Difference between revisions of "2019 AIME I Problems/Problem 13"
(Solution is incomplete. I just need to add the part about using similar triangles to find GF and BG) |
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<math>4 \cdot GD = BG \cdot 4\sqrt{2}</math> | <math>4 \cdot GD = BG \cdot 4\sqrt{2}</math> | ||
− | <math>\frac{ | + | <math>GD = BG \cdot \sqrt{2}</math> |
+ | |||
+ | Note that <math>\triangle GAC</math> is similar to <math>\triangle GFD</math>. <math>GF = \frac{BG + 4}{3}</math>. Also note that <math>\triangle GBC</math> is similar to <math>\triangle GFE</math>, which gives us <math>GF = \frac{7 \cdot BG}{5}</math>. Solving this system of linear equations, we get <math>BG = \frac{5}{4}</math>. Now, we can solve for <math>BE</math>, which is equal to <math>BG(\sqrt{2} + 1) + 4\sqrt{2}</math>. This simplifies to <math>\frac{5 + 21\sqrt{2}}{4}</math>, which means our answer is <math>\boxed{032}</math>. | ||
==See Also== | ==See Also== | ||
{{AIME box|year=2019|n=I|num-b=12|num-a=14}} | {{AIME box|year=2019|n=I|num-b=12|num-a=14}} | ||
{{MAA Notice}} | {{MAA Notice}} |
Revision as of 06:32, 15 March 2019
Problem 13
Triangle has side lengths
,
, and
. Points
and
are on ray
with
. The point
is a point of intersection of the circumcircles of
and
satisfying
and
. Then
can be expressed as
, where
,
,
, and
are positive integers such that
and
are relatively prime, and
is not divisible by the square of any prime. Find
.
Solution
Define to be the circumcircle of
and
to be the circumcircle of
.
Because of exterior angles,
But because
is cyclic. In addition,
because
is cyclic. Therefore,
. But
, so
. Using Law of Cosines on
, we can figure out that
. Since
,
. We are given that
and
, so we can use Law of Cosines on
to find that
.
Let be the intersection of segment
and
. Using Power of a Point with respect to
within
, we find that
. We can also apply Power of a Point with respect to
within
to find that
. Therefore,
.
Note that is similar to
.
. Also note that
is similar to
, which gives us
. Solving this system of linear equations, we get
. Now, we can solve for
, which is equal to
. This simplifies to
, which means our answer is
.
See Also
2019 AIME I (Problems • Answer Key • Resources) | ||
Preceded by Problem 12 |
Followed by Problem 14 | |
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 | ||
All AIME Problems and Solutions |
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