1994 AIME Problems/Problem 10
Problem
In triangle angle is a right angle and the altitude from meets at The lengths of the sides of are integers, and , where and are relatively prime positive integers. Find
Solution 1
Since , we have . It follows that and , so and are in the form and , respectively, where x is an integer.
By the Pythagorean Theorem, we find that , so . Letting , we obtain after dividing through by , . As , the pairs of factors of are ; clearly , so . Then, .
Thus, , and .
Solution 2
We will solve for using , which gives us . By the Pythagorean Theorem on , we have . Trying out factors of , we can either guess and check or just guess to find that and (The other pairs give answers over 999). Adding these, we have and , and our answer is .
Solution 3
Using similar right triangles, we identify that . Let be , to avoid too many radicals, getting . Next we know that and that . Applying the logic with the established values of k, we get and . Next we look to the integer requirement. Since is both outside and inside square roots, we know it must be an integer to keep all sides as integers. Let be , thus , and . Since is prime, and cannot be zero, we find and as the smallest integers satisfying this quadratic Diophantine equation. Then, since = . Plugging in we get , thus our answer is .
See also
1994 AIME (Problems • Answer Key • Resources) | ||
Preceded by Problem 9 |
Followed by Problem 11 | |
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