2011 AMC 12B Problems/Problem 10

Problem

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 $\angle AMD$?

$\textrm{(A)}\ 15 \qquad \textrm{(B)}\ 30 \qquad \textrm{(C)}\ 45 \qquad \textrm{(D)}\ 60 \qquad \textrm{(E)}\ 75$

Solution

Since $AB \parallel CD$, $\angle AMD = \angle CDM$, so $\angle AMD = \angle CMD = \angle CDM$, so $\bigtriangleup CMD$ is isosceles, and hence $CM=CD=6$. Therefore, $\triangle BMC$ is a 30-60-90 triangle with $\angle BMC = 30^\circ$. Therefore $\angle AMD=\boxed{\mathrm{(E)}\ 75^\circ}$

See also

2011 AMC 12B (ProblemsAnswer KeyResources)
Preceded by
Problem 9
Followed by
Problem 11
1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25
All AMC 12 Problems and Solutions

The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. AMC logo.png