Difference between revisions of "2019 AIME I Problems/Problem 11"
(→Solution) |
|||
Line 3: | Line 3: | ||
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> | ||
Line 13: | Line 13: | ||
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 | ||
+ | |||
+ | ==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 <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 | ||
+ | <cmath> | ||
+ | \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*} | ||
+ | </cmath> | ||
+ | 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
Contents
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 1
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
Solution 2
First assume that and , and scale up later. Notice that . Then, the height from is , so if , we know . Then, if denotes the -exradius for and denotes the semiperimeter, Then, if denotes the tangency point between the -excircle and , it is known that , so . Furthermore, . Then, It follows that whence . Then, since , the smallest possible perimeter is .
(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 |
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.