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Difference between revisions of "2011 AMC 10B Problems/Problem 18"

(trig subsolution.)
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draw(anglemark(A,M,D));
 
draw(anglemark(A,M,D));
 
draw(anglemark(D,M,C));
 
draw(anglemark(D,M,C));
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draw(anglemark(C,D,M));
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pair[] ps={A,B,C,D,M};
 
pair[] ps={A,B,C,D,M};
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\end{align*}</cmath>
 
\end{align*}</cmath>
  
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===Easier Way to Continue==
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===Easier Way to Continue===
 
After finding <math>MC = 6,</math> we can continue using trigonometry as follows.  
 
After finding <math>MC = 6,</math> we can continue using trigonometry as follows.  
  
−
We know that <math>\angle{BMC} = 180-2x</math> and so <math>\sin{180-2x} = \frac{3}{6} = \frac{1}{2}</math>
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We know that <math>\angle{BMC} = 180-2x</math> and so <math>\sin (180-2x) = \frac{3}{6} = \frac{1}{2}</math>
  
−
It is obvious that <math>\sin{30} = \frac{1}{2}</math> and so <math>180-2x=30.</math>
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It is obvious that <math>\sin (30) = \frac{1}{2}</math> and so <math>180-2x=30.</math>
  
 
Solving, we have <math>x = \boxed{75}</math>
 
Solving, we have <math>x = \boxed{75}</math>
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We have <math>DC=CM=6</math>. By the Pythagorean Theorem, <math>BM=\sqrt{6^2-3^2}=3\sqrt{3}</math>, and thus <math>AM=6-3\sqrt{3}</math>, We have <math>\tan(AMD)=\frac{6-3\sqrt{3}}{3}=2+\sqrt{3}</math>, or <math>\angle AMD=\boxed{75}</math>
 
We have <math>DC=CM=6</math>. By the Pythagorean Theorem, <math>BM=\sqrt{6^2-3^2}=3\sqrt{3}</math>, and thus <math>AM=6-3\sqrt{3}</math>, We have <math>\tan(AMD)=\frac{6-3\sqrt{3}}{3}=2+\sqrt{3}</math>, or <math>\angle AMD=\boxed{75}</math>
 
~awsomek
 
~awsomek
 +
 +
== Solution 4 (elimination) ==
 +
Let <math>\angle AMD=\angle DMC=\theta</math>. Thus, <math>\angle BMC=180-2\theta\implies\angle MCB=2\theta-90</math>. Since all angles should be positive, <math>\theta>45^\circ</math>, narrowing the options to D and E. Trying <math>60^\circ</math> (option D), <math>\Delta AMD</math> is a 30-60-90 triangle. <math>AD=3</math>, so it follows that <math>AM=\sqrt3</math>.
 +
Since <math>\angle BMC=180-2\theta</math>, <math>\angle BMC=60^\circ</math>, too. However, that would imply that <math>\Delta MBC</math> is also a <math>30-60-90</math> triangle, which would, in turn, imply that <math>MB=3\sqrt3</math>, since <math>BC=3</math>. We know that <math>AM+MB=AB</math> and <math>AB=6,</math> but we know that <math>AM=\sqrt3</math> and <math>MB=3\sqrt3</math>. <math>AM+MB</math> is clearly not <math>6</math>, so this is not possible.
 +
Thus, the answer must be <math>\boxed{\textbf{(E)}~75}</math>.
 +
~ Technodoggo
  
 
== See Also==
 
== See Also==

Revision as of 18:18, 2 September 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 $\angle AMD$?

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

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

See Also

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

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