constant term $a_0$ in $P(x)$ (when expanded) is $1$ , and the sum of the coefficients of $P(x)$ is $P(1)$ , we find the answer to be $P(1) - a_0 = \left(1 + 1 + 1^2 + \cdots 1^{14}\right)^2 - 1 = 15^2 - 1 = 224 \Rightarrow \boxed{224}$ | C | 224 |
773e5f6a995c8c85da6847bcaf18d904 | https://artofproblemsolving.com/wiki/index.php/2008_AMC_12A_Problems/Problem_19 | In the expansion of \[\left(1 + x + x^2 + \cdots + x^{27}\right)\left(1 + x + x^2 + \cdots + x^{14}\right)^2,\] what is the coefficient of $x^{28}$
$\mathrm{(A)}\ 195\qquad\mathrm{(B)}\ 196\qquad\mathrm{(C)}\ 224\qquad\mathrm{(D)}\ 378\qquad\mathrm{(E)}\ 405$ | We expand $(1 + x + x^2 + x^3 + \cdots + x^{14})^2$ to $(1 + x + x^2 + x^3 + \cdots + x^{14}) * (1 + x + x^2 + x^3 + \cdots + x^{14})$ and use FOIL to multiply. It expands out to:
$1 + x + x^2 + x^3 + x^4 + \cdots + x^{14} +$
$\qquad x + x^2 + x^3 + x^4 + \cdots + x^{14} + x^{15} +$
$\qquad \qquad x^2 + x^3 + x^4 + \cdots + x^{14} + x^{15} + x^{16} + \cdots$
It becomes apparent that
$(1 + x + x^2 + x^3 + \cdots + x^{14})^2 = 1 + 2x + 3x^2 + 4x^3 + \cdots + 15x^{14} + 14x^{15} + 13x^{16} + \cdots + x^{28}$
Now we have to find the coefficient of $x^{28}$ in the product:
$(1 + 2x + 3x^2 + 4x^3 + \cdots + 15x^{14} + 14x^{15} + 13x^{16} + \cdots + x^{28}) \cdot (1 + x + x^2 + x^3 + \cdots + x^{27})$
We quickly see that the we get $x^{28}$ terms from $x^{27} \cdot 2x$ $x^{26} \cdot 3x^2$ $x^{25} \cdot 4x^3$ , ... $15x^{14} \cdot x^{14}$ , ... $x^{28} \cdot 1$ . The coefficient of $x^{28}$ is just the sum of the coefficients of all these terms. $1 + 2 + 3 + 4 + \cdots + 15 + 14 + 13 + \cdots + 4 + 3 + 2 = 224$ , so the answer is $\boxed{224}$ | C | 224 |
773e5f6a995c8c85da6847bcaf18d904 | https://artofproblemsolving.com/wiki/index.php/2008_AMC_12A_Problems/Problem_19 | In the expansion of \[\left(1 + x + x^2 + \cdots + x^{27}\right)\left(1 + x + x^2 + \cdots + x^{14}\right)^2,\] what is the coefficient of $x^{28}$
$\mathrm{(A)}\ 195\qquad\mathrm{(B)}\ 196\qquad\mathrm{(C)}\ 224\qquad\mathrm{(D)}\ 378\qquad\mathrm{(E)}\ 405$ | Rewrite the product as $\frac{(x^{28} - 1)(x^{15} - 1)(x^{15} - 1)}{(x - 1)^3}$ . It is known that
\[\frac{1}{(1 - x)^n} = \binom{n - 1}{n - 1} + \binom{n}{n - 1}x + \binom{n + 1}{n - 1}x^2 + \binom{n + 2}{n - 1}x^3 + \cdots + \binom{n - 1 + k}{n - 1}x^k + \cdots .\]
Thus, our product becomes
\[-\left( x^{28} - 1 \right) \left( x^{15} - 1 \right) \left( x^{15} - 1 \right) \left( \binom{2}{2} + \binom{3}{2}x + \binom{4}{2}x^2 + \cdots \right).\]
\[= -\left( x^{28} - 1 \right) \left( x^{15} - 1 \right) \left( x^{15} - 1 \right) \left( 1 + 3x + 6x^2 + \cdots \right).\]
We determine the $x^{28}$ coefficient by doing casework on the first three terms in our product. We can obtain an $x^{28}$ term by choosing $x^{28}$ in the first term, $-1$ in the second and third terms, and $1$ in the fourth term. We can get two $x^{28}$ terms by choosing $x^{15}$ in either the second or third term, $-1$ in the first term, $-1$ in the second or third term from which $x^{15}$ has not been chosen, and the $\binom{15}{2}x^{13}$ in the fourth term. We get $\binom{15}{2} * 2 = 210$ $x^{28}$ terms this way. (We multiply by $2$ because the $x^{15}$ term could have been chosen from the second term or the third term). Lastly, we can get an $x^{28}$ term by choosing $-1$ in the first three terms and a $\binom{30}{2}x^{28}$ from the fourth term. We have a total of $1 + 210 - 435 = -224$ for the $x^{28}$ coefficient, but we recall that we have a negative sign in front of our product, so we obtain an answer of $224 \Rightarrow \boxed{224}$ | C | 224 |
c3bcc9dc1643bb23c7b30d4df10f577e | https://artofproblemsolving.com/wiki/index.php/2008_AMC_12A_Problems/Problem_21 | A permutation $(a_1,a_2,a_3,a_4,a_5)$ of $(1,2,3,4,5)$ is heavy-tailed if $a_1 + a_2 < a_4 + a_5$ . What is the number of heavy-tailed permutations?
$\mathrm{(A)}\ 36\qquad\mathrm{(B)}\ 40\qquad\textbf{(C)}\ 44\qquad\mathrm{(D)}\ 48\qquad\mathrm{(E)}\ 52$ | We use case work on the value of $a_3$
Case 1: $a_3 = 1$ . Since $a_1 + a_2 < a_4 + a_5$ $(a_1, a_2)$ can only be a permutation of $(2, 3)$ or $(2, 4)$ . The values of $a_1$ and $a_2$ , as well as the values of $a_4$ and $a_5$ , are interchangeable, so this case produces a total of $2(2 \cdot 2) = 8$ solutions.
Case 2: $a_3 = 2$ . Similarly, we have $(a_1, a_2)$ is a permutation of $(1, 3)$ $(1, 4)$ , or $(1, 5)$ , which gives a total of $3(2 \cdot 2) = 12$ solutions.
Case 3: $a_3 = 3$ $(a_1, a_2)$ is a permutation of $(1, 2)$ or $(1, 4)$ , which gives a total of $2(2 \cdot 2) = 8$ solutions.
Case 4: $a_3 = 4$ $(a_1, a_2)$ is a permutation of $(1, 2)$ $(1, 3)$ , or $(2, 3)$ , which gives a total of $3(2 \cdot 2) = 12$ solutions.
Case 5: $a_3 = 5$ $(a_1, a_2)$ is a permutation of $(1, 2)$ or $(1, 3)$ , which gives a total of $2(2 \cdot 2) = 8$ solutions.
Therefore, our answer is $8 + 12 + 8 + 12 + 8 = 48 \Rightarrow \boxed{48}$ | D | 48 |
6753ba46f9d601e2ab9d44d31e7b1c0e | https://artofproblemsolving.com/wiki/index.php/2008_AMC_12B_Problems/Problem_2 | $4\times 4$ block of calendar dates is shown. First, the order of the numbers in the second and the fourth rows are reversed. Then, the numbers on each diagonal are added. What will be the positive difference between the two diagonal sums?
$\begin{tabular}[t]{|c|c|c|c|} \multicolumn{4}{c}{}\\\hline 1&2&3&4\\\hline 8&9&10&11\\\hline 15&16&17&18\\\hline 22&23&24&25\\\hline \end{tabular}$
$\textbf{(A)}\ 2 \qquad \textbf{(B)}\ 4 \qquad \textbf{(C)}\ 6 \qquad \textbf{(D)}\ 8 \qquad \textbf{(E)}\ 10$ | After reversing the numbers on the second and fourth rows, the block will look like this:
$\begin{tabular}[t]{|c|c|c|c|} \multicolumn{4}{c}{}\\\hline 1&2&3&4\\\hline 11&10&9&8\\\hline 15&16&17&18\\\hline 25&24&23&22\\\hline \end{tabular}$
The positive difference between the two diagonal sums is then $(4+9+16+25)-(1+10+17+22)=3-1-1+3=\boxed{4}$ | B | 4 |
6753ba46f9d601e2ab9d44d31e7b1c0e | https://artofproblemsolving.com/wiki/index.php/2008_AMC_12B_Problems/Problem_2 | $4\times 4$ block of calendar dates is shown. First, the order of the numbers in the second and the fourth rows are reversed. Then, the numbers on each diagonal are added. What will be the positive difference between the two diagonal sums?
$\begin{tabular}[t]{|c|c|c|c|} \multicolumn{4}{c}{}\\\hline 1&2&3&4\\\hline 8&9&10&11\\\hline 15&16&17&18\\\hline 22&23&24&25\\\hline \end{tabular}$
$\textbf{(A)}\ 2 \qquad \textbf{(B)}\ 4 \qquad \textbf{(C)}\ 6 \qquad \textbf{(D)}\ 8 \qquad \textbf{(E)}\ 10$ | Notice that at baseline the diagonals sum to the same number ( $52$ ). Therefore we need only compute the effect of the swap. The positive difference between $9$ and $10$ is $1$ and the positive difference between $22$ and $25$ is $3$ . Adding gives $1+3=\boxed{4}$ | B | 4 |
45b6cf80fd8fd1b9b231f2984a8f2aa8 | https://artofproblemsolving.com/wiki/index.php/2008_AMC_12B_Problems/Problem_5 | A class collects $50$ dollars to buy flowers for a classmate who is in the hospital. Roses cost $3$ dollars each, and carnations cost $2$ dollars each. No other flowers are to be used. How many different bouquets could be purchased for exactly $50$ dollars?
$\textbf{(A)}\ 1 \qquad \textbf{(B)}\ 7 \qquad \textbf{(C)}\ 9 \qquad \textbf{(D)}\ 16 \qquad \textbf{(E)}\ 17$ | The class could send $25$ carnations and no roses, $22$ carnations and $2$ roses, $19$ carnations and $4$ roses, and so on, down to $1$ carnation and $16$ roses. There are 9 total possibilities (from 0 to 16 roses, incrementing by 2 at each step), $\Rightarrow \boxed{9}$ | C | 9 |
0a1f60fb305fd9d04d3fbb8b0a3623e5 | https://artofproblemsolving.com/wiki/index.php/2008_AMC_12B_Problems/Problem_6 | Postman Pete has a pedometer to count his steps. The pedometer records up to $99999$ steps, then flips over to $00000$ on the next step. Pete plans to determine his mileage for a year. On January $1$ Pete sets the pedometer to $00000$ . During the year, the pedometer flips from $99999$ to $00000$ forty-four times. On December $31$ the pedometer reads $50000$ . Pete takes $1800$ steps per mile. Which of the following is closest to the number of miles Pete walked during the year?
