AMC12 2012 B

AMC12 2012 B · Q25

AMC12 2012 B · Q25. It mainly tests Counting & probability misc, Coordinate geometry.

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 \[\prod_{t\in T} f(t)?\]

令 $S=\{(x,y) : x\in \{0,1,2,3,4\}, y\in \{0,1,2,3,4,5\},\text{ and } (x,y)\ne (0,0)\}$。令 $T$ 为所有顶点在 $S$ 中的直角三角形的集合。对每个直角三角形 $t=\triangle{ABC}$（顶点 $A$、$B$、$C$ 按逆时针顺序排列，且直角在 $A$），令 $f(t)=\tan(\angle{CBA})$。求 \[\prod_{t\in T} f(t)?\]

(A) 1 1

(B) $\frac{625}{144}$ $\frac{625}{144}$

(D) 6 6

(E) $\frac{625}{24}$ $\frac{625}{24}$

Answer

Correct choice: (B)

正确答案：（B）

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 $\subseteq T$ 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.

考虑反射。对任意满足题目所述标记方式的直角三角形 $ABC$，其任意一个按该方式标记的反射三角形 $A'B'C'$ 都满足 $\tan CBA\cdot \tan C'B'A'=1$。先考虑关于直线 $y=2.5$ 的反射。只有那些 $\subseteq T$ 且有一个顶点在 $(0,5)$ 的三角形，反射后不会得到一个仍属于 $\subseteq T$ 的三角形。在这些三角形中，再考虑关于直线 $y=5-x$ 的反射。此时只有那些 $\subseteq T$ 且有一个顶点在直线 $y=0$ 上的三角形，反射后不会得到一个仍属于 $\subseteq T$ 的三角形。因此我们只需考察顶点为 $(0,5),(*,0),(*,*)$ 的直角三角形。分三种情况：情形 1：$A=(0,5)$。则 $B=(*,0)$ 不可能。情形 2：$B=(0,5)$。此时寻找 $A=(x,y)$ 使得 $\angle BAC=90^{\circ}$ 且 $C=(*,0)$。它们为：$(A=(x,5), C=(x,0))$，$(A=(3,2), C=(1,0))$，以及 $(A=(4,1), C=(3,0))$。它们对应的 $\tan \angle CBA$ 的乘积为 $\frac{5}{1}\cdot \frac{5}{2} \cdot \frac{5}{3} \cdot \frac{5}{4} \cdot \frac{1}{4} \cdot \frac{2}{3} = \frac{625}{144}$。情形 3：$C=(0,5)$。则 $A=(*,0)$ 不可能。因此答案为 $\boxed{\textbf{(B)} \ \frac{625}{144}}$。

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