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

Revision as of 00:15, 26 May 2011

Problem 18

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

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

It is given that $\angle AMD \cong \angle CMD$ . 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 $\begin{align*} 2x + 30 &= 180\\ 2x &= 150\\ x &= \boxed{\textbf{(E)} 75} \end{align*}$

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

Revision as of 00:15, 26 May 2011

Problem 18

Solution

@@ Line 1: / Line 1: @@
-==Problem==
+== Problem 18 ==
-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>
 ==Solution==
-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("$6$",midpoint(A--B),N);
+label("$3$",midpoint(B--C),E);
+</asy>
+</center>
+It is given that <math>\angle AMD \cong \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*}
+x + 30 &= 180\\
+x &= 150\\
+x &= \boxed{\textbf{(E)} 75}
+\end{align*}</cmath>