Difference between revisions of "2019 AIME I Problems/Problem 11"

Latest revision as of 18:55, 10 October 2023

Problem

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 (Various Techniques)

Before we start thinking about the problem, let’s draw it out;

$[asy] unitsize(1cm); var x = 9; pair A = (0,sqrt(x^2-1)); pair B = (-1,0); pair C = (1,0); dot(Label("$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$",I,SE),I); draw(circle(I,r)); pair G = intersectionpoint(A--B,circle(I,r)); pair D = (0,0); dot(Label("$G$",G,W),G); dot(Label("$D$",D,SSE),D); draw(Label("$r$"),I--G,dashed); var rA = sqrt((x+1)/(x-1)); pair IA = (0,-rA); dot(Label("$I_A$",IA,SE),IA); draw(circle(IA,rA)); pair AB = (-1-2/(x-1),-2rA); pair AC = (1+2/(x-1),-2rA); draw(B--AB,EndArrow); draw(C--AC,EndArrow); pair H = intersectionpoint(B--AB,circle(IA,rA)); dot(Label("$H$",H,W),H); draw(Label("$r_{I_A}$"),IA--H,dashed); var rB = sqrt(x^2-1); pair IB = (x,rB); dot(Label("$I_B$",IB,SE),IB); pair BC = (x+1,0); pair BA = (2/x,sqrt(x^2-1)+2(sqrt(x^2-1)/x)); draw(C--BC,EndArrow); draw(A--BA,EndArrow); pair E = intersectionpoint(A--C,circle(IB,rB)); pair F = intersectionpoint(C--BC,circle(IB,rB)); dot(Label("$E$",E,SE),E); dot(Label("$F$",F,S),F); draw(Label("$r_{I_B}$"),IB--F,dashed); draw(circle(IB,rB)); draw(A--IA); draw(B--IB); pair J = intersectionpoint(B--BA,circle(IB,rB)); dot(Label("$J$",J,W),J); draw(circle(I,r+2rA)); pair W = intersectionpoint(B--F,circle(I,r+2rA)); dot(Label("$\omega$",W,SSE),W); [/asy]$

For the sake of space, I've drawn only 2 of the 3 excircles because the third one looks the same as the second large one because the triangle is isosceles. By the incenter-excenter lemma, $AII_A$ and $BII_B$ are collinear, $E$ is the tangent of circle $I_B$ to $AC$ , $F$ is the tangent of that circle to the extension of $BC$ , and $J$ is the tangent of the circle to the extension of $BA$ . The interesting part of the diagram is circle $\omega$ , which is internally tangent to circle $I_A$ yet externally tangent to circle $I_B$ . Therefore, perhaps we can relate the radius of this circle to the semiperimeter of triangle $ABC$ .

We can see that the radius of circle $\omega$ is $2r_{I_A}+r$ using the incenter and A-excenter of our main triangle. This radius is also equal to $BI_B - BI - r_{I_B}$ from the incenter and B-excenter of our triangle. Thus, we can solve for each of these separately in terms of the lengths of the triangle and set them equal to each other to form an equation.

To find the left hand side of the equation, we have to first find $r$ and $r_{I_A}$ . Let $a = AB = AC, b = BD = DC,$ and $h = AD$ . Then since the perimeter of the triangle is $2a+2b$ , the semiperimeter is $a+b$ .

Now let's take a look at triangle $BDI$ . Because $BI$ is the angle bisector of $\angle B$ , by the angle bisector theorem, $\frac{AI}{ID} = \frac{BA}{BD} \implies \frac{h-r}{r} = \frac{a}{b}$ . Rearranging, we get $r = \frac{hb}{a+b}$ .

