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 "2022 AIME I Problems/Problem 8"

(→‎Solution 1)
m (→‎Solution 3 (Simple Geometry))
 
(63 intermediate revisions by 10 users not shown)
Line 1: Line 1:
 
== Problem ==
 
== Problem ==
−
Equilateral triangle <math>\triangle ABC</math> is inscribed in circle <math>\omega</math> with radius <math>18.</math> Circle <math>\omega_A</math> is tangent to sides <math>\overline{AB}</math> and <math>\overline{AC}</math> and is internally tangent to <math>\omega</math>. Circles <math>\omega_B</math> and <math>\omega_C</math> are defined analogously. Circles <math>\omega_A</math>, <math>\omega_B</math>, and <math>\omega_C</math> meet in six points<math>-</math>two points for each pair of circles. The three intersection points closest to the vertices of <math>\triangle ABC</math> are the vertices of a large equilateral triangle in the interior of <math>\triangle ABC</math>, and the other three intersection points are the vertices of a smaller equilateral triangle in the interior of <math>\triangle ABC</math>. The side length of the smaller equilateral triangle can be written as <math>\sqrt{a}-\sqrt{b}</math>, where <math>a</math> and <math>b</math> are positive integers. Find <math>a+b</math>.
+
Equilateral triangle <math>\triangle ABC</math> is inscribed in circle <math>\omega</math> with radius <math>18.</math> Circle <math>\omega_A</math> is tangent to sides <math>\overline{AB}</math> and <math>\overline{AC}</math> and is internally tangent to <math>\omega.</math> Circles <math>\omega_B</math> and <math>\omega_C</math> are defined analogously. Circles <math>\omega_A,</math> <math>\omega_B,</math> and <math>\omega_C</math> meet in six points---two points for each pair of circles. The three intersection points closest to the vertices of <math>\triangle ABC</math> are the vertices of a large equilateral triangle in the interior of <math>\triangle ABC,</math> and the other three intersection points are the vertices of a smaller equilateral triangle in the interior of <math>\triangle ABC.</math> The side length of the smaller equilateral triangle can be written as <math>\sqrt{a} - \sqrt{b},</math> where <math>a</math> and <math>b</math> are positive integers. Find <math>a+b.</math>
  
 
== Diagram ==
 
== Diagram ==
 
<asy>
 
<asy>
−
unitsize(0.3cm);
+
/* Made by MRENTHUSIASM */
−
draw(circle((0,0),18));
+
size(250);
−
pair A = (9 * sqrt(3), -9);
+
pair A, B, C, W, WA, WB, WC, X, Y, Z;
−
pair B = (-9 * sqrt(3), -9);
+
A = 18*dir(90);
−
pair C = (0,18);
+
B = 18*dir(210);
−
draw(A--B--C--cycle);
+
C = 18*dir(330);
−
draw(circle((0,-6),12), gray);
+
W = (0,0);
−
draw(circle((3*sqrt(3),3),12), gray);
+
WA = 6*dir(270);
−
draw(circle((-3*sqrt(3),3),12), gray);
+
WB = 6*dir(30);
 +
WC = 6*dir(150);
 +
X = (sqrt(117)-3)*dir(270);
 +
Y = (sqrt(117)-3)*dir(30);
 +
Z = (sqrt(117)-3)*dir(150);
 +
filldraw(X--Y--Z--cycle,green,dashed);
 +
draw(Circle(WA,12)^^Circle(WB,12)^^Circle(WC,12),blue);
 +
draw(Circle(W,18)^^A--B--C--cycle);
 +
dot("$A$",A,1.5*dir(A),linewidth(4));
 +
dot("$B$",B,1.5*dir(B),linewidth(4));
 +
dot("$C$",C,1.5*dir(C),linewidth(4));
 +
dot("$\omega$",W,1.5*dir(270),linewidth(4));
 +
dot("$\omega_A$",WA,1.5*dir(-WA),linewidth(4));
 +
dot("$\omega_B$",WB,1.5*dir(-WB),linewidth(4));
 +
dot("$\omega_C$",WC,1.5*dir(-WC),linewidth(4));
 +
</asy>
 +
~MRENTHUSIASM ~ihatemath123
 +
 
