Difference between revisions of "2005 AMC 10B Problems/Problem 14"

Revision as of 13:45, 15 December 2021

Problem

Equilateral $\triangle ABC$ has side length $2$ , $M$ is the midpoint of $\overline{AC}$ , and $C$ is the midpoint of $\overline{BD}$ . What is the area of $\triangle CDM$ ? $[asy]defaultpen(linewidth(.8pt)+fontsize(8pt)); pair B = (0,0); pair A = 2*dir(60); pair C = (2,0); pair D = (4,0); pair M = midpoint(A--C); label("$A$",A,NW);label("$B$",B,SW);label("$C$",C, SE);label("$M$",M,NE);label("$D$",D,SE); draw(A--B--C--cycle); draw(C--D--M--cycle);[/asy]$

$\textbf{(A) }\ \frac {\sqrt {2}}{2}\qquad \textbf{(B) }\ \frac {3}{4}\qquad \textbf{(C) }\ \frac {\sqrt {3}}{2}\qquad \textbf{(D) }\ 1\qquad \textbf{(E) }\ \sqrt {2}$

Solutions

Solution 1 (simplest)

The area of a triangle can be given by $\frac12 ab \sin C$ . $MC=1$ because it is the midpoint of a side, and $CD=2$ because it is the same length as $BC$ . Each angle of an equilateral triangle is $60^\circ$ so $\angle MCD = 120^\circ$ . The area is $\frac12 (1)(2) \sin 120^\circ = \boxed{\textbf{(C)}\ \frac{\sqrt{3}}{2}}$ . Note: Even if you don't know the value of $\sin 120^\circ$ , you can use the fact that $\sin x = \sin (180^\circ - x)$ , so $\sin 120^\circ = \sin 60^\circ$ . You can easily calculate $\sin 60^\circ$ to be $\frac{\sqrt3}{2}$ using equilateral triangles.

Solution 2

In order to calculate the area of $\triangle CDM$ , we can use the formula $A=\dfrac{1}{2}bh$ , where $\overline{CD}$ is the base. We already know that $\overline{CD}=2$ , so the formula becomes $A=h$ . We can drop verticals down from $A$ and $M$ to points $E$ and $F$ , respectively. We can see that $\triangle AEC \sim \triangle MFC$ . Now, we establish the relationship that $\dfrac{AE}{MF}=\dfrac{AC}{MC}$ . We are given that $\overline{AC}=2$ , and $M$ is the midpoint of $\overline{AC}$ , so $\overline{MC}=1$ . Because $\triangle AEB$ is a $30-60-90$ triangle and the ratio of the sides opposite the angles are $1-\sqrt{3}-2$ $\overline{AE}$ is $\sqrt{3}$ . Plugging those numbers in, we have $\dfrac{\sqrt{3}}{MF}=\dfrac{2}{1}$ . Cross-multiplying, we see that $2\times\overline{MF}=\sqrt{3}\times1\implies \overline{MF}=\dfrac{\sqrt{3}}{2}$ Since $\overline{MF}$ is the height $\triangle CDM$ , the area is $\boxed{\mathrm{(C)}\ \dfrac{\sqrt{3}}{2}}$ .

Solution 3

Draw a line from $M$ to the midpoint of $\overline{BC}$ . Call the midpoint of $\overline{BC}$ $P$ . This is an equilateral triangle, since the two segments $\overline{PC}$ and $\overline{MC}$ are identical, and $\angle C$ is 60°. Using the Pythagorean Theorem, point $M$ to $\overline{BC}$ is $\dfrac{\sqrt{3}}{2}$ . Also, the length of $\overline{CD}$ is 2, since $C$ is the midpoint of $\overline{BD}$ . So, our final equation is $\frac{\sqrt{3}}{2}\times2\over2$ , which just leaves us with $\boxed{\mathrm{(C)}\ \dfrac{\sqrt{3}}{2}}$ .

Solution 4

Drop a vertical down from $M$ to $BC$ . Let us call the point of intersection $X$ and the midpoint of $BC$ , $Y$ . We can observe that $\triangle AYC$ and $\triangle MXC$ are similar. By the Pythagorean theorem, $AY$ is $\sqrt3$ . Since $AC:MC=2:1,$ we find $MX=\frac{\sqrt3}{2}.$ Because $C$ is the midpoint of $BD,$ and $BC=2,$ $CD=2.$ Using the area formula, $\frac{CD*MX}{2}=\frac{\sqrt3}{2},$ $\boxed{\mathrm{(C)}\ \dfrac{\sqrt{3}}{2}}.$

~ sdk652

@@ Line 14: / Line 14: @@
 draw(C--D--M--cycle);</asy>
-<math> \textrm{(A)}\ \frac {\sqrt {2}}{2}\qquad \textrm{(B)}\ \frac {3}{4}\qquad \textrm{(C)}\ \frac {\sqrt {3}}{2}\qquad \textrm{(D)}\ 1\qquad \textrm{(E)}\ \sqrt {2}</math>
+<math> \textbf{(A) }\ \frac {\sqrt {2}}{2}\qquad \textbf{(B) }\ \frac {3}{4}\qquad \textbf{(C) }\ \frac {\sqrt {3}}{2}\qquad \textbf{(D) }\ 1\qquad \textbf{(E) }\ \sqrt {2}</math>
 == Solutions ==

2005 AMC 10B (Problems • Answer Key • Resources)
Preceded by Problem 13	Followed by Problem 15
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