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> | ||
− | == Solution == | + | == 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 == | ||
+ | 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 15:01, 30 August 2017
Contents
Problem
In triangle angle is a right angle and the altitude from meets at The lengths of the sides of are integers, and , where and are relatively prime positive integers. Find
Solution 1
Since , we have . It follows that and , so and are in the form and , respectively.
By the Pythagorean Theorem, we find that , so . Letting , we obtain after dividing through by , . As , the pairs of factors of are ; clearly , so . Then, .
Thus, , and .
Solution 2
We will solve for using , which gives us . By the Pythagorean Theorem on , we have . Trying out factors of , we can either guess and check or just guess to find that and (The other pairs give answers over 999). Adding these, we have and , and our answer is .
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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