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Difference between revisions of "2013 AIME II Problems/Problem 13"
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==Problem 13==
 
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>.
 
==Solution==
 
==Solution==
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Then the area of <math>\triangle ABC</math> is <math>3\sqrt{7}</math>, so the answer is <math>\boxed{010}</math>
 
Then the area of <math>\triangle ABC</math> is <math>3\sqrt{7}</math>, so the answer is <math>\boxed{010}</math>
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==See Also==
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{{AIME box|year=2013|n=II|num-b=12|num-a=14}}

Revision as of 15:44, 6 April 2013

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

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 $\boxed{010}$

See Also

2013 AIME II (Problems • Answer Key • Resources)
Preceded by
Problem 12 		Followed by
Problem 14
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15
All AIME Problems and Solutions
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