2019 AIME I Problems/Problem 11

Problem 11

In $\triangle ABC$, the sides have integer 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 (Lots of Pythagorean Theorem)

First, assume $BC=2$ and $AB=AC=x$. The triangle can be scaled later if necessary. Let $I$ be the incenter and let $r$ be the inradius. Let the points at which the incircle intersects $AB$, $BC$, and $CA$ be denoted $M$, $N$, and $O$, respectively.

Next, we calculate $r$ in terms of $x$. Note the right triangle formed by $A$, $I$, and $M$. The length $IM$ is equal to $r$. Using the Pythagorean Theorem, the length $AN$ is $\sqrt{x^2-1}$, so the length $AI$ is $\sqrt{x^2-1}-r$. Note that $BN$ is half of $BC=2$, and by symmetry caused by the incircle, $BN=BM$ and $BM=1$, so $MA=x-1$. Applying the Pythagorean Theorem to $AIM$, we get \[r^2+(x-1)^2=\left(\sqrt{x^2-1}-r\right)^2.\] Expanding yields \[r^2+x^2-2x+1=x^2-1-2r\sqrt{x^2-1}+r^2,\] which can be simplified to \[2r\sqrt{x^2-1}=2x-2.\] Dividing by $2$ and then squaring results in \[r^2(x^2-1)=(x-1)^2,\] and isolating $r^2$ gets us \[r^2=\frac{(x-1)^2}{x^2-1}=\frac{(x-1)^2}{(x+1)(x-1)}=\frac{x-1}{x+1},\] so $r=\sqrt{\frac{x-1}{x+1}}$.

We then calculate the radius of the excircle tangent to $BC$. We denote the center of the excircle $E_N$ and the radius $r_N$.

Consider the quadrilateral formed by $M$, $I$, $E_N$, and the point at which the excircle intersects the extension of $AB$, which we denote $H$. By symmetry caused by the excircle, $BN=BH$, so $BH=1$.

Note that triangles $MBI$ and $NBI$ are congruent, and $HBE$ and $NBE$ are also congruent. Denoting the measure of angles $MBI$ and $NBI$ measure $\alpha$ and the measure of angles $HBE$ and $NBE$ measure $\beta$, straight angle $MBH=2\alpha+2\beta$, so $\alpha + \beta=90^\circ$. This means that angle $IBE$ is a right angle, so it forms a right triangle.

Setting the base of the right triangle to $IE$, the height is $BN=1$ and the base consists of $IN=r$ and $EN=r_N$. Triangles $INB$ and $BNE$ are similar to $IBE$, so $\frac{IN}{BN}=\frac{BN}{EN}$, or $\frac{r}{1}=\frac{1}{r_N}$. This makes $r_N$ the reciprocal of $r$, so $r_N=\sqrt{\frac{x+1}{x-1}}$.

Circle $\omega$'s radius can be expressed by the distance from the incenter $I$ to the bottom of the excircle with center $E_N$. This length is equal to $r+2r_N$, or $\sqrt{\frac{x-1}{x+1}}+2\sqrt{\frac{x+1}{x-1}}$. Denote this value $r_\omega$.

Finally, we calculate the distance from the incenter $I$ to the closest point on the excircle tangent to $AB$, which forms another radius of circle $\omega$ and is equal to $r_\omega$. We denote the center of the excircle $E_M$ and the radius $r_M$. We also denote the points where the excircle intersects $AB$ and the extension of $BC$ using $J$ and $K$, respectively. In order to calculate the distance, we must find the distance between $I$ and $E_M$ and subtract off the radius $r_M$.

We first must calculate the radius of the excircle. Because the excircle is tangent to both $AB$ and the extension of $AC$, its center must lie on the angle bisector formed by the two lines, which is parallel to $BC$. This means that the distance from $E_M$ to $K$ is equal to the length of $AN$, so the radius is also $\sqrt{x^2-1}$.

Next, we find the length of $IE_M$. We can do this by forming the right triangle $IAE_M$. The length of leg $AI$ is equal to $AN$ minus $r$, or $\sqrt{x^2-1}-\sqrt{\frac{x-1}{x+1}}$. In order to calculate the length of leg $AE_M$, note that right triangles $AJE_M$ and $BNA$ are congruent, as $JE_M$ and $NA$ share a length of $\sqrt{x^2-1}$, and angles $E_MAJ$ and $NAB$ add up to the right angle $NAE_M$. This means that $AE_M=BA=x$.

Using Pythagorean Theorem, we get \[IE_M=\sqrt{\left(\sqrt{x^2-1}-\sqrt{\frac{x-1}{x+1}}\right)^2+x^2}.\] Bringing back \[r_\omega=IE_M-r_M\] and substituting in some values, the equation becomes \[r_\omega=\sqrt{\left(\sqrt{x^2-1}-\sqrt{\frac{x-1}{x+1}}\right)^2+x^2}-\sqrt{x^2-1}.\] Rearranging and squaring both sides gets \[\left(r_\omega+\sqrt{x^2-1}\right)^2=\left(\sqrt{x^2-1}-\sqrt{\frac{x-1}{x+1}}\right)^2+x^2.\] Distributing both sides yields \[r_\omega^2+2r_\omega\sqrt{x^2-1}+x^2-1=x^2-1-2\sqrt{x^2-1}\sqrt{\frac{x-1}{x+1}}+\frac{x-1}{x+1}+x^2.\] Canceling terms results in \[r_\omega^2+2r_\omega\sqrt{x^2-1}=-2\sqrt{x^2-1}\sqrt{\frac{x-1}{x+1}}+\frac{x-1}{x+1}+x^2.\] Since \[-2\sqrt{x^2-1}\sqrt{\frac{x-1}{x+1}}=-2\sqrt{(x+1)(x-1)\frac{x-1}{x+1}}=-2(x-1),\] We can further simplify to \[r_\omega^2+2r_\omega\sqrt{x^2-1}=-2(x-1)+\frac{x-1}{x+1}+x^2.\] Substituting out $r_\omega$ gets \[\left(\sqrt{\frac{x-1}{x+1}}+2\sqrt{\frac{x+1}{x-1}}\right)^2+2\left(\sqrt{\frac{x-1}{x+1}}+2\sqrt{\frac{x+1}{x-1}}\right)\sqrt{x^2-1}=-2(x-1)+\frac{x-1}{x+1}+x^2\] which when distributed yields \[\frac{x-1}{x+1}+4+4\left(\frac{x+1}{x-1}\right)+2(x-1+2(x+1))=-2(x-1)+\frac{x-1}{x+1}+x^2.\] After some canceling, distributing, and rearranging, we obtain \[4\left(\frac{x+1}{x-1}\right)=x^2-8x-4.\] Multiplying both sides by $x-1$ results in \[4x+4=x^3-x^2-8x^2+8x-4x+4\] which can be rearranged into \[x^3-9x^2=0\] and factored into \[x^2(x-9)=0.\] This means that $x$ equals $0$ or $9$, and since a side length of $0$ cannot exist, $x=9$.

As a result, the triangle must have sides in the ratio of $9:2:9$. Since the triangle must have integer side lengths, and these values share no common factors greater than 1, the triangle with the smallest possible perimeter under these restrictions has a perimeter of $9+2+9=\boxed{020}$. ~emerald_block

Solution 3

Video Link Here: https://www.youtube.com/watch?v=zKHwTJBhKdM

See Also

2019 AIME I (ProblemsAnswer KeyResources)
Preceded by
Problem 10
Followed by
Problem 12
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All AIME Problems and Solutions

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