Difference between revisions of "2012 AMC 12B Problems/Problem 25"
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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. | ||
− | Case 2: <math>B=(0,5)</math>. Then we look for <math>A=(x,y)</math> such that <math>\angle BAC=90^{\circ}</math> and that <math>C=(*,0)</math>. They are: <math>(A=(x,5), C=(x,0))</math>, <math>(A=(2 | + | Case 2: <math>B=(0,5)</math>. Then we look for <math>A=(x,y)</math> such that <math>\angle BAC=90^{\circ}</math> and that <math>C=(*,0)</math>. They are: <math>(A=(x,5), C=(x,0))</math>, <math>(A=(3,2), C=(1,0))</math> and <math>(A=(4,1), C=(3,0))</math>. The product of their values of <math>\tan \angle CBA</math> is <math>\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}</math>. |
Case 3: <math>C=(0,5)</math>. Then <math>A=(*,0)</math> is impossible. | Case 3: <math>C=(0,5)</math>. Then <math>A=(*,0)</math> is impossible. |
Revision as of 18:23, 29 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 ' |
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All AMC 12 Problems and Solutions |
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