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Difference between revisions of "2013 AIME II Problems/Problem 13"

(Problem 13)
 
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In <math>\triangle ABC</math>, <math>AC = BC</math>, and point <math>D</math> is on <math>\overline{BC}</math> so that <math>CD = 3\cdot BD</math>. Let <math>E</math> be the midpoint of <math>\overline{AD}</math>. Given that <math>CE = \sqrt{7}</math> and <math>BE = 3</math>, the area of <math>\triangle ABC</math> can be expressed in the form <math>m\sqrt{n}</math>, where <math>m</math> and <math>n</math> are positive integers and <math>n</math> is not divisible by the square of any prime. Find <math>m+n</math>.
 
In <math>\triangle ABC</math>, <math>AC = BC</math>, and point <math>D</math> is on <math>\overline{BC}</math> so that <math>CD = 3\cdot BD</math>. Let <math>E</math> be the midpoint of <math>\overline{AD}</math>. Given that <math>CE = \sqrt{7}</math> and <math>BE = 3</math>, the area of <math>\triangle ABC</math> can be expressed in the form <math>m\sqrt{n}</math>, where <math>m</math> and <math>n</math> are positive integers and <math>n</math> is not divisible by the square of any prime. Find <math>m+n</math>.
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==Solution==
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After drawing the figure, we suppose <math>BD=a</math>, so that<math>CD=3a</math>,<math>AC=4a</math>, and <math>AE=ED=b</math>.
 +
 +
Using cosine law for <math>\triangle AEC</math> and <math>\triangle CED</math>,we get
 +
 +
<math>b^2+7-2\sqrt{7}\cdot cos(\angle CED)=9a^2</math> ... <math>(1)</math>
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<math>b^2+7+2\sqrt{7}\cdot cos(\angle CED)=16a^2</math> ...<math>(2)</math>
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So, <math>(1)+(2)</math>, we get<math>2b^2+14=25a^2</math>...<math>(3)</math>
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 +
Using cosine law in <math>\triangle ACD</math>,we get
 +
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<math>4b^2+9a^2-2\cdot 2b\cdot 3a\cdot cos(\angle ADC)=16a^2</math>
 +
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So, <math>cos(\angle ADC)=\frac{7a^2-4b^2}{12ab}</math>...<math>(4)</math>
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Using cosine law in <math>\triangle EDC</math> and <math>\triangle EDB</math>, we get
 +
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<math>b^2+9a^2-2\cdot 3a\cdot b\cdot cos(\angle ADC)=7</math>...<math>(5)</math>
 +
 +
<math>b^2+a^2+2\cdot a\cdot b\cdot cos(\angle ADC)=9</math>...<math>(6)</math>
 +
 +
<math>(5)+(6)</math>, and according to <math>(4)</math>, we can get <math>37a^2+2b^2=48</math>...<math>(7)</math>
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Using <math>(3)</math> and <math>(7)</math>, we can solve <math>a=1</math> and <math>b=\frac{\sqrt{22}}{2}</math>
 +
 +
Finally, we use cosine law for  <math>\triangle ADB</math>,
 +
 +
<math>4(\frac{\sqrt{22}}{2})^2+1+2\cdot\2(\frac{\sqrt{22}}{2})\cdot cos(ADC)=AB^2</math>
 +
 +
then <math>AB=2\sqrt{7}</math>
 +
 +
so the height of this <math>\triangle ABC</math> is <math>\sqrt{4^2-(\sqrt{7})^2}=3</math>
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 +
Then the area of <math>\triangle ABC</math> is <math>3\sqrt{7}</math>, so the answer is <math>\boxed{10}</math>

Revision as of 20:22, 4 April 2013

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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