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 |
Revision as of 22:48, 28 August 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
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.