Difference between revisions of "2018 AIME I Problems/Problem 13"
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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> | ||
+ | *Notice that we truly did minimize the area for <math>[A I_1 I_2]</math> because <math>b, c, \angle A, \angle B, \angle C</math> are all constants while only <math>\sin \alpha</math> is variable, so maximizing <math>\sin \alpha</math> would minimize the area. | ||
− | ==Solution 2 (A lengthier, but less trigonometric approach)== | + | ==Solution 2 (Similar to Official MAA)== |
+ | It's clear that <math>\angle I_{1}AI_{2}=\frac{1}{2}\angle BAX+\frac{1}{2}\angle CAX=\frac{1}{2}\angle A</math>. Thus <cmath>\begin{align*} AI_{1}I_{2}&=\frac{1}{2}\cdot AI_{1}\cdot AI_{2}\cdot\sin\angle I_{1}AI_{2} \\ &=\frac{1}{2}\cdot AI_{1}\cdot AI_{2}\cdot\sin\left(\frac{1}{2}\angle A\right).\end{align*}</cmath> By the Law of Sines on <math>\triangle AI_{1}B</math>, <cmath>\frac{AI_{1}}{\sin\left(\frac{1}{2}\angle B\right)}=\frac{AB}{\sin\angle AI_{1}B}.</cmath> Similarly, <cmath>\frac{AI_{2}}{\sin\left(\frac{1}{2}\angle C\right)}=\frac{AC}{\sin\angle AI_{2}C}.</cmath> It is well known that <cmath>\angle AI_{1}B=90+\frac{1}{2}\angle AXB~~~\text{and}~~~\angle AI_{2}C=90+\frac{1}{2}\angle AXC.</cmath> Denote <math>\alpha=\frac{1}{2}\angle AXB</math> and <math>\theta=\frac{1}{2}\angle AXC</math>, with <math>\alpha+\theta=90^{\circ}</math>. Thus <math>\sin\alpha=\cos\theta</math> and <cmath>\begin{align*}\frac{AI_{1}}{\sin\left(\frac{1}{2}\angle B\right)}=\frac{AB}{\sin\angle AI_{1}B}\Longrightarrow\frac{AB}{\sin\left(90^{\circ}+\alpha\right)}\Longrightarrow\frac{AB}{\cos\alpha} \\\frac{AI_{2}}{\sin\left(\frac{1}{2}\angle C\right)}=\frac{AC}{\sin\angle AI_{2}C}\Longrightarrow\frac{AC}{\sin\left(90^{\circ}+\theta\right)}\Longrightarrow\frac{AC}{\cos\theta}\Longrightarrow\frac{AC}{\sin\alpha}.\end{align*}</cmath> Thus <cmath>AI_{1}=\frac{AB\sin\left(\frac{1}{2}\angle B\right)}{\cos\alpha}~~~\text{and}~~~AI_{2}=\frac{AC\sin\left(\frac{1}{2}\angle C\right)}{\sin\alpha}</cmath> so <cmath>\begin{align*}[AI_{1}I_{2}]&=\frac{1}{2}\cdot AI_{1}\cdot AI_{2}\cdot\sin\left(\frac{1}{2}\angle A\right) \\ &=\frac{AB\sin\left(\frac{1}{2}\angle B\right)\cdot AC\sin\left(\frac{1}{2}\angle C\right)\cdot\sin\left(\frac{1}{2}\angle A\right)}{2\sin\alpha\cos\alpha} \\ &=\frac{AB\sin\left(\frac{1}{2}\angle B\right)\cdot AC\sin\left(\frac{1}{2}\angle C\right)\cdot\sin\left(\frac{1}{2}\angle A\right)}{\sin(2\alpha)}.\end{align*}</cmath> We intend to minimize this expression, which is equivalent to maximizing <math>\sin(2\alpha)</math>, and that occurs when <math>\alpha=45^{\circ}</math>, or <math>\angle AXB=90^{\circ}</math>. Ergo, <math>X</math> is the foot of the altitude from <math>A</math> to <math>\overline{BC}</math>. In that case, we intend to compute <cmath>AB\cdot AC\cdot\sin\left(\frac{1}{2}\angle B\right)\cdot\sin\left(\frac{1}{2}\angle C\right)\cdot\sin\left(\frac{1}{2}\angle A\right).</cmath> Recall that <cmath>\sin\left(\frac{1}{2}\angle B\right)=\sqrt{\frac{1-\cos\angle B}{2}}</cmath> and similarly for angles <math>C</math> and <math>A</math>. Applying the Law of Cosines to each angle of <math>\triangle ABC</math> gives <cmath>\begin{align*}\angle B&:\cos\angle B=\frac{30^{2}+32^{2}-34^{2}}{2\cdot 30\cdot 32}=\frac{2}{5} \\ \angle C&:\cos\angle C=\frac{32^{2}+34^{2}-30^{2}}{2\cdot 32\cdot 34}=\frac{10}{17} \\ \angle A&:\cos\angle A=\frac{30^{2}+34^{2}-32^{2}}{2\cdot 30\cdot 34}=\frac{43}{85}.