Difference between revisions of "2019 AIME I Problems/Problem 11"
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==Solution== | ==Solution== | ||
+ | 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> | ||
+ | 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. | ||
+ | Using this gives us the following equation: <cmath>\frac{2b*IB}{a}=r+\frac{2rs}{s-a}+\frac{rs}{s-b}</cmath> | ||
+ | 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> | ||
+ | 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> | ||
+ | Cancelling some terms, we have <cmath>\frac{r(x+4y)}{x}=IB</cmath> | ||
+ | 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> | ||
+ | -franchester | ||
+ | |||
==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 , the sides have integers lengths and . Circle has its center at the incenter of . An excircle of is a circle in the exterior of 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 is internally tangent to , and the other two excircles are both externally tangent to . Find the minimum possible value of the perimeter of .
Solution
Let the tangent circle be . Some notation first: let , , be the semiperimeter, , and be the inradius. Intuition tells us that the radius of is (using the exradius formula). However, the sum of the radius of and is equivalent to the distance between the incenter and the the excenter. Denote the B excenter as and the incenter as . Lemma: We draw the circumcircle of . Let the angle bisector of hit the circumcircle at a second point . By the incenter-excenter lemma, . Let this distance be . Ptolemy's theorem on gives us Again, by the incenter-excenter lemma, so as desired. Using this gives us the following equation: Motivated by the and , we make the following substitution: This changes things quite a bit. Here's what we can get from it: It is known (easily proved with Heron's and a=rs) that Using this, we can also find : let the midpoint of be . Using Pythagorean's Theorem on , We now look at the RHS of the main equation: Cancelling some terms, we have Squaring, Expanding and moving terms around gives Reverse substituting, Clearly the smallest solution is and , so our answer is -franchester
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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