Difference between revisions of "1979 AHSME Problems/Problem 12"
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Because <math>AB = OD</math>, triangles <math>ABO</math> and <math>BOE</math> are isosceles. Denote <math>\measuredangle BAO = \measuredangle AOB = \theta</math>. Then <math>\measuredangle ABO = 180^\circ-2\theta</math>, and <math>\measuredangle EBO = \measuredangle OEB = 2\theta</math>, so <math>\measuredangle BOE = 180^\circ-4\theta</math>. Notice that <math>\measuredangle AOB + \measuredangle BOE + 45^\circ = 180^\circ</math>. Therefore <math>\theta+180-4\theta = 135^\circ</math>, and <math>\theta = \boxed{\textbf{(B) } 15^\circ}</math>. | Because <math>AB = OD</math>, triangles <math>ABO</math> and <math>BOE</math> are isosceles. Denote <math>\measuredangle BAO = \measuredangle AOB = \theta</math>. Then <math>\measuredangle ABO = 180^\circ-2\theta</math>, and <math>\measuredangle EBO = \measuredangle OEB = 2\theta</math>, so <math>\measuredangle BOE = 180^\circ-4\theta</math>. Notice that <math>\measuredangle AOB + \measuredangle BOE + 45^\circ = 180^\circ</math>. Therefore <math>\theta+180-4\theta = 135^\circ</math>, and <math>\theta = \boxed{\textbf{(B) } 15^\circ}</math>. | ||
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+ | ==Another Solution== | ||
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+ | Draw <math>BO</math>. Let <math>y = \angle BAO</math>. Since <math>AB = OD = BO</math>, triangle <math>ABO</math> is isosceles, so <math>\angle BOA = \angle BAO = y</math>. Angle <math>\angle EBO</math> is exterior to triangle <math>ABO</math>, so <math>\angle EBO = \angle BAO + \angle BOA = y + y = 2y</math>. | ||
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+ | Triangle <math>BEO</math> is isosceles, so <math>\angle BEO = \angle EBO = 2y</math>. Then <math>\angle EOD</math> is external to triangle <math>AEO</math>, so <math>\angle EOD = \angle EAO + \angle AEO = y + 2y = 3y</math>. But <math>\angle EOD = 45^\circ</math>, so <math>\angle BAO = y = 45^\circ/3 = \boxed{15^\circ}</math>. That means the answer is <math>\boxed{\textbf{(B) } 15^\circ}</math>. | ||
== See also == | == See also == |
Revision as of 21:08, 14 March 2018
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 . Then , and , so . Notice that . Therefore , and .
Another Solution
Draw . Let . Since , triangle is isosceles, so . Angle is exterior to triangle , so .
Triangle is isosceles, so . Then is external to triangle , so . But , so . That means the answer is .
See also
1979 AHSME (Problems • Answer Key • Resources) | ||
Preceded by Problem 11 |
Followed by Problem 13 | |
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All AHSME Problems and Solutions |
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