Art of Problem Solving
AoPS Online
Math texts, online classes, and more
for students in grades 5-12.
Visit AoPS Online ‚
Books for Grades 5-12 Online Courses
Beast Academy
Engaging math books and online learning
for students ages 6-13.
Visit Beast Academy ‚
Books for Ages 6-13 Beast Academy Online
AoPS Academy
Small live classes for advanced math
and language arts learners in grades 2-12.
Visit AoPS Academy ‚
Find a Physical Campus Visit the Virtual Campus

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

(→‎Solution 2)
(→‎Solution 2)
Line 17: Line 17:
 
<asy>
 
<asy>
 
size(8cm);
 
size(8cm);
−
defaultpen(fontsize(8pt));
+
defaultpen(fontsize(80pt));
 
pair A, B, C, I, IA, IB, IC;
 
pair A, B, C, I, IA, IB, IC;
 
A=(0, 4sqrt(5));
 
A=(0, 4sqrt(5));

Revision as of 13:02, 29 March 2020

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, $AM=CM=IM$. 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

[asy] size(8cm); defaultpen(fontsize(80pt)); 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 $\overline{I_BAI_C}\parallel\overline{BC}$ ($\textbf{Note:}$ This is wrong. The two excircles B and C, by definition, must be externally tangent to AB and AC. However, this solution says that the two excircles are tangent at A, which means they are tangent to the same line. However, we know that AB and AC are distinct, a contradiction. Therefore, this solution is wrong.)

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)

Solution 3

On the Spot STEM solves this problem here: https://www.youtube.com/watch?v=zKHwTJBhKdM

Please like, share, and subscribe!

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. AMC logo.png

Retrieved from "https://artofproblemsolving.com/wiki/index.php?title=2019_AIME_I_Problems/Problem_11&oldid=120253"