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

(Solution 3)
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==Problem 11==
 
==Problem 11==
  
In <math>\triangle ABC</math>, the sides have integers lengths and <math>AB=AC</math>. Circle <math>\omega</math> has its center at the incenter of <math>\triangle ABC</math>. An ''excircle'' of <math>\triangle ABC</math> is a circle in the exterior of <math>\triangle ABC</math> 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 <math>\overline{BC}</math> is internally tangent to <math>\omega</math>, and the other two excircles are both externally tangent to <math>\omega</math>. Find the minimum possible value of the perimeter of <math>\triangle ABC</math>.
+
In <math>\triangle ABC</math>, the sides have integer lengths and <math>AB=AC</math>. Circle <math>\omega</math> has its center at the incenter of <math>\triangle ABC</math>. An ''excircle'' of <math>\triangle ABC</math> is a circle in the exterior of <math>\triangle ABC</math> 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 <math>\overline{BC}</math> is internally tangent to <math>\omega</math>, and the other two excircles are both externally tangent to <math>\omega</math>. Find the minimum possible value of the perimeter of <math>\triangle ABC</math>.
  
 
==Solution 1==
 
==Solution 1==
Line 14: Line 14:
 
-franchester
 
-franchester
  
==Solution 2==
+
==Solution 2 (Lots of Pythagorean Theorem)==
  
Video Link Here: https://www.youtube.com/watch?v=zKHwTJBhKdM
+
<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 <math>BC=2</math> and <math>AB=AC=x</math>. The triangle can be scaled later if necessary. Let <math>I</math> be the incenter and let <math>r</math> be the inradius. Let the points at which the incircle intersects <math>AB</math>, <math>BC</math>, and <math>CA</math> be denoted <math>M</math>, <math>N</math>, and <math>O</math>, respectively.
 +
 
 +
 
 +
Next, we calculate <math>r</math> in terms of <math>x</math>. Note the right triangle formed by <math>A</math>, <math>I</math>, and <math>M</math>. The length <math>IM</math> is equal to <math>r</math>. Using the Pythagorean Theorem, the length <math>AN</math> is <math>\sqrt{x^2-1}</math>, so the length <math>AI</math> is <math>\sqrt{x^2-1}-r</math>. Note that <math>BN</math> is half of <math>BC=2</math>, and by symmetry caused by the incircle, <math>BN=BM</math> and <math>BM=1</math>, so <math>MA=x-1</math>. Applying the Pythagorean Theorem to <math>AIM</math>, we get
 +
<cmath>r^2+(x-1)^2=\left(\sqrt{x^2-1}-r\right)^2.</cmath>
 +
Expanding yields
 +
<cmath>r^2+x^2-2x+1=x^2-1-2r\sqrt{x^2-1}+r^2,</cmath>
 +
which can be simplified to
 +
<cmath>2r\sqrt{x^2-1}=2x-2.</cmath>
 +
Dividing by <math>2</math> and then squaring results in
 +
<cmath>r^2(x^2-1)=(x-1)^2,</cmath>
 +
and isolating <math>r^2</math> gets us
 +
<cmath>r^2=\frac{(x-1)^2}{x^2-1}=\frac{(x-1)^2}{(x+1)(x-1)}=\frac{x-1}{x+1},</cmath>
 +
so <math>r=\sqrt{\frac{x-1}{x+1}}</math>.
 +
 
 +
 
 +
We then calculate the radius of the excircle tangent to <math>BC</math>. We denote the center of the excircle <math>E_N</math> and the radius <math>r_N</math>.
 +
 
 +
Consider the quadrilateral formed by <math>M</math>, <math>I</math>, <math>E_N</math>, and the point at which the excircle intersects the extension of <math>AB</math>, which we denote <math>H</math>. By symmetry caused by the excircle, <math>BN=BH</math>, so <math>BH=1</math>.
 +
 
 +
Note that triangles <math>MBI</math> and <math>NBI</math> are congruent, and <math>HBE</math> and <math>NBE</math> are also congruent. Denoting the measure of angles <math>MBI</math> and <math>NBI</math> measure <math>\alpha</math> and the measure of angles <math>HBE</math> and <math>NBE</math> measure <math>\beta</math>, straight angle <math>MBH=2\alpha+2\beta</math>, so <math>\alpha + \beta=90^\circ</math>. This means that angle <math>IBE</math> is a right angle, so it forms a right triangle.
 +
 
 +
Setting the base of the right triangle to <math>IE</math>, the height is <math>BN=1</math> and the base consists of <math>IN=r</math> and <math>EN=r_N</math>. Triangles <math>INB</math> and <math>BNE</math> are similar to <math>IBE</math>, so <math>\frac{IN}{BN}=\frac{BN}{EN}</math>, or <math>\frac{r}{1}=\frac{1}{r_N}</math>. This makes <math>r_N</math> the reciprocal of <math>r</math>, so <math>r_N=\sqrt{\frac{x+1}{x-1}}</math>.
 +
 
 +
 
 +
Circle <math>\omega</math>'s radius can be expressed by the distance from the incenter <math>I</math> to the bottom of the excircle with center <math>E_N</math>. This length is equal to <math>r+2r_N</math>, or <math>\sqrt{\frac{x-1}{x+1}}+2\sqrt{\frac{x+1}{x-1}}</math>. Denote this value <math>r_\omega</math>.
 +
 
 +
 
 +
Finally, we calculate the distance from the incenter <math>I</math> to the closest point on the excircle tangent to <math>AB</math>, which forms another radius of circle <math>\omega</math> and is equal to <math>r_\omega</math>. We denote the center of the excircle <math>E_M</math> and the radius <math>r_M</math>. We also denote the points where the excircle intersects <math>AB</math> and the extension of <math>BC</math> using <math>J</math> and <math>K</math>, respectively. In order to calculate the distance, we must find the distance between <math>I</math> and <math>E_M</math> and subtract off the radius <math>r_M</math>.
 +
 