\end{align*}</cmath> Thus <cmath>\begin{align*}\sin\left(\frac{1}{2}\angle B\right)=\sqrt{\frac{1-\tfrac{2}{5}}{2}}=\sqrt{\frac{3}{10}} \\ \sin\left(\frac{1}{2}\angle C\right)=\sqrt{\frac{1-\tfrac{10}{17}}{2}}=\sqrt{\frac{7}{34}} \\ \sin\left(\frac{1}{2}\angle A\right)=\sqrt{\frac{1-\tfrac{43}{85}}{2}}=\sqrt{\frac{21}{85}}.\end{align*}</cmath> Thus the answer is <cmath>\begin{align*} & 30\cdot 34\cdot\sqrt{\frac{3}{10}\cdot\frac{7}{34}\cdot\frac{21}{85}} \\ =&~30\cdot 34\cdot\sqrt{\frac{3}{2\cdot 5}\cdot\frac{7}{2\cdot 17}\cdot\frac{3\cdot 7}{5\cdot 17}} \\ =&~30\cdot 34\cdot\frac{3\cdot 7}{2\cdot 5\cdot 17} \\ =&~(2\cdot 3\cdot 5)\cdot(2\cdot 17)\cdot\frac{3\cdot 7}{2\cdot 5\cdot 17} \\ =&~2\cdot 3^{2}\cdot 7 \\ =&~\boxed{126}.\end{align*}</cmath> | ||
+ | |||
+ | |||
+ | ==Solution 3 (A lengthier, but less trigonometric approach)== | ||
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>. | ||
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https://www.youtube.com/watch?v=sT-wxV2rYqs | https://www.youtube.com/watch?v=sT-wxV2rYqs | ||
− | ==Solution | + | ==Solution 4 (Geometry only)== |
[[File:2018 AIME I 13.png|400px|right]] | [[File:2018 AIME I 13.png|400px|right]] | ||
Let <math>BC = a, s</math> be semiperimeter of <math>\triangle ABC, s = 48, h</math> be the height of <math>\triangle ABC</math> dropped from <math>A.</math> | Let <math>BC = a, s</math> be semiperimeter of <math>\triangle ABC, s = 48, h</math> be the height of <math>\triangle ABC</math> dropped from <math>A.</math> | ||
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Last is evident, the claim has been proven. | Last is evident, the claim has been proven. | ||
− | ''' | + | '''vladimir.shelomovskii@gmail.com, vvsss''' |
− | ==Solution | + | ==Solution 4a== |
[[File:2018 AIME I 13e.png|350px|right]] | [[File:2018 AIME I 13e.png|350px|right]] | ||
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Using diagrams, we can recall known facts and using those facts for making sequence of equations. | Using diagrams, we can recall known facts and using those facts for making sequence of equations. | ||
− | + | <cmath>\frac {(r-r_1)\cdot (r-r_2)}{r_1 \cdot r_2} =\tan\beta \tan\gamma.</cmath> | |
The twice area of <math>\triangle ABC</math> is | The twice area of <math>\triangle ABC</math> is | ||
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Therefore <cmath>r_1 + r_2 - \frac{2 r_1 r_2 }{h} = r.</cmath> | Therefore <cmath>r_1 + r_2 - \frac{2 r_1 r_2 }{h} = r.</cmath> | ||
− | [[File:2018 AIME I 13d.png| | + | |
− | ''' | + | [[File:2018 AIME I 13g.png|400px|right]] |
+ | [[File:2018 AIME I 13d.png|400px]] | ||
+ | [[File:2018 AIME I 13f.png|400px]] | ||
+ | |||
+ | '''vladimir.shelomovskii@gmail.com, vvsss''' | ||
+ | |||
+ | ==Video Solution by MOP 2024== | ||
+ | https://youtube.com/watch?v=ALzZA13PuZk | ||
==See Also== | ==See Also== |
Latest revision as of 21:36, 6 January 2024
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
- 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 | |
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 | ||
All AIME Problems and Solutions |
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.