2019 AIME I Problems/Problem 11
Contents
Problem 11
In , the sides have integers lengths and
. Circle
has its center at the incenter of
. An excircle of
is a circle in the exterior of
that is tangent to one side of the triangle and tangent to the extensions of the other two sides. Suppose that the excircle tangent to
is internally tangent to
, and the other two excircles are both externally tangent to
. Find the minimum possible value of the perimeter of
.
Solution 1
Let the tangent circle be . Some notation first: let
,
,
be the semiperimeter,
, and
be the inradius. Intuition tells us that the radius of
is
(using the exradius formula). However, the sum of the radius of
and
is equivalent to the distance between the incenter and the the
excenter. Denote the B excenter as
and the incenter as
.
Lemma:
We draw the circumcircle of
. Let the angle bisector of
hit the circumcircle at a second point
. By the incenter-excenter lemma,
. Let this distance be
. Ptolemy's theorem on
gives us
Again, by the incenter-excenter lemma,
so
as desired.
Using this gives us the following equation:
Motivated by the
and
, we make the following substitution:
This changes things quite a bit. Here's what we can get from it:
It is known (easily proved with Heron's and a=rs) that
Using this, we can also find
: let the midpoint of
be
. Using Pythagorean's Theorem on
,
We now look at the RHS of the main equation:
Cancelling some terms, we have
Squaring,
Expanding and moving terms around gives
Reverse substituting,
Clearly the smallest solution is
and
, so our answer is
-franchester
Solution 2
First assume that
and
, and scale up later. Notice that
. Then, the height from
is
, so if
, we know
. Then, if
denotes the
-exradius for
and
denotes the semiperimeter,
Then, if
denotes the tangency point between the
-excircle and
, it is known that
, so
. Furthermore,
. Then,
It follows that
whence
. Then, since
, the smallest possible perimeter is
.
(Solution by TheUltimate123)
See Also
2019 AIME I (Problems • Answer Key • Resources) | ||
Preceded by Problem 10 |
Followed by Problem 12 | |
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