 +
==Solution 1 (Coordinate Geometry)==
 +
 
 +
We can extend <math>AB</math> and <math>AC</math> to <math>B'</math> and <math>C'</math> respectively such that circle <math>\omega_A</math> is the incircle of <math>\triangle AB'C'</math>.
 +
<asy>
 +
/* Made by MRENTHUSIASM */
 +
size(300);
 +
pair A, B, C, B1, C1, W, WA, WB, WC, X, Y, Z;
 +
A = 18*dir(90);
 +
B = 18*dir(210);
 +
C = 18*dir(330);
 +
B1 = A+24*sqrt(3)*dir(B-A);
 +
C1 = A+24*sqrt(3)*dir(C-A);
 +
W = (0,0);
 +
WA = 6*dir(270);
 +
WB = 6*dir(30);
 +
WC = 6*dir(150);
 +
X = (sqrt(117)-3)*dir(270);
 +
Y = (sqrt(117)-3)*dir(30);
 +
Z = (sqrt(117)-3)*dir(150);
 +
filldraw(X--Y--Z--cycle,green,dashed);
 +
draw(Circle(WA,12)^^Circle(WB,12)^^Circle(WC,12),blue);
 +
draw(Circle(W,18)^^A--B--C--cycle);
 +
draw(B--B1--C1--C,dashed);
 +
dot("$A$",A,1.5*dir(A),linewidth(4));
 +
dot("$B$",B,1.5*(-1,0),linewidth(4));
 +
dot("$C$",C,1.5*(1,0),linewidth(4));
 +
dot("$B'$",B1,1.5*dir(B1),linewidth(4));
 +
dot("$C'$",C1,1.5*dir(C1),linewidth(4));
 +
dot("$O$",W,1.5*dir(90),linewidth(4));
 +
dot("$X$",X,1.5*dir(X),linewidth(4));
 +
dot("$Y$",Y,1.5*dir(Y),linewidth(4));
 +
dot("$Z$",Z,1.5*dir(Z),linewidth(4));
 +
</asy>
 +
Since the diameter of the circle is the height of this triangle, the height of this triangle is <math>36</math>. We can use inradius or equilateral triangle properties to get the inradius of this triangle is <math>12</math> (The incenter is also a centroid in an equilateral triangle, and the distance from a side to the centroid is a third of the height). Therefore, the radius of each of the smaller circles is <math>12</math>.
 +
 
 +
Let <math>O=\omega</math> be the center of the largest circle. We will set up a coordinate system with <math>O</math> as the origin. The center of <math>\omega_A</math> will be at <math>(0,-6)</math> because it is directly beneath <math>O</math> and is the length of the larger radius minus the smaller radius, or <math>18-12 = 6</math>. By rotating this point <math>120^{\circ}</math> around <math>O</math>, we get the center of <math>\omega_B</math>. This means that the magnitude of vector <math>\overrightarrow{O\omega_B}</math> is <math>6</math> and is at a <math>30</math> degree angle from the horizontal. Therefore, the coordinates of this point are <math>(3\sqrt{3},3)</math> and by symmetry the coordinates of the center of <math>\omega_C</math> is <math>(-3\sqrt{3},3)</math>.
  
−
pair X = (0, 3-sqrt(117));
+
The upper left and right circles intersect at two points, the lower of which is <math>X</math>. The equations of these two circles are:
−
pair Y = ( (sqrt(351)-sqrt(27))/2, (sqrt(117)-3)/2 );
+
<cmath>\begin{align*}
−
pair Z = ( (sqrt(27) - sqrt(351))/2, (sqrt(117)-3)/2 );
+
(x+3\sqrt3)^2 + (y-3)^2 &= 12^2, \\
−
dot(X);
+
(x-3\sqrt3)^2 + (y-3)^2 &= 12^2.
−
dot(Y);
+
\end{align*}</cmath>
−
dot(Z);
+
We solve this system by subtracting to get <math>x = 0</math>. Plugging back in to the first equation, we have <math>(3\sqrt{3})^2 + (y-3)^2 = 144 \implies (y-3)^2 = 117 \implies y-3 = \pm \sqrt{117} \implies y = 3 \pm \sqrt{117}</math>. Since we know <math>X</math> is the lower solution, we take the negative value to get <math>X = (0,3-\sqrt{117})</math>.
  
