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

(→‎Solution 3)
m (y^2-k^2 was written as (y+x)(y-x) instead of (y+k)(y-k))
 
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== Solution 3 ==
 
== Solution 3 ==
−
Using similar right triangles, we identify that <math>CD = \sqrt{AD \cdot BD}</math>. Let <math>AD</math> be <math>29 \cdot k^2</math>, to avoid too many radicals, getting <math>CD = k \cdot 29^2</math>. Next we know that <math>AC = \sqrt{AB \cdot AD}</math> and that <math>BC = \sqrt{AB \cdot BD}</math>. Applying the logic with the established values of k, we get <math>AC = 29k \cdot \sqrt{29^2 + k^2}</math> and <math>BC = 29^2 \cdot \sqrt{29^2 + k^2}</math>. Next we look to the integer requirement. Since <math>k</math> is both outside and inside square roots, we know it must be an integer to keep all sides as integers. Let <math>y</math> be <math>\sqrt{29^2 + k^2}</math>, thus <math>29^2 = y^2 - k^2</math>, and <math>29^2 = (y + x)(y - x)</math>. Since <math>29</math> is prime, and <math>k</math> cannot be zero, we find <math>k = 420</math> and <math>y = 421</math> as the smallest integers satisfying this quadratic Diophantine equation. Then, since <math>cos B</math> = <math>\frac{29}{\sqrt{29^2 + k^2}}</math>. Plugging in we get <math>cos B = \frac{29}{421}</math>, thus our answer is <math>\boxed{450}</math>.
+
Using similar right triangles, we identify that <math>CD = \sqrt{AD \cdot BD}</math>. Let <math>AD</math> be <math>29 \cdot k^2</math>, to avoid too many radicals, getting <math>CD = k \cdot 29^2</math>. Next we know that <math>AC = \sqrt{AB \cdot AD}</math> and that <math>BC = \sqrt{AB \cdot BD}</math>. Applying the logic with the established values of k, we get <math>AC = 29k \cdot \sqrt{29^2 + k^2}</math> and <math>BC = 29^2 \cdot \sqrt{29^2 + k^2}</math>. Next we look to the integer requirement. Since <math>k</math> is both outside and inside square roots, we know it must be an integer to keep all sides as integers. Let <math>y</math> be <math>\sqrt{29^2 + k^2}</math>, thus <math>29^2 = y^2 - k^2</math>, and <math>29^2 = (y + k)(y - k)</math>. Since <math>29</math> is prime, and <math>k</math> cannot be zero, we find <math>k = 420</math> and <math>y = 421</math> as the smallest integers satisfying this quadratic Diophantine equation. Then, since <math>cos B</math> = <math>\frac{29}{\sqrt{29^2 + k^2}}</math>. Plugging in we get <math>cos B = \frac{29}{421}</math>, thus our answer is <math>\boxed{450}</math>.
  
 
== See also ==
 
== See also ==

Latest revision as of 15:23, 26 August 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 $\boxed{450}$.

Solution 3

Using similar right triangles, we identify that $CD = \sqrt{AD \cdot BD}$. Let $AD$ be $29 \cdot k^2$, to avoid too many radicals, getting $CD = k \cdot 29^2$. Next we know that $AC = \sqrt{AB \cdot AD}$ and that $BC = \sqrt{AB \cdot BD}$. Applying the logic with the established values of k, we get $AC = 29k \cdot \sqrt{29^2 + k^2}$ and $BC = 29^2 \cdot \sqrt{29^2 + k^2}$. Next we look to the integer requirement. Since $k$ is both outside and inside square roots, we know it must be an integer to keep all sides as integers. Let $y$ be $\sqrt{29^2 + k^2}$, thus $29^2 = y^2 - k^2$, and $29^2 = (y + k)(y - k)$. Since $29$ is prime, and $k$ cannot be zero, we find $k = 420$ and $y = 421$ as the smallest integers satisfying this quadratic Diophantine equation. Then, since $cos B$ = $\frac{29}{\sqrt{29^2 + k^2}}$. Plugging in we get $cos B = \frac{29}{421}$, thus 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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