AIME 2018 I · 第 13 题

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

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

• 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

$$\begin{aligned} AI_{1}I_{2}&=\frac{1}{2}\cdot AI_{1}\cdot AI_{2}\cdot\sin\angle I_{1}AI_{2} \\ &=\frac{1}{2}\cdot AI_{1}\cdot AI_{2}\cdot\sin\left(\frac{1}{2}\angle A\right).\end{aligned}$$ 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

$$\begin{aligned}\frac{AI_{1}}{\sin\left(\frac{1}{2}\angle B\right)}=\frac{AB}{\sin\angle AI_{1}B}\Longrightarrow\frac{AB}{\sin\left(90^{\circ}+\alpha\right)}\Longrightarrow\frac{AB}{\cos\alpha} \\\frac{AI_{2}}{\sin\left(\frac{1}{2}\angle C\right)}=\frac{AC}{\sin\angle AI_{2}C}\Longrightarrow\frac{AC}{\sin\left(90^{\circ}+\theta\right)}\Longrightarrow\frac{AC}{\cos\theta}\Longrightarrow\frac{AC}{\sin\alpha}.\end{aligned}$$ 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

$$\begin{aligned}[AI_{1}I_{2}]&=\frac{1}{2}\cdot AI_{1}\cdot AI_{2}\cdot\sin\left(\frac{1}{2}\angle A\right) \\ &=\frac{AB\sin\left(\frac{1}{2}\angle B\right)\cdot AC\sin\left(\frac{1}{2}\angle C\right)\cdot\sin\left(\frac{1}{2}\angle A\right)}{2\sin\alpha\cos\alpha} \\ &=\frac{AB\sin\left(\frac{1}{2}\angle B\right)\cdot AC\sin\left(\frac{1}{2}\angle C\right)\cdot\sin\left(\frac{1}{2}\angle A\right)}{\sin(2\alpha)}.\end{aligned}$$ 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

$$\begin{aligned}\angle B&:\cos\angle B=\frac{30^{2}+32^{2}-34^{2}}{2\cdot 30\cdot 32}=\frac{2}{5} \\ \angle C&:\cos\angle C=\frac{32^{2}+34^{2}-30^{2}}{2\cdot 32\cdot 34}=\frac{10}{17} \\ \angle A&:\cos\angle A=\frac{30^{2}+34^{2}-32^{2}}{2\cdot 30\cdot 34}=\frac{43}{85}.\end{aligned}$$ Thus

$$\begin{aligned}\sin\left(\frac{1}{2}\angle B\right)=\sqrt{\frac{1-\tfrac{2}{5}}{2}}=\sqrt{\frac{3}{10}} \\ \sin\left(\frac{1}{2}\angle C\right)=\sqrt{\frac{1-\tfrac{10}{17}}{2}}=\sqrt{\frac{7}{34}} \\ \sin\left(\frac{1}{2}\angle A\right)=\sqrt{\frac{1-\tfrac{43}{85}}{2}}=\sqrt{\frac{21}{85}}.\end{aligned}$$ Thus the answer is

$$\begin{aligned} & 30\cdot 34\cdot\sqrt{\frac{3}{10}\cdot\frac{7}{34}\cdot\frac{21}{85}} \\ =&~30\cdot 34\cdot\sqrt{\frac{3}{2\cdot 5}\cdot\frac{7}{2\cdot 17}\cdot\frac{3\cdot 7}{5\cdot 17}} \\ =&~30\cdot 34\cdot\frac{3\cdot 7}{2\cdot 5\cdot 17} \\ =&~(2\cdot 3\cdot 5)\cdot(2\cdot 17)\cdot\frac{3\cdot 7}{2\cdot 5\cdot 17} \\ =&~2\cdot 3^{2}\cdot 7 \\ =&~\boxed{126}.\end{aligned}$$

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

Solution 5 (Barybash)