−
draw(X--Y--Z--cycle, dashed);
+
We can solve the problem two ways from here. We can find <math>Y</math> by rotation and use the distance formula to find the length, or we can be somewhat more clever. We notice that it is easier to find <math>OX</math> as they lie on the same vertical, <math>\angle XOY</math> is <math>120</math> degrees so we can make use of <math>30-60-90</math> triangles, and <math>OX = OY</math> because <math>O</math> is the center of triangle <math>XYZ</math>. We can draw the diagram as such:
 +
<asy>
 +
/* Made by MRENTHUSIASM */
 +
size(300);
 +
pair A, B, C, B1, C1, W, WA, WB, WC, X, Y, Z;
 +
A = 18*dir(90);
 +
B = 18*dir(210);
 +
C = 18*dir(330);
 +
B1 = A+24*sqrt(3)*dir(B-A);
 +
C1 = A+24*sqrt(3)*dir(C-A);
 +
W = (0,0);
 +
WA = 6*dir(270);
 +
WB = 6*dir(30);
 +
WC = 6*dir(150);
 +
X = (sqrt(117)-3)*dir(270);
 +
Y = (sqrt(117)-3)*dir(30);
 +
Z = (sqrt(117)-3)*dir(150);
 +
filldraw(X--Y--Z--cycle,green,dashed);
 +
draw(Circle(WA,12)^^Circle(WB,12)^^Circle(WC,12),blue);
 +
draw(Circle(W,18)^^A--B--C--cycle);
 +
draw(B--B1--C1--C^^W--X^^W--Y^^W--midpoint(X--Y),dashed);
 +
dot("$A$",A,1.5*dir(A),linewidth(4));
 +
dot("$B$",B,1.5*(-1,0),linewidth(4));
 +
dot("$C$",C,1.5*(1,0),linewidth(4));
 +
dot("$B'$",B1,1.5*dir(B1),linewidth(4));
 +
dot("$C'$",C1,1.5*dir(C1),linewidth(4));
 +
dot("$O$",W,1.5*dir(90),linewidth(4));
 +
dot("$X$",X,1.5*dir(X),linewidth(4));
 +
dot("$Y$",Y,1.5*dir(Y),linewidth(4));
 +
dot("$Z$",Z,1.5*dir(Z),linewidth(4));
 
</asy>
 
</asy>
 +
Note that <math>OX = OY = \sqrt{117} - 3</math>. It follows that
 +
<cmath>\begin{align*}
 +
XY &= 2 \cdot \frac{OX\cdot\sqrt{3}}{2} \\
 +
&= OX \cdot \sqrt{3} \\
 +
&= (\sqrt{117}-3) \cdot \sqrt{3} \\
 +
&= \sqrt{351}-\sqrt{27}.
 +
\end{align*}</cmath>
 +
Finally, the answer is <math>351+27 = \boxed{378}</math>.
  
−
==Solution 1==
+
~KingRavi
  
−
We can extend <math>AB</math> and <math>AC</math> to <math>B'</math> and <math>C'</math> respectively such that circle <math>\omega_A</math> is the incircle of <math>\triangle AB'C'</math>.
+
==Solution 2 (Euclidean Geometry)==
  
