2012 AMC 12B Problems/Problem 25
Problem 25
Let . Let be the set of all right triangles whose vertices are in . For every right triangle with vertices , , and in counter-clockwise order and right angle at , let . What is
Solution
Consider reflections. For any right triangle with the right labeling described in the problem, any reflection labeled that way will give us . First we consider the reflection about the line . Only those triangles that have one vertex at do not reflect to a traingle . Within those triangles, consider a reflection about the line . Then only those triangles that have one vertex on the line do not reflect to a triangle . So we only need to look at right triangles that have vertices . There are three cases:
Case 1: . Then is impossible.
Case 2: . Then we look for such that and that . They are: , and . The product of their values of is .
Case 3: . Then is impossible.
Therefore is the answer.
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.