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

(Created page with '==Problem== There is a rectangle ABCD with AB = 6 and BC = 3. A point M lies on AB such that angles CMD and AMD are congruent. What is the measure of angle CMD? (A) 15 (B) 30 (…')
 
(→‎Solution 3(Easier Trig))
(24 intermediate revisions by 13 users not shown)
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==Problem==
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== Problem==
  
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There is a rectangle ABCD with AB = 6 and BC = 3. A point M lies on AB such that angles CMD and AMD are congruent. What is the measure of angle CMD?
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Rectangle <math>ABCD</math> has <math>AB = 6</math> and <math>BC = 3</math>. Point <math>M</math> is chosen on side <math>AB</math> so that <math>\angle AMD = \angle CMD</math>. What is the degree measure of <math>\angle AMD</math>?
  
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(A) 15 (B) 30 (C) 45 (D) 60 (E) 75
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<math> \textbf{(A)}\ 15 \qquad\textbf{(B)}\ 30 \qquad\textbf{(C)}\ 45 \qquad\textbf{(D)}\ 60 \qquad\textbf{(E)}\ 75</math>
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[[Category: Introductory Geometry Problems]]
  
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==Solution==
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==Solution 1==
  
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Using alternate interior angles and the fact that AB and CD are parallel because it is a rectangle, the angles AMD and MDC are congruent. Using the transitive property and the given angular congruence, MDC and CMD are congruent, and so CD = CM = 6. So sin(BMC) = 3/6 = 0.5, and angle BMC is 30 degrees. Then AMD = 75 degrees (E).
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<center><asy>
 +
unitsize(10mm);
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defaultpen(linewidth(.5pt)+fontsize(10pt));
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dotfactor=3;
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pair A=(0,3), B=(6,3), C=(6,0), D=(0,0);
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pair M=(0.80385,3);
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draw(A--B--C--D--cycle);
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draw(M--C);
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draw(M--D);
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draw(anglemark(A,M,D));
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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};
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dot(ps);
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label("$A$",A,NW);
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label("$B$",B,NE);
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label("$C$",C,SE);
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label("$D$",D,SW);
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label("$M$",M,N);
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label("$6$",midpoint(C--M),SW);
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label("$3$",midpoint(B--C),E);
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label("$6$",midpoint(C--D),S);
 +
 
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</asy>
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</center>
 +
 
 +
It is given that <math>\angle AMD \sim \angle CMD</math>. Since <math>\angle AMD</math> and <math>\angle CDM</math> are alternate interior angles and <math>\overline{AB} \parallel \overline{DC}</math>, <math>\angle AMD \cong \angle CDM \longrightarrow \angle CMD \cong \angle CDM</math>. Use the [[Base Angle Theorem]] to show <math>\overline{DC} \cong \overline{MC}</math>. We know that <math>ABCD</math> is a [[rectangle]], so it follows that <math>\overline{MC} = 6</math>. We notice that <math>\triangle BMC</math> is a <math>30-60-90</math> triangle, and <math>\angle BMC = 30^{\circ}</math>. If we let <math>x</math> be the measure of <math>\angle AMD,</math> then
 +
<cmath>\begin{align*}
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2x + 30 &= 180\\
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2x &= 150\\
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x &= \boxed{\textbf{(E)} 75}
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\end{align*}</cmath>
 +
 
 +
==Solution 2 (with trig, not recommended)==
 +
 
 +
Let <math>\angle{DMC} = \angle{AMD} = \theta</math>. If we let <math>AM = x</math>, we have that <math>MD = \sqrt{x^2 + 9}</math>, by the Pythagorean Theorem, and similarily, <math>MC = \sqrt{x^2 - 12x + 45}</math>. Applying the law of cosine, we see that
 +
<cmath>2x^2 + 54 - 12x - 2 \sqrt{x^4 - 12x^3 + 54x^2 - 108x + 405} \cdot \cos (\theta) = 36</cmath> and <cmath>\tan (\theta) = \frac{3}{x}</cmath> 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 <math>\sin (\theta) = \frac{3}{\sqrt{x^2+9}}</math>, so solving for <math>\cos (\theta)</math> in terms of <math>x</math>, we get that <math>\cos (\theta) = \frac{x \cdot \sqrt{x^2 + 9}}{x^2 + 9}</math>. The equation now becomes
 +
 
 +
<cmath>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</cmath>
 +
Simplifying, we get
 +
 
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<cmath>4x^4 - 48x^3 + 216x^2 - 432x + 324</cmath>
 +
 
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Now, we apply the quartic formula to get
 +
 
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<cmath>x = 6 \pm 3 \sqrt{3}</cmath>
 +
 
 +
We can easily see that <math>x = 6 + 3 \sqrt{3}</math> is an invalid solution. Thus, <math>x = 6 - 3 \sqrt{3}</math>.  
 +
 
 +
Finally, since <math>\tan (\theta) = \frac{3}{6 - 3 \sqrt{3}} = 2 + \sqrt{3}</math>, <math>\theta = \frac{5 + 12n}{12} \pi</math>, where <math>n</math> is any integer. Converting to degrees, we have that <math>\theta = 75 + 180n</math>. Since <math>0 < \theta < 90</math>, we have that <math>\theta = \boxed{75}</math>. <math>\square</math>
 +
 
 +
~ilovepi3.14
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==Solution 3(Easier Trig)==
 +
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
 +
 
 +
== See Also==
 +
 
 +
{{AMC10 box|year=2011|ab=B|num-b=17|num-a=19}}
 +
{{MAA Notice}}

Revision as of 11:13, 13 July 2021

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*}

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

See Also

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

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