2011 AMC 12B Problem 10

Below is the professionally curated solution for Problem 10 of the 2011 AMC 12B, 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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Concepts:parallel lines isosceles triangle special right triangle

Difficulty rating: 1450

10.

Rectangle $ABCD$ has $AB=6$ and $BC=3.$ Point $M$ is chosen on side $AB$ so that $\angle AMD=\angle CMD.$ What is the degree measure of $\angle AMD?$

$15$

$30$

$45$

$60$

$75$

Solution:

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

Then $\triangle MBC$ is right-angled at $B$ with hypotenuse $CM=6$ and leg $BC=3,$ so it is a $30$ - $60$ - $90^\circ$ triangle with $\angle BMC=30^\circ.$

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

Thus, the correct answer is E.

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