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 1
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.
See Also
2012 AMC 12B (Problems • Answer Key • Resources) | |
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