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>\ | + | 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>. |
== See also == | == See also == |
Revision as of 12:58, 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 . Then , and , so . Notice that . Therefore , and .
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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