Difference between revisions of "1994 AIME Problems/Problem 10"
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== Solution 3 == | == Solution 3 == | ||
− | Using similar right triangles, we identify that <math>CD = \sqrt{AD \cdot BD}</math>. Let <math>AD</math> be <math>29 \cdot k^2</math>, to avoid too many radicals, getting <math>CD = k \cdot 29^2</math>. Next we know that <math>AC = \sqrt{AB \cdot AD}</math> and that <math>BC = \sqrt{AB \cdot BD}</math>. Applying the logic with the established values of k, we get <math>AC = 29k \cdot \sqrt{29^2 + k^2}</math> and <math>BC = 29^2 \cdot \sqrt{29^2 + k^2}</math>. Next we look to the integer requirement. Since <math>k</math> is both outside and inside square roots, we know it must be an integer to keep all sides as integers. Let <math>y</math> be <math>\sqrt{29^2 + k^2}</math>, thus <math>29^2 = y^2 - k^2</math>, and <math>29^2 = (y + | + | Using similar right triangles, we identify that <math>CD = \sqrt{AD \cdot BD}</math>. Let <math>AD</math> be <math>29 \cdot k^2</math>, to avoid too many radicals, getting <math>CD = k \cdot 29^2</math>. Next we know that <math>AC = \sqrt{AB \cdot AD}</math> and that <math>BC = \sqrt{AB \cdot BD}</math>. Applying the logic with the established values of k, we get <math>AC = 29k \cdot \sqrt{29^2 + k^2}</math> and <math>BC = 29^2 \cdot \sqrt{29^2 + k^2}</math>. Next we look to the integer requirement. Since <math>k</math> is both outside and inside square roots, we know it must be an integer to keep all sides as integers. Let <math>y</math> be <math>\sqrt{29^2 + k^2}</math>, thus <math>29^2 = y^2 - k^2</math>, and <math>29^2 = (y + k)(y - k)</math>. Since <math>29</math> is prime, and <math>k</math> cannot be zero, we find <math>k = 420</math> and <math>y = 421</math> as the smallest integers satisfying this quadratic Diophantine equation. Then, since <math>cos B</math> = <math>\frac{29}{\sqrt{29^2 + k^2}}</math>. Plugging in we get <math>cos B = \frac{29}{421}</math>, thus our answer is <math>\boxed{450}</math>. |
== See also == | == See also == |
Revision as of 15:23, 26 August 2018
Contents
[hide]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 | |
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 | ||
All AIME Problems and Solutions |
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.