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

(→‎Solution)
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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 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>.
  
−
==Solution==
+
==Solution 1==
 
Let the tangent circle be <math>\omega</math>. Some notation first: let <math>BC=a</math>, <math>AB=b</math>, <math>s</math> be the semiperimeter, <math>\theta=\angle ABC</math>, and <math>r</math> be the inradius. Intuition tells us that the radius of <math>\omega</math> is <math>r+\frac{2rs}{s-a}</math> (using the exradius formula). However, the sum of the radius of <math>\omega</math> and <math>\frac{rs}{s-b}</math> is equivalent to the distance between the incenter and the the <math>B/C</math> excenter. Denote the B excenter as <math>I_B</math> and the incenter as <math>I</math>.  
 
Let the tangent circle be <math>\omega</math>. Some notation first: let <math>BC=a</math>, <math>AB=b</math>, <math>s</math> be the semiperimeter, <math>\theta=\angle ABC</math>, and <math>r</math> be the inradius. Intuition tells us that the radius of <math>\omega</math> is <math>r+\frac{2rs}{s-a}</math> (using the exradius formula). However, the sum of the radius of <math>\omega</math> and <math>\frac{rs}{s-b}</math> is equivalent to the distance between the incenter and the the <math>B/C</math> excenter. Denote the B excenter as <math>I_B</math> and the incenter as <math>I</math>.  
 
Lemma: <math>I_BI=\frac{2b*IB}{a}</math>
 
Lemma: <math>I_BI=\frac{2b*IB}{a}</math>
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Squaring, <cmath>\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)</cmath> Expanding and moving terms around gives <cmath>(x-8y)(x+2y)=0\to x=8y</cmath> Reverse substituting, <cmath>s-a=8s-8b\to b=\frac{9}{2}a</cmath> Clearly the smallest solution is <math>a=2</math> and <math>b=9</math>, so our answer is <math>2+9+9=\boxed{020}</math>
 
Squaring, <cmath>\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)</cmath> Expanding and moving terms around gives <cmath>(x-8y)(x+2y)=0\to x=8y</cmath> Reverse substituting, <cmath>s-a=8s-8b\to b=\frac{9}{2}a</cmath> Clearly the smallest solution is <math>a=2</math> and <math>b=9</math>, so our answer is <math>2+9+9=\boxed{020}</math>
 
-franchester
 
-franchester
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 +
==Solution 2==
 +
<asy>
 +
size(8cm);
 +
defaultpen(fontsize(8pt));
 +
pair A, B, C, I, IA, IB, IC;
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A=(0, 4sqrt(5));
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B=(-1, 0);
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C=(1, 0);
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I=incenter(A, B, C);
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IA=2*circumcenter(I,B,C)-I;
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IB=2*circumcenter(I,C,A)-I;
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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));
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draw(circle(IA, length(IA-foot(IA, B, C))));
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draw(arc(IB, IB-(4sqrt(5), 0), IB-(0, 4sqrt(5))));
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draw(arc(IC, IC-(0, 4sqrt(5)), IC+(4sqrt(5), 0)));
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draw(circle(I, 2/sqrt(5)+sqrt(5)));
 +
 +
dot("$A$", A, N);
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dot("$B$", B, SW);
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dot("$C$", C, SE);
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dot("$I$", I, N);
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dot("$I_A$", IA, S);
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dot("$I_B$", IB, NE);
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dot("$I_C$", IC, NW);
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</asy>
 +
First assume that <math>BC=2</math> and <math>AB=AC=x</math>, and scale up later. Notice that <math>\overline{I_BAI_C}\parallel\overline{BC}</math>. Then, the height from <math>A</math> is <math>\sqrt{x^2-1}</math>, so if <math>K=[ABC]</math>, we know <math>K=\sqrt{x^2-1}</math>. Then, if <math>r_D</math> denotes the <math>D</math>-exradius for <math>D\in\{A,B,C\}</math> and <math>s=x+1</math> denotes the semiperimeter, <cmath>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}.</cmath>Then, if <math>X</math> denotes the tangency point between the <math>B</math>-excircle and <math>\overline{BC}</math>, it is known that <math>BX=s</math>, so <math>AI_B=s-1=x</math>. Furthermore, <math>AI=\sqrt{(s-2)^2+r^2}=\sqrt{(x-1)^2+(K/(x+1))^2}</math>. Then, <cmath>r+2r_A=II_A=II_B-r_B.</cmath>It follows that
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<cmath>
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\begin{align*}
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II_B&=r+2r_A+r_B\\
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\sqrt{AI^2+AI_B^2}&=\frac{K}{x+1}+\frac{2K}{x-1}+K\\
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\sqrt{x^2+(x-1)^2+\left(\frac{K}{x+1}\right)^2}&=K\left(\frac1{x+1}+\frac2{x-1}+1\right)\\
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\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}}\\
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2x(x-1)&=x^2+6x^2+9\\
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0&=x^2-8x-9\\
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&=(x+1)(x-9),
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\end{align*}
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</cmath>
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whence <math>x=9</math>. Then, since <math>\gcd(2,9,9)=1</math>, the smallest possible perimeter is <math>2+9+9=\boxed{020}</math>.
 +
 +
(Solution by TheUltimate123)
  
 
==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 19:05, 15 March 2019

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, $BM=CM=IM$. 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)

See Also

2019 AIME I (Problems • Answer Key • Resources)
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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