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 | + | ==Solution 2== |
<asy> | <asy> | ||
import math; | import math; | ||
Revision as of 20:01, 4 June 2020
Problem 10
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 2
![[asy] import math; import olympiad; import graph; pair A, B, K, L; B = (sqrt(13), 0); A=(4+sqrt(13), 0); dot(B); dot(A); draw(Circle((0,0), sqrt(13))); label("$O$", (0,0), S);label("$B$", B, SW);label("$A$", A, S); dot((0,0)); [/asy]](http://latex.artofproblemsolving.com/f/4/1/f419c223647108f3ec0e7539c755b1fef842ed2d.png)
Draw 
perpendicular to 
at 
. Draw 
perpendicular to at 
.
![\[\frac{\triangle OKL}{\triangle BKL} = \frac{OC}{BD} = \frac{AO}{AB} = \frac{4+\sqrt{13}}{4}\]](http://latex.artofproblemsolving.com/f/0/0/f00d16a122c17afcb2e71e4dc6dc03fda0210c17.png)
Therefore, to maximize area of , we need to maximize area of 
.
![\[\triangle OKL = \frac12 r^2 \sin{\angle KOL}\]](http://latex.artofproblemsolving.com/4/2/7/4273517b375629983d5a281e55038084fe3eb77b.png)
So when area of is maximized, 
.
Eventually, we get ![\[\triangle BKL= (\frac12 \cdot \sqrt{13}^2)\cdot(\frac{4}{4+\sqrt{13}})=\frac{104-26\sqrt{13}}{3}\]](http://latex.artofproblemsolving.com/0/0/d/00dbbb05a48deb9830cc84f09f6fb5762d9d0fe9.png)
So the answer is 
.
See Also
http://girlsangle.wordpress.com/2013/11/26/2013-aime-2-problem-10/
| 2013 AIME II (Problems • Answer Key • Resources) | ||
| 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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