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2012 AMC 12B Problems/Problem 25

Revision as of 13:07, 6 December 2012 by Lightest (talk | contribs) (→‎Solution 1)

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 1

Four points in a rectangular arrangement allow 4 possible right angle triangles. These four congruent triangles will form two pairs of different vertice labelling - two ABC and two ACB. These pairs will multiple to equal 1 due to the fact that $\tan x \tan(90^{\circ}-x) = \tan x \cot x = 1.$

Due to the missing point (0,0) in the grid then in the rectangular arrangement (0,0), (x,0), (x,y), (0,y) there is only one triangle which will not cancel out to zero with a reflected version.


So we need to consider all triangles of the form A(x,y), B(0,y), C(x,0). For these triangle tanB = y/x. Multiplying them all together gives: 1/1 * 1/2 * 1/3 * 1/4 * 2/1 * 2/2 * 2/3 * 2/4 * 3/1 * 3/2 * 3/3 * 3/4 * 4/1 * 4/2 * 4/3 * 4/4 * 5/1 * 5/2 * 5/3 * 5/4 = 625/24

Solution 2

Consider reflection. 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 $y=2.5$. Then only those triangle $\subseteq S$ that has one vertex at $(0,5)$ does not $\subseteq S$. Within those triangles, consider a reflection about the line $y=5-x$. Then only those triangle $\subseteq S$ that has one vertex on the line $y=0$. 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 tangents of $\angle CBA$ is $5/1 \cdot 5/2 \cdot 5/3 \cdot 5/4 \cdot 1/4 \cdot 2/3 = 625/144$.

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

Therefore $\framebox{B}$ is the answer.

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