# 2013 AIME II Problems/Problem 13

## Problem 13

In $\triangle ABC$, $AC = BC$, and point $D$ is on $\overline{BC}$ so that $CD = 3\cdot BD$. Let $E$ be the midpoint of $\overline{AD}$. Given that $CE = \sqrt{7}$ and $BE = 3$, the area of $\triangle ABC$ can be expressed in the form $m\sqrt{n}$, where $m$ and $n$ are positive integers and $n$ is not divisible by the square of any prime. Find $m+n$.

## Solution

### Solution 1

After drawing the figure, we suppose $BD=a$, so that $CD=3a$, $AC=4a$, and $AE=ED=b$.

Using cosine law for $\triangle AEC$ and $\triangle CED$,we get

$$b^2+7-2\sqrt{7}\cdot \cos(\angle CED)=9a^2\qquad (1)$$ $$b^2+7+2\sqrt{7}\cdot \cos(\angle CED)=16a^2\qquad (2)$$ So, $(1)+(2)$, we get$$2b^2+14=25a^2. \qquad (3)$$

Using cosine law in $\triangle ACD$, we get

$$b^2+9a^2-2\cdot 2b\cdot 3a\cdot \cos(\angle ADC)=16a^2$$

So, $$\cos(\angle ADC)=\frac{7a^2-4b^2}{12ab}.\qquad (4)$$

Using cosine law in $\triangle EDC$ and $\triangle EDB$, we get

$$b^2+9a^2-2\cdot 3a\cdot b\cdot \cos(\angle ADC)=7\qquad (5)$$

$$b^2+a^2+2\cdot a\cdot b\cdot \cos(\angle ADC)=9.\qquad (6)$$

$(5)+(6)$, and according to $(4)$, we can get $$37a^2+2b^2=48. \qquad (7)$$

Using $(3)$ and $(7)$, we can solve $a=1$ and $b=\frac{\sqrt{22}}{2}$.

Finally, we use cosine law for $\triangle ADB$,

$$4(\frac{\sqrt{22}}{2})^2+1+2\cdot\2(\frac{\sqrt{22}}{2})\cdot \cos(ADC)=AB^2$$

then $AB=2\sqrt{7}$, so the height of this $\triangle ABC$ is $\sqrt{4^2-(\sqrt{7})^2}=3$.

Then the area of $\triangle ABC$ is $3\sqrt{7}$, so the answer is $\boxed{010}$.

### Solution 2

Let $X$ be the foot of the altitude from $C$ with other points labelled as shown below. $[asy] size(200); pair A=(0,0),B=(2*sqrt(7),0),C=(sqrt(7),3),D=(3*B+C)/4,L=C/5,M=3*B/7; draw(A--B--C--cycle);draw(A--D^^B--L^^C--M); label("A",A,SW);label("B",B,SE);label("C",C,N);label("D",D,NE);label("L",L,NW);label("M",M,S); pair X=foot(C,A,B), Y=foot(L,A,B); pair EE=D/2; label("X",X,S);label("E",EE,NW);label("Y",Y,S); draw(C--X^^L--Y,dotted); draw(rightanglemark(B,X,C)^^rightanglemark(B,Y,L)); [/asy]$ Now we proceed using mass points. To balance along the segment $BC$, we assign $B$ a mass of $3$ and $C$ a mass of $1$. Therefore, $D$ has a mass of $4$. As $E$ is the midpoint of $AD$, we must assign $A$ a mass of $4$ as well. This gives $L$ a mass of $5$ and $M$ a mass of $7$.

Now let $AB=b$ be the base of the triangle, and let $CX=h$ be the height. Then as $AM:MB=3:4$, and as $AX=\frac{b}{2}$, we know that $$MX=\frac{b}{2}-\frac{3b}{7}=\frac{b}{14}.$$ Also, as $CE:EM=7:1$, we know that $EM=\frac{1}{\sqrt{7}}$. Therefore, by the Pythagorean Theorem on $\triangle {XCM}$, we know that $$\frac{b^2}{196}+h^2=\left(\sqrt{7}+\frac{1}{\sqrt{7}}\right)^2=\frac{64}{7}.$$

Also, as $LE:BE=5:3$, we know that $BL=\frac{8}{5}\cdot 3=\frac{24}{5}$. Furthermore, as $\triangle YLA\sim \triangle XCA$, and as $AL:LC=1:4$, we know that $LY=\frac{h}{5}$ and $AY=\frac{b}{10}$, so $YB=\frac{9b}{10}$. Therefore, by the Pythagorean Theorem on $\triangle BLY$, we get $$\frac{81b^2}{100}+\frac{h^2}{25}=\frac{576}{25}.$$ Solving this system of equations yields $b=2\sqrt{7}$ and $h=3$. Therefore, the area of the triangle is $3\sqrt{7}$, giving us an answer of $\boxed{010}$.