$\textbf{(A)}\ 2500 \qquad \textbf{(B)}\ 3000 \qquad \textbf{(C)}\ 3500 \qquad \textbf{(D)}\ 4000 \qquad \textbf{(E)}\ 4500$ | Every time the pedometer flips, Pete has walked $100,000$ steps. Therefore, Pete has walked a total of $100,000 \cdot 44 + 50,000 = 4,450,000$ steps, which is $4,450,000/1,800 = 2472.2$ miles, which is the closest to the answer choice $\boxed{2500}$ | A | 2500 |
dff3409cbb0d8e6720188e280ce986fc | https://artofproblemsolving.com/wiki/index.php/2008_AMC_12B_Problems/Problem_12 | For each positive integer $n$ , the mean of the first $n$ terms of a sequence is $n$ . What is the $2008$ th term of the sequence?
$\textbf{(A)}\ 2008 \qquad \textbf{(B)}\ 4015 \qquad \textbf{(C)}\ 4016 \qquad \textbf{(D)}\ 4030056 \qquad \textbf{(E)}\ 4032064$ | Letting the sum of the sequence equal $a_1+a_2+\cdots+a_n$ yields the following two equations:
$\frac{a_1+a_2+\cdots+a_{2008}}{2008}=2008$ and
$\frac{a_1+a_2+\cdots+a_{2007}}{2007}=2007$
Therefore:
$a_1+a_2+\cdots+a_{2008}=2008^2$ and $a_1+a_2+\cdots+a_{2007}=2007^2$
Hence, by substitution, $a_{2008}=2008^2-2007^2=(2008+2007)(2008-2007)=4015(1)=4015\implies\boxed{4015}$ | B | 4015 |
e13c7abfc823d707440a9edbac62f5fb | https://artofproblemsolving.com/wiki/index.php/2008_AMC_12B_Problems/Problem_18 | A pyramid has a square base $ABCD$ and vertex $E$ . The area of square $ABCD$ is $196$ , and the areas of $\triangle ABE$ and $\triangle CDE$ are $105$ and $91$ , respectively. What is the volume of the pyramid?
$\textbf{(A)}\ 392 \qquad \textbf{(B)}\ 196\sqrt {6} \qquad \textbf{(C)}\ 392\sqrt {2} \qquad \textbf{(D)}\ 392\sqrt {3} \qquad \textbf{(E)}\ 784$ | Let $h$ be the height of the pyramid and $a$ be the distance from $h$ to $CD$ . The side length of the base is $14$ . The heights of $\triangle ABE$ and $\triangle CDE$ are $2\cdot105\div14=15$ and $2\cdot91\div14=13$ , respectively. Consider a side view of the pyramid from $\triangle BCE$ . We have a systems of equations through the Pythagorean Theorem:
$13^2-(14-a)^2=h^2 \\ 15^2-a^2=h^2$
Setting them equal to each other and simplifying gives $-27+28a=225 \implies a=9$
Therefore, $h=\sqrt{15^2-9^2}=12$ , and the volume of the pyramid is $\frac{bh}{3}=\frac{12\cdot 196}{3}=\boxed{784}$ | E | 784 |
3796b52a3c81159bcb3cd0b6c7894358 | https://artofproblemsolving.com/wiki/index.php/2008_AMC_12B_Problems/Problem_20 | Michael walks at the rate of $5$ feet per second on a long straight path. Trash pails are located every $200$ feet along the path. A garbage truck traveling at $10$ feet per second in the same direction as Michael stops for $30$ seconds at each pail. As Michael passes a pail, he notices the truck ahead of him just leaving the next pail. How many times will Michael and the truck meet?
$\mathrm{(A)}\ 4\qquad\mathrm{(B)}\ 5\qquad\mathrm{(C)}\ 6\qquad\mathrm{(D)}\ 7\qquad\mathrm{(E)}\ 8$ | Pick a coordinate system where Michael's starting pail is $0$ and the one where the truck starts is $200$ .
Let $M(t)$ and $T(t)$ be the coordinates of Michael and the truck after $t$ seconds.
Let $D(t)=T(t)-M(t)$ be their (signed) distance after $t$ seconds.
Meetings occur whenever $D(t)=0$ .
We have $D(0)=200$
The truck always moves for $20$ seconds, then stands still for $30$ . During the first $20$ seconds of the cycle the truck moves by $200$ feet and Michael by $100$ , hence during the first $20$ seconds of the cycle $D(t)$ increases by $100$ .
During the remaining $30$ seconds $D(t)$ decreases by $150$
From this observation it is obvious that after four full cycles, i.e. at $t=200$ , we will have $D(t)=0$ for the first time.
During the fifth cycle, $D(t)$ will first grow from $0$ to $100$ , then fall from $100$ to $-50$ . Hence Michael overtakes the truck while it is standing at the pail.
During the sixth cycle, $D(t)$ will first grow from $-50$ to $50$ , then fall from $50$ to $-100$ . Hence the truck starts moving, overtakes Michael on their way to the next pail, and then Michael overtakes the truck while it is standing at the pail.
During the seventh cycle, $D(t)$ will first grow from $-100$ to $0$ , then fall from $0$ to $-150$ . Hence the truck meets Michael at the moment when it arrives to the next pail.
Obviously, from this point on $D(t)$ will always be negative, meaning that Michael is already too far ahead. Hence we found all $\boxed{5}$ meetings. | B | 5 |
3796b52a3c81159bcb3cd0b6c7894358 | https://artofproblemsolving.com/wiki/index.php/2008_AMC_12B_Problems/Problem_20 | Michael walks at the rate of $5$ feet per second on a long straight path. Trash pails are located every $200$ feet along the path. A garbage truck traveling at $10$ feet per second in the same direction as Michael stops for $30$ seconds at each pail. As Michael passes a pail, he notices the truck ahead of him just leaving the next pail. How many times will Michael and the truck meet?
$\mathrm{(A)}\ 4\qquad\mathrm{(B)}\ 5\qquad\mathrm{(C)}\ 6\qquad\mathrm{(D)}\ 7\qquad\mathrm{(E)}\ 8$ | The truck takes $20$ seconds to go from one pail to the next and then stops for $30$ seconds at the new pail. Thus it sets off from a pail every 50 sec. Let $t$ denote the time elapsed and write $t=50k + \Delta$ , where $\Delta \in [0,50)$ . In this time Michael has traveled $5t = 250k+5\Delta$ feet. What about the truck? In the first $50k$ seconds the truck covers $k$ pails, i.e. $200k$ feet so it moves $200+200k$ feet from Michael's starting point. Then we have two cases:
(a) if $\Delta < 20$ , then the truck travels an additional $10\Delta$ feet. For them to intersect we must have $200+200k + 10\Delta = 250k + 5\Delta$ . Solving, we get $\Delta = 10k - 40$ . Since $\Delta$ must lie in the interval $[0,20)$ we get $k \in \{4, 5\}$
(b) if $\Delta \in [20, 50)$ , then the truck travels an additional $200$ feet. For them to intersect we must have $200+200k + 200 = 250k + 5\Delta$ . Solving, we get $\Delta = 10(8-k)$ . Since $\Delta$ must lie in the interval $[20,50)$ we get $k \in \{4, 5, 6\}$
Thus Michael intersects with the truck $5$ times, which is option $\boxed{5}$ | B | 5 |
3796b52a3c81159bcb3cd0b6c7894358 | https://artofproblemsolving.com/wiki/index.php/2008_AMC_12B_Problems/Problem_20 | Michael walks at the rate of $5$ feet per second on a long straight path. Trash pails are located every $200$ feet along the path. A garbage truck traveling at $10$ feet per second in the same direction as Michael stops for $30$ seconds at each pail. As Michael passes a pail, he notices the truck ahead of him just leaving the next pail. How many times will Michael and the truck meet?
$\mathrm{(A)}\ 4\qquad\mathrm{(B)}\ 5\qquad\mathrm{(C)}\ 6\qquad\mathrm{(D)}\ 7\qquad\mathrm{(E)}\ 8$ | We make a chart by seconds in increments of ten.
$\begin{tabular}{c|c|c}Time (s) &Distance of Michael&Distance of Garbage\\ \hline 0&0&200\\ 10&50&300\\ 20&100&400\\ 30&150&400\\ 40&200&400\\ 50&250&400\\ 60&300&500\\ 70&350&600\\ 80&400&600\\ 90&450&600\\ 100&500&600\\ 110&550&700\\ 120&600&800\\ 130&650&800\\ 140&700&800\\ 150&750&800\\ 160&800&900\\ 170&850&1000\\ 180&900&1000\\ 190&950&1000\\ 200&1000&1000\\ 210&1050&1100\\ 220&1100&1200\\ 230&1150&1200\\ 240&1200&1200\\ 250&1250&1200\\ 260&1300&1300\\ 270&1350&1400\\ 280&1400&1400\\ 290&1450&1400\\ 300&1500&1400\\ 310&1550&1500\\ 320&1600&1600\\ 330&1650&1600\\ 340&1700&1600\\ 350&1750&1600\\ 360&1800&1700\\ 370&1850&1800\\ 380&1900&1800\\ 390&1950&1800\\ 400&2000&1800\\ \end{tabular}$
Notice at 200, 240, 260, 280, and 320 seconds, Michael and the garbage truck meet. It is clear that they met at these times, and will meet no more. Thus the answer is $\boxed{5}$ | B | 5 |
3796b52a3c81159bcb3cd0b6c7894358 | https://artofproblemsolving.com/wiki/index.php/2008_AMC_12B_Problems/Problem_20 | Michael walks at the rate of $5$ feet per second on a long straight path. Trash pails are located every $200$ feet along the path. A garbage truck traveling at $10$ feet per second in the same direction as Michael stops for $30$ seconds at each pail. As Michael passes a pail, he notices the truck ahead of him just leaving the next pail. How many times will Michael and the truck meet?
$\mathrm{(A)}\ 4\qquad\mathrm{(B)}\ 5\qquad\mathrm{(C)}\ 6\qquad\mathrm{(D)}\ 7\qquad\mathrm{(E)}\ 8$ | This solution might be time consuming, but it is pretty rigorous. Also, throughout the solution refer to the graph in solution 1 to understand this one more. Lets first start off by defining the position function for Michael. We let $M(t) = 5t$ , where $t$ is the amount of seconds passed. Now, lets define the position function of the truck as two independent functions. It is obvious, graphically, that the position function of the truck is a piecewise function alternating between linear lines and constant lines. Lets focus on the linear pieces of the truck's position function. Let the $kth$ linear part of the truck's position function be denoted as $L_k(t)$ . Then through algebra, it is found that $L_k(t) = 10t + 500 - 300k$ ${50k - 50 <= t <= 50k - 30}$ . Now, lets move on to the constant pieces, which is a lot easier in terms of algebra. Let the $ith$ constant part of the truck's position function be denoted as $C_i(t)$ . Then, again, through algebra we obtain $C_i(t) = 200 * (i + 1)$ ${50i - 30 <= t <= 50i}$ . Now, let me stress that $i$ and $k$ are disjoint, which is why I used different variable names. We are interested in where $C_i(t)$ and $L_k(t)$ intersect $M(t)$ or $5t$ $L_k(t)$ intersects $M$ at $(60k - 100, 300k - 500)$ . However, the only $k$ values that actually work are $5 <= k <= 7$ because of the domain restrictions on $L_k(t)$ . Similarly, we also see that for $C_i(t)$ it intersects at $(40i + 40, 200i + 200)$ . The only $i$ values that work is $4 <= i <= 7$ . However, some pair $(i, k)$ might yield the same exact intersection point. Checking this through simple algebra we see that $(4,5)$ and $(7,7)$ do indeed yield the same intersection point. Thus, our answer is $(7 - 5 + 1) + (7 - 4 + 1) - 2 = 7 - 2 = 5$ or $\boxed{5}$ | B | 5 |
85d8f5f2573240810fe0a911684c39ad | https://artofproblemsolving.com/wiki/index.php/2008_AMC_12B_Problems/Problem_23 | The sum of the base- $10$ logarithms of the divisors of $10^n$ is $792$ . What is $n$
$\text{(A)}\ 11\qquad \text{(B)}\ 12\qquad \text{(C)}\ 13\qquad \text{(D)}\ 14\qquad \text{(E)}\ 15$ | Every factor of $10^n$ will be of the form $2^a \times 5^b , a\leq n , b\leq n$ . Not all of these base ten logarithms will be rational, but we can add them together in a certain way to make it rational. Recall the logarithmic property $\log(a \times b) = \log(a)+\log(b)$ . For any factor $2^a \times 5^b$ , there will be another factor $2^{n-a} \times 5^{n-b}$ . Note this is not true if $10^n$ is a perfect square. When these are multiplied, they equal $2^{a+n-a} \times 5^{b+n-b} = 10^n$ $\log 10^n=n$ so the number of factors divided by 2 times n equals the sum of all the factors, 792.
