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

Revision as of 15:51, 26 June 2022

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 lengthier, but less trigonometric approach)

First, instead of using angles to find $[AI_1I_2]$ , let's try to find the area of other, 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 $[AI_1I_2]$ . (Proof needed here).

We need to minimize $AX$ . 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, $s=12$ whereby we conclude that $d=6\sqrt{21}$ .

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 which is equal to $\frac{(9+3\sqrt{21})(r_1)}{2} + \frac{(7+3\sqrt{21})(r_2)}{2}.$

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 $[ABC]-\left(\frac{(35+3\sqrt{21})(r_1)}{2}+\frac{(45+3r_2\sqrt{21})(r_2)}{2}\right).$

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

Doing so gets us that the minimum area of $AI_1I_2=\boxed{126}.$

-Azeem H.(Mathislife52) ~edited by phoenixfire

Video Solution by Osman Nal

https://www.youtube.com/watch?v=sT-wxV2rYqs

Solution 3 (Geometry only)

Let $BC = a, s$ be semiperimeter of $\triangle ABC, s = 48, h$ be the height of $\triangle ABC$ dropped from $A.$

Let $r, r_1, r_2$ be inradius of the $\triangle ABC, \triangle ABX,$ and $\triangle ACX,$ respectively.

Using the Lemma (below), we get the area $[ AI_1 I_2] = \frac{AX \cdot r }{2} \ge \frac { hr}{2} =\frac{ r^2 s}{a},$ $[ AI_1 I_2]= \frac{(s-a)(s-b)(s-c)}{a} = \frac{18 \cdot 16 \cdot 14}{32} = 9 \cdot 14 = \boxed {126}.$ Lemma

$[AI_1 I_2] = \frac{AX \cdot r }{2}$

Proof $\angle AXI_1 = \angle BXI_1, \angle AXI_2 = \angle CXI_2, \angle BXC = 180^\circ \implies \angle I_1 X I_2 = 90^\circ.$ WLOG $\hspace{70mm} \sin \angle AXB = \frac {h}{AX}.$ $[AI_1 X] =\frac{ AX\cdot r_1}{2}, \hspace{20mm} [AI_2 X] =\frac{ AX\cdot r_2}{2},\hspace{20mm}[XI_1 I_2] =\frac{XI_1 \cdot XI_2}{2},$ $XI_1 \cdot XI_2 = \frac{r_1}{\sin \angle AXI_1} \cdot \frac{r_2}{\sin \angle AXI_2} = \frac{2 r_1 r_2}{\sin \angle AXB} = \frac{2 AX \cdot r_1 \cdot r_2 }{h}.$ $\hspace{20mm} [AI_1 I_2] = [AI_1 X] + [AI_2 X] - [XI_1 I_2] = AX (r_1 + r_2 - \frac{2 r_1 r_2 }{h}) = \frac{AX \cdot r }{2}$ if and only if

Claim $r_1 + r_2 - \frac{2 r_1 r_2 }{h} = r.$ Proof

Let $\hspace{30mm} AX = t, AC = b, AB = c, x_1 = BX, x_2 = CX, \frac{c+t}{x_1} = u, \frac{b+t}{x_2} = v.$ $r_1 + r_2 - \frac{2 r_1 r_2 }{h} = r \iff \frac{h}{r_2} + \frac{h}{r_1}- 2 = \frac{h}{r_2} \frac{h}{r_1} \frac{r}{h}.$

$2[ABX] = r_1 (c + t + x_1) = h x_1 \implies \frac{h}{r_1} = 1 + u,$ $2[ACX] = r_2 (b + t + x_2) = h x_2 \implies \frac{h}{r_2} = 1 + v,$ $\frac{h}{r_2} + \frac{h}{r_1}- 2 = \frac{h}{r_2} \frac{h}{r_1} \frac{r}{h} \iff (u+v)(a+b+c)=(u+1)(v+1)a \iff (u+v)(b+c)=(uv+1)a,$ $c^2 x_2 + b^2 x_1 = t^2 x_1 + t^2 x_2 + x_1^2 x_2 + x_1 x_2^2,$ $(b^2 -t^2 -x_2^2)x_1 + (c^2 –t^2 -x_1^2)x_2 = 0,$ We use Cosine Law for $\triangle ABX$ and $\triangle ACX$ and get $2 t x_1 x_2 \cos \angle AXB + 2 t x_1 x_2 \cos \angle AXC = 0 \iff \cos \angle AXB + \cos \angle AXC = 0.$ Last is evident, the claim has been proven.

Shelomovskii, vvsss, www.deoma-cmd.ru

Solution 3a

Geometry proof of the equation $r_1 + r_2 - \frac{2 r_1 r_2 }{h} = r.$ $\frac{r^2}{r_1 r_2}-\frac{r}{r_1} -\frac{r}{r_2} +1 = 1 - \frac{2r}{h} \implies \frac {(r-r_1)\cdot (r-r_2)}{r_1 \cdot r_2} = 1-\frac{2r}{h}.$

Using diagrams, we can recall known facts and using those facts for making sequence of equations.

$\frac {(r-r_1)\cdot (r-r_2)}{r_1 \cdot r_2} =\tan\beta \tan\gamma.$

The twice area of $\triangle ABC$ is $r(a+b+c) = ha = r_a (b+c-a)\implies$

$1 -\frac{2r}{h} = \frac {b+c-a}{b+c+a} = \frac {r}{r_a} = \tan\beta \tan\gamma .$

Therefore $r_1 + r_2 - \frac{2 r_1 r_2 }{h} = r.$

Shelomovskii, vvsss, www.deoma-cmd.ru

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 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
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 [[File:2018 AIME I 13g.png|350px|right]]
-[[File:2018 AIME I 13d.png|350px]]
+[[File:2018 AIME I 13d.png|400px]]
-[[File:2018 AIME I 13f.png|300px]]
+[[File:2018 AIME I 13f.png|350px]]
 '''Shelomovskii, vvsss,  www.deoma-cmd.ru'''

2018 AIME I (Problems • Answer Key • Resources)
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