 
<asy>
 
<asy>
−
unitsize(0.3cm);
+
/* Made by MRENTHUSIASM */
−
draw(circle((0,0),18));
+
/* Modified by isabelchen */
−
pair C = (9 * sqrt(3), -9);
+
size(250);
−
pair B = (-9 * sqrt(3), -9);
+
pair A, B, C, W, WA, WB, WC, X, Y, Z, D, E;
−
pair B2 = (-12 * sqrt(3), -18);
+
A = 18*dir(90);
−
pair C2 = (12 * sqrt(3), -18);
+
B = 18*dir(210);
 +
C = 18*dir(330);
 +
W = (0,0);
 +
WA = 6*dir(270);
 +
WB = 6*dir(30);
 +
WC = 6*dir(150);
 +
X = (sqrt(117)-3)*dir(270);
 +
Y = (sqrt(117)-3)*dir(30);
 +
Z = (sqrt(117)-3)*dir(150);
 +
D = intersectionpoint(Circle(WA,12),A--C);
 +
E = intersectionpoints(Circle(WB,12),Circle(WC,12))[0];
 +
filldraw(X--Y--Z--cycle,green,dashed);
 +
draw(Circle(WA,12)^^Circle(WB,12)^^Circle(WC,12),blue);
 +
draw(Circle(W,18)^^A--B--C--cycle);
 +
dot("$A$",A,1.5*dir(A),linewidth(4));
 +
dot("$B$",B,1.5*dir(B),linewidth(4));
 +
dot("$C$",C,1.5*dir(C),linewidth(4));
 +
dot("$\omega$",W,1.5*dir(270),linewidth(4));
 +
dot("$\omega_A$",WA,1.5*dir(-WA),linewidth(4));
 +
dot("$\omega_B$",WB,1.5*dir(-WB),linewidth(4));
 +
dot("$\omega_C$",WC,1.5*dir(-WC),linewidth(4));
 +
dot("$X$",X,1.5*dir(X),linewidth(4));
 +
dot("$Y$",Y,1.5*dir(Y),linewidth(4));
 +
dot("$Z$",Z,1.5*dir(Z),linewidth(4));
 +
dot("$E$",E,1.5*dir(E),linewidth(4));
 +
dot("$D$",D,1.5*dir(D),linewidth(4));
 +
draw(WC--WB^^WC--X^^WC--E^^WA--D^^A--X);
 +
</asy>
 +
For equilateral triangle with side length <math>l</math>, height <math>h</math>, and circumradius <math>r</math>, there are relationships: <math>h = \frac{\sqrt{3}}{2} l</math>, <math>r = \frac{2}{3} h = \frac{\sqrt{3}}{3} l</math>, and <math>l = \sqrt{3}r</math>.
  
−
pair A = (0,18);
+
There is a lot of symmetry in the figure. The radius of the big circle <math>\odot \omega</math> is <math>R = 18</math>, let the radius of the small circles <math>\odot \omega_A</math>, <math>\odot \omega_B</math>, <math>\odot \omega_C</math> be <math>r</math>.
−
draw(A--B--C--cycle);
 
−
draw(circle((0,-6),12), gray);
 
−
draw(circle((3*sqrt(3),3),12), gray);
 
−
draw(circle((-3*sqrt(3),3),12), gray);
 
  
−
draw(B--B2,dashed);
+
We are going to solve this problem in <math>3</math> steps:
−
draw(C--C2,dashed);
 
−
draw(B2--C2,dashed);
 
  
−
dot(B2);
+
<math>\textbf{Step 1:}</math>
−
dot(C2);
 
  
−
pair X = (0, 3-sqrt(117));
+
We have <math>\triangle A \omega_A D</math> is a <math>30-60-90</math> triangle, and <math>A \omega_A = 2 \cdot \omega_A D</math>, <math>A \omega_A = 2R-r</math> (<math>\odot \omega</math> and <math>\odot \omega_A</math> are tangent), and <math>\omega_A D = r</math>. So, we get <math>2R-r = 2r</math> and <math>r = \frac{2}{3} \cdot R = 12</math>.
−
pair Y = ( (sqrt(351)-sqrt(27))/2, (sqrt(117)-3)/2 );
 
−
pair Z = ( (sqrt(27) - sqrt(351))/2, (sqrt(117)-3)/2 );
 
−
dot(X);
 
−
dot(Y);
 
−
dot(Z);
 
