2011 AMC 10B Problem 18

Below is the professionally curated solution for Problem 18 of the 2011 AMC 10B, from LIVE by Po-Shen Loh. You can also try the full timed exam, view all 2011 AMC 10B solutions, or check the answer key.

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Concepts:rectangleisosceles triangletrigonometry

Difficulty rating: 1540

18.

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?

1515

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7575

Solution:

The angles AMD\angle AMD and MDC\angle MDC are equal since ABDC.AB \mid \mid 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 (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 .

Problem 18 in Other Years