# Difference between revisions of "2013 AIME II Problems/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>. | ||

+ | |||

+ | ==Solution== | ||

+ | 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> | ||

+ | |||

+ | 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> | ||

+ | |||

+ | 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} | ||

+ | |||

+ | 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> | ||

+ | |||

+ | 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> | ||

+ | |||

+ | So the answer is <math>104+26+13+3=\boxed{146}</math> |

## Revision as of 04:00, 5 April 2013

Given a circle of radius , let be a point at a distance from the center of the circle. Let be the point on the circle nearest to point . A line passing through the point intersects the circle at points and . The maximum possible area for can be written in the form , where , , , and are positive integers, and are relatively prime, and is not divisible by the square of any prime. Find .

## Solution

Now we put the figure in the Cartesian plane, let the center of the circle , then , and

The equation for Circle O is , and let the slope of the line be , then the equation for line is

Then we get , according to Vieta's formulas, we get

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

So,$ (Error compiling LaTeX. ! Missing $ inserted.)LK=\sqrt{1+k^2}\cdot \sqrt{(x1+x2)^2-4x1x2}OLK\frac{k\times \sqrt{13}-(4+\sqrt{13})\cdot k}{\sqrt{1+k^2}}=\frac{-4k}{\sqrt{1+k^2}}S=0.5ah=\frac{-4k\sqrt{(16-8\sqrt{13})k^2-13}}{k^2+1}

Then the maximum value of is

So the answer is