There are $n+1$ choices for the exponent of 5 in each factor, and for each of those choices, there are $n+1$ factors (each corresponding to a different exponent of 2), yielding $(n+1)^2$ total factors. $\frac{(n+1)^2}{2}*n = 792$ . We then plug in answer choices and arrive at the answer $\boxed{11}$ | null | 11 |
88e6295832a05bba99a2a06177871713 | https://artofproblemsolving.com/wiki/index.php/2008_AMC_12B_Problems/Problem_24 | Let $A_0=(0,0)$ . Distinct points $A_1,A_2,\dots$ lie on the $x$ -axis, and distinct points $B_1,B_2,\dots$ lie on the graph of $y=\sqrt{x}$ . For every positive integer $n,\ A_{n-1}B_nA_n$ is an equilateral triangle. What is the least $n$ for which the length $A_0A_n\geq100$
$\textbf{(A)}\ 13\qquad \textbf{(B)}\ 15\qquad \textbf{(C)}\ 17\qquad \textbf{(D)}\ 19\qquad \textbf{(E)}\ 21$ | Let $a_n=|A_{n-1}A_n|$ . We need to rewrite the recursion into something manageable. The two strange conditions, $B$ 's lie on the graph of $y=\sqrt{x}$ and $A_{n-1}B_nA_n$ is an equilateral triangle, can be compacted as follows: \[\left(a_n\frac{\sqrt{3}}{2}\right)^2=\frac{a_n}{2}+a_{n-1}+a_{n-2}+\cdots+a_1\] which uses $y^2=x$ , where $x$ is the height of the equilateral triangle and therefore $\frac{\sqrt{3}}{2}$ times its base.
The relation above holds for $n=k$ and for $n=k-1$ $(k>1)$ , so \[\left(a_k\frac{\sqrt{3}}{2}\right)^2-\left(a_{k-1}\frac{\sqrt{3}}{2}\right)^2=\] \[=\left(\frac{a_k}{2}+a_{k-1}+a_{k-2}+\cdots+a_1\right)-\left(\frac{a_{k-1}}{2}+a_{k-2}+a_{k-3}+\cdots+a_1\right)\] Or, \[a_k-a_{k-1}=\frac23\] This implies that each segment of a successive triangle is $\frac23$ more than the last triangle. To find $a_{1}$ , we merely have to plug in $k=1$ into the aforementioned recursion and we have $a_{1} - a_{0} = \frac23$ . Knowing that $a_{0}$ is $0$ , we can deduce that $a_{1} = 2/3$ .Thus, $a_n=\frac{2n}{3}$ , so $A_0A_n=a_n+a_{n-1}+\cdots+a_1=\frac{2}{3} \cdot \frac{n(n+1)}{2} = \frac{n(n+1)}{3}$ . We want to find $n$ so that $n^2<300<(n+1)^2$ $n=\boxed{17}$ is our answer. | null | 17 |
88e6295832a05bba99a2a06177871713 | https://artofproblemsolving.com/wiki/index.php/2008_AMC_12B_Problems/Problem_24 | Let $A_0=(0,0)$ . Distinct points $A_1,A_2,\dots$ lie on the $x$ -axis, and distinct points $B_1,B_2,\dots$ lie on the graph of $y=\sqrt{x}$ . For every positive integer $n,\ A_{n-1}B_nA_n$ is an equilateral triangle. What is the least $n$ for which the length $A_0A_n\geq100$
$\textbf{(A)}\ 13\qquad \textbf{(B)}\ 15\qquad \textbf{(C)}\ 17\qquad \textbf{(D)}\ 19\qquad \textbf{(E)}\ 21$ | Consider two adjacent equilateral triangles obeying the problem statement. For each, drop an altitude to the $x$ axis and denote the resulting heights $h_n$ and $h_{n+1}$ . From 30-60-90 rules, the distance between the points where these altitudes meet the x-axis is \[\frac{h_{n+1}}{\sqrt{3}}+\frac{h_n}{\sqrt{3}} = \frac{h_{n+1}+h_n}{\sqrt{3}}\]
But the square root curve means that this distance is also expressible as $h_{n+1}^2-h_n^2$ (the $x$ coordinates are the squares of the heights). Setting these expressions equal and dividing throughout by $h_{n+1}+h_n$ leaves $h_{n+1}-h_n=\frac{1}{\sqrt{3}}$ . So the difference in height of successive triangles is $\frac{1}{\sqrt{3}}$ , meaning their bases are wider by $2/3$ units each time. From here, one can proceed as in Solution 1 to arrive at $n=\boxed{17}$ | null | 17 |
88e6295832a05bba99a2a06177871713 | https://artofproblemsolving.com/wiki/index.php/2008_AMC_12B_Problems/Problem_24 | Let $A_0=(0,0)$ . Distinct points $A_1,A_2,\dots$ lie on the $x$ -axis, and distinct points $B_1,B_2,\dots$ lie on the graph of $y=\sqrt{x}$ . For every positive integer $n,\ A_{n-1}B_nA_n$ is an equilateral triangle. What is the least $n$ for which the length $A_0A_n\geq100$
$\textbf{(A)}\ 13\qquad \textbf{(B)}\ 15\qquad \textbf{(C)}\ 17\qquad \textbf{(D)}\ 19\qquad \textbf{(E)}\ 21$ | Note that $A_{1}$ is of the form $(2x,0)$ for some $x$ , and thus $B_{1}$ is of the form $(x, x \sqrt{3}).$ Then, we are told that $B_{1}$ lies on the graph of $y = \sqrt{x}$ , so \[(x \sqrt{3})^{2} = x.\] Solving for x, we get that $x = \frac{1}{3},$ and so $A_{1} = (2/3,0)$
Now, similarly to before, let $|A_{1}A_{2}|=2y.$ Then, $B_{2} = (y+\frac{2}{3},y \sqrt{3})$ , and so \[(y \sqrt{3})^{2} = (y+\frac{2}{3}).\] Solving using the quadratic formula gives \[y = \frac{-(-1) \pm \sqrt{1^{2}-4(3)(-\frac{2}{3})}}{6} = \frac{1 \pm \sqrt{1^{2}+8}}{6} = \frac{2}{3}.\] Then, $2y = \frac{4}{3},$ so $A_{2} = (2,0).$
In general, if $A_{n} = (a_{n},0)$ for all integers $n$ , and $|A_{n}A_{n+1}| = 2x$ for some real number $x$ , we have the following equation for $x$ \[(x \sqrt{3})^{2} = x+a_{n},\] which give us when plugged into the quadratic formula gives \[x = \frac{1 \pm \sqrt{1+12a_{n}}}{6},\] but since $x$ must be positive, we have that \[x = \frac{1 + \sqrt{1+12a_{n}}}{6},\] and so \[a_{n+1} = a_{n}+2x = a_{n}+\frac{1 + \sqrt{1+12a_{n}}}{3}.\] Computing a few terms of $a_{n}$ using this method gives $a_{3} = 4,$ $a_{4} = \frac{20}{3},$ and $a_{5} = 10$
Notice how all our $a_{n}$ terms so far are rational, even though there is an abundance of radicals in the recurrence. This motivates us to look at our discriminants in the quadratic formula that is solved for $x$
The discriminant of \[(x \sqrt{3})^{2} = x+a_{2}\] is \[1+12 \cdot \frac{2}{3} = 9.\] Similarly, the discriminant of \[(x \sqrt{3})^{2} = x+a_{3}\] is $1+12 \cdot a_{3} = 25,$ and $1+12 \cdot a_{4} = 49$
Note how our results keep coming out as the squares of the odd integers. Moreover, it seems that \[1+12a_{n} = (2n+1)^{2}.\] We will prove this with induction. The base case, $n=2$ , we have already verified.
Now, for the Inductive step, assume that \[1+12a_{n-1} = (2n-1)^{2}.\] for some integer $n-1$ . We will prove that this is true for $n$ as well.
Plugging this into our recurrence formula gives us \[a_{n} = a_{n}+2x = a_{n-1}+\frac{1 + \sqrt{1+12a_{n-1}}}{3}\] \[a_{n} = \frac{(2n-1)^{2}-1}{12} + \frac{1 + (2n-1)}{3}\] \[a_{n} = \frac{4n^{2}-4n+1-1}{12} + \frac{8n}{12}\] \[a_{n} = \frac{4n^{2}+4n+1-1}{12}\] \[a_{n} = \frac{(2n+1)^{2}-1}{12}.\] Therefore, we have proved our claim. Now, we have that $|A_{0}A_{n}|=a_{n},$ so we just need the least integer $n$ so that \[a_{n} > 100,\] or \[(2n+1)^{2} > 1201.\] Then, we see that $35^{2} = 1225$ is the smallest odd square larger than $1201.$ Therefore, we have $2n+1=35$ , so $n = \boxed{17}.$ | null | 17 |
88e6295832a05bba99a2a06177871713 | https://artofproblemsolving.com/wiki/index.php/2008_AMC_12B_Problems/Problem_24 | Let $A_0=(0,0)$ . Distinct points $A_1,A_2,\dots$ lie on the $x$ -axis, and distinct points $B_1,B_2,\dots$ lie on the graph of $y=\sqrt{x}$ . For every positive integer $n,\ A_{n-1}B_nA_n$ is an equilateral triangle. What is the least $n$ for which the length $A_0A_n\geq100$
$\textbf{(A)}\ 13\qquad \textbf{(B)}\ 15\qquad \textbf{(C)}\ 17\qquad \textbf{(D)}\ 19\qquad \textbf{(E)}\ 21$ | We can iteratively calculate out the first few $A_i$ and $B_i$ . We know that $A_0 = (0,0)$ and the line through $A_0$ and $B_1$ needs to make a $60^{\circ}$ angle with the x-axis (because the triangle is equilateral). The equation of a line that makes an angle $\theta$ with the x-axis and passes through the origin has equation $\tan(\theta)x$ , so the line through $A_0$ and $B_1$ has equation $\sqrt{3}x$ . We set this equal to $\sqrt{x}$ (to find where the two curves intersect) and, when solving, find that $x = \frac{1}{3}$ . Therefore, $B_1 = (\frac{1}{3}, \frac{\sqrt{3}}{3})$ . By equilateral triangle properties, we must then have that $A_1 = (\frac{2}{3}, 0)$ . To find $B_2$ , we find the equation of the line through $A_1$ that makes a $60^{\circ}$ angle with the $x$ -axis. This is the same line as the one through the origin (which we already found) shifted $\frac{2}{3}$ to the right, so it has equation $y = \sqrt{3}(x - \frac{2}{3})$ . Setting this equal to $\sqrt{x}$ and solving for $x$ , we get $x = \frac{4}{3}$ , so $B_2 = (\frac{4}{3}, \frac{2\sqrt{3}}{3})$ . By equilateral triangle properties, we have that $A_2 = (2, 0)$ . Repeating this process, we find that $B_3 = (3, \sqrt{3})$ and $A_3 = (4,0)$ . At this point, we notice that the $y$ -coordinate of each $B_i$ is $\frac{\sqrt{3}}{3}$ more than that of $B_{i-1}$ , so $B_4 = (\frac{16}{3}, \frac{4\sqrt{3}}{3})$ and $A_4 = (\frac{20}{3}, 0)$ . It may be helpful to continue calculating out the $A_i$ and $B_i$ if the pattern in the $x$ -coordinates of the $A_i$ isn't visible yet. At this point, it can be noticed that the differences between the x coordinates of consecutive $A_i$ form an arithmetic sequence ( $x_{A_1} - x_{A_0} = \frac{2}{3}$ $x_{A_2} - x_{A_1} = \frac{4}{3}$ , etc.). Additionally, the numerator of each of the fractions is a consecutive even number, and the sum of the first $n$ even numbers is $n(n+1)$ . Thus, we must find the least integral solution to $\frac{n(n+1)}{3} \geq 100$ . This least solution is $\boxed{17}$ | C | 17 |
a6149367de9ad3f338ea3cb01ff7dc1a | https://artofproblemsolving.com/wiki/index.php/2007_AMC_12A_Problems/Problem_4 | Kate rode her bicycle for 30 minutes at a speed of 16 mph, then walked for 90 minutes at a speed of 4 mph. What was her overall average speed in miles per hour?
