Difference between revisions of "2012 AMC 12B Problems/Problem 25"
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− | Consider reflections. For any right triangle <math>ABC</math> with the right labeling described in the problem, any reflection <math>A'B'C'</math> labeled that way will give us <math>\tan CBA \cdot \tan C'B'A' = 1</math>. First we consider the reflection about the line <math>y=2.5</math>. Only those triangles <math>\subseteq T</math> that have one vertex at <math>(0,5)</math> do not reflect to a traingle <math>\subseteq | + | Consider reflections. For any right triangle <math>ABC</math> with the right labeling described in the problem, any reflection <math>A'B'C'</math> labeled that way will give us <math>\tan CBA \cdot \tan C'B'A' = 1</math>. First we consider the reflection about the line <math>y=2.5</math>. Only those triangles <math>\subseteq T</math> that have one vertex at <math>(0,5)</math> do not reflect to a traingle <math>\subseteq T</math>. Within those triangles, consider a reflection about the line <math>y=5-x</math>. Then only those triangles <math>\subseteq S</math> that have one vertex on the line <math>y=0</math> do not reflect to a triangle <math>\subseteq S</math>. So we only need to look at right triangles that have vertices <math>(0,5), (*,0), (*,*)</math>. There are three cases: |
Case 1: <math>A=(0,5)</math>. Then <math>B=(*,0)</math> is impossible. | Case 1: <math>A=(0,5)</math>. Then <math>B=(*,0)</math> is impossible. |
Revision as of 22:28, 28 December 2015
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.
See Also
2012 AMC 12B (Problems • Answer Key • Resources) | |
Preceded by Problem 24 |
Followed by ' |
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 |
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