2005 AMC 10B Problems/Problem 14

Revision as of 16:59, 7 January 2022 by Doulai1 (talk | contribs) (Solution 1 (trig))
(diff) ← Older revision | Latest revision (diff) | Newer revision → (diff)


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]

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


Solution 1 (trig)

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.

~Minor Edits by doulai1

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 $\boxed{\textbf{(C) }\frac{\sqrt{3}}{2}}$.

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 $60^{\circ}$. 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 $\boxed{\textbf{(C) }\dfrac{\sqrt{3}}{2}}$.

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 $\bigtriangleup MXC$ 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, $\frac{CD*MX}{2}=\boxed{\textbf{(C) }\dfrac{\sqrt{3}}{2}}.$

~ sdk652

See Also

2005 AMC 10B (ProblemsAnswer KeyResources)
Preceded by
Problem 13
Followed by
Problem 15
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