$\mathrm{(A)}\ 7\qquad \mathrm{(B)}\ 9\qquad \mathrm{(C)}\ 10\qquad \mathrm{(D)}\ 12\qquad \mathrm{(E)}\ 14$ | \[16 \cdot \frac{30}{60}+4\cdot\frac{90}{60}=14\] \[\frac{14}2=7\Rightarrow\boxed{7}\] | A | 7 |
614e723e5f64ed464c336ca976221e26 | https://artofproblemsolving.com/wiki/index.php/2007_AMC_12A_Problems/Problem_9 | Yan is somewhere between his home and the stadium. To get to the stadium he can walk directly to the stadium, or else he can walk home and then ride his bicycle to the stadium. He rides 7 times as fast as he walks, and both choices require the same amount of time. What is the ratio of Yan's distance from his home to his distance from the stadium?
$\mathrm{(A)}\ \frac 23\qquad \mathrm{(B)}\ \frac 34\qquad \mathrm{(C)}\ \frac 45\qquad \mathrm{(D)}\ \frac 56\qquad \mathrm{(E)}\ \frac 78$ | Let $x$ represent the distance from home to the stadium, and let $r$ represent the distance from Yan to home. Our goal is to find $\frac{r}{x-r}$ . If Yan walks directly to the stadium, then assuming he walks at a rate of $1$ , it will take him $x-r$ units of time. Similarly, if he walks back home it will take him $r + \frac{x}{7}$ units of time. Because the two times are equal, we can create the following equation: $x-r = r + \frac{x}{7}$ . We get $x-2r=\frac{x}{7}$ , so $\frac{6}{7}x = 2r$ , and $\frac{x}{r} = \frac{7}{3}$ . This minus one is the reciprocal of what we want to find: $\frac{7}{3}-1 = \frac{4}{3}$ , so the answer is $\boxed{34}$ | B | 34 |
614e723e5f64ed464c336ca976221e26 | https://artofproblemsolving.com/wiki/index.php/2007_AMC_12A_Problems/Problem_9 | Yan is somewhere between his home and the stadium. To get to the stadium he can walk directly to the stadium, or else he can walk home and then ride his bicycle to the stadium. He rides 7 times as fast as he walks, and both choices require the same amount of time. What is the ratio of Yan's distance from his home to his distance from the stadium?
$\mathrm{(A)}\ \frac 23\qquad \mathrm{(B)}\ \frac 34\qquad \mathrm{(C)}\ \frac 45\qquad \mathrm{(D)}\ \frac 56\qquad \mathrm{(E)}\ \frac 78$ | [asy] draw((0,0)--(7,0)); dot((0,0)); dot((3,0)); dot((6,0)); dot((7,0)); label("$H$",(0,0),S); label("$Y$",(3,0),S); label("$P$",(6,0),S); label("$S$",(7,0),S); [/asy] Let $H$ represent Yan's home, $S$ represent the stadium, and $Y$ represent Yan's current position. If Yan walks directly to the stadium, he will reach Point $P$ the same time he will reach $H$ if he is walking home.
Since he bikes $7$ times as fast as he walks and the time is the same, the distance from his home to the stadium must be $7$ times the distance from $P$ to the stadium. If $PS=x$ , then $HS=7x$ and $HP=6x$ . Since Y is the midpoint of $\overline{HP}$ $HY=YP=3x$ . Therefore, the ratio of Yan's distance from his home to his distance from the stadium is $\frac{YH}{YS}=\frac{3x}{4x}=\boxed{34}$ | B | 34 |
672eaf84f366f95023b522fae340d0a9 | https://artofproblemsolving.com/wiki/index.php/2007_AMC_12A_Problems/Problem_11 | A finite sequence of three-digit integers has the property that the tens and units digits of each term are, respectively, the hundreds and tens digits of the next term, and the tens and units digits of the last term are, respectively, the hundreds and tens digits of the first term. For example, such a sequence might begin with the terms 247, 475, and 756 and end with the term 824. Let $S$ be the sum of all the terms in the sequence. What is the largest prime factor that always divides $S$
$\mathrm{(A)}\ 3\qquad \mathrm{(B)}\ 7\qquad \mathrm{(C)}\ 13\qquad \mathrm{(D)}\ 37\qquad \mathrm{(E)}\ 43$ | A given digit appears as the hundreds digit, the tens digit, and the units digit of a term the same number of times. Let $k$ be the sum of the units digits in all the terms. Then $S=111k=3 \cdot 37k$ , so $S$ must be divisible by $37\ \mathrm{(D)}$ . To see that it need not be divisible by any larger prime, the sequence $123, 231, 312$ gives $S=666=2 \cdot 3^2 \cdot 37\Rightarrow \mathrm{\boxed{37}$ | D | 37 |
11404e601c27b039d000554fd1d20078 | https://artofproblemsolving.com/wiki/index.php/2007_AMC_12A_Problems/Problem_12 | Integers $a, b, c,$ and $d$ , not necessarily distinct, are chosen independently and at random from 0 to 2007, inclusive. What is the probability that $ad-bc$ is even
$\mathrm{(A)}\ \frac 38\qquad \mathrm{(B)}\ \frac 7{16}\qquad \mathrm{(C)}\ \frac 12\qquad \mathrm{(D)}\ \frac 9{16}\qquad \mathrm{(E)}\ \frac 58$ | The only time when $ad-bc$ is even is when $ad$ and $bc$ are of the same parity . The chance of $ad$ being odd is $\frac 12 \cdot \frac 12 = \frac 14$ , since the only way to have $ad$ be odd is to have both $a$ and $d$ be odd. As a result, $ad$ has a $\frac 34$ probability of being even. $bc$ also has a $\frac 14$ chance of being odd and a $\frac34$ chance of being even. Therefore, the probability that $ad-bc$ will be even is $\left(\frac 14\right)^2+\left(\frac 34\right)^2=\boxed{58}$ | E | 58 |
11404e601c27b039d000554fd1d20078 | https://artofproblemsolving.com/wiki/index.php/2007_AMC_12A_Problems/Problem_12 | Integers $a, b, c,$ and $d$ , not necessarily distinct, are chosen independently and at random from 0 to 2007, inclusive. What is the probability that $ad-bc$ is even
$\mathrm{(A)}\ \frac 38\qquad \mathrm{(B)}\ \frac 7{16}\qquad \mathrm{(C)}\ \frac 12\qquad \mathrm{(D)}\ \frac 9{16}\qquad \mathrm{(E)}\ \frac 58$ | If we don't know our parity rules, we can check and see that $ad-bc$ is only even when $ad$ and $bc$ are of the same parity (as stated above). From here, we have two cases.
Case 1: $odd-odd$ (which must be $o \cdot o-o \cdot o$ ). The probability for this to occur is $\left(\frac 12\right)^4 = \frac 1{16}$ , because each integer has a $\frac 12$ chance of being odd.
Case 2: $even-even$ (which occurs in 4 cases: $(e \cdot e-e \cdot e$ ), ( $o \cdot e-o \cdot e$ ) (alternating of any kind), and ( $e \cdot e-o \cdot e$ ) with its reverse, ( $o \cdot e-e \cdot e$ ).
Our first subcase of case 2 has a chance of $\frac 1{16}$ (same reasoning as above).
Our second subcase of case 2 has a $\frac 14$ chance, since only the 2nd and 4th flip matter (or 1st and 3rd).
Our third subcase of case 2 has a $\frac 18$ chance, because the 1st, 2nd, and either 3rd or 4th flip matter.
Our fourth subcase of case 2 has a $\frac 18$ chance, because it's the same, just reversed.
