Difference between revisions of "2013 AIME II Problems/Problem 10"

(Solution)
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Given a circle of radius <math>\sqrt{13}</math>, let <math>A</math> be a point at a distance <math>4 + \sqrt{13}</math> from the center <math>O</math> of the circle. Let <math>B</math> be the point on the circle nearest to point <math>A</math>. A line passing through the point <math>A</math> intersects the circle at points <math>K</math> and <math>L</math>. The maximum possible area for <math>\triangle BKL</math> can be written in the form <math>\frac{a - b\sqrt{c}}{d}</math>, where <math>a</math>, <math>b</math>, <math>c</math>, and <math>d</math> are positive integers, <math>a</math> and <math>d</math> are relatively prime, and <math>c</math> is not divisible by the square of any prime. Find <math>a+b+c+d</math>.
 
Given a circle of radius <math>\sqrt{13}</math>, let <math>A</math> be a point at a distance <math>4 + \sqrt{13}</math> from the center <math>O</math> of the circle. Let <math>B</math> be the point on the circle nearest to point <math>A</math>. A line passing through the point <math>A</math> intersects the circle at points <math>K</math> and <math>L</math>. The maximum possible area for <math>\triangle BKL</math> can be written in the form <math>\frac{a - b\sqrt{c}}{d}</math>, where <math>a</math>, <math>b</math>, <math>c</math>, and <math>d</math> are positive integers, <math>a</math> and <math>d</math> are relatively prime, and <math>c</math> is not divisible by the square of any prime. Find <math>a+b+c+d</math>.
  
==Solution==
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==Solution 1==
 
Now we put the figure in the Cartesian plane, let the center of the circle <math>O (0,0)</math>, then <math>B (\sqrt{13},0)</math>, and <math>A(4+\sqrt{13},0)</math>
 
Now we put the figure in the Cartesian plane, let the center of the circle <math>O (0,0)</math>, then <math>B (\sqrt{13},0)</math>, and <math>A(4+\sqrt{13},0)</math>
  
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So, <math>LK=\sqrt{1+k^2}\cdot \sqrt{(x_1+x_2)^2-4x_1x_2}</math>
 
So, <math>LK=\sqrt{1+k^2}\cdot \sqrt{(x_1+x_2)^2-4x_1x_2}</math>
  
Also, the distance between <math>O</math> and <math>LK</math> is <math>\frac{k\times \sqrt{13}-(4+\sqrt{13})\cdot k}{\sqrt{1+k^2}}=\frac{-4k}{\sqrt{1+k^2}}</math>
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Also, the distance between <math>B</math> and <math>LK</math> is <math>\frac{k\times \sqrt{13}-(4+\sqrt{13})\cdot k}{\sqrt{1+k^2}}=\frac{-4k}{\sqrt{1+k^2}}</math>
 
   
 
   
 
So the ares <math>S=0.5ah=\frac{-4k\sqrt{(16-8\sqrt{13})k^2-13}}{k^2+1}</math>
 
So the ares <math>S=0.5ah=\frac{-4k\sqrt{(16-8\sqrt{13})k^2-13}}{k^2+1}</math>
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So the answer is <math>104+26+13+3=\boxed{146}</math>.
 
So the answer is <math>104+26+13+3=\boxed{146}</math>.
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 +
==Solution 2==
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Draw <math>OC</math> perpendicular to <math>KL</math> at <math>C</math>. Draw <math>BD</math> perpendicular to <math>KL</math> at <math>D</math>.
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 +
<cmath>\frac{\triangle OKL}{\triangle BKL} = \frac{OC}{BD} = \frac{AO}{AB} = \frac{4+\sqrt{13}}{4}</cmath>
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 +
Therefore, to maximize area of <math>\triangle BKL</math>, we need to maximize area of <math>\triangle OKL</math>.
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 +
<cmath>\triangle OKL = \frac12 r^2 \sin{\angle KOL}</cmath>
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So when area of <math>\triangle OKL</math> is maximized, <math>\angle KOL = \frac{\pi}{2}</math>.
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 +
Eventually, we get <cmath>\triangle BKL=  (\frac12 \cdot \sqrt{13}^2)\cdot(\frac{4}{4+\sqrt{13}})=\frac{104-26\sqrt{13}}{3}</cmath>
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So the answer is <math>104+26+13+3=\boxed{146}</math>.
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So the answer is
  
 
==See Also==
 
==See Also==
 
{{AIME box|year=2013|n=II|num-b=9|num-a=11}}
 
{{AIME box|year=2013|n=II|num-b=9|num-a=11}}
 
{{MAA Notice}}
 
{{MAA Notice}}

Revision as of 08:56, 31 January 2014

Problem 10

Given a circle of radius $\sqrt{13}$, let $A$ be a point at a distance $4 + \sqrt{13}$ from the center $O$ of the circle. Let $B$ be the point on the circle nearest to point $A$. A line passing through the point $A$ intersects the circle at points $K$ and $L$. The maximum possible area for $\triangle BKL$ can be written in the form $\frac{a - b\sqrt{c}}{d}$, where $a$, $b$, $c$, and $d$ are positive integers, $a$ and $d$ are relatively prime, and $c$ is not divisible by the square of any prime. Find $a+b+c+d$.

Solution 1

Now we put the figure in the Cartesian plane, let the center of the circle $O (0,0)$, then $B (\sqrt{13},0)$, and $A(4+\sqrt{13},0)$

The equation for Circle O is $x^2+y^2=13$, and let the slope of the line$AKL$ be $k$, then the equation for line$AKL$ is $y=k(x-4-\sqrt{13})$

Then we get $(k^2+1)x^2-2k^2(4+\sqrt{13})x+k^2\cdot (4+\sqrt{13})^2-13=0$, according to Vieta's Formulas, we get

$x_1+x_2=\frac{2k^2(4+\sqrt{13})}{k^2+1}$, and $x_1x_2=\frac{(4+\sqrt{13})^2\cdot k^2-13}{k^2+1}$

So, $LK=\sqrt{1+k^2}\cdot \sqrt{(x_1+x_2)^2-4x_1x_2}$

Also, the distance between $B$ and $LK$ is $\frac{k\times \sqrt{13}-(4+\sqrt{13})\cdot k}{\sqrt{1+k^2}}=\frac{-4k}{\sqrt{1+k^2}}$

So the ares $S=0.5ah=\frac{-4k\sqrt{(16-8\sqrt{13})k^2-13}}{k^2+1}$

Then the maximum value of $S$ is $\frac{104-26\sqrt{13}}{3}$

So the answer is $104+26+13+3=\boxed{146}$.

Solution 2

Draw $OC$ perpendicular to $KL$ at $C$. Draw $BD$ perpendicular to $KL$ at $D$.

\[\frac{\triangle OKL}{\triangle BKL} = \frac{OC}{BD} = \frac{AO}{AB} = \frac{4+\sqrt{13}}{4}\]

Therefore, to maximize area of $\triangle BKL$, we need to maximize area of $\triangle OKL$.

\[\triangle OKL = \frac12 r^2 \sin{\angle KOL}\]

So when area of $\triangle OKL$ is maximized, $\angle KOL = \frac{\pi}{2}$.

Eventually, we get \[\triangle BKL=  (\frac12 \cdot \sqrt{13}^2)\cdot(\frac{4}{4+\sqrt{13}})=\frac{104-26\sqrt{13}}{3}\]

So the answer is $104+26+13+3=\boxed{146}$. So the answer is

See Also

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

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