Difference between revisions of "2018 AIME I Problems/Problem 13"
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==Solution 2 (A more elegant, but lengthy, approach)== | ==Solution 2 (A more elegant, but lengthy, approach)== | ||
− | First, instead of using angles to find <math>[AI_1I_2]</math>, let's try to find the area of other, | + | 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>. To minimize <math>[AI_1I_2]</math>, intuitively, we should try to minimize the length of <math>AX</math>, since, after using the <math>rs=A</math> formula for the area of a triangle, we'll be able to minimize the inradii lengths, and thus, eventually minimize the area of <math>[AI_1I_2]</math> (Proof needed here). |
− | To minimize <math>AX</math>, | + | To minimize <math>AX</math>, use Stewart's Theorem. Let <math>AX=d</math>, <math>BX=s</math>, and <math>CX=32-s</math>. After an application of Stewart's Theorem, we will get that <cmath>d=\sqrt{s^2-24s+900}</cmath> To minimize this quadratic, we need to let <math>s=12</math> whereby we conclude that <math>d=6\sqrt{21}</math>. |
− | From here, draw perpendiculars down from <math>I_1</math> and <math>I_2</math> to <math>AB</math> and <math>AC</math> respectively, and label the foot of these perpendiculars <math>D</math> and <math>E</math> respectively. After, draw the inradii from <math>I_1</math> to <math>BX</math>, and from <math>I_2</math> to <math>CX</math>, and draw in <math>I_1I_2 | + | From here, draw perpendiculars down from <math>I_1</math> and <math>I_2</math> to <math>AB</math> and <math>AC</math> respectively, and label the foot of these perpendiculars <math>D</math> and <math>E</math> respectively. After, draw the inradii from <math>I_1</math> to <math>BX</math>, and from <math>I_2</math> to <math>CX</math>, and draw in <math>I_1I_2</math>. |
− | <math> | + | Label the foot of the inradii to <math>BX</math> and <math>CX</math>, <math>F</math> and <math>G</math>, respectively. From here, we see that to find <math>[AI_1I_2]</math>, we need to find <math>[ABC]</math>, and subtract off the sum of <math>[DBCEI_2I_1], [ADI_1],</math> and <math>[AEI_2]</math>. |
− | + | <math>[DBCEI_2I_1]</math> can be found by finding the area of two quadrilaterals <math>[DBFI_1]+[ECGI_2]</math> as well as the area of a trapezoid <math>[FGI_2I_1]</math>. If we let the inradius of <math>ABX</math> be <math>r_1</math> and if we let the inradius of <math>ACX</math> be <math>r_2</math>, we'll find, after an application of basic geometry and careful calculations on paper, that <math>[DBCEI_2I_1]=13r_1+19r_2</math>. | |
− | + | The area of two triangles can be found in a similar fashion, however, we must use <math>XYZ</math> substitution to solve for <math>AD</math> as well as <math>AE</math>. After doing this, we'll get a similar sum in terms of <math>r_1</math> and <math>r_2</math> for the area of those two triangles which is equal to <math>\frac{(9+3\sqrt{21})(r_1)}{2} + \frac{(7+3\sqrt{21})(r_2)}{2}</math>. | |
− | + | Now we're set. Summing up the area of the Hexagon and the two triangles and simplifying, we get that the formula for <math>[AI_1I_2]</math> is just <math>[ABC]-\left(\frac{(35+3\sqrt{21})(r_1)}{2}+\frac{(45+3r_2\sqrt{21})(r_2)}{2}\right)</math>. | |
− | -Mathislife52 | + | Using Heron's formula, <math>[ABC]=96\sqrt{21}</math>. Solving for <math>r_1</math> and <math>r_2</math> using Heron's in <math>ABX</math> and <math>ACX</math>, we get that <math>r_1=3\sqrt{21}-9</math> and <math>r_2=3\sqrt{21}-7</math>. From here, we just have to plug into our above equation and solve. Doing so gets us that the minimum area of <math>AI_1I_2=\boxed{126}</math>. |
+ | |||
+ | -Mathislife52 ~edited by phoenixfire | ||
==See Also== | ==See Also== |
Revision as of 03:08, 9 November 2020
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 more elegant, but lengthy, 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).
To minimize , 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 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
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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