We sum these, and get our answer of $\frac 1{16} + \frac 1{16} + \frac 14 + \frac 18 + \frac 18 = \boxed{58}.$ | E | 58 |
052f48df539941530df5acba92a63026 | https://artofproblemsolving.com/wiki/index.php/2007_AMC_12A_Problems/Problem_13 | A piece of cheese is located at $(12,10)$ in a coordinate plane . A mouse is at $(4,-2)$ and is running up the line $y=-5x+18$ . At the point $(a,b)$ the mouse starts getting farther from the cheese rather than closer to it. What is $a+b$
$\mathrm{(A)}\ 6\qquad \mathrm{(B)}\ 10\qquad \mathrm{(C)}\ 14\qquad \mathrm{(D)}\ 18\qquad \mathrm{(E)}\ 22$ | The point $(a,b)$ is the foot of the perpendicular from $(12,10)$ to the line $y=-5x+18$ . The perpendicular has slope $\frac{1}{5}$ , so its equation is $y=10+\frac{1}{5}(x-12)=\frac{1}{5}x+\frac{38}{5}$ . The $x$ -coordinate at the foot of the perpendicular satisfies the equation $\frac{1}{5}x+\frac{38}{5}=-5x+18$ , so $x=2$ and $y=-5\cdot2+18=8$ . Thus $(a,b) = (2,8)$ , and $a+b = \boxed{10}$ | null | 10 |
052f48df539941530df5acba92a63026 | https://artofproblemsolving.com/wiki/index.php/2007_AMC_12A_Problems/Problem_13 | A piece of cheese is located at $(12,10)$ in a coordinate plane . A mouse is at $(4,-2)$ and is running up the line $y=-5x+18$ . At the point $(a,b)$ the mouse starts getting farther from the cheese rather than closer to it. What is $a+b$
$\mathrm{(A)}\ 6\qquad \mathrm{(B)}\ 10\qquad \mathrm{(C)}\ 14\qquad \mathrm{(D)}\ 18\qquad \mathrm{(E)}\ 22$ | If the mouse is at $(x, y) = (x, 18 - 5x)$ , then the square of the distance from the mouse to the cheese is $(x - 12)^2 + (8 - 5x)^2 = 26(x^2 - 4x + 8) = 26((x - 2)^2 + 4).$ The value of this expression is smallest when $x = 2$ , so the mouse is closest to the cheese at the point $(2, 8)$ , and $a+b=2+8 = \boxed{10}$ | null | 10 |
052f48df539941530df5acba92a63026 | https://artofproblemsolving.com/wiki/index.php/2007_AMC_12A_Problems/Problem_13 | A piece of cheese is located at $(12,10)$ in a coordinate plane . A mouse is at $(4,-2)$ and is running up the line $y=-5x+18$ . At the point $(a,b)$ the mouse starts getting farther from the cheese rather than closer to it. What is $a+b$
$\mathrm{(A)}\ 6\qquad \mathrm{(B)}\ 10\qquad \mathrm{(C)}\ 14\qquad \mathrm{(D)}\ 18\qquad \mathrm{(E)}\ 22$ | We are trying to find the point where distance between the mouse and $(12, 10)$ is minimized. This point is where the line that passes through $(12, 10)$ and is perpendicular to $y=-5x+18$ intersects $y=-5x+18$ . By basic knowledge of perpendicular lines, this line is $y=\frac{x}{5}+\frac{38}{5}$ . This line intersects $y=-5x+18$ at $(2,8)$ . So $a+b=\boxed{10}$ . - MegaLucario1001 | null | 10 |
052f48df539941530df5acba92a63026 | https://artofproblemsolving.com/wiki/index.php/2007_AMC_12A_Problems/Problem_13 | A piece of cheese is located at $(12,10)$ in a coordinate plane . A mouse is at $(4,-2)$ and is running up the line $y=-5x+18$ . At the point $(a,b)$ the mouse starts getting farther from the cheese rather than closer to it. What is $a+b$
$\mathrm{(A)}\ 6\qquad \mathrm{(B)}\ 10\qquad \mathrm{(C)}\ 14\qquad \mathrm{(D)}\ 18\qquad \mathrm{(E)}\ 22$ | If the mouse is at $(x, y) = (x, 18 - 5x)$ , then the square of the distance from the mouse to the cheese is $(x - 12)^2 + (8 - 5x)^2 = 26(x^2 - 4x + 8) = 26((x - 2)^2 + 4)$ .
The value of this expression is smallest when $x = 2$ , so the mouse is closest to the cheese at the point $(2, 8)$ , and $a+b=2+8 = \boxed{10}$ .
-Paixiao | null | 10 |
9e5222aa39dd7ffe0f424bc6b502823e | https://artofproblemsolving.com/wiki/index.php/2007_AMC_12A_Problems/Problem_22 | For each positive integer $n$ , let $S(n)$ denote the sum of the digits of $n.$ For how many values of $n$ is $n + S(n) + S(S(n)) = 2007?$
$\mathrm{(A)}\ 1 \qquad \mathrm{(B)}\ 2 \qquad \mathrm{(C)}\ 3 \qquad \mathrm{(D)}\ 4 \qquad \mathrm{(E)}\ 5$ | It is well-known that $n \equiv S(n)\equiv S(S(n)) \pmod{9}.$ Substituting, we have that \[n+n+n \equiv 2007 \pmod{9} \implies n \equiv 0 \pmod{3}.\] Since $n \leq 2007,$ we must have that $\max S(n)=1+9+9+9=28.$ Now, we list out the possible vales for $S(n)$ in a table, noting that it is a multiple of $3$ because $n$ is a multiple of $3.$
Then, we compute the corresponding values of $S(S(n)).$
Finally, we may compute the corresponding values of $n$ using the fact that $n=2007-S(n)-S(S(n)).$
Notice how all conditions are designed to be satisfied except whether $S(n)$ is accurate with respect to $n.$ So, the only thing that remains is to check this. We may eliminate, for example, when $n=2007$ we have $S(n)=9$ while the table states that it is $0.$ Proceeding similarly, we obtain the following table.
It follows that there are $\boxed{4}$ possible values for $n.$ ~samrocksnature | D | 4 |
9e5222aa39dd7ffe0f424bc6b502823e | https://artofproblemsolving.com/wiki/index.php/2007_AMC_12A_Problems/Problem_22 | For each positive integer $n$ , let $S(n)$ denote the sum of the digits of $n.$ For how many values of $n$ is $n + S(n) + S(S(n)) = 2007?$
$\mathrm{(A)}\ 1 \qquad \mathrm{(B)}\ 2 \qquad \mathrm{(C)}\ 3 \qquad \mathrm{(D)}\ 4 \qquad \mathrm{(E)}\ 5$ | Claim. The only positive integers $n$ that satisfy the condition are perfect multiples of $3$
Proof of claim:
We examine the positive integers mod $9$ . Here are the cases.
Case 1. $n \equiv 1 \pmod 9$ . Now, we examine $S(n)$ modulo $9$ .
Case 1.1. The tens digit of $n$ is different from the tens digit of the largest multiple of $9$ under $n$ . (In other words, this means we will carry when adding from the perfect multiple of $9$ under $n$ .)
Observe that when we carry, i.e. Add $1$ onto $1989$ to obtain $1990$ , the units digit decreases by $9$ while the tens digit increases by $1$ . This means that the sum of the digits decreases by $8$ in total, and we have $-8 \equiv 1 \pmod 9$ , so the "mod 9" of the sum increases by $1$ . This means that, regardless of whether the sum carries or not, the modulo 9 of the sum of the digits always increases by $1$
Case 1.2. The tens digits are the same, which is trivial since the units digit just increases by $1$ which means that the sum is also equivalent to $1 \pmod 9$
This means that $S(n) \equiv 1 \pmod 9$ and similarly letting the next $n=S(n)$ $S(S(n)) \equiv 1 \pmod 9$ . Summing these, we have $n+S(n)+S(S(n)) \equiv 3 \pmod 9$ . Clearly, no integers of this form will satisfy the condition because $2007$ is a perfect multiple of $9$
Case 2. $n \equiv 2 \pmod 9$
In this case, we apply exactly the same argument. There is at most one carry, which means that the sum of the digits will always be congruent to $2$ mod $9$ . Then we can apply similar arguments to get $S(n) \equiv 2 \pmod 9$ and $S(S(n)) \equiv 2 \pmod 9$ , so adding gives $n+S(n)+S(S(n)) \equiv 6 \pmod 9$
It is trivial to see that for $n \equiv k \pmod 9$ , for $0 \leq k \leq 8$ , we must have $n+S(n)+S(S(n)) \equiv 3k \pmod 9$ . Only when $k=0, 3, 6$ is $3k$ a multiple of $9$ , which means that $n$ must be a multiple of $3$
Now, we find the integers. Again, consider two cases: Integers that are direct multiples of $9$ and integers that are multiples of $3$ but not $9$
Case 1. $n$ is a multiple of $9$ . An integer of the form $\overline{20ab}$ will not work since the least such integer is $2007$ which already exceeds our bounds. Thus, we need only consider the integers of the form $\overline{19ab}$ . The valid sums of the digits of $n$ are $18$ and $27$ in this case.
Case 1.1. The sum of the digits is $18$ . This means that $S(n)=18, S(S(n))=9$ , so $n=2007-18-9=1980$ . Clearly this number satisfies our constraints.
Case 1.2. The sum of the digits is $27$ . This means that $S(n)=27, S(S(n))=9$ , ,so $n=2007-27-9=1971$ . Since the sum of the digits of $1971$ is not $27$ , this does not work.
This means that there is $1$ integer in this case.
Case 2. $n$ is a multiple of $3$ , not $9$ .
.
Case 2.1. Integers of the form $\overline{20ab}$ . Then $S(n)=3$ or $S(n)=6$ ; it is trivial to see that $S(n)=6$ exceeds our bounds, so $S(n)=3$ and $n=2007-6=2001$
Case 2.2. Integers of the form $\overline{19ab}$ . Then $S(n)=12, 15, 21, 24$ and we consider each case separately.
Case 2.2.1. Integers with $S(n)=12$ . That means $n=2007-12-3=1992$ which clearly does not work.
Case 2.2.2. Integers with $S(n)=15$ . That means $n=2007-15-6=1986$ which also does not work
Case 2.2.3. Integers with $S(n)=21$ . That means $n=2007-21-3=1983$ which is valid.
Case 2.2.4. Integers with $S(n)=24$ . That means $n=2007-24-6=1977$ which is also valid.
We have considered every case, so there are $\boxed{4}$ integers that satisfy the given condition. | null | 4 |
9e5222aa39dd7ffe0f424bc6b502823e | https://artofproblemsolving.com/wiki/index.php/2007_AMC_12A_Problems/Problem_22 | For each positive integer $n$ , let $S(n)$ denote the sum of the digits of $n.$ For how many values of $n$ is $n + S(n) + S(S(n)) = 2007?$
$\mathrm{(A)}\ 1 \qquad \mathrm{(B)}\ 2 \qquad \mathrm{(C)}\ 3 \qquad \mathrm{(D)}\ 4 \qquad \mathrm{(E)}\ 5$ | Let the number of digits of $n$ be $m$ . If $m = 5$ $n$ will already be greater than $2007$ . Notice that $S(n)$ is always at most $9m$ . Then if $m = 3$ $n$ will be at most $999$ $S(n)$ will be at most $27$ , and $S(S(n))$ will be even smaller than $27$ . Clearly we cannot reach a sum of $2007$ , unless $m = 4$ (i.e. $n$ has $4$ digits).
Then, let $n$ be a four digit number in the form $1000a + 100b + 10c + d$ . Then $S(n) = a + b + c + d$
$S(S(n))$ is the sum of the digits of $a + b + c + d$ . We can represent $S(S(n))$ as the sum of the tens digit and the ones digit of $S(n)$ . The tens digit in the form of a decimal is
$\frac{a + b + c + d}{10}$
To remove the decimal portion, we can simply take the floor of the expression,
$\lfloor\frac{a + b + c + d}{10}\rfloor$
Now that we have expressed the tens digit, we can express the ones digit as $S(n) -10$ times the above expression, or
$a + b + c + d - 10\lfloor\frac{a + b + c + d}{10}\rfloor$
Adding the two expressions yields the value of $S(S(n))$
$= a + b + c + d - 9\lfloor\frac{a + b + c + d}{10}\rfloor$
Combining this expression to the ones for $n$ and $S(n)$ yields
$1002a + 102b + 12c + 3d - 9\lfloor\frac{a + b + c + d}{10}\rfloor$
Setting this equal to $2007$ and rearranging a bit yields
$12c + 3d = 2007 - 1002a - 102b + 9\lfloor\frac{a + b + c + d}{10}\rfloor$
$\Rightarrow$ $4c + d = 669 - 334a - 34b + 3\lfloor\frac{a + b + c + d}{10}\rfloor$
(The reason for this slightly weird arrangement will soon become evident)
Now we examine the possible values of $a$ . If $a \ge 3$ $n$ is already too large. $a$ must also be greater than $0$ , or $n$ would be a $3$ -digit number. Therefore, $a = 1 \, \text{or} \, 2$ . Now we examine by case.
