Art of Problem Solving
AoPS Online
Math texts, online classes, and more
for students in grades 5-12.
Visit AoPS Online ‚
Books for Grades 5-12 Online Courses
Beast Academy
Engaging math books and online learning
for students ages 6-13.
Visit Beast Academy ‚
Books for Ages 6-13 Beast Academy Online
AoPS Academy
Small live classes for advanced math
and language arts learners in grades 2-12.
Visit AoPS Academy ‚
Find a Physical Campus Visit the Virtual Campus

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 (…')
 
 
(31 intermediate revisions by 16 users not shown)
Line 1: Line 1:
−
==Problem==
+
== 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?
+
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>?
  
−
(A) 15 (B) 30 (C) 45 (D) 60 (E) 75
+
<math> \textbf{(A)}\ 15 \qquad\textbf{(B)}\ 30 \qquad\textbf{(C)}\ 45 \qquad\textbf{(D)}\ 60 \qquad\textbf{(E)}\ 75</math>
 +
[[Category: Introductory Geometry Problems]]
  
−
==Solution==
+
==Solution 1==
  
−
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).
+
<center><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>
 +
</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*}
 +
2x + 30 &= 180\\
 +
2x &= 150\\
 +
x &= \boxed{\textbf{(E)} 75}
 +
\end{align*}</cmath>
 +
 
 +
===Easier Way to Continue===
 +
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>
 +
 
 +
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>
 +
 
 +
~mathboy282
 +
 
 +
==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
 +
 
 +
<cmath>4x^4 - 48x^3 + 216x^2 - 432x + 324</cmath>
 +
 
 +
Now, we apply the quartic formula to get
 +
 
 +
<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
 +
 
 +
==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
 +
 
 +
== 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==
 +
 
 +
{{AMC10 box|year=2011|ab=B|num-b=17|num-a=19}}
 +
{{MAA Notice}}

Latest revision as of 14:29, 1 April 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)); 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

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

The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. AMC logo.png

Retrieved from "https://artofproblemsolving.com/wiki/index.php?title=2011_AMC_10B_Problems/Problem_18&oldid=217647"