Difference between revisions of "1994 AIME Problems/Problem 10"
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− | Since <math>\triangle ABC \sim \triangle CBD</math>, we have <math>\frac{BC}{AB} = \frac{29^3}{BC} \Longrightarrow BC^2 = 29^3 AB</math>. It follows that <math>29^2 | BC</math> and <math>29 | AB</math>, so <math>BC</math> and <math>AB</math> are in the form <math>29^2 x</math> and <math>29 x^2</math>, respectively. | + | Since <math>\triangle ABC \sim \triangle CBD</math>, we have <math>\frac{BC}{AB} = \frac{29^3}{BC} \Longrightarrow BC^2 = 29^3 AB</math>. It follows that <math>29^2 | BC</math> and <math>29 | AB</math>, so <math>BC</math> and <math>AB</math> are in the form <math>29^2 x</math> and <math>29 x^2</math>, respectively, where x is an integer. |
− | By the [[Pythagorean Theorem]], we find that <math>AC^2 + BC^2 = AB^2 \Longrightarrow (29^2x)^2 + AC^2 = (29 x^2)^2</math>, so <math>29x | AC</math>. Letting <math>y = AC / | + | By the [[Pythagorean Theorem]], we find that <math>AC^2 + BC^2 = AB^2 \Longrightarrow (29^2x)^2 + AC^2 = (29 x^2)^2</math>, so <math>29x | AC</math>. Letting <math>y = AC / 29x</math>, we obtain after dividing through by <math>(29x)^2</math>, <math>29^2 = x^2 - y^2 = (x-y)(x+y)</math>. As <math>x,y \in \mathbb{Z}</math>, the pairs of factors of <math>29^2</math> are <math>(1,29^2)(29,29)</math>; clearly <math>y = \frac{AC}{29x} \neq 0</math>, so <math>x-y = 1, x+y= 29^2</math>. Then, <math>x = \frac{1+29^2}{2} = 421</math>. |
Thus, <math>\cos B = \frac{BC}{AB} = \frac{29^2 x}{29x^2} = \frac{29}{421}</math>, and <math>m+n = \boxed{450}</math>. | Thus, <math>\cos B = \frac{BC}{AB} = \frac{29^2 x}{29x^2} = \frac{29}{421}</math>, and <math>m+n = \boxed{450}</math>. |
Revision as of 16:49, 12 March 2018
Contents
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 .
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.