Difference between revisions of "2013 AIME II Problems/Problem 10"
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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 | + | 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>. |
Revision as of 21:00, 4 June 2020
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
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 | ||
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