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

Revision as of 18:41, 4 September 2021

Problem

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

$[asy] import math; import olympiad; import graph; pair A, B, K, L; B = (sqrt(13), 0); A=(4+sqrt(13), 0); dot(B); dot(A); draw(Circle((0,0), sqrt(13))); label("$O$", (0,0), S);label("$B$", B, SW);label("$A$", A, S); dot((0,0)); [/asy]$

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

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

$[asy] import math; import olympiad; import graph; pair A, B, K, L; B = (sqrt(13), 0); A=(4+sqrt(13), 0); dot(B); dot(A); draw(Circle((0,0), sqrt(13))); label("$O$", (0,0), S);label("$B$", B, SW);label("$A$", A, S); dot((0,0)); [/asy]$

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 $104+26+13+3=\boxed{146}$ .

Solution 3 (simplest solution)

A rather easier solution is presented in the Girls Angle WordPress:

http://girlsangle.wordpress.com/2013/11/26/2013-aime-2-problem-10/

@@ Line 71: / Line 71: @@
 So the answer is <math>104+26+13+3=\boxed{146}</math>.
+==Solution 3 (simplest solution)==
+A rather easier solution is presented in the Girls Angle WordPress:
+http://girlsangle.wordpress.com/2013/11/26/2013-aime-2-problem-10/
 ==See Also==
-http://girlsangle.wordpress.com/2013/11/26/2013-aime-2-problem-10/
 {{AIME box|year=2013|n=II|num-b=9|num-a=11}}
 {{MAA Notice}}

2013 AIME II (Problems • Answer Key • Resources)
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