Difference between revisions of "1979 AHSME Problems/Problem 12"
(Created page with "== Problem 12 == <asy> size(200); pair A=(-2,0),B,C=(-1,0),D=(1,0),EE,O=(0,0); draw(arc(O,1, 0, 180)); EE=midpoint(arc(O,1, 0, 90)); draw(A--EE); draw(A--D); B=intersection...") |
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==Solution== | ==Solution== | ||
+ | Solution by e_power_pi_times_i | ||
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+ | Because <math>AB = OD</math>, triangles <math>ABO</math> and <math>BOE</math> are isosceles. Denote <math>\measuredangleBAO = \measuredangleAOB = \theta</math>. Then <math>\measuredangleABO = 180^\circ-2\theta</math>, and <math>\measuredangleEBO = \measuredangleOEB = 2\theta</math>, so <math>\measuredangleBOE = 180^\circ-4\theta</math>. Notice that <math>\measuredangleAOB + \measuredangleBOE + 45^\circ = 180^\circ</math>. Therefore <math>\theta+180-4\theta = 135^\circ</math>, and <math>\theta = \boxed{\textbf{(B) } 15^\circ}</math>. | ||
== See also == | == See also == |
Revision as of 11:57, 6 January 2017
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 | |
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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