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2012 AMC 12B Problems/Problem 25

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 1

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 $\subseteq T$ that have one vertex at $(0,5)$ do not reflect to a traingle $\subseteq T$. Within those triangles, consider a reflection about the line $y=5-x$. Then only those triangles $\subseteq T$ that have one vertex on the line $y=0$ do not reflect to a triangle $\subseteq T$. 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: $(A=(x,5), C=(x,0))$, $(A=(3,2), C=(1,0))$ 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.

Solution 2

This is just another way for the reasoning of solution 1. Picture the question to be a grid of unit squares instead of a coordinate system. Note that the restriction, (This is just to make visualization easier.) Define a "cell" to be a rectangle in the set of $S.$ For example, a cell can be (labeled in a red):

[asy] unitsize(0.5 cm); draw((0,1)--(0,5),black); draw((1,0)--(1,5),black); draw((2,0)--(2,5),black); draw((3,0)--(3,5),black); draw((4,0)--(4,5),black); draw((0,1)--(5,1),black); draw((5,0)--(5,5),black); draw((1,0)--(5,0),black); draw((0,2)--(5,2),black); draw((0,3)--(5,3),black); draw((0,4)--(5,4),black); draw((0,5)--(5,5),black); draw((0,1)--(5,1),red); draw((0,1)--(0,5),red); draw((0,5)--(5,5),red); draw((5,5)--(5,1),red); [/asy]

Note that choosing any three points and labeling them according to the problem will result in a product of one. For example, with the cell we just labeled, the four triangles we can create are:

[asy] unitsize(0.5 cm); draw((0,1)--(5,1),red); draw((0,1)--(0,5),red); draw((0,5)--(5,5),red); draw((5,5)--(5,1),red); draw((0,1)--(5,1),black); draw((0,1)--(0,5),black); draw((0,5)--(5,1),black); pair A, B, C; A = (0,1); B = (0,5); C = (5,1); [/asy]

[asy] unitsize(0.5 cm); draw((0,1)--(5,1),red); draw((0,1)--(0,5),red); draw((0,5)--(5,5),red); draw((5,5)--(5,1),red); draw((0,1)--(5,1),black); draw((5,1)--(5,5),black); draw((5,5)--(0,1),black); [/asy]

[asy] unitsize(0.5 cm); draw((0,1)--(5,1),red); draw((0,1)--(0,5),red); draw((0,5)--(5,5),red); draw((5,5)--(5,1),red); draw((0,5)--(5,5),black); draw((5,1)--(5,5),black); draw((0,5)--(5,1),black); [/asy]

[asy] unitsize(0.5 cm); draw((0,1)--(5,1),red); draw((0,1)--(0,5),red); draw((0,5)--(5,5),red); draw((5,5)--(5,1),red); draw((0,5)--(5,5),black); draw((0,1)--(0,5),black); draw((0,1)--(5,5),black); [/asy]

If we define the longer side to be $x$ and the shorter side to be $y,$ then the product will be $\frac{x}{y} \cdot \frac{y}{x} \cdot \frac{x}{y} \cdot \frac{y}{x}=1,$ and we are done.

Otherwise, the three points are not contained in a "cell." This will result in the solution 1 path as described before. Our three points must take the form $(0,5), (*,0), (*,*),$ where $*$ is a number defined by the boundaries of $S.$ Thus, by the three cases, our answer is $\boxed{\textbf{(B)} \ \frac{625}{144}}.$

~wesserwes7254

Video Solution by Richard Rusczyk

https://artofproblemsolving.com/videos/amc/2012amc12b/279

~dolphin7

See Also

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

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