Take a look at triangle $AGI$ . $AG = a - GB = a - BD = a-b$ , $AI = h-r = \frac{ha}{a+b}$ (angle bisector theorem), and $GI = r = \frac{hb}{a+b}$ . Now let's analyze triangle $AHI_A$ . $AH = AB + BH = AB+ BD = a+b$ , $AI_A = h+r_{I_A}$ , and $HI_A = r_{I_A}$ . Since $\angle GAI = \angle HAI_A$ and $\angle IGA = \angle I_AHA = 90^{\circ}$ , triangle $AGI$ and $AHI_A$ are similar by AA. Then $\frac{r_{I_A}}{r} = \frac{h+r_{I_A}}{h-r} \implies r_{I_A} = r \cdot \frac{h+r_{I_A}}{h-r} = \frac{hb}{a+b} \cdot \frac{h+r_{I_A}}{\frac{ha}{a+b}} = \frac{b(h+r_{I_A})}{a}$ . Now, solving yields $r_{I_A} = \frac{hb}{a-b}$ .

Finally, the left hand side of our equation is $\frac{2hb}{a-b} + \frac{hb}{a+b}$

Now let's look at triangle $BFI_B$ . How will we find $BI_B$ ? Let's first try to find $BF$ and $I_BF$ in terms of the lengths of the triangle. We recognize:

$BF = BC + CF = BC + CK$ . We really want to have $CA$ instead of $CK$ , and $AK$ looks very similar in length to $DC$ , so let's try to prove that they are equal.

$BJ = BF$ , so we can try to add these two and see if we get anything interesting. We have: $BJ + BF = BA + AJ + BC + CF = BA + AE + BC + CE = BA + BC + CA$ , which is our perimeter. Thus, $BF = a+b$ .

Triangle $BDI$ is similar to triangle $BFI_B$ by AA, and we know that $BD = b$ , and $ID = r = \frac{hb}{a+b}$ , so thus $I_BF = BF \cdot \frac{ID}{BD} = (a+b) \cdot \frac{\frac{hb}{a+b}}{b} = \frac{hb}{b} = h$ . Thus, the height of this triangle is $h$ by similarity ratios, the same height as vertex $A$ . By the Pythagorean Theorem, $BI_B = \sqrt{(a+b)^2 + h^2}$ and by similarity ratios, $BI = \frac{b}{a+b} \cdot \sqrt{(a+b)^2 + h^2}$ . Finally, $r_{I_B} = I_BF = h$ , and thus the right hand side of our equation is $\sqrt{(a+b)^2 + h^2} - \frac{b}{a+b} \cdot \sqrt{(a+b)^2 + h^2} - h = \sqrt{(a+b)^2 + h^2}(1 - \frac{b}{a+b}) - h = \sqrt{(a+b)^2 + h^2} \cdot \frac{a}{a+b} - h$ .

Setting the two equal, we have $\frac{2hb}{a-b} + \frac{hb}{a+b} = \sqrt{(a+b)^2 + h^2} \cdot \frac{a}{a+b} - h$

Multiplying both sides by $(a+b)(a-b)$ we have $2hb(a+b) + hb(a-b) = \sqrt{(a+b)^2 + h^2} \cdot a(a-b) - h(a^2 - b^2)$

From here, let $b = 1$ arbitrarily; note that we can always scale this value to fit the requirements later. Thus our equation is $2h(a+1) + h(a-1) = \sqrt{(a+1)^2 + h^2} \cdot a(a-1) - h(a^2 - 1)$ . Now since $h = \sqrt{a^2 - b^2}$ , we can plug into our equation:

$2h(a+1) + h(a-1) = \sqrt{a² + 2a + 1 + a^2 - b^2 } \cdot a(a-1) - h(a^2 - 1)$ . Remembering $b = 1$ ;