  
−
label("$A$",A,N);
+
Since <math>\odot \omega</math> and <math>\odot \omega_A</math> are tangent, we get <math>\omega \omega_A = R - r = \frac{1}{3} \cdot R = 6</math>.
−
label("$B$",B,W);
 
−
label("$C$",C,E);
 
−
label("$B'$",B2,W);
 
−
label("$C'$",C2,E);
 
−
label("$X$",X,S);
 
−
label("$Y$",Y,E);
 
−
label("$Z$",Z,W);
 
  
 +
Note that <math>\triangle \omega_A \omega_B \omega_C</math> is an equilateral triangle, and <math>\omega</math> is its center, so <math>\omega_B \omega_C = \sqrt{3} \cdot \omega \omega_A = 6 \sqrt{3}</math>.
  
−
draw(X--Y--Z--cycle, dashed);
+
<math>\textbf{Step 2:}</math>
−
</asy>
+
 
 +
Note that <math>\triangle \omega_C E X</math> is an isosceles triangle, so <cmath>EX = 2 \sqrt{(\omega_C E)^2 - \left(\frac{\omega_B \omega_C}{2}\right)^2} = 2 \sqrt{r^2 - \left(\frac{\omega_B \omega_C}{2}\right)^2} = 2 \sqrt{12^2 - (3 \sqrt{3})^2} = 2 \sqrt{117}.</cmath>
 +
 
 +
<math>\textbf{Step 3:}</math>
 +
 
 +
In <math>\odot \omega_C</math>, Power of a Point gives <math>\omega X \cdot \omega E = r^2 - (\omega_C \omega)^2</math> and <math>\omega E = EX - \omega X = 2\sqrt{117} - \omega X</math>.
 +
 
 +
It follows that <math>\omega X \cdot (2\sqrt{117} - \omega X) = 12^2 - 6^2</math>. We solve this quadratic equation: <math>\omega X = \sqrt{117} - 3</math>.
 +
 
 +
Since <math>\omega X</math> is the circumradius of equilateral <math>\triangle XYZ</math>, we have <math>XY = \sqrt{3} \cdot \omega X = \sqrt{3} \cdot (\sqrt{117} - 3) = \sqrt{351}-\sqrt{27}</math>.
 +
 
 +
Therefore, the answer is <math>351+27 = \boxed{378}</math>.
 +
 
 +
~[https://artofproblemsolving.com/wiki/index.php/User:Isabelchen isabelchen]
 +
 
 +
==Solution 3 (Simple Geometry)==
 +
[[File:AIME 2022 I 7.png|500px|right]]
 +
Let <math>O</math> be the center, <math>R = 18</math> be the radius, and <math>CC'</math> be the diameter of <math>\omega.</math>
 +
Let <math>r</math>  be the radius, <math>E,D,F</math> are the centers of <math>\omega_A, \omega_B,\omega_C.</math>
 +
Let <math>KGH</math> be the desired triangle with side  <math>x.</math>
 +
We find <math>r</math> using
 +
<cmath>CC' = 2R = C'K + KC = r + \frac{r}{\sin 30^\circ} = 3r.</cmath>
 +
<cmath>r = \frac{2R}{3} = 12.</cmath>
 +
<cmath>OE = R – r = 6.</cmath>
 +
Triangles <math>\triangle DEF</math> and <math>\triangle KGH</math> – are equilateral triangles with a common center <math>O,</math> therefore in the triangle  <math>OEH</math> <math>OE = 6, \angle EOH = 120^\circ, OH = \frac{x}{\sqrt3}.</math>
 +
 +
We apply the Law of Cosines to <math>\triangle OEH</math> and get
 +
<cmath>OE^2 + OH^2 + OE \cdot OH = EH^2.</cmath>
 +
<cmath>6^2 + \frac{x^2}{3} + \frac{6x}{\sqrt3} = 12^2.</cmath>
 +
<cmath>x^2 + 6x \sqrt{3} = 324</cmath>
 +
<cmath>x= \sqrt{351} - \sqrt{27}  \implies 351 + 27 = \boxed {378}</cmath>
 +
 
