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Difference between revisions of "2012 AMC 12B Problems/Problem 25"

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[[2012 AMC 12B Problems/Problem 25|Solution]]
 
[[2012 AMC 12B Problems/Problem 25|Solution]]
  
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==Solution 1==
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==Solution==
 
Consider reflections. For any right triangle <math>ABC</math> with the right labeling described in the problem, any reflection <math>A'B'C'</math> labeled that way will give us <math>\tan CBA \cdot \tan C'B'A' = 1</math>. First we consider the reflection about the line <math>y=2.5</math>. Only those triangles <math>\subseteq S</math> that have one vertex at <math>(0,5)</math> do not reflect to a traingle <math>\subseteq S</math>. Within those triangles, consider a reflection about the line <math>y=5-x</math>. Then only those triangles <math>\subseteq S</math> that have one vertex on the line <math>y=0</math> do not reflect to a triangle <math>\subseteq S</math>. So we only need to look at right triangles that have vertices <math>(0,5), (*,0), (*,*)</math>. There are three cases:
 
Consider reflections. For any right triangle <math>ABC</math> with the right labeling described in the problem, any reflection <math>A'B'C'</math> labeled that way will give us <math>\tan CBA \cdot \tan C'B'A' = 1</math>. First we consider the reflection about the line <math>y=2.5</math>. Only those triangles <math>\subseteq S</math> that have one vertex at <math>(0,5)</math> do not reflect to a traingle <math>\subseteq S</math>. Within those triangles, consider a reflection about the line <math>y=5-x</math>. Then only those triangles <math>\subseteq S</math> that have one vertex on the line <math>y=0</math> do not reflect to a triangle <math>\subseteq S</math>. So we only need to look at right triangles that have vertices <math>(0,5), (*,0), (*,*)</math>. There are three cases:
  

Revision as of 16:47, 11 February 2013

Problem 25

Let $S=\{(x,y) : x\in \{0,1,2,3,4\}, y\in \{0,1,2,3,4,5\},\text{ and } (x,y)\ne (0,0)\}$. Let $T$ be the set of all right triangles whose vertices are in $S$. For every right triangle $t=\triangle{ABC}$ with vertices $A$, $B$, and $C$ in counter-clockwise order and right angle at $A$, let $f(t)=\tan(\angle{CBA})$. What is \[\prod_{t\in T} f(t)?\]

$\textbf{(A)}\ 1\qquad\textbf{(B)}\ \frac{625}{144}\qquad\textbf{(C)}\ \frac{125}{24}\qquad\textbf{(D)}\ 6\qquad\textbf{(E)}\ \frac{625}{24}$

Solution

Solution

Consider reflections. For any right triangle $ABC$ with the right labeling described in the problem, any reflection $A'B'C'$ labeled that way will give us $\tan CBA \cdot \tan C'B'A' = 1$. First we consider the reflection about the line $y=2.5$. Only those triangles $\subseteq S$ that have one vertex at $(0,5)$ do not reflect to a traingle $\subseteq S$. Within those triangles, consider a reflection about the line $y=5-x$. Then only those triangles $\subseteq S$ that have one vertex on the line $y=0$ do not reflect to a triangle $\subseteq S$. So we only need to look at right triangles that have vertices $(0,5), (*,0), (*,*)$. There are three cases:

Case 1: $A=(0,5)$. Then $B=(*,0)$ is impossible.

Case 2: $B=(0,5)$. Then we look for $A=(x,y)$ such that $\angle BAC=90^{\circ}$ and that $C=(*,0)$. They are: $(A=(x,5), C=(x,0))$, $(A=(2,4), C=(1,0))$ and $(A=(4,1), C=(3,0))$. The product of their values of $\tan \angle CBA$ is $\frac{5}{1}\cdot \frac{5}{2} \cdot \frac{5}{3} \cdot \frac{5}{4} \cdot \frac{1}{4} \cdot \frac{2}{3} = \frac{625}{144}$.

Case 3: $C=(0,5)$. Then $A=(*,0)$ is impossible.

Therefore $\boxed{\textbf{(B)} \ \frac{625}{144}}$ is the answer.

See Also

2012 AMC 12B (Problems • Answer Key • Resources)
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All AMC 12 Problems and Solutions
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