2007 AMC 12B Problems/Problem 23
Problem
How many non-congruent right triangles with positive integer leg lengths have areas that are numerically equal to times their perimeters?
Solution 1
Let and be the two legs of the triangle.
We have .
Then .
We can complete the square under the root, and we get, .
Let and , we have .
After rearranging, squaring both sides, and simplifying, we have .
Putting back and , and after factoring using Simon's Favorite Factoring Trick, we've got .
Factoring 72, we get 6 pairs of and
And this gives us solutions .
Alternatively, note that . Then 72 has factors. However, half of these are repeats, so we have solutions.
Solution 2
We will proceed by using the fact that , where is the radius of the incircle and is the semiperimeter .
We are given .
The incircle of breaks the triangle's sides into segments such that , and . Since ABC is a right triangle, one of , and is equal to its radius, 6. Let's assume .
The side lengths then become , and . Plugging into Pythagorean's theorem:
We can factor to arrive with pairs of solutions: and .
Solution 3
Let and be the two legs of the triangle, and be the hypotenuse.
By using , where is the in-radius, we get:
In right triangle,
By the triangle's area we get:
By substituting in:
As , there are solutions, .
See Also
2007 AMC 12B (Problems • Answer Key • Resources) | |
Preceded by Problem 22 |
Followed by Problem 24 |
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