1991 AHSME Problems/Problem 19
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
See also
1991 AHSME (Problems • Answer Key • Resources) | ||
Preceded by Problem 18 |
Followed by Problem 20 | |
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 • 16 • 17 • 18 • 19 • 20 • 21 • 22 • 23 • 24 • 25 • 26 • 27 • 28 • 29 • 30 | ||
All AHSME Problems and Solutions |
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.