# Difference between revisions of "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