 +
'''vladimir.shelomovskii@gmail.com, vvsss'''
 +
 
 +
==Solution 4 (Mixtilinear Incircles)==
 +
Let <math>O</math> be the center of <math>\omega</math>, <math>X</math> be the intersection of <math>\omega_B,\omega_C</math> further from <math>A</math>, and <math>O_A</math> be the center of <math>\omega_A</math>. Define <math>Y, Z, O_B, O_C</math> similarly. It is well-known that the <math>A</math>-mixtilinear inradius <math>R_A</math> is <math>\tfrac{r}{\cos^2\left(\frac{\angle A}{2}\right)} = \tfrac{9}{\cos^2\left(30^{\circ}\right)} = 12</math>, so in particular this means that <math>OO_B = 18 - R_B = 6 = OO_C</math>. Since <math>\angle O_BOO_C = \angle BOC = 120^\circ</math>, it follows by Law of Cosines on <math>\triangle OO_BO_C</math> that <math>O_BO_C = 6\sqrt{3}</math>. Then the Pythagorean theorem gives that the altitude of <math>O_BO_CX</math> is <math>\sqrt{117}</math>, so <math>OY = OX = \text{dist}(X, YZ) - \text{dist}(O, YZ) = \sqrt{117} - 3</math> and <math>YZ = \tfrac{O_BO_C\cdot OY}{OO_B} = \tfrac{6\sqrt{3}(\sqrt{117} - 3)}{6}=\sqrt{351} - \sqrt{27}</math> so the answer is <math>351 + 27 = \boxed{378}</math>.
 +
 
 +
~Kagebaka
  
 +
==Video Solution==
  
−
Since the diameter of the circle is the height of this triangle, the height of this triangle is 36. We can use inradius or equilateral triangle properties to get the inradius of this trianglee is 12 (The incenter is also a centroid in an equilateral triangle, and the distance from a side to the centroid is a third of the height). Therefore, the radius of each of the smaller circles is 12.
+
https://youtu.be/q6_LslAfFpI
  
−
==Solution 2==
+
~MathProblemSolvingSkills.com
  
−
Let bottom left point as the origin, the radius of each circle is <math>36/3=12</math>, note that three centers for circles are <math>(9\sqrt{3},3),(12\sqrt{3},12),(6\sqrt{3},12)</math>
+
==Video Solution==
  
−
It is not hard to find that one intersection point lies on <math>\frac{\sqrt{3}x}{3}</math> since the intersection must lie on the angle bisector of the bigger triangle, plug it into equation
+
https://youtu.be/NTbdG4IiCRY
−
<math>(x-9\sqrt{3})^2+(\frac{\sqrt{3}x}{3}-3)^2=12^2</math>, getting that <math>x=\frac{15\sqrt{3}+3\sqrt{39}}{2}</math>, the length is <math>2*(\frac{15\sqrt{3}+3\sqrt{39}-18\sqrt{3}}{2}=3\sqrt{39}-3\sqrt{3}</math>, leads to the answer <math>378</math>
 
  
−
~bluesoul
+
~AMC & AIME Training
  
 
==See Also==
 
==See Also==
 
{{AIME box|year=2022|n=I|num-b=7|num-a=9}}
 
{{AIME box|year=2022|n=I|num-b=7|num-a=9}}
 +
 +
[[Category:Intermediate Geometry Problems]]
 
{{MAA Notice}}
 
{{MAA Notice}}

Latest revision as of 20:11, 31 January 2024

Problem

Equilateral triangle $\triangle ABC$ is inscribed in circle $\omega$ with radius $18.$ Circle $\omega_A$ is tangent to sides $\overline{AB}$ and $\overline{AC}$ and is internally tangent to $\omega.$ Circles $\omega_B$ and $\omega_C$ are defined analogously. Circles $\omega_A,$ $\omega_B,$ and $\omega_C$ meet in six points---two points for each pair of circles. The three intersection points closest to the vertices of $\triangle ABC$ are the vertices of a large equilateral triangle in the interior of $\triangle ABC,$ and the other three intersection points are the vertices of a smaller equilateral triangle in the interior of $\triangle ABC.$ The side length of the smaller equilateral triangle can be written as $\sqrt{a} - \sqrt{b},$ where $a$ and $b$ are positive integers. Find $a+b.$

