2019 AIME I Problems/Problem 11

Problem 11

In $\triangle ABC$ , the sides have integers lengths and $AB=AC$ . Circle $\omega$ has its center at the incenter of $\triangle ABC$ . An excircle of $\triangle ABC$ is a circle in the exterior of $\triangle ABC$ that is tangent to one side of the triangle and tangent to the extensions of the other two sides. Suppose that the excircle tangent to $\overline{BC}$ is internally tangent to $\omega$ , and the other two excircles are both externally tangent to $\omega$ . Find the minimum possible value of the perimeter of $\triangle ABC$ .

Solution 1

Let the tangent circle be $\omega$ . Some notation first: let $BC=a$ , $AB=b$ , $s$ be the semiperimeter, $\theta=\angle ABC$ , and $r$ be the inradius. Intuition tells us that the radius of $\omega$ is $r+\frac{2rs}{s-a}$ (using the exradius formula). However, the sum of the radius of $\omega$ and $\frac{rs}{s-b}$ is equivalent to the distance between the incenter and the the $B/C$ excenter. Denote the B excenter as $I_B$ and the incenter as $I$ . Lemma: $I_BI=\frac{2b*IB}{a}$ We draw the circumcircle of $\triangle ABC$ . Let the angle bisector of $\angle ABC$ hit the circumcircle at a second point $M$ . By the incenter-excenter lemma, $AM=CM=IM$ . Let this distance be $\alpha$ . Ptolemy's theorem on $ABCM$ gives us $a\alpha+b\alpha=b(\alpha+IB)\to \alpha=\frac{b*IB}{a}$ Again, by the incenter-excenter lemma, $II_B=2IM$ so $II_b=\frac{2b*IB}{a}$ as desired. Using this gives us the following equation: $\frac{2b*IB}{a}=r+\frac{2rs}{s-a}+\frac{rs}{s-b}$ Motivated by the $s-a$ and $s-b$ , we make the following substitution: $x=s-a, y=s-b$ This changes things quite a bit. Here's what we can get from it: $a=2y, b=x+y, s=x+2y$ It is known (easily proved with Heron's and a=rs) that $r=\sqrt{\frac{(s-a)(s-b)(s-b)}{s}}=\sqrt{\frac{xy^2}{x+2y}}$ Using this, we can also find $IB$ : let the midpoint of $BC$ be $N$ . Using Pythagorean's Theorem on $\triangle INB$ , $IB^2=r^2+(\frac{a}{2})^2=\frac{xy^2}{x+2y}+y^2=\frac{2xy^2+2y^3}{x+2y}=\frac{2y^2(x+y)}{x+2y}$ We now look at the RHS of the main equation: $r+\frac{2rs}{s-a}+\frac{rs}{s-b}=r(1+\frac{2(x+2y)}{x}+\frac{x+2y}{y})=r(\frac{x^2+5xy+4y^2}{xy})=\frac{r(x+4y)(x+y)}{xy}=\frac{2(x+y)IB}{2y}$ Cancelling some terms, we have $\frac{r(x+4y)}{x}=IB$ Squaring, $\frac{2y^2(x+y)}{x+2y}=\frac{(x+4y)^2*xy^2}{x^2(x+2y)}\to \frac{(x+4y)^2}{x}=2(x+y)$ Expanding and moving terms around gives $(x-8y)(x+2y)=0\to x=8y$ Reverse substituting, $s-a=8s-8b\to b=\frac{9}{2}a$ Clearly the smallest solution is $a=2$ and $b=9$ , so our answer is $2+9+9=\boxed{020}$ -franchester

Solution 2

$[asy] size(8cm); defaultpen(fontsize(8pt)); pair A, B, C, I, IA, IB, IC; A=(0, 4sqrt(5)); B=(-1, 0); C=(1, 0); I=incenter(A, B, C); IA=2*circumcenter(I,B,C)-I; IB=2*circumcenter(I,C,A)-I; IC=2*circumcenter(I,A,B)-I; draw(B -- A -- C); draw(IB -- IC); draw(incircle(A, B, C)); draw(foot(IB, B, C) -- foot(IC, B, C)); draw(circle(IA, length(IA-foot(IA, B, C)))); draw(arc(IB, IB-(4sqrt(5), 0), IB-(0, 4sqrt(5)))); draw(arc(IC, IC-(0, 4sqrt(5)), IC+(4sqrt(5), 0))); draw(circle(I, 2/sqrt(5)+sqrt(5))); dot("$A$", A, N); dot("$B$", B, SW); dot("$C$", C, SE); dot("$I$", I, N); dot("$I_A$", IA, S); dot("$I_B$", IB, NE); dot("$I_C$", IC, NW); [/asy]$ First assume that $BC=2$ and $AB=AC=x$ , and scale up later. Notice that $\overline{I_BAI_C}\parallel\overline{BC}$ . Then, the height from $A$ is $\sqrt{x^2-1}$ , so if $K=[ABC]$ , we know $K=\sqrt{x^2-1}$ . Then, if $r_D$ denotes the $D$ -exradius for $D\in\{A,B,C\}$ and $s=x+1$ denotes the semiperimeter, $r_A=\frac{K}{s-2}=\frac{K}{x-1},\;r_b=r_C=\frac{K}{s-x}=K,\text{ and }r=\frac{K}{s}=\frac{K}{x+1}.$ Then, if $X$ denotes the tangency point between the $B$ -excircle and $\overline{BC}$ , it is known that $BX=s$ , so $AI_B=s-1=x$ . Furthermore, $AI=\sqrt{(s-2)^2+r^2}=\sqrt{(x-1)^2+(K/(x+1))^2}$ . Then, $r+2r_A=II_A=II_B-r_B.$ It follows that $\begin{align*} II_B&=r+2r_A+r_B\\ \sqrt{AI^2+AI_B^2}&=\frac{K}{x+1}+\frac{2K}{x-1}+K\\ \sqrt{x^2+(x-1)^2+\left(\frac{K}{x+1}\right)^2}&=K\left(\frac1{x+1}+\frac2{x-1}+1\right)\\ \frac{\sqrt{(x^2+(x-1)^2)(x+1)^2+x^2-1}}{x+1}&=K\left(\frac{x^2+3x}{x^2-1}\right)\\ \frac{\sqrt{2x^3(x+1)}}{x+1}&=\frac{x(x+3)}{\sqrt{x^2-1}}\\ 2x(x-1)&=x^2+6x^2+9\\ 0&=x^2-8x-9\\ &=(x+1)(x-9), \end{align*}$ whence $x=9$ . Then, since $\gcd(2,9,9)=1$ , the smallest possible perimeter is $2+9+9=\boxed{020}$ .

(Solution by TheUltimate123)

Solution 3

On the Spot STEM solves this problem here: https://www.youtube.com/watch?v=zKHwTJBhKdM

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Solution 4

Guess it right. That's what I did.

2019 AIME I (Problems • Answer Key • Resources)
Preceded by Problem 10	Followed by Problem 12
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