$\implies 2h(a+1) + h(a-1) = \sqrt{2a^2 + 2a} \cdot a(a-1) - h(a^2 - 1)$

$\implies 3ha + h + ha^2 - h = \sqrt{2a^2 + 2a} \cdot a(a-1)$

$\implies ah(3+a) = a(a-1) \cdot \sqrt{2a^2 + 2a}$

$\implies h^2(3+a)^2 = (a-1)^2 \cdot 2a(a+1)$

$\implies (a^2 - 1) (3 + a)^2 = 2a(a+1)(a - 1)^2$

$\implies (3+ a)^2 = 2a(a-1)$

$\implies 9 + 6a + a^2 = 2a^2 - 2a$

$\implies a^2 - 8a - 9 = 0$

$\implies (a-9)(a+1) = 0$

$\implies a = 9$ because the side lengths have to be positive numbers. Furthermore, because our values for $a$ and $b$ are relatively prime, we don't have to scale down our triangle further, and we are done. Therefore, our answer is $2a + 2b = 18 + 2 = \boxed{020}$

~KingRavi

Solution 4 (Not that hard construction)

Notice that the $A$ -excircle would have to be very small to fit the property that it is internally tangent to $\omega$ and the other two excircles are both externally tangent, given that circle $\omega$ 's centre is at the incenter of $\triangle ABC$ . If $BC=2$ , we see that $AB=AC$ must be somewhere in the $6$ to $13$ range. If we test $6$ by construction, we notice the $A$ -excircle is too big for it to be internally tangent to $\omega$ while the other two are externally tangent. This means we should test $8$ or $9$ next. I actually did this and found that $9$ worked, so the answer is $2+9+9=\boxed{20}$ . Note that $BC$ cannot be $1$ because then $AB=AC$ would have to be $4.5$ which is not an integer.

~icematrix2

Solution 5 (Standard geometry)

Let $M$ be the midpoint $BC, BM = a, AB= BC = b,$ $s = b+a$ be the semiperimeter, $r$ be the inradius. Let $I_A, I_B$ be excenters, $r_A, r_B$ be exradius, $R$ be radius $\omega.$ Then $R = r + 2 r_A,$ $r = \sqrt{\frac{(s-2a)(s-b)^2}{s}} = a \sqrt{\frac{b-a}{b+a}},$ $r \cdot s = r_A \cdot (s-2a) \implies r_A = a \sqrt{\frac{b+a}{b–a}},$ $r \cdot s = r_B \cdot (s-b) \implies r_B = \sqrt{b^2 – a^2} = AM.$ $II_B =R + r_B = r + 2r_A + r_B \implies$ $II_B= a \sqrt{\frac{b-a}{b+ a}}+ 2a \sqrt{\frac{b+a}{b – a}} + \sqrt{b^2-a^2} = b\frac{3a+b}{\sqrt{b^2-a^2}}.$ $\overline{AI_B}= \frac{\overline{A} \cdot 2a - \overline{B} \cdot b + \overline{C}\cdot b}{2a -b + b}-\overline{A} =b \frac{\overline{C}- \overline{B}}{2a},$ $AI_B = b, \overline{AI_B}\perp AI \implies II_B = \sqrt{AI_B^2 + (AM – r)^2},$ $II_B = \sqrt{b^2 +\left(\sqrt{b^2-a^2}- a \sqrt{\frac{b-a}{b+ a}}\right)^2} = b\sqrt{\frac{2b}{b+a}}$ Therefore we get problem’s condition in the form of $b\frac{b +3a}{\sqrt{b^2-a^2}} = b \sqrt{\frac {2b}{b+a}} \implies b + 3a = \sqrt{2b(b-a)} \implies (b-9a)(b+a) = 0 \implies b = 9a.$

We use $a = 1$ an get $b = 9, 2s = 18+2 = \boxed{020}$ .

vladimir.shelomovskii@gmail.com, vvsss

Video Solution (On the Spot STEM)

This solution is the video solution for Solution 3 - not posted by ~KingRavi

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

Video Solution 2 (More concise)

https://www.youtube.com/watch?v=ldr4yi3t6hQ

@@ Line 316: / Line 316: @@
 ==Solution 5 (Standard geometry)==
-[[File:2019 AIME I 11a.png|500px|right]]
+[[File:AIME-I-2019-11.png|450px|right]]
 Let <math>M</math> be the midpoint <math>BC, BM = a, AB= BC = b,</math>

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