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

(Problem 10)
 
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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>.
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==Solution==
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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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The equation for Circle O is <math>x^2+y^2=13</math>, and let the slope of the line<math>AKL</math> be <math>k</math>, then the equation for line<math>AKL</math> is <math>y=k(x-4-\sqrt{13})</math>
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Then we get <math>(k^2+1)x^2-2k^2(4+\sqrt{13})x+k^2\cdot (4+\sqrt{13})^2-13=0</math>, according to Vieta's formulas, we get
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<math>x1+x2=\frac{2k^2(4+\sqrt{13})}{k^2+1}</math>, and <math>x1x2=\frac{(4+\sqrt{13})^2\cdot k^2-13}{k^2+1}
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So, </math>LK=\sqrt{1+k^2}\cdot \sqrt{(x1+x2)^2-4x1x2}<math>
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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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So the ares </math>S=0.5ah=\frac{-4k\sqrt{(16-8\sqrt{13})k^2-13}}{k^2+1}
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Then the maximum value of <math>S</math> is <math>\frac{104-26\sqrt{13}}{3}</math>
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So the answer is <math>104+26+13+3=\boxed{146}</math>

Revision as of 03:00, 5 April 2013

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

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

$x1+x2=\frac{2k^2(4+\sqrt{13})}{k^2+1}$, and $x1x2=\frac{(4+\sqrt{13})^2\cdot k^2-13}{k^2+1}

So,$ (Error compiling LaTeX. Unknown error_msg)LK=\sqrt{1+k^2}\cdot \sqrt{(x1+x2)^2-4x1x2}$Also, the distance between$O$and$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 $104+26+13+3=\boxed{146}$

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