If $a = 2$ , then $b$ and $c$ must both be $0$ (otherwise $n$ would already be greater than $2007$ ). Substituting these values into the equation yields
$d = 1 + 3\lfloor\frac{2 + d}{10}\rfloor$
$\Rightarrow$ $d=1$
Sure enough, $2001 + (2+1) + 3=2007$
Now we move onto the case where $a = 1$ . Then our initial equation simplifies to
$4c + d = 335 - 34b + 3\lfloor\frac{1 + b + c + d}{10}\rfloor$
Since $c$ and $d$ can each be at most $9$ , we substitute that value to find the lower bound of $b$ . Doing so yields
$34b \ge 290 + 3\lfloor\frac{19 + b}{10}\rfloor$
The floor expression is at least $3\lfloor\frac{19}{10}\rfloor=3$ , so the right-hand side is at least $293$ . Solving for $b$ , we see that $b \ge 9$ $\Rightarrow$ $b=9$ . Again, we substitute for $b$ and the equation becomes
$4c + d = 29 + 3\lfloor\frac{10 + c + d}{10}\rfloor$
$\Rightarrow$ $4c + d = 32 + 3\lfloor\frac{c + d}{10}\rfloor$
Just like we did for $b$ , we can find the lower bound of $c$ by assuming $d = 9$ and solving:
$4c + 9 \ge 29 + 3\lfloor\frac{c + 9}{10}\rfloor$
$\Rightarrow$ $4c \ge 20 + 3\lfloor\frac{c + 9}{10}\rfloor$
The right hand side is $20$ for $c=0$ and $23$ for $c \ge 1$ . Solving for c yields $c \ge 6$ . Looking back at the previous equation, the floor expression is $0$ for $c+d \le 9$ and $3$ for $c+d \ge 10$ . Thus, the right-hand side is $32$ for $c+d \le 9$ and $35$ for $c+d \ge 10$ . We can solve these two scenarios as systems of equations/inequalities:
$4c+d = 32$
$c+d \le 9$
and
$4c+d=35$
$c+d \ge 10$
Solving yields three pairs $(c, d):$ $(8, 0)$ $(8, 3)$ ; and $(7, 7)$ . Checking the numbers $1980$ $1983$ , and $1977$ ; we find that all three work. Therefore there are a total of $4$ possibilities for $n$ $\Rightarrow$ $\boxed{4}$ | D | 4 |
0f312a868ce3a8d148b1fff2605e1349 | https://artofproblemsolving.com/wiki/index.php/2007_AMC_12A_Problems/Problem_25 | Call a set of integers spacy if it contains no more than one out of any three consecutive integers. How many subsets of $\{1,2,3,\ldots,12\},$ including the empty set , are spacy?
$\mathrm{(A)}\ 121 \qquad \mathrm{(B)}\ 123 \qquad \mathrm{(C)}\ 125 \qquad \mathrm{(D)}\ 127 \qquad \mathrm{(E)}\ 129$ | Let $S_{n}$ denote the number of spacy subsets of $\{ 1, 2, ... n \}$ . We have $S_{0} = 1, S_{1} = 2, S_{2} = 3$
The spacy subsets of $S_{n + 1}$ can be divided into two groups:
Hence,
From this recursion , we find that
And so the answer is $\boxed{129}$ | E | 129 |
0f312a868ce3a8d148b1fff2605e1349 | https://artofproblemsolving.com/wiki/index.php/2007_AMC_12A_Problems/Problem_25 | Call a set of integers spacy if it contains no more than one out of any three consecutive integers. How many subsets of $\{1,2,3,\ldots,12\},$ including the empty set , are spacy?
$\mathrm{(A)}\ 121 \qquad \mathrm{(B)}\ 123 \qquad \mathrm{(C)}\ 125 \qquad \mathrm{(D)}\ 127 \qquad \mathrm{(E)}\ 129$ | Since each of the elements of the subsets must be spaced at least two apart, a divider counting argument can be used.
From the set $\{1,2,3,4,5,6,7,8,9,10,11,12\}$ we choose at most four numbers. Let those numbers be represented by balls. Between each of the balls there are at least two dividers. So for example, o | | o | | o | | o | | represents ${1,4,7,10}$
For subsets of size $k$ there must be $2(k - 1)$ dividers between the balls, leaving $12 - k - 2(k - 1) = 12 - 3k + 2$ dividers to be be placed in $k + 1$ spots between the balls. The number of way this can be done is $\binom{(12 - 3k + 2) + (k + 1) - 1}k = \binom{12 - 2k + 2}k$
Therefore, the number of spacy subsets is $\binom 64 + \binom 83 + \binom{10}2 + \binom{12}1 + \binom{14}0 = \boxed{129}$ | null | 129 |
0f312a868ce3a8d148b1fff2605e1349 | https://artofproblemsolving.com/wiki/index.php/2007_AMC_12A_Problems/Problem_25 | Call a set of integers spacy if it contains no more than one out of any three consecutive integers. How many subsets of $\{1,2,3,\ldots,12\},$ including the empty set , are spacy?
$\mathrm{(A)}\ 121 \qquad \mathrm{(B)}\ 123 \qquad \mathrm{(C)}\ 125 \qquad \mathrm{(D)}\ 127 \qquad \mathrm{(E)}\ 129$ | A shifting argument is also possible, and is similar in spirit to Solution 2. Clearly we can have at most $4$ elements. Given any arrangment, we subract $2i-2$ from the $i-th$ element in our subset, when the elements are arranged in increasing order. This creates a bijection with the number of size $k$ subsets of the set of the first $14-2k$ positive integers. For instance, the arrangment o | | o | | o | | | o | corresponds to the arrangment o o o | o |. Notice that there is no longer any restriction on consectutive numbers. Therefore, we can easily plug in the possible integers 0, 1, 2, 3, 4, 5 for $k$ ${14 \choose 0} + {12 \choose 1} + {10 \choose 2} + {8 \choose 3} + {6 \choose 4} = \boxed{129}$ | null | 129 |
0f312a868ce3a8d148b1fff2605e1349 | https://artofproblemsolving.com/wiki/index.php/2007_AMC_12A_Problems/Problem_25 | Call a set of integers spacy if it contains no more than one out of any three consecutive integers. How many subsets of $\{1,2,3,\ldots,12\},$ including the empty set , are spacy?
$\mathrm{(A)}\ 121 \qquad \mathrm{(B)}\ 123 \qquad \mathrm{(C)}\ 125 \qquad \mathrm{(D)}\ 127 \qquad \mathrm{(E)}\ 129$ | Let us consider each size of subset individually. Since each integer in the subset must be at least $3$ away from any other integer in the subset, the largest spacy subset contains $4$ elements.
First, it is clear that there is $1$ spacy set with $0$ elements in it, the empty set. Next, there are $12$ spacy subsets with $1$ element in them, one for each integer $1$ through $12$
Now, let us consider the spacy subsets with $2$ elements in them. If the smaller integer is $1$ , the larger integer is any of the $9$ integers from $4$ to $12$ . If the smaller integer is $2$ , the larger integer is any of the $8$ integers from $5$ to $12$ . This continues, up to a smaller integer of $9$ and $1$ choice for the larger integer, $12$ . This means that there are $9 + 8 + \cdots + 1 = 45$ spacy subsets with $2$ elements.
For spacy subsets with $3$ elements, we first consider the middle integer. The smallest such integer is $4$ , and it allows for $1$ possible value for the smaller integer ( $1$ ) and possible $6$ values for the larger integer ( $7$ through $12$ ), for a total of $1 \cdot 6 = 6$ possible subsets. The next middle integer, $5$ , allows for $2$ smaller integers and $5$ larger integers, and this pattern continues up until the middle integer of $9$ , which has $6$ values for the smaller integer and $1$ value for the larger integer. This means that there are $1 \cdot 6 + 2 \cdot 5 + \cdots + 6 \cdot 1 = 56$ spacy subsets with $3$ elements.
Lastly, there are $3$ main categories for spacy subsets with $4$ elements, defined by the difference between their smallest and largest values. The difference ranges from $9$ to $11$ . If it is $9$ , there is only $1$ set of places to put the two middle values ( $n + 3$ and $n + 6$ , where $n$ is the smallest value). Since there are $3$ possible sets of smallest and largest values, there are $1 \cdot 3 = 3$ sets in this category. If the difference is $10$ , there are now $3$ sets of places to put the two middle values ( $n + 3$ and $n + 6$ or $7$ , and $n + 4$ and $n + 7$ ). There are $2$ possible sets of smallest and largest values, so there are $3 \cdot 2 = 6$ sets in this category. Finally, if the difference is $11$ , there are $6$ possible sets of places to put the two middle values ( $n + 3$ and $n + 6$ $7$ , or $8$ $n + 4$ and $n + 7$ or $8$ , and $n + 5$ and $n + 8$ ) and one possible set of smallest and largest values, meaning that there are $6 \cdot 1 = 6$ sets in this category. Adding them up, there are $3 + 6 + 6 = 15$ spacy subsets with $4$ elements.
Adding these all up, we have a total of $1 + 12 + 45 + 56 + 15 = \boxed{129}$ spacy subsets. ~ emerald_block | E | 129 |
0f312a868ce3a8d148b1fff2605e1349 | https://artofproblemsolving.com/wiki/index.php/2007_AMC_12A_Problems/Problem_25 | Call a set of integers spacy if it contains no more than one out of any three consecutive integers. How many subsets of $\{1,2,3,\ldots,12\},$ including the empty set , are spacy?
$\mathrm{(A)}\ 121 \qquad \mathrm{(B)}\ 123 \qquad \mathrm{(C)}\ 125 \qquad \mathrm{(D)}\ 127 \qquad \mathrm{(E)}\ 129$ | We build a modified version of Markov's chain to solve this problem. First, start off by representing each element of the chosen subset with a $1$ , and every other element $0$ . The possible states are as follows; if the restriction was broken previously or not. For the sake of not over counting, we let the former be its own state (call it "C"), and the latter will be split up into more states. The previous two digits can either be $00$ $10$ , or $01$ . If a previous digit doesn't exist, it is $0$ for obvious reasons. We know that we start off at state " $00$ ". We have two options; include the first digit ( $1$ ) or not ( $0$ ). Thus, state $00$ transitions to state $01$ in $1$ case and state $00$ in $1$ case. State $01$ transitions to $10$ in $1$ case and the "C" state in $1$ case. State $10$ transitions to $00$ in $1$ case and state "C" in $1$ case. Since it doesn't matter where or how many times the restriction is broken for the subset to fail, state "C" can transition to itself in $2$ cases.
Notice how this corresponds to a transition matrix of $\begin{bmatrix} 1 & 1 & 0 & 0\\ 0 & 0 & 1 & 1\\ 1 & 0 & 0 & 1 \\ 0 & 0 & 0 & 2 \\ \end{bmatrix}$ . Since we start off on state $00$ , our starting state matrix is $\begin{bmatrix} 1 & 0 & 0 & 0\\ \end{bmatrix}$ . We wish to compute the value $a+b+c = 4096-d$ when given $f^{12}(\begin{bmatrix} 1 & 0 & 0 & 0\\ \end{bmatrix})=\begin{bmatrix} a & b & c & d\\ \end{bmatrix}$ , where $f(x)$ is $x\times\begin{bmatrix} 1 & 1 & 0 & 0\\ 0 & 0 & 1 & 1\\ 1 & 0 & 0 & 1 \\ 0 & 0 & 0 & 2 \\ \end{bmatrix}$ Through repeated applications of matrix multiplication, we obtain the output matrix of $\begin{bmatrix} 60 & 41 & 28 & 3967\\ \end{bmatrix}$ . Our answer is thus $60+41+28=\boxed{129}$ .
Remark: Markov's Chain is more commonly used to calculate probabilities, but it also can be very useful for counting problems with a uniform requirement.
