2019 AIME I Problems/Problem 13
Contents
Problem
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 1
Notice that By the Law of Cosines,
Then,
Let
,
, and
. Then,
However, since
,
, but since
,
and the requested sum is
.
(Solution by TheUltimate123)
Solution 2
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
.
Solution 3
Construct and let
. Let
. Using
,
Using
, it can be found that
This also means that
. It suffices to find
. It is easy to see the following:
Using reverse Law of Cosines on
,
. Using Law of Cosines on
gives
, so
.
-franchester
Solution 4 (No <C = <DFE, no LoC)
Let . Let
and
; from
we have
and
. From
we have
giving
. So
and
. These similar triangles also gives us
so
. Now, Stewart's Theorem on
and cevian
tells us that
so
. Then
so the answer is
as desired. (Solution by Trumpeter, but not added to the Wiki by Trumpeter)
Solution 5
Connect meeting
at
. We can observe that
Getting that
. We can also observe that
, getting that
Assume that , since
, we can get that
, getting that
Using Power of Point, we can get that Assume that
, getting that
Now applying Law of Cosine on two triangles, separately, we can get two equations
Since , we can use
to eliminate the
term
Then we can get that , getting
, so the desired answer is
, which leads to the answer
~bluesoul
Solution 6
Nice problem!
First, let and
intersect at
. Our motivation here is to introduce cyclic quadrilaterals and find useful relationships in terms of angles. Observe that
By the so-called "Reverse Law of Cosines" on
we have
Applying on
gives
So
, now by our cyclic quadrilaterals again, we are motivated by the multiple appearances of similar triangles throughout the figure. We want some that are related to
and
, which are crucial lengths in the problem. Suppose
for simplicity. We have:
So
So
. The requested sum is
.
~CoolJupiter
See Also
2019 AIME I (Problems • Answer Key • Resources) | ||
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Followed by Problem 14 | |
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