2018 AIME I Problems/Problem 13
Problem
Let have side lengths
,
, and
. Point
lies in the interior of
, and points
and
are the incenters of
and
, respectively. Find the minimum possible area of
as
varies along
.
Solution 1 (Official MAA)
First note that is a constant not depending on
, so by
it suffices to minimize
. Let
,
,
, and
. Remark that
Applying the Law of Sines to
gives
Analogously one can derive
, and so
with equality when
, that is, when
is the foot of the perpendicular from
to
. In this case the desired area is
. To make this feasible to compute, note that
Applying similar logic to
and
and simplifying yields a final answer of
Solution 2 (A more elegant approach)
First, instead of using angles to find , let's try to find the area of other, more simpler figures, and subtract that from
. However, to do this, we need to be able to figure out the length of the inradii, and so, we need to find
. To minimize
, intuitively, we should try to minimize the length of
, since, after using the
formula for the area of a triangle, we'll be able to minimize the inradii lengths, and thus, eventually minimize the area of
(I did not come up with a solid proof to back this intuition, if any does, don't hesitate to add a note at the bottom).
To minimize , let's use Stewart's Theorem. Let
,
, and
. After an application of Stewart's Theorem, we will get that
To minimize this quadratic, we need to let
whereby we conclude that
.
From here, draw perpendiculars down from and
to
and
respectively, and label the foot of these perpendiculars
and
respectively. After, draw the inradii from
to
, and from
to
, and draw in
. Label the foot of the inradii to
and
,
and
, respectively. From here, we see that to find
, we need to find
, and subtract off the sum of
,
, and
.
can be found by finding the area of two quadrilaterals (
+
) as well as the area of a trapezoid (
). If we let the inradius of
be
and if we let the inradius of
be
, we'll find, after an application of basic geometry and careful calculations on paper, that
.
The area of two triangles can be found in a similar fashion, however, we must use substitution to solve for
as well as
. After doing this, we'll get a similar sum in terms of
and
for the area of those two triangles (I'm not very good at LaTeX, so I can't typeset the area, but the area of the two triangles is equal to (9+3sqrt(21)(r_1)/2 + (7+3sqrt(21)(r_2)/2).
Now we're set. Summing up the area of the Hexagon and the two triangles and simplifying, we get that the formula for is just [ABC]-(35+3sqrt(21)(r_1)/2 -(45+3sqrt(21)(r_2)/2.
Using Heron's formula, . Solving for
and
using Heron's in
and
, we get that
and
. From here, we just have to plug into our above equation and solve. Doing so gets us that the minimum area of
.
-Mathislife52
See Also
2018 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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