-SigmaPiE | null | 129 |
b688e0ffe9a1ba9d35b5eeb52d8c89be | https://artofproblemsolving.com/wiki/index.php/2007_AMC_12B_Problems/Problem_1 | Isabella's house has $3$ bedrooms. Each bedroom is $12$ feet long, $10$ feet wide, and $8$ feet high. Isabella must paint the walls of all the bedrooms. Doorways and windows, which will not be painted, occupy $60$ square feet in each bedroom. How many square feet of walls must be painted?
$\mathrm{(A)}\ 678 \qquad \mathrm{(B)}\ 768 \qquad \mathrm{(C)}\ 786 \qquad \mathrm{(D)}\ 867 \qquad \mathrm{(E)}\ 876$ | There are four walls in each bedroom (she can't paint floors or ceilings). Therefore, we calculate the number of square feet of walls there is in one bedroom: \[2\cdot(12\cdot8+10\cdot8)-60=2\cdot176-60=292\] We have three bedrooms, so she must paint $292\cdot3=\boxed{876}$ square feet of walls. | E | 876 |
1d84e503d01e7e34c2a19ed9cee36ef6 | https://artofproblemsolving.com/wiki/index.php/2007_AMC_12B_Problems/Problem_2 | A college student drove his compact car $120$ miles home for the weekend and averaged $30$ miles per gallon. On the return trip the student drove his parents' SUV and averaged only $20$ miles per gallon. What was the average gas mileage, in miles per gallon, for the round trip?
$\textbf{(A) } 22 \qquad\textbf{(B) } 24 \qquad\textbf{(C) } 25 \qquad\textbf{(D) } 26 \qquad\textbf{(E) } 28$ | The trip was $240$ miles long and took $\dfrac{120}{30}+\dfrac{120}{20}=4+6=10$ gallons. Therefore, the average mileage was $\dfrac{240}{10}= \boxed{24}$ | B | 24 |
1d84e503d01e7e34c2a19ed9cee36ef6 | https://artofproblemsolving.com/wiki/index.php/2007_AMC_12B_Problems/Problem_2 | A college student drove his compact car $120$ miles home for the weekend and averaged $30$ miles per gallon. On the return trip the student drove his parents' SUV and averaged only $20$ miles per gallon. What was the average gas mileage, in miles per gallon, for the round trip?
$\textbf{(A) } 22 \qquad\textbf{(B) } 24 \qquad\textbf{(C) } 25 \qquad\textbf{(D) } 26 \qquad\textbf{(E) } 28$ | Alternatively, we can use the harmonic mean to get $\frac{2}{\frac{1}{20} + \frac{1}{30}} = \frac{2}{\frac{1}{12}} = \boxed{24}$ | B | 24 |
b60320f4f485ad4edb6d3373cbf5a11d | https://artofproblemsolving.com/wiki/index.php/2007_AMC_12B_Problems/Problem_8 | Tom's age is $T$ years, which is also the sum of the ages of his three children. His age $N$ years ago was twice the sum of their ages then. What is $T/N$
$\textbf{(A) } 2 \qquad\textbf{(B) } 3 \qquad\textbf{(C) } 4 \qquad\textbf{(D) } 5 \qquad\textbf{(E) } 6$ | Tom's age $N$ years ago was $T-N$ . The sum of the ages of his three children $N$ years ago was $T-3N,$ since there are three children. If his age $N$ years ago was twice the sum of the children's ages then, \begin{align*}T-N&=2(T-3N)\\ T-N&=2T-6N\\ T&=5N\\ T/N&=\boxed{5} Note that actual values were not found. | D | 5 |
13e6c5ea0496e122e8202385f761f7b2 | https://artofproblemsolving.com/wiki/index.php/2007_AMC_12B_Problems/Problem_10 | Some boys and girls are having a car wash to raise money for a class trip to China. Initially $40\%$ of the group are girls. Shortly thereafter two girls leave and two boys arrive, and then $30\%$ of the group are girls. How many girls were initially in the group?
$\textbf{(A) } 4 \qquad\textbf{(B) } 6 \qquad\textbf{(C) } 8 \qquad\textbf{(D) } 10 \qquad\textbf{(E) } 12$ | If we let $p$ be the number of people initially in the group, then $0.4p$ is the number of girls. If two girls leave and two boys arrive, the number of people in the group is still $p$ , but the number of girls is $0.4p-2$ . Since only $30\%$ of the group are girls, \begin{align*} \frac{0.4p-2}{p}&=\frac{3}{10}\\ 4p-20&=3p\\ p&=20\end{align*} The number of girls initially in the group is $0.4p=0.4(20)=\boxed{8}$ | C | 8 |
13e6c5ea0496e122e8202385f761f7b2 | https://artofproblemsolving.com/wiki/index.php/2007_AMC_12B_Problems/Problem_10 | Some boys and girls are having a car wash to raise money for a class trip to China. Initially $40\%$ of the group are girls. Shortly thereafter two girls leave and two boys arrive, and then $30\%$ of the group are girls. How many girls were initially in the group?
$\textbf{(A) } 4 \qquad\textbf{(B) } 6 \qquad\textbf{(C) } 8 \qquad\textbf{(D) } 10 \qquad\textbf{(E) } 12$ | Let $x$ be the number of people initially in the group and $g$ the number of girls. $\frac{2}{5}x = g$ , so $x = \frac{5}{2}g$ . Also, the problem states $\frac{3}{10}x = g-2$ . Substituting $x$ in terms of $g$ into the second equation yields that $g = \boxed{8}$ | C | 8 |
9f28aacdb7e70977ee78e860e95e6a39 | https://artofproblemsolving.com/wiki/index.php/2007_AMC_12B_Problems/Problem_11 | The angles of quadrilateral $ABCD$ satisfy $\angle A=2 \angle B=3 \angle C=4 \angle D.$ What is the degree measure of $\angle A,$ rounded to the nearest whole number?
$\textbf{(A) } 125 \qquad\textbf{(B) } 144 \qquad\textbf{(C) } 153 \qquad\textbf{(D) } 173 \qquad\textbf{(E) } 180$ | The sum of the interior angles of any quadrilateral is $360^\circ.$ \begin{align*} 360 &= \angle A + \angle B + \angle C + \angle D\\ &= \angle A + \frac{1}{2}A + \frac{1}{3}A + \frac{1}{4}A\\ &= \frac{12}{12}A + \frac{6}{12}A + \frac{4}{12}A + \frac{3}{12}A\\ &= \frac{25}{12}A \end{align*} \[\angle A = 360 \cdot \frac{12}{25} = 172.8 \approx \boxed{173}\] | D | 173 |
604dadb1759f1b1c452462891feac98f | https://artofproblemsolving.com/wiki/index.php/2007_AMC_12B_Problems/Problem_12 | A teacher gave a test to a class in which $10\%$ of the students are juniors and $90\%$ are seniors. The average score on the test was $84.$ The juniors all received the same score, and the average score of the seniors was $83.$ What score did each of the juniors receive on the test?
$\textbf{(A) } 85 \qquad\textbf{(B) } 88 \qquad\textbf{(C) } 93 \qquad\textbf{(D) } 94 \qquad\textbf{(E) } 98$ | We can assume there are $10$ people in the class. Then there will be $1$ junior and $9$ seniors. The sum of everyone's scores is $10 \cdot 84 = 840$ . Since the average score of the seniors was $83$ , the sum of all the senior's scores is $9 \cdot 83 = 747$ . The only score that has not been added to that is the junior's score, which is $840 - 747 = \boxed{93}$ | C | 93 |
604dadb1759f1b1c452462891feac98f | https://artofproblemsolving.com/wiki/index.php/2007_AMC_12B_Problems/Problem_12 | A teacher gave a test to a class in which $10\%$ of the students are juniors and $90\%$ are seniors. The average score on the test was $84.$ The juniors all received the same score, and the average score of the seniors was $83.$ What score did each of the juniors receive on the test?
$\textbf{(A) } 85 \qquad\textbf{(B) } 88 \qquad\textbf{(C) } 93 \qquad\textbf{(D) } 94 \qquad\textbf{(E) } 98$ | Let the average score of the juniors be $j$ . The problem states the average score of the seniors is $83$ . The equation for the average score of the class (juniors and seniors combined) is $\frac{j}{10} + \frac{83 \cdot 9}{10} = 84$ . Simplifying this equation yields $j = \boxed{93}$ | C | 93 |
c64167d60b45f76aff5753eb1ab192be | https://artofproblemsolving.com/wiki/index.php/2007_AMC_12B_Problems/Problem_18 | Let $a$ $b$ , and $c$ be digits with $a\ne 0$ . The three-digit integer $abc$ lies one third of the way from the square of a positive integer to the square of the next larger integer. The integer $acb$ lies two thirds of the way between the same two squares. What is $a+b+c$
$\mathrm{(A)}\ 10 \qquad \mathrm{(B)}\ 13 \qquad \mathrm{(C)}\ 16 \qquad \mathrm{(D)}\ 18 \qquad \mathrm{(E)}\ 21$ | Let $k$ be the lesser of the two integers. Then the squares of the integers are $k^2$ and $k^2+2k+1$ , and the distance between them is $2k+1$ . Let this be equivalent to $3d$ , so that the one-third of the distance between the squares is equivalent to $d$ . The numbers $abc$ and $acb$ are one-third and two-thirds of the way between $k^2$ and $(k+1)^2$ . Therefore, the distance between these two numbers is also one-third the distance between the squares, or $d$ . Setting these equal to each other, we have
$\frac{2k+1}{3} = 9(c-b)$
$\Rightarrow 2k+1 = 27(c-b)$
Notice that since $c$ and $b$ are digits, their difference is at most $9$ and at least $0$ . Also notice that since $acb$ is greater than $abc$ $c > b$ . Representing this as an inequality, we have
$27 \le 27(c-b) \le 243$
Substituting $2k+1$ , we have
$27 \le 2k+1 \le 243$
$\Rightarrow 13 \le k \le 121$
However, we know that $abc$ is a $3$ -digit number, and since $k^2$ is less than $abc$ $k^2$ must be at most $961$ , or $31^2$ . Therefore $k \le 31$ . Plugging this back into our inequality, we have
$13 \le k \le 31$
$\Rightarrow 27 \le 2k+1 \le63$
$\Rightarrow 27 \le 27(c-b) \le 63$
$\Rightarrow 1 \le (c-b) \le \frac{7}{3}$
But (c-b) must be an integer, so now we have
$1 \le (c-b) \le 2$
$\Rightarrow 27 \le 27(c-b) \le 54$
$\Rightarrow 27 \le 2k+1 \le 54$
$\Rightarrow 13 \le k \le\frac{53}{2}$
$k$ is also an integer, so now we have
$\Rightarrow 13 \le k \le 26$
$\Rightarrow 27 \le 2k+1 \le 53$
$\Rightarrow 27 \le 27(c-b) \le 53$
$\Rightarrow 1 \le (c-b) \le \frac{53}{27}$
