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

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== Solution 1 ==
 
== Solution 1 ==
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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 / 29a</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>.
+
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 15:49, 12 March 2018

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 $29^2 x$ and $29 x^2$, respectively, where x is an integer.

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

Thus, $\cos B = \frac{BC}{AB} = \frac{29^2 x}{29x^2} = \frac{29}{421}$, 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
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15
All AIME Problems and Solutions

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