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Difference between revisions of "2013 AIME II Problems/Problem 10"
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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
 
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
  
−
<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}
+
<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}</math>
  
−
So, </math>LK=\sqrt{1+k^2}\cdot \sqrt{(x1+x2)^2-4x1x2}<math>
+
So, <math>LK=\sqrt{1+k^2}\cdot \sqrt{(x1+x2)^2-4x1x2}</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>
+
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>
 
   
 
   
−
So the ares </math>S=0.5ah=\frac{-4k\sqrt{(16-8\sqrt{13})k^2-13}}{k^2+1}
+
So the ares <math>S=0.5ah=\frac{-4k\sqrt{(16-8\sqrt{13})k^2-13}}{k^2+1}
  
−
Then the maximum value of <math>S</math> is <math>\frac{104-26\sqrt{13}}{3}</math>
+
Then the maximum value of </math>S<math> is </math>\frac{104-26\sqrt{13}}{3}<math>
  
−
So the answer is <math>104+26+13+3=\boxed{146}</math>
+
So the answer is </math>104+26+13+3=\boxed{146}$

Revision as of 03:01, 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, $LK=\sqrt{1+k^2}\cdot \sqrt{(x1+x2)^2-4x1x2}$

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$ (Error compiling LaTeX. Unknown error_msg)S$is$\frac{104-26\sqrt{13}}{3}$So the answer is$104+26+13+3=\boxed{146}$

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