Once again, $(c-b)$ must be an integer, so we have
$1 \le (c-b) \le 1$
$\Rightarrow (c-b) = 1$
$\Rightarrow 27(c-b) = 27$
$\Rightarrow 2k+1 = 27$
$\Rightarrow k = 13$
The two squares are $13^2$ and $14^2$ , or $169$ and $196$ . A third of the distance between them is $9$ , and $169 +9 = 178$ $1 + 7 + 8 = 16 \Rightarrow \boxed{16}$ | C | 16 |
5dba9a0b30bb787f260c13b44f288a73 | https://artofproblemsolving.com/wiki/index.php/2007_AMC_12B_Problems/Problem_20 | The parallelogram bounded by the lines $y=ax+c$ $y=ax+d$ $y=bx+c$ , and $y=bx+d$ has area $18$ . The parallelogram bounded by the lines $y=ax+c$ $y=ax-d$ $y=bx+c$ , and $y=bx-d$ has area $72$ . Given that $a$ $b$ $c$ , and $d$ are positive integers, what is the smallest possible value of $a+b+c+d$
$\mathrm {(A)} 13\qquad \mathrm {(B)} 14\qquad \mathrm {(C)} 15\qquad \mathrm {(D)} 16\qquad \mathrm {(E)} 17$ | Plotting the parallelogram on the coordinate plane, the 4 corners are at $(0,c),(0,d),\left(\frac{d-c}{a-b},\frac{ad-bc}{a-b}\right),\left(\frac{c-d}{a-b},\frac{bc-ad}{a-b}\right)$ . Because $72= 4\cdot 18$ , we have that $4(c-d)\left(\frac{c-d}{a-b}\right) = (c+d)\left(\frac{c+d}{a-b}\right)$ or that $2(c-d)=c+d$ , which gives $c=3d$ (consider a homothety , or dilation, that carries the first parallelogram to the second parallelogram; because the area increases by $4\times$ , it follows that the stretch along the diagonal, or the ratio of side lengths, is $2\times$ ). The area of the triangular half of the parallelogram on the right side of the y-axis is given by $9 = \frac{1}{2} (c-d)\left(\frac{d-c}{a-b}\right)$ , so substituting $c = 3d$
Thus $3|d$ , and we verify that $d = 3$ $a-b = 2 \Longrightarrow a = 3, b = 1$ will give us a minimum value for $a+b+c+d$ . Then $a+b+c+d = 3 + 1 + 9 + 3 = \boxed{16}$ | D | 16 |
5dba9a0b30bb787f260c13b44f288a73 | https://artofproblemsolving.com/wiki/index.php/2007_AMC_12B_Problems/Problem_20 | The parallelogram bounded by the lines $y=ax+c$ $y=ax+d$ $y=bx+c$ , and $y=bx+d$ has area $18$ . The parallelogram bounded by the lines $y=ax+c$ $y=ax-d$ $y=bx+c$ , and $y=bx-d$ has area $72$ . Given that $a$ $b$ $c$ , and $d$ are positive integers, what is the smallest possible value of $a+b+c+d$
$\mathrm {(A)} 13\qquad \mathrm {(B)} 14\qquad \mathrm {(C)} 15\qquad \mathrm {(D)} 16\qquad \mathrm {(E)} 17$ | The key to this solution is that area is invariant under translation. By suitably shifting the plane, the problem is mapped to the lines $c,d,(b-a)x+c,(b-a)x+d$ and $c,-d,(b-a)x+c,(b-a)x-d$ . Now, the area of the parallelogram contained by is the former is equal to the area of a rectangle with sides $d-c$ and $\frac{d-c}{b-a}$ $\frac{(d-c)^2}{b-a}=18$ , and the area contained by the latter is $\frac{(c+d)^2}{b-a}=72$ . Thus, $d=3c$ and $b-a$ must be even if the former quantity is to equal $18$ $c^2=18(b-a)$ so $c$ is a multiple of $3$ . Putting this all together, the minimal solution for $(a,b,c,d)=(3,1,3,9)$ , so the sum is $\boxed{16}$ | D | 16 |
5dba9a0b30bb787f260c13b44f288a73 | https://artofproblemsolving.com/wiki/index.php/2007_AMC_12B_Problems/Problem_20 | The parallelogram bounded by the lines $y=ax+c$ $y=ax+d$ $y=bx+c$ , and $y=bx+d$ has area $18$ . The parallelogram bounded by the lines $y=ax+c$ $y=ax-d$ $y=bx+c$ , and $y=bx-d$ has area $72$ . Given that $a$ $b$ $c$ , and $d$ are positive integers, what is the smallest possible value of $a+b+c+d$
$\mathrm {(A)} 13\qquad \mathrm {(B)} 14\qquad \mathrm {(C)} 15\qquad \mathrm {(D)} 16\qquad \mathrm {(E)} 17$ | Let $a$ and $b$ be the slopes of the lines such that $b > a$ (i.e. the line $bx+c$ is steeper than $ax+c$ ) and $c > d$ (i.e. the point $(0, c)$ is higher than the point $(0, d)$ . Upon drawing a diagram, we see that both the smaller and the larger parallelogram can be split along the x-axis, such that both of their areas are the combinations of two corresponding triangles. The area of a triangle is $\frac{bh}{2}$ , but since a parallelogram is two such triangles, the area becomes $bh$
Let $b_1$ and $h_1$ denote the base length and height, respectively, of the triangles pertaining to the smaller parallelogram and $b_2$ and $h_2$ denote those of the larger parallelogram. Notice that $b_1$ is simple the distance from $(0, d)$ to $(0, c)$ , or $(c-d)$ . Also notice that $h_1$ is the distance from the $x$ -axis to the intersection of lines $ax+c$ and $bx+d$ . This is equivalent to the value of the $x$ -coordinate of intersection, so we solve for $x$
$ax+c=bx+d$
$\Rightarrow bx-ax = c-d$
$\Rightarrow (b-a)x = c-d$
$\Rightarrow x = \frac{c-d}{b-a}$
The area of the smaller parallelogram is $b_1*h_1$ , or
$(c-d) * \frac{c-d}{b-a}$
$\Rightarrow \frac{(c-d)^2}{b-a}$
$b_2$ is the distance from $(0, -d)$ to $(0, c)$ , or $(c+d)$ $h_2$ is the $x$ -coordinate of the intersection of the lines $ax+c$ and $bx-d$ . Again, we solve for x:
$ax+c=bx-d$
$\Rightarrow bx-ax = c+d$
$\Rightarrow (b-a)x = c+d$
$\Rightarrow x = \frac{c+d}{b-a}$
The area of the larger parallelogram is $b-1*h_1$ , or
$(c+d) * \frac{c+d}{b-a}$
$\Rightarrow \frac{(c+d)^2}{b-a}$
The areas of the parallelograms are given to us: $18$ and $72$ . Therefore we can set up a ratio:
$\frac{18}{72} = \frac{\frac{(c-d)^2}{b-a}}{\frac{(c+d)^2}{b-a}}$
$\Rightarrow 18(c+d)^2 = 72(c-d)^2$
$\Rightarrow (c+d)^2 = 4(c-d)^2$
$\Rightarrow c^2 + 2cd + d^2 = 4c^2 - 8cd + 4d^2$
$\Rightarrow 3c^2 - 10cd +3d^2 = 0$
$\Rightarrow (3c-d)(c-3d)=0$
$\Rightarrow c=3d, c=\frac{d}{3}$
We established earlier that $c>d$ , so $c=3d$ . Plugging this into the intial equations yields
$\frac{16d^2}{b-a} = 72$
and
$\frac{4d^2}{b-a} = 18$
Solving for $d$ , we get
$d = 3\sqrt{\frac{b-a}{2}}$
We want the sum of $a$ $b$ $c$ , and $d$ . We can now rewrite this
$a + b + 12\sqrt{\frac{b-a}{2}}$
We are told that $a$ $b$ $c$ , and $d$ are all positive integers. Therefore the value under the radical must be a perfect square greater than 0. We can rewrite this
$\frac{b-a}{2} = k^2$
where $k$ is some positive integer.
Rearranging, we get
$b = a + 2k^2$
Now we can rewrite the sum as
$a + a + 2k^2 +12k$
Since both $a$ and $k$ must be at least $1$ , the minimum value is
$1 + 1 + 2(1)^2 + 12(1) = 16 \Rightarrow \boxed{16}$ | D | 16 |
fbce6bda07d66372129e3393674093cd | https://artofproblemsolving.com/wiki/index.php/2007_AMC_12B_Problems/Problem_23 | How many non-congruent right triangles with positive integer leg lengths have areas that are numerically equal to $3$ times their perimeters?
$\mathrm {(A)} 6\qquad \mathrm {(B)} 7\qquad \mathrm {(C)} 8\qquad \mathrm {(D)} 10\qquad \mathrm {(E)} 12$ | Let $a$ and $b$ be the two legs of the triangle, and $c$ be the hypotenuse.
By using $Area = \frac{r}{2} (a+b+c)$ , where $r$ is the in-radius, we get:
\[3(a+b+c) = \frac{r}{2} (a+b+c)\] \[r=6\]
In right triangle, $r = \frac{a+b-c}{2}$ \[a+b-c = 12\] \[c = a + b - 12\]
By the triangle's area we get:
\[\frac{ab}{2} = 6 \cdot \frac{a+b+c}{2}\] \[ab = 6(a+b+c)\]
By substituting $c$ in:
\[ab = 6(a+b+a + b - 12)\] \[ab - 12a - 12b + 72 = 0\] \[(a - 12)(b - 12) = 72\]
As $72 = 2^3 \cdot 3^2$ , there are $\frac{(3+1)(2+1)}{2} = 6$ solutions, $\boxed{6}$ | A | 6 |
fbce6bda07d66372129e3393674093cd | https://artofproblemsolving.com/wiki/index.php/2007_AMC_12B_Problems/Problem_23 | How many non-congruent right triangles with positive integer leg lengths have areas that are numerically equal to $3$ times their perimeters?
$\mathrm {(A)} 6\qquad \mathrm {(B)} 7\qquad \mathrm {(C)} 8\qquad \mathrm {(D)} 10\qquad \mathrm {(E)} 12$ | Well, obviously MAA would try to make the answer choices trap some people. One way they could do that is by thinking "non-congruent" would be ignored, so the answer would be multiplied by 2. The only answer choice that can be divided by 2 to create an existing answer is 12, so the answer is $\boxed{6}$ | A | 6 |
c84575ab947ea92b2fa92f55386966ac | https://artofproblemsolving.com/wiki/index.php/2007_AMC_12B_Problems/Problem_24 | problem_id
c84575ab947ea92b2fa92f55386966ac Also refer to the 2007 AMC 10B #25 (same problem)
c84575ab947ea92b2fa92f55386966ac How many pairs of positive integers $(a,b)$ ar...
Name: Text, dtype: object | Rewrite the equation\[ $\frac{a}{b}+\frac{14b}{9a}=k$ \]in two different forms. First, multiply both sides by $b$ and subtract $a$ to obtain\[ $\frac{14b^2}{9a}=bk-a.$ \]Because $a$ $b$ , and $k$ are integers, $14b^2$ must be a multiple of $a$ , and because $a$ and $b$ have no common factors greater than 1, it follows that 14 is divisible by $a$ . Next, multiply both sides of the original equation by $9a$ and subtract $14b$ to obtain\[ $\frac{9a^2}{b}=9ak-14b.$ \]This shows that $9a^2$ is a multiple of $b$ , so 9 must be divisible by $b$ . Thus if $(a,b)$ is a solution, then $b=1$ $3$ , or $9$ , and $a=1$ , 2, 7, or 14. This gives a total of twelve possible solutions $(a,b)$ , each of which can be checked quickly. $\[
\frac{a}{b}+\frac{14b}{9a}\]$ (Error compiling LaTeX. Unknown error_msg) will only be an integer when $(a,b) \in \{(1,3),(2,3), (7,3), (14,3)\},$ for a total of $\boxed{4}$ pairs. | null | 4 |
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