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

Latest revision as of 20:36, 6 January 2024

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 $\begin{align*}bc\sin\frac A2\sin\frac B2\sin\frac C2&=bc\cdot\dfrac{(a-b+c)(b-c+a)(c-a+b)}{8abc}\\&=\dfrac{(30-32+34)(32-34+30)(34-30+32)}{8\cdot 32}=\boxed{126}.\end{align*}$

Notice that we truly did minimize the area for $[A I_1 I_2]$ because $b, c, \angle A, \angle B, \angle C$ are all constants while only $\sin \alpha$ is variable, so maximizing $\sin \alpha$ would minimize the area.

Solution 2 (Similar to Official MAA)

It's clear that $\angle I_{1}AI_{2}=\frac{1}{2}\angle BAX+\frac{1}{2}\angle CAX=\frac{1}{2}\angle A$ . Thus $\begin{align*} AI_{1}I_{2}&=\frac{1}{2}\cdot AI_{1}\cdot AI_{2}\cdot\sin\angle I_{1}AI_{2} \\ &=\frac{1}{2}\cdot AI_{1}\cdot AI_{2}\cdot\sin\left(\frac{1}{2}\angle A\right).\end{align*}$ By the Law of Sines on $\triangle AI_{1}B$ , $\frac{AI_{1}}{\sin\left(\frac{1}{2}\angle B\right)}=\frac{AB}{\sin\angle AI_{1}B}.$ Similarly, $\frac{AI_{2}}{\sin\left(\frac{1}{2}\angle C\right)}=\frac{AC}{\sin\angle AI_{2}C}.$ It is well known that $\angle AI_{1}B=90+\frac{1}{2}\angle AXB~~~\text{and}~~~\angle AI_{2}C=90+\frac{1}{2}\angle AXC.$ Denote $\alpha=\frac{1}{2}\angle AXB$ and $\theta=\frac{1}{2}\angle AXC$ , with $\alpha+\theta=90^{\circ}$ . Thus $\sin\alpha=\cos\theta$ and $\begin{align*}\frac{AI_{1}}{\sin\left(\frac{1}{2}\angle B\right)}=\frac{AB}{\sin\angle AI_{1}B}\Longrightarrow\frac{AB}{\sin\left(90^{\circ}+\alpha\right)}\Longrightarrow\frac{AB}{\cos\alpha} \\\frac{AI_{2}}{\sin\left(\frac{1}{2}\angle C\right)}=\frac{AC}{\sin\angle AI_{2}C}\Longrightarrow\frac{AC}{\sin\left(90^{\circ}+\theta\right)}\Longrightarrow\frac{AC}{\cos\theta}\Longrightarrow\frac{AC}{\sin\alpha}.\end{align*}$ Thus $AI_{1}=\frac{AB\sin\left(\frac{1}{2}\angle B\right)}{\cos\alpha}~~~\text{and}~~~AI_{2}=\frac{AC\sin\left(\frac{1}{2}\angle C\right)}{\sin\alpha}$ so $\begin{align*}[AI_{1}I_{2}]&=\frac{1}{2}\cdot AI_{1}\cdot AI_{2}\cdot\sin\left(\frac{1}{2}\angle A\right) \\ &=\frac{AB\sin\left(\frac{1}{2}\angle B\right)\cdot AC\sin\left(\frac{1}{2}\angle C\right)\cdot\sin\left(\frac{1}{2}\angle A\right)}{2\sin\alpha\cos\alpha} \\ &=\frac{AB\sin\left(\frac{1}{2}\angle B\right)\cdot AC\sin\left(\frac{1}{2}\angle C\right)\cdot\sin\left(\frac{1}{2}\angle A\right)}{\sin(2\alpha)}.\end{align*}$ We intend to minimize this expression, which is equivalent to maximizing $\sin(2\alpha)$ , and that occurs when $\alpha=45^{\circ}$ , or $\angle AXB=90^{\circ}$ . Ergo, $X$ is the foot of the altitude from $A$ to $\overline{BC}$ . In that case, we intend to compute $AB\cdot AC\cdot\sin\left(\frac{1}{2}\angle B\right)\cdot\sin\left(\frac{1}{2}\angle C\right)\cdot\sin\left(\frac{1}{2}\angle A\right).$ Recall that $\sin\left(\frac{1}{2}\angle B\right)=\sqrt{\frac{1-\cos\angle B}{2}}$ and similarly for angles $C$ and $A$ . Applying the Law of Cosines to each angle of $\triangle ABC$ gives $\begin{align*}\angle B&:\cos\angle B=\frac{30^{2}+32^{2}-34^{2}}{2\cdot 30\cdot 32}=\frac{2}{5} \\ \angle C&:\cos\angle C=\frac{32^{2}+34^{2}-30^{2}}{2\cdot 32\cdot 34}=\frac{10}{17} \\ \angle A&:\cos\angle A=\frac{30^{2}+34^{2}-32^{2}}{2\cdot 30\cdot 34}=\frac{43}{85}.\end{align*}$ Thus $\begin{align*}\sin\left(\frac{1}{2}\angle B\right)=\sqrt{\frac{1-\tfrac{2}{5}}{2}}=\sqrt{\frac{3}{10}} \\ \sin\left(\frac{1}{2}\angle C\right)=\sqrt{\frac{1-\tfrac{10}{17}}{2}}=\sqrt{\frac{7}{34}} \\ \sin\left(\frac{1}{2}\angle A\right)=\sqrt{\frac{1-\tfrac{43}{85}}{2}}=\sqrt{\frac{21}{85}}.\end{align*}$ Thus the answer is $\begin{align*} & 30\cdot 34\cdot\sqrt{\frac{3}{10}\cdot\frac{7}{34}\cdot\frac{21}{85}} \\ =&~30\cdot 34\cdot\sqrt{\frac{3}{2\cdot 5}\cdot\frac{7}{2\cdot 17}\cdot\frac{3\cdot 7}{5\cdot 17}} \\ =&~30\cdot 34\cdot\frac{3\cdot 7}{2\cdot 5\cdot 17} \\ =&~(2\cdot 3\cdot 5)\cdot(2\cdot 17)\cdot\frac{3\cdot 7}{2\cdot 5\cdot 17} \\ =&~2\cdot 3^{2}\cdot 7 \\ =&~\boxed{126}.\end{align*}$