We use barycentric coordinates with $A=(1,\,0,\,0)$, $B=(0,\,1,\,0)$, $C=(0,\,0,\,1)$, $a:=BC$, $b:=CA$, $C:=AB$. Let $X=:(0,\,t,\,1-t)$ and $d:=AX$. Then $\overline{XA}=(1,\,-t,\,t-1)$ so

$$d^2=-a^2t(1-t)+b^2(1-t)+c^2t.$$ By the angle bisector theorem, the angle bisector of $\angle BAX$ intersects side $BC$ at

$$\frac{d}{c+d}B+\frac{c}{c+d}X=\left(0,\,\frac{d+ct}{c+d},\,\frac{c(1-t)}{c+d}\right).$$ Thus

$$I_1=(a(1-t):d+ct:c(1-t))=\left(\frac{a(1-t)}{a(1-t)+c+d},\,\frac{d+ct}{a(1-t)+c+d},\,\frac{c(1-t)}{a(1-t)+c+d}\right).$$ Similarly,

$$I_2=\left(\frac{at}{at+b+d},\,\frac{bt}{at+b+d},\,\frac{d+b(1-t)}{at+b+d}\right).$$ Hence, \begin{align*} \frac{[AI_1I_2]}{[ABC]}&= \begin{vmatrix} 1 & 0 & 0\\ \frac{a(1-t)}{a(1-t)+c+d} & \frac{d+ct}{a(1-t)+c+d} & \frac{c(1-t)}{a(1-t)+c+d}\\ \frac{at}{at+b+d} & \frac{bt}{at+b+d} & \frac{d+b(1-t)}{at+b+d} \end{vmatrix}\\ &=\frac{(d+ct)(d+b(1-t))-bct(1-t)}{(a(1-t)+c+d)(at+b+d)}\\ &=\frac{d^2+d(b(1-t)+ct)}{(a(1-t)+c+d)(at+b+d)}. \end{align*} The denominator equals \begin{align*} (a(1-t)+c+d)(at+b+d)&=a^2t(1-t)+ab(1-t)+ad(1-t)+act+bc+cd+adt+bd+d^2\\ &=a^2t(1-t)+ab(1-t)+ad+act+bc+cd+bd-a^2t(1-t)+b^2(1-t)+c^2t\\ &=b(a+b)(1-t)+c(a+c)t+bc+d(a+b+c). \end{align*} Note that $(b(1-t)+ct)(a+b+c)=b(a+b)(1-t)+c(a+c)t+bc$ so

$$\frac{[AI_1I_2]}{[ABC]}=\frac{d^2+d(b(1-t)+ct)}{(a(1-t)+c+d)(at+b+d)}=\frac{d}{a+b+c},$$ which is minimized when $AX$ is an altitude. By Heron, we get $[ABC]=96\sqrt{21}$ so $AX=6\sqrt{21}$. Thus $[AI_1I_2]=\frac{96\sqrt{21}\cdot 6\sqrt{21}}{30+32+34}=\boxed{126}$.

- KevinYang2.71

Solution 6 (Similar Triangles)

Draw in the incenter of $\triangle ABC$, $I$, as well as $AI$, $BI$, and $CI$. Through some angle chasing, $\angle I_1 X A = \angle I I_2 A$ and $\angle I_1 A X = \angle I A I_2$. Thus, $\frac{AI_1}{AX}=\frac{AX}{AI_2}$. The area of $\triangle AI_1 I_2$ can then be expressed as $\frac{1}{2}\cdot\overline{AI_1}\cdot\overline{AI_2}\cdot\sin{(\angle I_1 AI_2)}=\frac{1}{2}\cdot\overline{AX}\cdot\overline{AI}\cdot\sin{\angle I_1 AI_2}$. Since $\overline{AI}$ and $\sin{\angle I_1 AI_2}$ are constant, $\overline{AX}$ should be minimized. Thus, $\overline{AX}$ should be an altitude. Assuming so and noticing $\overline{AI}\cdot\sin{(\angle BAI)}=\overline{AI}\cdot\sin{(\angle I_1 AI_2)}=r$, $\frac{1}{2}\cdot\overline{AX}\cdot\overline{AI}\cdot\sin{(\angle I_1 AI_2)}=\frac{1}{2}\cdot \frac{2[ABC]}{\overline{BC}}\cdot r=\frac{[ABC]}{\overline{BC}}\cdot\frac{[ABC]}{s}=\frac{[ABC]^2}{s\cdot\overline{BC}}=\frac{s(s-32)(s-30)(s-34)}{32s}=\frac{(16)(18)(14)}{32}=\boxed{126}$.

- `\rlap{\Delta}{\rlap{\mathtt{7}}\!\!\!\!\!\mathtt{\text{ }\text{ }7}}` (first time writing a solution :D)

Video Solution by MOP 2024

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