Difference between revisions of "2012 AMC 12B Problems/Problem 25"
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<math>\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} </math> | <math>\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} </math> | ||
− | + | ==Solution== | |
+ | 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 T</math> that have one vertex on the line <math>y=0</math> do not reflect to a triangle <math>\subseteq T</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 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. | |
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− | + | Therefore <math>\boxed{\textbf{(B)} \ \frac{625}{144}}</math> is the answer. | |
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− | + | ==Video Solution by Richard Rusczyk== | |
+ | https://artofproblemsolving.com/videos/amc/2012amc12b/279 | ||
− | + | ~dolphin7 | |
− | + | == See Also == | |
− | + | {{AMC12 box|year=2012|ab=B|num-b=24|after=Last Problem}} | |
+ | {{MAA Notice}} |
Latest revision as of 16:10, 4 April 2020
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 |
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