2018 AIME I Problems/Problem 13
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[hide]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
- Notice that we truly did minimize the area for because are all constants while only is variable, so maximizing would minimize the area.
Solution 2 (Similar to Official MAA)
It's clear that . Thus By the Law of Sines on , Similarly, It is well known that Denote and , with . Thus and Thus so We intend to minimize this expression, which is equivalent to maximizing , and that occurs when , or . Ergo, is the foot of the altitude from to . In that case, we intend to compute Recall that and similarly for angles and . Applying the Law of Cosines to each angle of gives Thus Thus the answer is
Solution 3 (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
-Azeem H.(Mathislife52) ~edited by phoenixfire
Video Solution by Osman Nal
https://www.youtube.com/watch?v=sT-wxV2rYqs
Solution 4 (Geometry only)
Let be semiperimeter of be the height of dropped from
Let be inradius of the and respectively.
Using the Lemma (below), we get the area Lemma
Proof WLOG if and only if
Claim Proof
Let
We use Cosine Law for and and get Last is evident, the claim has been proven.
vladimir.shelomovskii@gmail.com, vvsss
Solution 4a
Geometry proof of the equation
Using diagrams, we can recall known facts and using those facts for making sequence of equations.
The twice area of is
Therefore
vladimir.shelomovskii@gmail.com, vvsss
Video Solution by MOP 2024
https://youtube.com/watch?v=ALzZA13PuZk
See Also
2018 AIME I (Problems • Answer Key • Resources) | ||
Preceded by Problem 12 |
Followed by Problem 14 | |
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