# 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

$[asy] unitsize(10mm); defaultpen(linewidth(.5pt)+fontsize(10pt)); dotfactor=3; pair A=(0,3), B=(6,3), C=(6,0), D=(0,0); pair M=(0.80385,3); draw(A--B--C--D--cycle); draw(M--C); draw(M--D); draw(anglemark(A,M,D)); draw(anglemark(D,M,C)); draw(anglemark(C,D,M)); pair[] ps={A,B,C,D,M}; dot(ps); label("A",A,NW); label("B",B,NE); label("C",C,SE); label("D",D,SW); label("M",M,N); label("6",midpoint(C--M),SW); label("3",midpoint(B--C),E); label("6",midpoint(C--D),S); [/asy]$

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