 +
We first must calculate the radius of the excircle. Because the excircle is tangent to both <math>AB</math> and the extension of <math>AC</math>, its center must lie on the angle bisector formed by the two lines, which is parallel to <math>BC</math>. This means that the distance from <math>E_M</math> to <math>K</math> is equal to the length of <math>AN</math>, so the radius is also <math>\sqrt{x^2-1}</math>.
 +
 
 +
Next, we find the length of <math>IE_M</math>. We can do this by forming the right triangle <math>IAE_M</math>. The length of leg <math>AI</math> is equal to <math>AN</math> minus <math>r</math>, or <math>\sqrt{x^2-1}-\sqrt{\frac{x-1}{x+1}}</math>. In order to calculate the length of leg <math>AE_M</math>, note that right triangles <math>AJE_M</math> and <math>BNA</math> are congruent, as <math>JE_M</math> and <math>NA</math> share a length of <math>\sqrt{x^2-1}</math>, and angles <math>E_MAJ</math> and <math>NAB</math> add up to the right angle <math>NAE_M</math>. This means that <math>AE_M=BA=x</math>.
 +
 
 +
Using Pythagorean Theorem, we get
 +
<cmath>IE_M=\sqrt{\left(\sqrt{x^2-1}-\sqrt{\frac{x-1}{x+1}}\right)^2+x^2}.</cmath>
 +
Bringing back
 +
<cmath>r_\omega=IE_M-r_M</cmath>
 +
and substituting in some values, the equation becomes
 +
<cmath>r_\omega=\sqrt{\left(\sqrt{x^2-1}-\sqrt{\frac{x-1}{x+1}}\right)^2+x^2}-\sqrt{x^2-1}.</cmath>
 +
Rearranging and squaring both sides gets
 +
<cmath>\left(r_\omega+\sqrt{x^2-1}\right)^2=\left(\sqrt{x^2-1}-\sqrt{\frac{x-1}{x+1}}\right)^2+x^2.</cmath>
 +
Distributing both sides yields
 +
<cmath>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.</cmath>
 +
Canceling terms results in
 +
<cmath>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.</cmath>
 +
Since
 +
<cmath>-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),</cmath>
 +
We can further simplify to
 +
<cmath>r_\omega^2+2r_\omega\sqrt{x^2-1}=-2(x-1)+\frac{x-1}{x+1}+x^2.</cmath>
 +
Substituting out <math>r_\omega</math> gets
 +
<cmath>\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</cmath>
 +
which when distributed yields
 +
<cmath>\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.</cmath>
 +
After some canceling, distributing, and rearranging, we obtain
 +
<cmath>4\left(\frac{x+1}{x-1}\right)=x^2-8x-4.</cmath>
 +
Multiplying both sides by <math>x-1</math> results in
 +
<cmath>4x+4=x^3-x^2-8x^2+8x-4x+4,</cmath>
 +
which can be rearranged into
 +
<cmath>x^3-9x^2=0</cmath>
 +
and factored into
 +
<cmath>x^2(x-9)=0.</cmath>
 +
This means that <math>x</math> equals <math>0</math> or <math>9</math>, and since a side length of <math>0</math> cannot exist, <math>x=9</math>.
 +
 
 +
 
 +
As a result, the triangle must have sides in the ratio of <math>9:2:9</math>. Since the triangle must have integer side lengths, and these values share no common factors greater than <math>1</math>, the triangle with the smallest possible perimeter under these restrictions has a perimeter of <math>9+2+9=\boxed{020}</math>. ~[[User:emerald_block|emerald_block]]
 +
 
 +
==Solution 3 (Not that hard construction)==
 +
Notice that the <math>A</math>-excircle would have to be very small to fit the property that it is internally tangent to <math>\omega</math>, given that circle <math>\omega</math>'s centre is at the incenter of <math>\triangle ABC</math>. If <math>BC=2</math>, we see that <math>AB=AC</math> must be somewhere in the <math>6</math> to <math>13</math> range. If we test <math>6</math> by construction, we notice the <math>A</math>-excircle is too big for it to be internally tangent to <math>\omega</math>. This means we should test <math>8</math> or <math>9</math> next. I actually did this and found that <math>9</math> worked, so the answer is <math>2+9+9=\boxed{20}</math>. Note that <math>BC</math> cannot be <math>1</math> because then <math>AB=AC</math> would have to be <math>4.5</math> which is not an integer.
 +
 
 +
~[[User:Icematrix2|icematrix2]]
 +
 
 +
==Video Solution (On the Spot STEM)==
 +
 
 +
https://www.youtube.com/watch?v=zKHwTJBhKdM
 +
 
 +
==Video Solution 2 (I would recommend this one because it's more concise)==
 +
 
 +
https://www.youtube.com/watch?v=ldr4yi3t6hQ
  
 
==See Also==
 
==See Also==
 
{{AIME box|year=2019|n=I|num-b=10|num-a=12}}
 
{{AIME box|year=2019|n=I|num-b=10|num-a=12}}
 
{{MAA Notice}}
 
{{MAA Notice}}

Revision as of 17:05, 19 September 2020

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 (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 $\omega$, 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 $\omega$. 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

Video Solution (On the Spot STEM)

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

Video Solution 2 (I would recommend this one because it's more concise)

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

See Also

2019 AIME I (ProblemsAnswer KeyResources)
Preceded by
Problem 10
Followed by
Problem 12
1 2 3 4 5 6 7 8 9 10 11 12 13 14 15
All AIME Problems and Solutions

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