# 2005 AMC 10B Problems/Problem 14

## 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]$$\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}$

## Solution

### Solution 1

The area of a circle can be given by $\frac12 ab \text{sin} C$. $MC=1$ because it is the midpoint of a side, and $CD=2$ because it is twice the length of $BC$. Each angle of an equilateral triangle is $60^\circ$ so $\angle MCD = 120^\circ$. The area is $\frac12 (1)(2) \text{sin} 120^\circ = \boxed{\textbf{(C)}\ \frac{\sqrt{3}}{2}}$.

### 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 now 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 $\boxed{\mathrm{(C)}\ \dfrac{\sqrt{3}}{2}}$.