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2013 AIME II Problems/Problem 13

Revision as of 20:22, 4 April 2013 by Aux770 (talk | contribs)

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

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

$b^2+7-2\sqrt{7}\cdot cos(\angle CED)=9a^2$ ... $(1)$

$b^2+7+2\sqrt{7}\cdot cos(\angle CED)=16a^2$ ...$(2)$

So, $(1)+(2)$, we get$2b^2+14=25a^2$...$(3)$

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

$4b^2+9a^2-2\cdot 2b\cdot 3a\cdot cos(\angle ADC)=16a^2$

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

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

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

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

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

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

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

$4(\frac{\sqrt{22}}{2})^2+1+2\cdot\2(\frac{\sqrt{22}}{2})\cdot cos(ADC)=AB^2$ (Error compiling LaTeX. Unknown error_msg)

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 $\boxed{10}$

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