2011 AMC 10B Problems/Problem 17
In the given circle, the diameter is parallel to , and is parallel to . The angles and are in the ratio . What is the degree measure of angle ?
We can let be and be because they are in the ratio . When an inscribed angle contains the diameter, the inscribed angle is a right angle. Therefore by triangle sum theorem, and .
because they are alternate interior angles and . Opposite angles in a cyclic quadrilateral are supplementary, so . Use substitution to get
We could also tell that quadrilateral is an isosceles trapezoid because for and to be parallel, the line going through the center of the circle and perpendicular to must fall through the center of .
Note as before. The sum of the interior angles for quadrilateral is . Denote the center of the circle as . . Denote and . We wish to find . Our equation is . Our final equation becomes . After subtracting and dividing by , our answer becomes
Solution 3 (Circle Geometry)
Note that intercepts . Since, , thus (courtesy of the Inscribed Angles Theorem).
Since we know that , then , (courtesy of the Triangle Sum Theorem) and also . By solving this variation, and . After that, due to the Alternate Interior Angles Theorem, , which means .
After doing some angle chasing, then these following facts should be true, .
Note that the arcs have to equal 360, so,
Notice how intercepts and that .
According to the Inscribed Angles Theorem, , therefore the answer is
Video Solution by OmegaLearn
|2011 AMC 10B (Problems • Answer Key • Resources)|
|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|
|All AMC 10 Problems and Solutions|
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.