2012 AMC 12B Problems/Problem 25
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[hide]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.
Solution 2
This is just another way for the reasoning of solution 1. Picture the question to be a grid of unit squares instead of a coordinate system. Note that the restriction, (This is just to make visualization easier.) Define a "cell" to be a rectangle in the set of For example, a cell can be (labeled in a red):
Note that choosing any three points and labeling them according to the problem will result in a product of one. For example, with the cell we just labeled, the four triangles we can create are:
If we define the longer side to be and the shorter side to be then the product will be and we are done.
Otherwise, the three points are not contained in a "cell." This will result in the solution 1 path as described before. Our three points must take the form where is a number defined by the boundaries of Thus, by the three cases, our answer is
~wesserwes7254
Video Solution by Richard Rusczyk
https://artofproblemsolving.com/videos/amc/2012amc12b/279
~dolphin7
See Also
2012 AMC 12B (Problems • Answer Key • Resources) | |
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