Difference between revisions of "1991 AHSME Problems/Problem 19"
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Extend lines <math>AD</math> and <math>CE</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{\textbf{(B) } 128}</math>. | Extend lines <math>AD</math> and <math>CE</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{\textbf{(B) } 128}</math>. | ||
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+ | == Solution 3 (Trig)== | ||
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+ | We have <math>\angle ABC = \arcsin\left(\frac{3}{5}\right)</math> and <math>\angle DBA=\arcsin\left(\frac{12}{13}\right).</math> Now we are trying to find <math>\sin(\angle DBE)=\sin\left(180^{\circ}-\angle DBC=\sin(180^{\circ}-\arcsin\left(\frac{3}{5}\right)\right)-\arcsin\left(\frac{12}{13}\right)=\sin(\arcsin\left(\frac{3}{5}\right)+\arcsin\left(\frac{12}{13}\right)).</math> Now we use the <math>\sin</math> angle sum identity, which states <math>\sin(a+b)=\sin(a)\cos(b)+\cos(a)\sin(b).</math> Using this identity yields <math>\sin\left(\arcsin\left(\frac{3}{5}\right)\right)\cos\left(\arcsin\left(\frac{12}{13}\right)\right)+\cos\left(\arcsin\left(\frac{3}{5}\right)\right)+\sin\left(\arcsin\left(\frac{12}{13}\right)\right).</math> | ||
== See also == | == See also == |
Latest revision as of 21:41, 23 November 2020
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 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 .
Solution 3 (Trig)
We have and Now we are trying to find Now we use the angle sum identity, which states Using this identity yields
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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