Difference between revisions of "2018 AIME I Problems/Problem 13"
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First, instead of using angles to find <math>[AI_1I_2]</math>, let's try to find the area of other, simpler figures, and subtract that from <math>[ABC]</math>. However, to do this, we need to be able to figure out the length of the inradii, and so, we need to find <math>AX</math>. | First, instead of using angles to find <math>[AI_1I_2]</math>, let's try to find the area of other, simpler figures, and subtract that from <math>[ABC]</math>. However, to do this, we need to be able to figure out the length of the inradii, and so, we need to find <math>AX</math>. |
Revision as of 23:32, 18 July 2021
Contents
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 lengthier, but less trigonometric approach)
First, instead of using angles to find , let's try to find the area of other, 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
. (Proof needed here).
We need to minimize . Let
,
, and
. After an application of Stewart's Theorem, we will get that
To minimize this quadratic,
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 which is equal to
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
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 ~edited by phoenixfire
Solution 3 (Cheap, if you are running out of time or if you are confused)
What 2 kinds of cevians would yield clean values for angles BAX and CAX? The angle bisector and the altitude! However, the angle bisector clearly seems dirtier, and angle BXA is not clean then. One can just test the altitude out then. The area of triangle ABC, by Heron's formula, is ; one can easily find the altitude length as
, with BX=12 and CX=20. The area of ABC can also tell us that our answer should be a multiple of 21 most likely. One can find AI1 by keeping in mind that in a right triangle, the Incenter, the right angle vertex, and the two contact points of the incircle adjacent to the right angle vertex form a square. This means that one can use the pythagorean theorem to get AI1^2 = r^2 + (3*sqrt(21)+9) using the well known theorems (involving s-a, s-b, s-c) about the length of the tangent segments to the incircle in a triangle. r=s-a = 3*sqrt(21)-9 due to the square, so one can get AI1 = sqrt(540). One can perform analagous calculations to obtain that AI2 = sqrt(476). Sin(A/2) = sqrt(21/85) by the decently well known sin(a/2) = sqrt((s-b)*(s-c)/bc) formula. One can easily use the sine area formula then; the cancellation is almost like magic, giving a clean multiple of 21; 126! This must be the answer, so mark it and move on.
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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