Diagram

[asy] /* Made by MRENTHUSIASM */ size(250); pair A, B, C, W, WA, WB, WC, X, Y, Z; A = 18*dir(90); B = 18*dir(210); C = 18*dir(330); W = (0,0); WA = 6*dir(270); WB = 6*dir(30); WC = 6*dir(150); X = (sqrt(117)-3)*dir(270); Y = (sqrt(117)-3)*dir(30); Z = (sqrt(117)-3)*dir(150); filldraw(X--Y--Z--cycle,green,dashed); draw(Circle(WA,12)^^Circle(WB,12)^^Circle(WC,12),blue); draw(Circle(W,18)^^A--B--C--cycle); dot("$A$",A,1.5*dir(A),linewidth(4)); dot("$B$",B,1.5*dir(B),linewidth(4)); dot("$C$",C,1.5*dir(C),linewidth(4)); dot("$\omega$",W,1.5*dir(270),linewidth(4)); dot("$\omega_A$",WA,1.5*dir(-WA),linewidth(4)); dot("$\omega_B$",WB,1.5*dir(-WB),linewidth(4)); dot("$\omega_C$",WC,1.5*dir(-WC),linewidth(4)); [/asy] ~MRENTHUSIASM ~ihatemath123

Solution 1 (Coordinate Geometry)

We can extend $AB$ and $AC$ to $B'$ and $C'$ respectively such that circle $\omega_A$ is the incircle of $\triangle AB'C'$. [asy] /* Made by MRENTHUSIASM */ size(300); pair A, B, C, B1, C1, W, WA, WB, WC, X, Y, Z; A = 18*dir(90); B = 18*dir(210); C = 18*dir(330); B1 = A+24*sqrt(3)*dir(B-A); C1 = A+24*sqrt(3)*dir(C-A); W = (0,0); WA = 6*dir(270); WB = 6*dir(30); WC = 6*dir(150); X = (sqrt(117)-3)*dir(270); Y = (sqrt(117)-3)*dir(30); Z = (sqrt(117)-3)*dir(150); filldraw(X--Y--Z--cycle,green,dashed); draw(Circle(WA,12)^^Circle(WB,12)^^Circle(WC,12),blue); draw(Circle(W,18)^^A--B--C--cycle); draw(B--B1--C1--C,dashed); dot("$A$",A,1.5*dir(A),linewidth(4)); dot("$B$",B,1.5*(-1,0),linewidth(4)); dot("$C$",C,1.5*(1,0),linewidth(4)); dot("$B'$",B1,1.5*dir(B1),linewidth(4)); dot("$C'$",C1,1.5*dir(C1),linewidth(4)); dot("$O$",W,1.5*dir(90),linewidth(4)); dot("$X$",X,1.5*dir(X),linewidth(4)); dot("$Y$",Y,1.5*dir(Y),linewidth(4)); dot("$Z$",Z,1.5*dir(Z),linewidth(4)); [/asy] Since the diameter of the circle is the height of this triangle, the height of this triangle is $36$. We can use inradius or equilateral triangle properties to get the inradius of this triangle is $12$ (The incenter is also a centroid in an equilateral triangle, and the distance from a side to the centroid is a third of the height). Therefore, the radius of each of the smaller circles is $12$.

Let $O=\omega$ be the center of the largest circle. We will set up a coordinate system with $O$ as the origin. The center of $\omega_A$ will be at $(0,-6)$ because it is directly beneath $O$ and is the length of the larger radius minus the smaller radius, or $18-12 = 6$. By rotating this point $120^{\circ}$ around $O$, we get the center of $\omega_B$. This means that the magnitude of vector $\overrightarrow{O\omega_B}$ is $6$ and is at a $30$ degree angle from the horizontal. Therefore, the coordinates of this point are $(3\sqrt{3},3)$ and by symmetry the coordinates of the center of $\omega_C$ is $(-3\sqrt{3},3)$.

