2011 AMC 10B Problem 18

Attempt Problem 18 of the 2011 AMC 10B 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 10B solutions, or check the answer key.

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18.

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

Answer: E

Concepts:rectangle isosceles triangle trigonometry

Difficulty rating: 1540

Small hint:

Use

AB\parallel CD

to transfer one angle to

\angle CDM

Big hint:

Triangle

CMD

becomes isosceles

Solution:

The angles

\angle AMD

and

\angle MDC

are equal since

AB \parallel DC.

As such,

\angle MDC = \angle DMC ,

making

MDC

isosceles and

MC = DC = 6.

As we can see,

\sin (\angle CMB) = \frac 12,

making

\angle CMB = 30^\circ .

Therefore,

\angle AMC = 150^\circ .

Since

\angle AMD

is half of that,

\angle AMD = 75^\circ .

Thus, the correct answer is E .