2011 AMC 10B Problems/Problem 18

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 $\angle AMD$ ?

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

Solution 1

unitsize(10mm);
defaultpen(linewidth(.5pt)+fontsize(10pt));
dotfactor=3;

pair A=(0,3), B=(6,3), C=(6,0), D=(0,0);
pair M=(0.80385,3);

draw(A--B--C--D--cycle);
draw(M--C);
draw(M--D);
draw(anglemark(A,M,D));
draw(anglemark(C,D,M);

pair[] ps={A,B,C,D,M};
dot(ps);
label("$A$",A,NW);
label("$B$",B,NE);
label("$C$",C,SE);
label("$D$",D,SW);
label("$M$",M,N);
label("$6$",midpoint(C--M),SW);
label("$3$",midpoint(B--C),E);
label("$6$",midpoint(C--D),S);

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It is given that $\angle AMD \sim \angle CMD$ . Since $\angle AMD$ and $\angle CDM$ are alternate interior angles and $\overline{AB} \parallel \overline{DC}$ , $\angle AMD \cong \angle CDM \longrightarrow \angle CMD \cong \angle CDM$ . Use the Base Angle Theorem to show $\overline{DC} \cong \overline{MC}$ . We know that $ABCD$ is a rectangle, so it follows that $\overline{MC} = 6$ . We notice that $\triangle BMC$ is a $30-60-90$ triangle, and $\angle BMC = 30^{\circ}$ . If we let $x$ be the measure of $\angle AMD,$ then $\begin{align*} 2x + 30 &= 180\\ 2x &= 150\\ x &= \boxed{\textbf{(E)} 75} \end{align*}$

Easier Way to Continue

After finding $MC = 6,$ we can continue using trigonometry as follows.

We know that $\angle{BMC} = 180-2x$ and so $\sin (180-2x) = \frac{3}{6} = \frac{1}{2}$

It is obvious that $\sin (30) = \frac{1}{2}$ and so $180-2x=30.$

Solving, we have $x = \boxed{75}$

~mathboy282

Solution 2 (with trig, not recommended)

Let $\angle{DMC} = \angle{AMD} = \theta$ . If we let $AM = x$ , we have that $MD = \sqrt{x^2 + 9}$ , by the Pythagorean Theorem, and similarily, $MC = \sqrt{x^2 - 12x + 45}$ . Applying the law of cosine, we see that $2x^2 + 54 - 12x - 2 \sqrt{x^4 - 12x^3 + 54x^2 - 108x + 405} \cdot \cos (\theta) = 36$ and $\tan (\theta) = \frac{3}{x}$ YAY!!! We have two equations for two variables... that are relatively difficult to deal with. Well, we'll try to solve it. First of all, note that $\sin (\theta) = \frac{3}{\sqrt{x^2+9}}$ , so solving for $\cos (\theta)$ in terms of $x$ , we get that $\cos (\theta) = \frac{x \cdot \sqrt{x^2 + 9}}{x^2 + 9}$ . The equation now becomes

$2x^2 + 54 - 12x - 2 \sqrt{x^4 - 12x^3 + 54x^2 - 108x + 405} \cdot \frac{x \cdot \sqrt{x^2 + 9}}{x^2 + 9} = 36$ Simplifying, we get

$4x^4 - 48x^3 + 216x^2 - 432x + 324$

Now, we apply the quartic formula to get

$x = 6 \pm 3 \sqrt{3}$

We can easily see that $x = 6 + 3 \sqrt{3}$ is an invalid solution. Thus, $x = 6 - 3 \sqrt{3}$ .

Finally, since $\tan (\theta) = \frac{3}{6 - 3 \sqrt{3}} = 2 + \sqrt{3}$ , $\theta = \frac{5 + 12n}{12} \pi$ , where $n$ is any integer. Converting to degrees, we have that $\theta = 75 + 180n$ . Since $0 < \theta < 90$ , we have that $\theta = \boxed{75}$ . $\square$

~ilovepi3.14

Solution 3(Easier Trig)

We have $DC=CM=6$ . By the Pythagorean Theorem, $BM=\sqrt{6^2-3^2}=3\sqrt{3}$ , and thus $AM=6-3\sqrt{3}$ , We have $\tan(AMD)=\frac{6-3\sqrt{3}}{3}=2+\sqrt{3}$ , or $\angle AMD=\boxed{75}$ ~awsomek

Solution 4 (elimination)

Let $\angle AMD=\angle DMC=\theta$ . Thus, $\angle BMC=180-2\theta\implies\angle MCB=2\theta-90$ . Since all angles should be positive, $\theta>45^\circ$ , narrowing the options to D and E. Trying $60^\circ$ (option D), $\Delta AMD$ is a 30-60-90 triangle. $AD=3$ , so it follows that $AM=\sqrt3$ . Since $\angle BMC=180-2\theta$ , $\angle BMC=60^\circ$ , too. However, that would imply that $\Delta MBC$ is also a $30-60-90$ triangle, which would, in turn, imply that $MB=3\sqrt3$ , since $BC=3$ . We know that $AM+MB=AB$ and $AB=6,$ but we know that $AM=\sqrt3$ and $MB=3\sqrt3$ . $AM+MB$ is clearly not $6$ , so this is not possible. Thus, the answer must be $\boxed{\textbf{(E)}~75}$ . ~ Technodoggo

2011 AMC 10B (Problems • Answer Key • Resources)
Preceded by Problem 17	Followed by Problem 19
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