Difference between revisions of "1991 AHSME Problems/Problem 19"
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Let <math>F</math> be the point such that <math>DF</math> and <math>CF</math> are parallel to <math>CE</math> and <math>DE</math>, respectively, and let <math>DE = x</math> and <math>BE^2 = 169-x^2</math>. Then, <math>[FDEC] = x(4+\sqrt{169-x^2}) = [ABC] + [BED] + [ABD] + [AFD] = 6 + \dfrac{x\sqrt{169-x^2}}{2} + 30 + \dfrac{(x-3)(4+\sqrt{169-x^2})}{2}</math>. So, <math>4x+x\sqrt{169-x^2} = 60 + x\sqrt{169-x^2} - 3\sqrt{169-x^2}</math>. Simplifying <math>3\sqrt{169-x^2} = 60 - 4x</math>, and <math>1521 - 9x^2 = 16x^2 - 480x + 3600</math>. Therefore <math>25x^2 - 480x + 2079 = 0</math>, and <math>x = \dfrac{48\pm15}{5}</math>. Checking, <math>x = \dfrac{63}{5}</math> is the answer, so <math>\dfrac{DE}{DB} = \dfrac{\dfrac{63}{5}}{13} = \dfrac{63}{65}</math>. The answer is <math>\boxed{\textbf{(B) } 128}</math>. | Let <math>F</math> be the point such that <math>DF</math> and <math>CF</math> are parallel to <math>CE</math> and <math>DE</math>, respectively, and let <math>DE = x</math> and <math>BE^2 = 169-x^2</math>. Then, <math>[FDEC] = x(4+\sqrt{169-x^2}) = [ABC] + [BED] + [ABD] + [AFD] = 6 + \dfrac{x\sqrt{169-x^2}}{2} + 30 + \dfrac{(x-3)(4+\sqrt{169-x^2})}{2}</math>. So, <math>4x+x\sqrt{169-x^2} = 60 + x\sqrt{169-x^2} - 3\sqrt{169-x^2}</math>. Simplifying <math>3\sqrt{169-x^2} = 60 - 4x</math>, and <math>1521 - 9x^2 = 16x^2 - 480x + 3600</math>. Therefore <math>25x^2 - 480x + 2079 = 0</math>, and <math>x = \dfrac{48\pm15}{5}</math>. Checking, <math>x = \dfrac{63}{5}</math> is the answer, so <math>\dfrac{DE}{DB} = \dfrac{\dfrac{63}{5}}{13} = \dfrac{63}{65}</math>. The answer is <math>\boxed{\textbf{(B) } 128}</math>. | ||
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== Solution 2 == | == Solution 2 == |
Revision as of 06:28, 1 August 2019
Contents
[hide]Problem
Triangle has a right angle at and . Triangle has a right angle at and . Points and are on opposite sides of . The line through parallel to meets extended at . If where and are relatively prime positive integers, then
Solution 1
Solution by e_power_pi_times_i
Let be the point such that and are parallel to and , respectively, and let and . Then, . So, . Simplifying , and . Therefore , and . Checking, is the answer, so . The answer is .
Solution 2
Solution by arjvik (Arjun Vikram)
Extend lines and to meet at a new point . Now, we see that . Using this relationship, we can see that , (so ), and the ratio of similarity between and is . This ratio gives us that . By the Pythagorean Theorem, . Thus, , and the answer is .
See also
1991 AHSME (Problems • Answer Key • Resources) | ||
Preceded by Problem 18 |
Followed by Problem 20 | |
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All AHSME Problems and Solutions |
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