2018 AIME I Problems/Problem 13

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 (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, ater 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 $XYZ$ substituion 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 $AI_1I_2=126$ .

2018 AIME I (Problems • Answer Key • Resources)
Preceded by Problem 12	Followed by Problem 14
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2018 AIME I Problems/Problem 13

Contents

Problem

Solution (Official MAA)

Solution 2 (A more elegant approach)

See Also