Difference between revisions of "2018 AIME I Problems/Problem 13"
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Let <math>\triangle ABC</math> have side lengths <math>AB=30</math>, <math>BC=32</math>, and <math>AC=34</math>. Point <math>X</math> lies in the interior of <math>\overline{BC}</math>, and points <math>I_1</math> and <math>I_2</math> are the incenters of <math>\triangle ABX</math> and <math>\triangle ACX</math>, respectively. Find the minimum possible area of <math>\triangle AI_1I_2</math> as <math>X</math> varies along <math>\overline{BC}</math>. | Let <math>\triangle ABC</math> have side lengths <math>AB=30</math>, <math>BC=32</math>, and <math>AC=34</math>. Point <math>X</math> lies in the interior of <math>\overline{BC}</math>, and points <math>I_1</math> and <math>I_2</math> are the incenters of <math>\triangle ABX</math> and <math>\triangle ACX</math>, respectively. Find the minimum possible area of <math>\triangle AI_1I_2</math> as <math>X</math> varies along <math>\overline{BC}</math>. | ||
− | ==Solution (Official MAA)== | + | ==Solution 1 (Official MAA)== |
First note that <cmath>\angle I_1AI_2 = \angle I_1AX + \angle XAI_2 = \frac{\angle BAX}2 + \frac{\angle CAX}2 = \frac{\angle A}2</cmath> is a constant not depending on <math>X</math>, so by <math>[AI_1I_2] = \tfrac12(AI_1)(AI_2)\sin\angle I_1AI_2</math> it suffices to minimize <math>(AI_1)(AI_2)</math>. Let <math>a = BC</math>, <math>b = AC</math>, <math>c = AB</math>, and <math>\alpha = \angle AXB</math>. Remark that <cmath>\angle AI_1B = 180^\circ - (\angle I_1AB + \angle I_1BA) = 180^\circ - \tfrac12(180^\circ - \alpha) = 90^\circ + \tfrac\alpha 2.</cmath> Applying the Law of Sines to <math>\triangle ABI_1</math> gives <cmath>\frac{AI_1}{AB} = \frac{\sin\angle ABI_1}{\sin\angle AI_1B}\qquad\Rightarrow\qquad AI_1 = \frac{c\sin\frac B2}{\cos\frac\alpha 2}.</cmath> Analogously one can derive <math>AI_2 = \tfrac{b\sin\frac C2}{\sin\frac\alpha 2}</math>, and so <cmath>[AI_1I_2] = \frac{bc\sin\frac A2 \sin\frac B2\sin\frac C2}{2\cos\frac\alpha 2\sin\frac\alpha 2} = \frac{bc\sin\frac A2 \sin\frac B2\sin\frac C2}{\sin\alpha}\geq bc\sin\frac A2 \sin\frac B2\sin\frac C2,</cmath> with equality when <math>\alpha = 90^\circ</math>, that is, when <math>X</math> is the foot of the perpendicular from <math>A</math> to <math>\overline{BC}</math>. In this case the desired area is <math>bc\sin\tfrac A2\sin\tfrac B2\sin\tfrac C2</math>. To make this feasible to compute, note that <cmath>\sin\frac A2=\sqrt{\frac{1-\cos A}2}=\sqrt{\frac{1-\frac{b^2+c^2-a^2}{2bc}}2} = \sqrt{\dfrac{(a-b+c)(a+b-c)}{4bc}}.</cmath> Applying similar logic to <math>\sin \tfrac B2</math> and <math>\sin\tfrac C2</math> and simplifying yields a final answer of <cmath>\begin{align*}bc\sin\frac A2\sin\frac B2\sin\frac C2&=bc\cdot\dfrac{(a-b+c)(b-c+a)(c-a+b)}{8abc}\\&=\dfrac{(30-32+34)(32-34+30)(34-30+32)}{8\cdot 32}=\boxed{126}.\end{align*}</cmath> | First note that <cmath>\angle I_1AI_2 = \angle I_1AX + \angle XAI_2 = \frac{\angle BAX}2 + \frac{\angle CAX}2 = \frac{\angle A}2</cmath> is a constant not depending on <math>X</math>, so by <math>[AI_1I_2] = \tfrac12(AI_1)(AI_2)\sin\angle I_1AI_2</math> it suffices to minimize <math>(AI_1)(AI_2)</math>. Let <math>a = BC</math>, <math>b = AC</math>, <math>c = AB</math>, and <math>\alpha = \angle AXB</math>. Remark that <cmath>\angle AI_1B = 180^\circ - (\angle I_1AB + \angle I_1BA) = 180^\circ - \tfrac12(180^\circ - \alpha) = 90^\circ + \tfrac\alpha 2.</cmath> Applying the Law of Sines to <math>\triangle ABI_1</math> gives <cmath>\frac{AI_1}{AB} = \frac{\sin\angle ABI_1}{\sin\angle AI_1B}\qquad\Rightarrow\qquad AI_1 = \frac{c\sin\frac B2}{\cos\frac\alpha 2}.</cmath> Analogously one can derive <math>AI_2 = \tfrac{b\sin\frac C2}{\sin\frac\alpha 2}</math>, and so <cmath>[AI_1I_2] = \frac{bc\sin\frac A2 \sin\frac B2\sin\frac C2}{2\cos\frac\alpha 2\sin\frac\alpha 2} = \frac{bc\sin\frac A2 \sin\frac B2\sin\frac C2}{\sin\alpha}\geq bc\sin\frac A2 \sin\frac B2\sin\frac C2,</cmath> with equality when <math>\alpha = 90^\circ</math>, that is, when <math>X</math> is the foot of the perpendicular from <math>A</math> to <math>\overline{BC}</math>. In this case the desired area is <math>bc\sin\tfrac A2\sin\tfrac B2\sin\tfrac C2</math>. To make this feasible to compute, note that <cmath>\sin\frac A2=\sqrt{\frac{1-\cos A}2}=\sqrt{\frac{1-\frac{b^2+c^2-a^2}{2bc}}2} = \sqrt{\dfrac{(a-b+c)(a+b-c)}{4bc}}.</cmath> Applying similar logic to <math>\sin \tfrac B2</math> and <math>\sin\tfrac C2</math> and simplifying yields a final answer of <cmath>\begin{align*}bc\sin\frac A2\sin\frac B2\sin\frac C2&=bc\cdot\dfrac{(a-b+c)(b-c+a)(c-a+b)}{8abc}\\&=\dfrac{(30-32+34)(32-34+30)(34-30+32)}{8\cdot 32}=\boxed{126}.\end{align*}</cmath> | ||
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==See Also== | ==See Also== | ||
{{AIME box|year=2018|n=I|num-b=12|num-a=14}} | {{AIME box|year=2018|n=I|num-b=12|num-a=14}} | ||
+ | |||
+ | [[Category:Intermediate Geometry Problems]] | ||
{{MAA Notice}} | {{MAA Notice}} |
Revision as of 17:15, 7 October 2020
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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