1979 AHSME Problems/Problem 12
Problem 12
In the adjoining figure, is the diameter of a semi-circle with center . Point lies on the extension of past ; point lies on the semi-circle, and is the point of intersection (distinct from ) of line segment with the semi-circle. If length equals length , and the measure of is , then the measure of is
Solution
Solution by e_power_pi_times_i
Because , triangles and are isosceles. Denote $\measuredangleBAO = \measuredangleAOB = \theta$ (Error compiling LaTeX. ! Undefined control sequence.). Then $\measuredangleABO = 180^\circ-2\theta$ (Error compiling LaTeX. ! Undefined control sequence.), and $\measuredangleEBO = \measuredangleOEB = 2\theta$ (Error compiling LaTeX. ! Undefined control sequence.), so $\measuredangleBOE = 180^\circ-4\theta$ (Error compiling LaTeX. ! Undefined control sequence.). Notice that $\measuredangleAOB + \measuredangleBOE + 45^\circ = 180^\circ$ (Error compiling LaTeX. ! Undefined control sequence.). Therefore , and .
See also
1979 AHSME (Problems • Answer Key • Resources) | ||
Preceded by Problem 11 |
Followed by Problem 13 | |
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