Difference between revisions of "1984 AHSME Problems/Problem 4"
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Revision as of 11:49, 5 July 2013
A rectangle intersects a circle as shown: , , and . Then equals:
Solution
Draw and , forming a trapezoid. Since it's cyclic, this trapezoid must be isosceles. Also, drop altitudes from to , to , and to , and let the feet of these altitudes be , , and respectively. is a rectangle since it has right angles. Therefore, , and . By the same logic, is also a rectangle, and . since they're both altitudes to a trapezoid, and since the trapezoid is isosceles. Therefore, $\triangleBHE\congruent\triangleCIF$ (Error compiling LaTeX. Unknown error_msg) by HL congruence, so . Also, is a rectangle from right angles, and . Therefore, .
See Also
1984 AHSME (Problems • Answer Key • Resources) | ||
Preceded by Problem 3 |
Followed by Problem 5 | |
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 |
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