Difference between revisions of "1991 AHSME Problems/Problem 19"
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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> | 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> | ||
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+ | == Solution 4 (Really simple similar triangles)== | ||
+ | Triangle <math>ABC</math> has a right angle at <math>C</math>, <math>AC = 3</math> and <math>BC = 4</math>. Triangle <math>ABD</math> has a right angle at <math>A</math> and <math>AD = 12</math>. Points <math>C</math> and <math>D</math> are on opposite sides of <math>\overline{AB}</math>. The line through <math>D</math> parallel to <math>\overline{AC}</math> meets <math>\overline{CB}</math> extended at <math>E</math>. If <math>DE/DB = m/n</math>, where <math>m</math> and <math>n</math> are relatively prime positive integers, then <math>m + n =</math> | ||
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+ | ~ math31415926535 | ||
== See also == | == See also == |
Revision as of 21:40, 23 November 2020
Contents
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
Solution 4 (Really simple similar triangles)
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
~ math31415926535
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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