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== | + | ==Solution (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> | ||
Revision as of 12:55, 25 September 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 (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
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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