Art of Problem Solving
AoPS Online
Math texts, online classes, and more
for students in grades 5-12.
Visit AoPS Online ‚
Books for Grades 5-12 Online Courses
Beast Academy
Engaging math books and online learning
for students ages 6-13.
Visit Beast Academy ‚
Books for Ages 6-13 Beast Academy Online
AoPS Academy
Small live classes for advanced math
and language arts learners in grades 2-12.
Visit AoPS Academy ‚
Find a Physical Campus Visit the Virtual Campus

2012 AMC 12B Problems/Problem 25

Revision as of 18:21, 2 August 2024 by Wes (talk | contribs) (→‎Solution 2)

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] 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] 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); 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); 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),red); draw((0,1)--(0,5),red); draw((0,5)--(5,5),red); draw((5,5)--(5,1),red); [/asy]

Video Solution by Richard Rusczyk

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

~dolphin7

See Also

2012 AMC 12B (Problems • Answer Key • Resources)
Preceded by
Problem 24 		Followed by
Last Problem
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 • 16 • 17 • 18 • 19 • 20 • 21 • 22 • 23 • 24 • 25
All AMC 12 Problems and Solutions

The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. AMC logo.png

Retrieved from "https://artofproblemsolving.com/wiki/index.php?title=2012_AMC_12B_Problems/Problem_25&oldid=224278"