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

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== Solution ==
 
== Solution ==
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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 a</math> and <math>29 a^2</math>, respectively.
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By the [[Pythagorean Theorem]], we find that <math>AC^2 + BC^2 = AB^2 \Longrightarrow (29^2a)^2 + AC^2 = (29 a^2)^2</math>, so <math>29a | AC</math>. Letting <math>b = AC / 29a</math>, we obtain after dividing through by <math>(29a)^2</math>, <math>29^2 = a^2 - b^2 = (a-b)(a+b)</math>. As <math>a,b \in \mathbb{Z}</math>, the pairs of factors of <math>29^2</math> are <math>(1,29^2)(29,29)</math>; clearly <math>b = \frac{AC}{29a} \neq 0</math>, so <math>a-b = 1, a+b= 29^2</math>. Then, <math>a = \frac{1+29^2}{2} = 421</math>.
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Thus, <math>\cos B = \frac{BC}{AB} = \frac{29^2 a}{29a^2} = \frac{29}{421}</math>, and <math>m+n = \boxed{450}</math>.
  
 
== See also ==
 
== See also ==
 
{{AIME box|year=1994|num-b=9|num-a=11}}
 
{{AIME box|year=1994|num-b=9|num-a=11}}
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[[Category:Intermediate Geometry Problems]]
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[[Category:Intermediate Number Theory Problems]]

Revision as of 23:58, 6 November 2008

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 $m+n = \boxed{450}$.

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
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