Difference between revisions of "2013 AIME II Problems/Problem 10"
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− | ==Problem | + | ==Problem== |
+ | |||
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 1 (Coordbash)== | ||
+ | <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 <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>x_1+x_2=\frac{2k^2(4+\sqrt{13})}{k^2+1}</math>, and <math>x_1x_2=\frac{(4+\sqrt{13})^2\cdot k^2-13}{k^2+1}</math> | ||
+ | |||
+ | So, <math>LK=\sqrt{1+k^2}\cdot \sqrt{(x_1+x_2)^2-4x_1x_2}</math> | ||
+ | |||
+ | Also, the distance between <math>B</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 area <math>S=0.5ah=\frac{-4k\sqrt{(16+8\sqrt{13})k^2-13}}{k^2+1}</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>. | ||
==Solution 2== | ==Solution 2== | ||
Line 30: | Line 68: | ||
So when area of <math>\triangle OKL</math> is maximized, <math>\angle KOL = \frac{\pi}{2}</math>. | So when area of <math>\triangle OKL</math> is maximized, <math>\angle KOL = \frac{\pi}{2}</math>. | ||
− | Eventually, we get <cmath>\triangle BKL= | + | Eventually, we get <cmath>\triangle BKL= \frac12 \cdot (\sqrt{13})^2\cdot(\frac{4}{4+\sqrt{13}})=\frac{104-26\sqrt{13}}{3}</cmath> |
So the answer is <math>104+26+13+3=\boxed{146}</math>. | So the answer is <math>104+26+13+3=\boxed{146}</math>. | ||
+ | |||
+ | ==Solution 3 (simpler solution)== | ||
+ | A rather easier solution is presented in the Girls' Angle WordPress: | ||
+ | |||
+ | http://girlsangle.wordpress.com/2013/11/26/2013-aime-2-problem-10/ | ||
+ | |||
+ | |||
+ | ==Solution 4== | ||
+ | |||
+ | Let <math>N,M</math> les on <math>AL</math> such that <math>BM\bot AL, ON\bot AL</math>, call <math>BM=h, ON=k,LN=KN=d</math> We call <math>\angle{LON}=\alpha</math> By similar triangle, we have <math>\frac{h}{k}=\frac{4}{4+\sqrt{13}}, h=\frac{4k}{4+\sqrt{13}}</math>. Then, we realize the area is just <math>dh=d\cdot \frac{4K}{4+\sqrt{13}}</math> As <math>\sin \alpha=\frac{d}{\sqrt{13}}, \cos \alpha=\frac{k}{\sqrt{13}}</math>. Now, we have to maximize <math>\frac{52\sin \alpha \cos \alpha}{4+\sqrt{13}}=\frac{26\sin 2\alpha}{4+\sqrt{13}}</math>, which is obviously reached when <math>\alpha=45^{\circ}</math>, the answer is <math>\frac{104-26\sqrt{13}}{3}</math> leads to <math>\boxed{146}</math> | ||
+ | |||
+ | ~bluesoul | ||
+ | |||
+ | ==Solution 5== | ||
+ | [[File:AIME-II-2013-10.png|400px|right]] | ||
+ | Let C and D be the base of perpendiculars dropped from points O and B to AK. Denote BD = h, OC = H. | ||
+ | <cmath>\triangle ABD \sim \triangle AOC \implies \frac {h}{H} = \frac {4}{4 + \sqrt{13}}.</cmath> | ||
+ | <math>KL</math> is the base of triangles <math>\triangle OKL</math> and <math>\triangle BKL \implies \frac {[BKL]}{[OKL]} = \frac{h}{H} =</math> const <math>\implies</math> | ||
+ | The maximum possible area for <math>\triangle BKL</math> and <math>\triangle OKL</math> are at the same position of point <math>K</math>. | ||
+ | |||
+ | <math>\triangle OKL</math> has sides <math>OK = OL = \sqrt{13}\implies \max[\triangle OKL] = \frac {OK^2}{2} = \frac {13}{2}</math> | ||
+ | |||
+ | in the case <math>\angle KOL = 90^\circ.</math> It is possible – if we rotate such triangle, we can find position when <math>A</math> lies on <math>KL.</math> | ||
+ | <cmath>\max[\triangle BKL] = \max[\triangle OKL] \cdot \frac {4}{4+\sqrt{13}} = \frac {26}{4+\sqrt{13}} \implies \boxed{\textbf{146}}</cmath> | ||
+ | '''vladimir.shelomovskii@gmail.com, vvsss''' | ||
==See Also== | ==See Also== | ||
− | |||
{{AIME box|year=2013|n=II|num-b=9|num-a=11}} | {{AIME box|year=2013|n=II|num-b=9|num-a=11}} | ||
{{MAA Notice}} | {{MAA Notice}} |
Latest revision as of 22:42, 15 June 2024
Contents
Problem
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 1 (Coordbash)
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
So,
Also, the distance between and
is
So the area
Then the maximum value of is
So the answer is .
Solution 2
Draw perpendicular to
at
. Draw
perpendicular to
at
.
Therefore, to maximize area of , we need to maximize area of
.
So when area of is maximized,
.
Eventually, we get
So the answer is .
Solution 3 (simpler solution)
A rather easier solution is presented in the Girls' Angle WordPress:
http://girlsangle.wordpress.com/2013/11/26/2013-aime-2-problem-10/
Solution 4
Let les on
such that
, call
We call
By similar triangle, we have
. Then, we realize the area is just
As
. Now, we have to maximize
, which is obviously reached when
, the answer is
leads to
~bluesoul
Solution 5
Let C and D be the base of perpendiculars dropped from points O and B to AK. Denote BD = h, OC = H.
is the base of triangles
and
const
The maximum possible area for
and
are at the same position of point
.
has sides
in the case It is possible – if we rotate such triangle, we can find position when
lies on
vladimir.shelomovskii@gmail.com, vvsss
See Also
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 |
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.