Solution 3 (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 4 (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.

vladimir.shelomovskii@gmail.com, vvsss

Solution 4a

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

vladimir.shelomovskii@gmail.com, vvsss

Video Solution by MOP 2024

https://youtube.com/watch?v=ALzZA13PuZk

@@ Line 5: / Line 5: @@
 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 more elegant, but lengthy, 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>.
@@ Line 35: / Line 40: @@
 Doing so gets us that the minimum area of <math>AI_1I_2=\boxed{126}.</math>
--Mathislife52 ~edited by phoenixfire
+-Azeem H.(Mathislife52) ~edited by phoenixfire
-==Solution 3 (Cheap, if you are running out of time or if you are confused)==
-What 2 kinds of cevians would yield clean values for angles BAX and CAX? The angle bisector and the altitude! However, the angle bisector clearly seems dirtier, and angle BXA is not clean then. One can just test the altitude out then. The area of triangle ABC, by Heron's formula, is <math>96\sqrt{21}</math>; one can easily find the altitude length as <math>6\sqrt{21}</math>, with BX=12 and CX=20. The area of ABC can also tell us that our answer should be a multiple of 21 most likely. One can find AI1 by keeping in mind that in a right triangle, the Incenter, the right angle vertex, and the two contact points of the incircle adjacent to the right angle vertex form a square. This means that one can use the pythagorean theorem to get AI1^2 = r^2 + (3*sqrt(21)+9) using the well known theorems (involving s-a, s-b, s-c) about the length of the tangent segments to the incircle in a triangle. r=s-a = 3*sqrt(21)-9 due to the square, so one can get AI1 = sqrt(540). One can perform analagous calculations to obtain that AI2 = sqrt(476). Sin(A/2) = sqrt(21/85) by the decently well known sin(a/2) = sqrt((s-b)*(s-c)/bc) formula. One can easily use the sine area formula then; the cancellation is almost like magic, giving a clean multiple of 21; 126! This must be the answer, so mark it and move on.
+==Video Solution by Osman Nal==
+https://www.youtube.com/watch?v=sT-wxV2rYqs
+==Solution 4 (Geometry only)==
+[[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>r, r_1, r_2</math> be inradius of the <math>\triangle ABC, \triangle ABX,</math> and <math>\triangle ACX,</math> respectively.
+Using the Lemma (below), we get the area
+<cmath>[ AI_1 I_2] = \frac{AX \cdot r }{2} \ge \frac { hr}{2} =\frac{ r^2 s}{a},</cmath>
+<cmath> [ AI_1 I_2]=  \frac{(s-a)(s-b)(s-c)}{a} =   \frac{18 \cdot 16 \cdot 14}{32} = 9 \cdot 14 = \boxed {126}. </cmath>
+<i><b>Lemma </b></i>
+<cmath>[AI_1 I_2] = \frac{AX \cdot r }{2}</cmath>
+<i><b>Proof </b></i>
+<cmath>\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.</cmath>
+WLOG <math>\hspace{70mm} \sin \angle AXB = \frac {h}{AX}.</math>
+<cmath>[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},</cmath>
+<cmath>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}.</cmath>
+<math> \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}</math> if and only if
+<i><b>Claim </b></i>
+<cmath>r_1 + r_2 -  \frac{2 r_1 r_2 }{h} = r.</cmath>
+<i><b>Proof </b></i>
+Let <math> \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.</math>
+<cmath>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}.</cmath>
+<cmath>2[ABX] = r_1 (c + t + x_1) = h x_1 \implies \frac{h}{r_1} = 1 + u,</cmath>
+<cmath>2[ACX] = r_2 (b + t + x_2) = h x_2 \implies \frac{h}{r_2} = 1 + v,</cmath>
+<cmath>\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,</cmath>
+<cmath>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,</cmath>
+<cmath>(b^2 -t^2 -x_2^2)x_1 + (c^2 –t^2 -x_1^2)x_2 = 0,</cmath>
+We use Cosine Law for <math>\triangle ABX</math> and <math>\triangle ACX</math> and get
+<cmath>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.</cmath>
+Last is evident, the claim has been proven.
+'''vladimir.shelomovskii@gmail.com, vvsss'''
+==Solution 4a==
+[[File:2018 AIME I 13e.png|350px|right]]
+[[File:2018 AIME I 13c.png|350px|right]]
+Geometry proof of the equation <math>r_1 + r_2 -  \frac{2 r_1 r_2 }{h} = r.</math>
+<math>\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}.</math>
+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
+<math>r(a+b+c) = ha = r_a (b+c-a)\implies </math>
+<cmath>1 -\frac{2r}{h} = \frac {b+c-a}{b+c+a} = \frac {r}{r_a} = \tan\beta \tan\gamma .</cmath>
+Therefore <cmath>r_1 + r_2 -  \frac{2 r_1 r_2 }{h} = r.</cmath>
+[[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==

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