1963 AHSME Problems/Problem 18
Problem
Chord is the perpendicular bisector of chord , intersecting it in . Between and point is taken, and extended meets the circle in . Then, for any selection of , as described, is similar to:
Solution
Note that is a right triangle with one of the angles being . This leads to prediction that is the similar triangle as it shares an angle, and to prove this, we need to show that is a diameter.
If we let be on so that , then by SAS Congruency, , so . Since three points define a circle and point is equidistant from three points, is the center, so is a diameter. Therefore, is a right angle, and by AA Similarity, we can confirm that . The answer is .
See Also
1963 AHSC (Problems • Answer Key • Resources) | ||
Preceded by Problem 17 |
Followed by Problem 19 | |
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 • 31 • 32 • 33 • 34 • 35 • 36 • 37 • 38 • 39 • 40 | ||
All AHSME Problems and Solutions |
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.