Difference between revisions of "2019 AIME II Problems/Problem 15"
Thebeast5520 (talk | contribs) (→Solution) |
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Then | Then | ||
− | <math>AB | + | <math>AB\cdot AC= \frac{a}{k} \frac{b}{k} = \frac{ab}{\frac{ab}{u} \cdot\frac{ab}{u} } = \frac{u^2}{ab} |
− | = \frac{1400 | + | = \frac{1400 \cdot 1400}{ \sqrt{ 1000\cdot 875 }} = 560 \sqrt{14}</math> |
So the final answer is <math>560 + 14 = \boxed{574} </math> | So the final answer is <math>560 + 14 = \boxed{574} </math> |
Revision as of 19:14, 22 June 2019
Problem
In acute triangle points
and
are the feet of the perpendiculars from
to
and from
to
, respectively. Line
intersects the circumcircle of
in two distinct points,
and
. Suppose
,
, and
. The value of
can be written in the form
where
and
are positive integers, and
is not divisible by the square of any prime. Find
.
Solution
Let
Therefore
By power of point, we have
Which are simplified to
Or
(1)
Or
Let
Then,
In triangle , by law of cosine
Pluging (1)
Or
Substitute everything by
The quadratic term is cancelled out after simplified
Which gives
Plug back in,
Then
So the final answer is
By SpecialBeing2017
See Also
2019 AIME II (Problems • Answer Key • Resources) | ||
Preceded by Problem 14 |
Followed by Last Question | |
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.