Difference between revisions of "1991 AHSME Problems/Problem 19"
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\text{(E) } 256</math> | \text{(E) } 256</math> | ||
− | == Solution == | + | == Solution 1 == |
Solution by e_power_pi_times_i | Solution by e_power_pi_times_i | ||
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 == | ||
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+ | Solution by arjvik | ||
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+ | Extend lines <math>AD</math> and <math>CF</math> to meet at a new point <math>F</math>. Now, we see that <math>FAC\sim FDE \sim ACB</math>. Using this relationship, we can see that <math>AF=\frac{15}4</math>, (so <math>FD=\frac{63}4</math>), and the ratio of similarity between <math>FDE</math> and <math>FAC</math> is <math>\frac{63}{15}</math>. This ratio gives us that <math>\frac{63}5</math>. By the Pythagorean Theorem, <math>DB=13</math>. Thus, <math>\frac{DE}{DB}=\frac{63}{65}</math>, and the answer is <math>63+65=\boxed{128}</math>. | ||
== See also == | == See also == |
Revision as of 20:06, 6 January 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
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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