Difference between revisions of "2013 AIME II Problems/Problem 10"
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==Solution 4== | ==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}}{ | + | 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 | ~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== |
Revision as of 11:41, 24 December 2022
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
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.