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

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==Solution==
 
==Solution==
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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>.
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Lemma: <math>I_BI=\frac{2b*IB}{a}</math>
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We draw the circumcircle of <math>\triangle ABC</math>. Let the angle bisector of <math>\angle ABC</math> hit the circumcircle at a second point <math>M</math>. By the incenter-excenter lemma, <math>BM=CM=IM</math>. Let this distance be <math>\alpha</math>. Ptolemy's theorem on <math>ABCM</math> gives us <cmath>a\alpha+b\alpha=b(\alpha+IB)\to \alpha=\frac{b*IB}{a}</cmath> Again, by the incenter-excenter lemma, <math>II_B=2IM</math> so <math>II_b=\frac{2b*IB}{a}</math> as desired.
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Using this gives us the following equation: <cmath>\frac{2b*IB}{a}=r+\frac{2rs}{s-a}+\frac{rs}{s-b}</cmath>
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Motivated by the <math>s-a</math> and <math>s-b</math>, we make the following substitution: <math>x=s-a, y=s-b</math>
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This changes things quite a bit. Here's what we can get from it: <cmath>a=2y, b=x+y, s=x+2y</cmath> It is known (easily proved with Heron's and a=rs) that <cmath>r=\sqrt{\frac{(s-a)(s-b)(s-b)}{s}}=\sqrt{\frac{xy^2}{x+2y}}</cmath> Using this, we can also find <math>IB</math>: let the midpoint of <math>BC</math> be <math>N</math>. Using Pythagorean's Theorem on <math>\triangle INB</math>, <cmath>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}  </cmath> We now look at the RHS of the main equation: <cmath>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}</cmath>
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Cancelling some terms, we have <cmath>\frac{r(x+4y)}{x}=IB</cmath>
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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=\textbf{ 020}</math>
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-franchester
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==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 16:25, 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

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 $2+9+9=\textbf{ 020}$ -franchester

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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