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)

$[asy] unitsize(1cm); var x = 9; pair A = (0,sqrt(x^2-1)); pair B = (-1,0); pair C = (1,0); dot(Label("$A$",A,NE),A); dot(Label("$B$",B,SW),B); dot(Label("$C$",C,SE),C); draw(A--B--C--cycle); var r = sqrt((x-1)/(x+1)); pair I = (0,r); dot(Label("$I$",I,SE),I); draw(circle(I,r)); draw(Label("$r$"),I--I+r*SSW,dashed); pair M = intersectionpoint(A--B,circle(I,r)); pair N = (0,0); pair O = intersectionpoint(A--C,circle(I,r)); dot(Label("$M$",M,W),M); dot(Label("$N$",N,S),N); dot(Label("$O$",O,E),O); var rN = sqrt((x+1)/(x-1)); pair EN = (0,-rN); dot(Label("$E_N$",EN,SE),EN); draw(circle(EN,rN)); draw(Label("$r_N$"),EN--EN+rN*SSW,dashed); pair AB = (-1-2/(x-1),-2rN); pair AC = (1+2/(x-1),-2rN); draw(B--AB,EndArrow); draw(C--AC,EndArrow); pair H = intersectionpoint(B--AB,circle(EN,rN)); dot(Label("$H$",H,W),H); var rM = sqrt(x^2-1); pair EM = (-x,rM); dot(Label("$E_M$",EM,SW),EM); draw(Label("$r_M$"),EM--EM+rM*SSE,dashed); pair CB = (-x-1,0); pair CA = (-2/x,sqrt(x^2-1)+2(sqrt(x^2-1)/x)); draw(B--CB,EndArrow); draw(A--CA,EndArrow); pair J = intersectionpoint(A--B,circle(EM,rM)); pair K = intersectionpoint(B--CB,circle(EM,rM)); dot(Label("$J$",J,W),J); dot(Label("$K$",K,S),K); draw(arc(EM,rM,-100,15),Arrows); [/asy]$

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

2019 AIME I (Problems • Answer Key • Resources)
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2019 AIME I Problems/Problem 11

Contents

Problem 11

Solution 1

Solution 2 (Lots of Pythagorean Theorem)

Solution 3

See Also