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2011 AMC 10B Problem 18

Problem 18 of 25IntermediateGeometry

Rectangle ABCDABCD has AB=6AB = 6 and BC=3.BC = 3. Point MM is chosen on side ABAB so that ∠AMD=∠CMD.\angle AMD = \angle CMD. What is the degree measure of ∠AMD?\angle AMD?

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Solution

The angles ∠AMD\angle AMD and ∠MDC\angle MDC are equal since AB∥DC.AB \parallel DC. As such, ∠MDC=∠DMC,\angle MDC = \angle DMC , making MDCMDC isosceles and MC=DC=6.MC = DC = 6. As we can see, sin⁡(∠CMB)=12,\sin (\angle CMB) = \frac 12, making ∠CMB=30∘.\angle CMB = 30^\circ . Therefore, ∠AMC=150∘.\angle AMC = 150^\circ . Since ∠AMD\angle AMD is half of that, ∠AMD=75∘.\angle AMD = 75^\circ . Thus, the correct answer is E .

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Concepts: rectangle · isosceles triangle · trigonometry

Problem text and solution from the LIVE past-contest archive. See also the AoPS wiki page for community solutions.