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2011 AMC 12B Problems/Problem 10: Difference between revisions

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m →Solution: should mention that BMC is a 30-60-90 in order to conclude BMC = 30 deg
 
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== Solution ==
== Solution ==
Since <math>AB \parallel CD</math>, <math>\angle AMD = \angle CDM</math>, so <math>\angle AMD = \angle CMD = \angle CDM</math>, so <math>\bigtriangleup CMD</math> is isosceles, and hence <math>CM=CD=6</math>.  Therefore, <math>\angle BMC = 30^\circ</math>.  Therefore <math>\angle AMD=\boxed{\mathrm{(E)}\ 75^\circ}</math>
Since <math>AB \parallel CD</math>, <math>\angle AMD = \angle CDM</math>, so <math>\angle AMD = \angle CMD = \angle CDM</math>, so <math>\bigtriangleup CMD</math> is isosceles, and hence <math>CM=CD=6</math>.  Therefore, <math>\triangle BMC</math> is a 30-60-90 triangle with <math>\angle BMC = 30^\circ</math>.  Therefore <math>\angle AMD=\boxed{\mathrm{(E)}\ 75^\circ}</math>


== See also ==
== See also ==
{{AMC12 box|year=2011|ab=B|num-b=9|num-a=11}}
{{AMC12 box|year=2011|ab=B|num-b=9|num-a=11}}
{{MAA Notice}}
{{MAA Notice}}

Latest revision as of 16:39, 20 June 2024

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 (Problems • Answer Key • Resources)
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

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