Art of Problem Solving
AoPS Online
Math texts, online classes, and more
for students in grades 5-12.
Visit AoPS Online ‚
Books for Grades 5-12 Online Courses
Beast Academy
Engaging math books and online learning
for students ages 6-13.
Visit Beast Academy ‚
Books for Ages 6-13 Beast Academy Online
AoPS Academy
Small live classes for advanced math
and language arts learners in grades 2-12.
Visit AoPS Academy ‚
Find a Physical Campus Visit the Virtual Campus

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

(→‎Solution 1)
m (→‎Solution 1 (Coordbash))
 
(21 intermediate revisions by 11 users not shown)
Line 1: Line 1:
−
==Problem 10==
+
==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==
+
==Solution 1 (Coordbash)==
 
<asy>
 
<asy>
 
import math;
 
import math;
Line 13: Line 14:
  
 
draw(Circle((0,0), sqrt(13)));
 
draw(Circle((0,0), sqrt(13)));
−
label("$O$", (0,0), S);label("$B$", B, W);label("$A$", A, S);
+
label("$O$", (0,0), S);label("$B$", B, SW);label("$A$", A, S);
 
dot((0,0));
 
dot((0,0));
  
Line 23: Line 24:
 
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>
 
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>
+
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
+
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>
 
<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>
Line 33: Line 34:
 
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>
 
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>
+
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>
 
Then the maximum value of <math>S</math> is <math>\frac{104-26\sqrt{13}}{3}</math>
Line 40: Line 41:
  
 
==Solution 2==
 
==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 <math>OC</math> perpendicular to <math>KL</math> at <math>C</math>. Draw <math>BD</math> perpendicular to <math>KL</math> at <math>D</math>.  
 
Draw <math>OC</math> perpendicular to <math>KL</math> at <math>C</math>. Draw <math>BD</math> perpendicular to <math>KL</math> at <math>D</math>.  
  
Line 50: 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=  (\frac12 \cdot \sqrt{13}^2)\cdot(\frac{4}{4+\sqrt{13}})=\frac{104-26\sqrt{13}}{3}</cmath>
+
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==
−
http://girlsangle.wordpress.com/2013/11/26/2013-aime-2-problem-10/
 
  
 
{{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 21:42, 15 June 2024

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 (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 $O (0,0)$, 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 (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 $N,M$ les on $AL$ such that $BM\bot AL, ON\bot AL$, call $BM=h, ON=k,LN=KN=d$ We call $\angle{LON}=\alpha$ By similar triangle, we have $\frac{h}{k}=\frac{4}{4+\sqrt{13}}, h=\frac{4k}{4+\sqrt{13}}$. Then, we realize the area is just $dh=d\cdot \frac{4K}{4+\sqrt{13}}$ As $\sin \alpha=\frac{d}{\sqrt{13}}, \cos \alpha=\frac{k}{\sqrt{13}}$. Now, we have to maximize $\frac{52\sin \alpha \cos \alpha}{4+\sqrt{13}}=\frac{26\sin 2\alpha}{4+\sqrt{13}}$, which is obviously reached when $\alpha=45^{\circ}$, the answer is $\frac{104-26\sqrt{13}}{3}$ leads to $\boxed{146}$

~bluesoul

Solution 5

AIME-II-2013-10.png

Let C and D be the base of perpendiculars dropped from points O and B to AK. Denote BD = h, OC = H. \[\triangle ABD \sim \triangle AOC \implies \frac {h}{H} = \frac {4}{4 + \sqrt{13}}.\] $KL$ is the base of triangles $\triangle OKL$ and $\triangle BKL \implies \frac {[BKL]}{[OKL]} = \frac{h}{H} =$ const $\implies$ The maximum possible area for $\triangle BKL$ and $\triangle OKL$ are at the same position of point $K$.

$\triangle OKL$ has sides $OK = OL = \sqrt{13}\implies \max[\triangle OKL] = \frac {OK^2}{2} = \frac {13}{2}$

in the case $\angle KOL = 90^\circ.$ It is possible – if we rotate such triangle, we can find position when $A$ lies on $KL.$ \[\max[\triangle BKL] = \max[\triangle OKL] \cdot \frac {4}{4+\sqrt{13}} = \frac {26}{4+\sqrt{13}} \implies \boxed{\textbf{146}}\] 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. AMC logo.png

Retrieved from "https://artofproblemsolving.com/wiki/index.php?title=2013_AIME_II_Problems/Problem_10&oldid=220591"