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Difference between revisions of "1994 AIME Problems/Problem 10"
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In triangle <math>ABC,\,</math> angle <math>C</math> is a right angle and the altitude from <math>C\,</math> meets <math>\overline{AB}\,</math> at <math>D.\,</math>  The lengths of the sides of <math>\triangle ABC\,</math> are integers, <math>BD=29^3,\,</math> and <math>\cos B=m/n\,</math>, where <math>m\,</math> and <math>n\,</math> are relatively prime positive integers.  Find <math>m+n.\,</math>
 
In triangle <math>ABC,\,</math> angle <math>C</math> is a right angle and the altitude from <math>C\,</math> meets <math>\overline{AB}\,</math> at <math>D.\,</math>  The lengths of the sides of <math>\triangle ABC\,</math> are integers, <math>BD=29^3,\,</math> and <math>\cos B=m/n\,</math>, where <math>m\,</math> and <math>n\,</math> are relatively prime positive integers.  Find <math>m+n.\,</math>
  
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== Solution ==
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== Solution 1 ==
 
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.  
 
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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Thus, <math>\cos B = \frac{BC}{AB} = \frac{29^2 a}{29a^2} = \frac{29}{421}</math>, and <math>m+n = \boxed{450}</math>.
 
Thus, <math>\cos B = \frac{BC}{AB} = \frac{29^2 a}{29a^2} = \frac{29}{421}</math>, and <math>m+n = \boxed{450}</math>.
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== Solution 2 ==
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We will solve for <math>\cos B</math> using <math>\triangle CBD</math>, which gives us <math>\cos B = \frac{29^3}{BC}</math>. By the Pythagorean Theorem on <math>\triangle CBD</math>, we have <math>BC^2 - DC^2 = (BC + DC)(BC - DC) = 29^6</math>. Trying out factors of <math>29^6</math>, we can either guess and check or just guess to find that <math>BC + DC = 29^4</math> and <math>BC - DC = 29^2</math> (The other pairs give answers over 999). Adding these, we have <math>2BC = 29^4 + 29^2</math> and <math>\frac{29^3}{BC} = \frac{2*29^3}{29^2 (29^2 +1)} = \frac{58}{842} = \frac{29}{421}</math>, and our answer is <math>\boxed{450}</math>.
  
 
== See also ==
 
== See also ==

Revision as of 16:01, 30 August 2017

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 1

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

Solution 2

We will solve for $\cos B$ using $\triangle CBD$, which gives us $\cos B = \frac{29^3}{BC}$. By the Pythagorean Theorem on $\triangle CBD$, we have $BC^2 - DC^2 = (BC + DC)(BC - DC) = 29^6$. Trying out factors of $29^6$, we can either guess and check or just guess to find that $BC + DC = 29^4$ and $BC - DC = 29^2$ (The other pairs give answers over 999). Adding these, we have $2BC = 29^4 + 29^2$ and $\frac{29^3}{BC} = \frac{2*29^3}{29^2 (29^2 +1)} = \frac{58}{842} = \frac{29}{421}$, and our answer is $\boxed{450}$.

See also

1994 AIME (Problems • Answer Key • Resources)
Preceded by
Problem 9 		Followed by
Problem 11
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All AIME Problems and Solutions

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