Difference between revisions of "2013 AIME II Problems/Problem 10"
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==Solution 1== | ==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, W);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> | 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> | ||
Revision as of 20:01, 2 November 2014
Contents
Problem 10
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 .
See Also
http://girlsangle.wordpress.com/2013/11/26/2013-aime-2-problem-10/
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.