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

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

Solution 5 (guesstimation)

We draw a diagram. It seems that they are larger than 60 degrees. We try 75, and with the knowledge that $\text{sin}(75^\circ)\approx 0.96$ , and $\text{cos}(75^\circ)\approx 0.26$ , so we have $\text{cot}(75^circ)\approx \frac{4}{15}$ , and this gives us missing side lengths of 0.8, 5.2, 5.9, and 3.1, which happens to satisfy all equations. So, $\boxed{\textbf{E}}$

Solution 6 (using trig to express related sides)

Let $\angle AMD=\angle CMD=x$. Then, $\angle BMC=180-2x$. $\tan (x)=\frac{AD}{AM}=\frac{3}{AM}$. Thus, we have $AM=\frac{3}{\tan (x)}$. Similarly, $\tan (180-2x)=\frac{BC}{BM}=\frac{3}{BM}$. Thus, we have $BM=\frac{3}{\tan (180-2x)}$. Using the fact that $\tan (180-2x)=-\tan (2x)$, we can write $BM$ as $\frac{-3}{\tan (2x)}$. Since $AM+BM=6$, we have $\frac{3}{\tan (x)}-\frac{3}{\tan (2x)}=6$. $\tan (2x)=\frac{2\tan x}{1-\tan ^{2}x}$, so $\frac{3}{\tan (x)}-\frac{3(1-\tan ^{2}x)}{2\tan (x)}=6$. Multiplying both sides by $2\tan (x)$, we have $6-3+3\tan ^{2}(x)=12\tan (x)$. Therefore, $\tan ^{2}(x)-4\tan (x)+1=0$. Solving the quadratic, we have $\tan (x)=2\pm \sqrt{3}$. If $\tan (x)=2-\sqrt{3}$, then $AM=\frac{3}{2-\sqrt{3}}=6+3\sqrt{3}$, which is larger than the side length of $6$. Therefore, $\tan (x)=2+\sqrt{3}$, and since $x$ is between $0$ and $180$ degrees, $x=75$ degrees. ~Spiritmoon_OSU

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