Art of Problem Solving
AoPS Online
Math texts, online classes, and more
for students in grades 5-12.
Visit AoPS Online ‚
Books for Grades 5-12 Online Courses
Beast Academy
Engaging math books and online learning
for students ages 6-13.
Visit Beast Academy ‚
Books for Ages 6-13 Beast Academy Online
AoPS Academy
Small live classes for advanced math
and language arts learners in grades 2-12.
Visit AoPS Academy ‚
Find a Physical Campus Visit the Virtual Campus

Difference between revisions of "2018 AIME I Problems/Problem 13"

(→‎Solution 2 (A more elegant approach))
Line 2: Line 2:
 
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>
  
Line 26: Line 26:
 
==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 $\triangle ABC$ have side lengths $AB=30$, $BC=32$, and $AC=34$. Point $X$ lies in the interior of $\overline{BC}$, and points $I_1$ and $I_2$ are the incenters of $\triangle ABX$ and $\triangle ACX$, respectively. Find the minimum possible area of $\triangle AI_1I_2$ as $X$ varies along $\overline{BC}$.

Solution 1 (Official MAA)

First note that \[\angle I_1AI_2 = \angle I_1AX + \angle XAI_2 = \frac{\angle BAX}2 + \frac{\angle CAX}2 = \frac{\angle A}2\] is a constant not depending on $X$, so by $[AI_1I_2] = \tfrac12(AI_1)(AI_2)\sin\angle I_1AI_2$ it suffices to minimize $(AI_1)(AI_2)$. Let $a = BC$, $b = AC$, $c = AB$, and $\alpha = \angle AXB$. Remark that \[\angle AI_1B = 180^\circ - (\angle I_1AB + \angle I_1BA) = 180^\circ - \tfrac12(180^\circ - \alpha) = 90^\circ + \tfrac\alpha 2.\] Applying the Law of Sines to $\triangle ABI_1$ gives \[\frac{AI_1}{AB} = \frac{\sin\angle ABI_1}{\sin\angle AI_1B}\qquad\Rightarrow\qquad AI_1 = \frac{c\sin\frac B2}{\cos\frac\alpha 2}.\] Analogously one can derive $AI_2 = \tfrac{b\sin\frac C2}{\sin\frac\alpha 2}$, and so \[[AI_1I_2] = \frac{bc\sin\frac A2 \sin\frac B2\sin\frac C2}{2\cos\frac\alpha 2\sin\frac\alpha 2} = \frac{bc\sin\frac A2 \sin\frac B2\sin\frac C2}{\sin\alpha}\geq bc\sin\frac A2 \sin\frac B2\sin\frac C2,\] with equality when $\alpha = 90^\circ$, that is, when $X$ is the foot of the perpendicular from $A$ to $\overline{BC}$. In this case the desired area is $bc\sin\tfrac A2\sin\tfrac B2\sin\tfrac C2$. To make this feasible to compute, note that \[\sin\frac A2=\sqrt{\frac{1-\cos A}2}=\sqrt{\frac{1-\frac{b^2+c^2-a^2}{2bc}}2} = \sqrt{\dfrac{(a-b+c)(a+b-c)}{4bc}}.\] Applying similar logic to $\sin \tfrac B2$ and $\sin\tfrac C2$ and simplifying yields a final answer of \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*}


Solution 2 (A more elegant approach)

First, instead of using angles to find $[AI_1I_2]$, let's try to find the area of other, more simpler figures, and subtract that from $[ABC]$. However, to do this, we need to be able to figure out the length of the inradii, and so, we need to find $AX$. To minimize $[AI_1I_2]$, intuitively, we should try to minimize the length of $AX$, since, after using the $rs=A$ formula for the area of a triangle, we'll be able to minimize the inradii lengths, and thus, eventually minimize the area of $[AI_1I_2]$ (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 $AX$, let's use Stewart's Theorem. Let $AX=d$, $BX=s$, and $CX=32-s$. After an application of Stewart's Theorem, we will get that \[d=sqrt(s^2-24s+900)\] To minimize this quadratic, we need to let $s=12$ whereby we conclude that $d=6sqrt{21}$.

From here, draw perpendiculars down from $I_1$ and $I_2$ to $AB$ and $AC$ respectively, and label the foot of these perpendiculars $D$ and $E$ respectively. After, draw the inradii from $I_1$ to $BX$, and from $I_2$ to $CX$, and draw in $I_1I_2$. Label the foot of the inradii to $BX$ and $CX$, $F$ and $G$, respectively. From here, we see that to find $[AI_1I_2]$, we need to find $[ABC]$, and subtract off the sum of $[DBCEI_2I_1]$, $[ADI_1]$, and $[AEI_2]$.

$[DBCEI_2I_1]$ can be found by finding the area of two quadrilaterals ($[DBFI_1]$+$[ECGI_2]$) as well as the area of a trapezoid ($[FGI_2I_1]$). If we let the inradius of $ABX$ be $r_1$ and if we let the inradius of $ACX$ be $r_2$, we'll find, after an application of basic geometry and careful calculations on paper, that $[DBCEI_2I_1]=13r_1+19r_2$.

The area of two triangles can be found in a similar fashion, however, we must use $XYZ$ substitution to solve for $AD$ as well as $AE$. After doing this, we'll get a similar sum in terms of $r_1$ and $r_2$ 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 $[AI_1I_2]$ is just [ABC]-(35+3sqrt(21)(r_1)/2 -(45+3sqrt(21)(r_2)/2.

Using Heron's formula, $[ABC]=96sqrt{21}$. Solving for $r_1$ and $r_2$ using Heron's in $ABX$ and $ACX$, we get that $r_1=3sqrt{21}-9$ and $r_2=3sqrt{21}-7$. From here, we just have to plug into our above equation and solve. Doing so gets us that the minimum area of $AI_1I_2=126$.

-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

These problems are copyrighted © by the Mathematical Association of America, as part of the American Mathematics Competitions. AMC Logo.png

Retrieved from "https://artofproblemsolving.com/wiki/index.php?title=2018_AIME_I_Problems/Problem_13&oldid=134763"