Difference between revisions of "1994 AIME Problems/Problem 10"

Revision as of 18:28, 4 July 2013

Problem

In triangle $ABC,\,$ angle $C$ is a right angle and the altitude from $C\,$ meets $\overline{AB}\,$ at $D.\,$ The lengths of the sides of $\triangle ABC\,$ are integers, $BD=29^3,\,$ and $\cos B=m/n\,$ , where $m\,$ and $n\,$ are relatively prime positive integers. Find $m+n.\,$

Solution

Since $\triangle ABC \sim \triangle CBD$ , we have $\frac{BC}{AB} = \frac{29^3}{BC} \Longrightarrow BC^2 = 29^3 AB$ . It follows that $29^2 | BC$ and $29 | AB$ , so $BC$ and $AB$ are in the form $29^2 a$ and $29 a^2$ , respectively.

By the Pythagorean Theorem, we find that $AC^2 + BC^2 = AB^2 \Longrightarrow (29^2a)^2 + AC^2 = (29 a^2)^2$ , so $29a | AC$ . Letting $b = AC / 29a$ , we obtain after dividing through by $(29a)^2$ , $29^2 = a^2 - b^2 = (a-b)(a+b)$ . As $a,b \in \mathbb{Z}$ , the pairs of factors of $29^2$ are $(1,29^2)(29,29)$ ; clearly $b = \frac{AC}{29a} \neq 0$ , so $a-b = 1, a+b= 29^2$ . Then, $a = \frac{1+29^2}{2} = 421$ .

Thus, $\cos B = \frac{BC}{AB} = \frac{29^2 a}{29a^2} = \frac{29}{421}$ , and $m+n = \boxed{450}$ .

@@ Line 14: / Line 14: @@
 [[Category:Intermediate Geometry Problems]]
 [[Category:Intermediate Number Theory Problems]]
+{{MAA Notice}}

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

Page

Toolbox

Search

Difference between revisions of "1994 AIME Problems/Problem 10"

Revision as of 18:28, 4 July 2013

Problem

Solution

See also