2011 AMC 12B Problem 10

Attempt Problem 10 of the 2011 AMC 12B below, then check your answer against the professionally curated solution from LIVE by Po-Shen Loh. You can also try the full timed exam, view all 2011 AMC 12B solutions, or check the answer key.

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10.

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

3030

4545

6060

7575

Answer: E
Concepts:parallel linesisosceles trianglespecial right triangle
Difficulty rating: 1450
Solution:

Because ABCD,AB\parallel CD, we have CDM=AMD.\angle CDM=\angle AMD. Combined with AMD=CMD,\angle AMD=\angle CMD, this gives CDM=CMD,\angle CDM=\angle CMD, so CMD\triangle CMD is isosceles with CM=CD=6.CM=CD=6.

Then MBC\triangle MBC is right-angled at BB with hypotenuse CM=6CM=6 and leg BC=3,BC=3, so it is a 3030-6060-9090^\circ triangle with BMC=30.\angle BMC=30^\circ.

Finally, AMD+CMD\angle AMD+\angle CMD +BMC=180,+\angle BMC=180^\circ, so 2AMD+30=180,2\angle AMD+30^\circ=180^\circ, giving AMD=75.\angle AMD=75^\circ.

Thus, the correct answer is E.

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