Problem 396

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Problem 396: (2012 AMC12B, 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

Let $G=\{0,\dots,4\}\times\{0,\dots,5\}$ be the full grid, so $S=G\setminus\{O\}$ with $O=(0,0)$. Write $f(t)=\dfrac{AC}{AB}$, the ratio of the leg toward $C$ to the leg toward $B$ (going $A\to B\to C$ counterclockwise).
\emph{Mirror pairing.} Reflect a right triangle $t$ of $G$ in the line $x=2$. The mirror image is again a right triangle of $G$, but its orientation is reversed, so we must swap the names $B$ and $C$ to keep them counterclockwise. This changes $f$ into $\frac{AB}{AC}=\frac1{f(t)}$. Hence the triangles of $G$ split into pairs with product $1$, and a triangle equal to its own mirror image is isosceles with $f=1$. Therefore $\prod_{t\subset G}f(t)=1$, and \[\prod_{t\in T}f(t)=\frac1{P},\qquad P=\prod_{\text{right triangles of $G$ with a vertex at $O$}}f(t).\]
\emph{Diagonal pairing.} The square $\{0,\dots,4\}^2\subset G$ is symmetric about $y=x$, which fixes $O$. The same argument shows that the triangles with vertex $O$ lying inside this square pair off with product $1$. So $P$ is the product over triangles with vertex $O$ having a vertex on the line $y=5$.
\emph{Right angle at $O$.} The legs lie along the axes: $B=(a,0)$, $C=(0,5)$, $a=1,\dots,4$, so $f=\frac5a$ and the product is $\frac{5^4}{4!}=\frac{625}{24}$.
\emph{Right angle at $A=(0,5)$.} Then $AO$ is vertical, so the other vertex is $X=(k,5)$, $k=1,\dots,4$; here $C=X$ and $B=O$, so $f=\frac{k}{5}$ and the product is $\frac{4!}{5^4}=\frac{24}{625}$. These two families cancel.
\emph{Right angle at $A=(k,0)$, $k=1,\dots,4$, with the other vertex $X=(k,5)$.} Here $B=X$ and $C=O$, so $f=\frac{AO}{AX}=\frac k5$. The product is $\frac{4!}{5^4}=\frac{24}{625}$.
\emph{The remaining triangles} have the right angle at a point $A$ with $AX\perp AO$, $X$ on the line $y=5$ (the perpendicular to $AO$ at $A$ must reach the top row). Checking the perpendiculars to $AO$ gives exactly these, each with $B=X$, $C=O$ and $f=\frac{AO}{AX}$: \begin{center} \begin{tabular}{c|cccccc} $A$ & $(2,1)$ & $(2,4)$ & $(3,2)$ & $(3,3)$ & $(4,1)$ & $(4,4)$\\\hline $X$ & $(0,5)$ & $(0,5)$ & $(1,5)$ & $(1,5)$ & $(3,5)$ & $(3,5)$\\ $f$ & $\frac12$ & $2$ & $1$ & $\frac32$ & $1$ & $4$ \end{tabular} \end{center} whose product is $\frac12\cdot2\cdot1\cdot\frac32\cdot1\cdot4=6$.
Therefore $P=\frac{24}{625}\cdot6=\frac{144}{625}$ and $\prod_{t\in T}f(t)=\dfrac{625}{144}$, answer (B).

Exercise 396: Solution

In $\triangle ABC$, $AB=13$, $BC=14$, and $CA=15$. What is $\tan\dfrac A2$?

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