The upper left and right circles intersect at two points, the lower of which is $X$. The equations of these two circles are: \begin{align*} (x+3\sqrt3)^2 + (y-3)^2 &= 12^2, \\ (x-3\sqrt3)^2 + (y-3)^2 &= 12^2. \end{align*} We solve this system by subtracting to get $x = 0$. Plugging back in to the first equation, we have $(3\sqrt{3})^2 + (y-3)^2 = 144 \implies (y-3)^2 = 117 \implies y-3 = \pm \sqrt{117} \implies y = 3 \pm \sqrt{117}$. Since we know $X$ is the lower solution, we take the negative value to get $X = (0,3-\sqrt{117})$.

We can solve the problem two ways from here. We can find $Y$ by rotation and use the distance formula to find the length, or we can be somewhat more clever. We notice that it is easier to find $OX$ as they lie on the same vertical, $\angle XOY$ is $120$ degrees so we can make use of $30-60-90$ triangles, and $OX = OY$ because $O$ is the center of triangle $XYZ$. We can draw the diagram as such: [asy] /* Made by MRENTHUSIASM */ size(300); pair A, B, C, B1, C1, W, WA, WB, WC, X, Y, Z; A = 18*dir(90); B = 18*dir(210); C = 18*dir(330); B1 = A+24*sqrt(3)*dir(B-A); C1 = A+24*sqrt(3)*dir(C-A); W = (0,0); WA = 6*dir(270); WB = 6*dir(30); WC = 6*dir(150); X = (sqrt(117)-3)*dir(270); Y = (sqrt(117)-3)*dir(30); Z = (sqrt(117)-3)*dir(150); filldraw(X--Y--Z--cycle,green,dashed); draw(Circle(WA,12)^^Circle(WB,12)^^Circle(WC,12),blue); draw(Circle(W,18)^^A--B--C--cycle); draw(B--B1--C1--C^^W--X^^W--Y^^W--midpoint(X--Y),dashed); dot("$A$",A,1.5*dir(A),linewidth(4)); dot("$B$",B,1.5*(-1,0),linewidth(4)); dot("$C$",C,1.5*(1,0),linewidth(4)); dot("$B'$",B1,1.5*dir(B1),linewidth(4)); dot("$C'$",C1,1.5*dir(C1),linewidth(4)); dot("$O$",W,1.5*dir(90),linewidth(4)); dot("$X$",X,1.5*dir(X),linewidth(4)); dot("$Y$",Y,1.5*dir(Y),linewidth(4)); dot("$Z$",Z,1.5*dir(Z),linewidth(4)); [/asy] Note that $OX = OY = \sqrt{117} - 3$. It follows that \begin{align*} XY &= 2 \cdot \frac{OX\cdot\sqrt{3}}{2} \\ &= OX \cdot \sqrt{3} \\ &= (\sqrt{117}-3) \cdot \sqrt{3} \\ &= \sqrt{351}-\sqrt{27}. \end{align*} Finally, the answer is $351+27 = \boxed{378}$.

~KingRavi

Solution 2 (Euclidean Geometry)

[asy] /* Made by MRENTHUSIASM */ /* Modified by isabelchen */ size(250); pair A, B, C, W, WA, WB, WC, X, Y, Z, D, E; A = 18*dir(90); B = 18*dir(210); C = 18*dir(330); W = (0,0); WA = 6*dir(270); WB = 6*dir(30); WC = 6*dir(150); X = (sqrt(117)-3)*dir(270); Y = (sqrt(117)-3)*dir(30); Z = (sqrt(117)-3)*dir(150); D = intersectionpoint(Circle(WA,12),A--C); E = intersectionpoints(Circle(WB,12),Circle(WC,12))[0]; filldraw(X--Y--Z--cycle,green,dashed); draw(Circle(WA,12)^^Circle(WB,12)^^Circle(WC,12),blue); draw(Circle(W,18)^^A--B--C--cycle); dot("$A$",A,1.5*dir(A),linewidth(4)); dot("$B$",B,1.5*dir(B),linewidth(4)); dot("$C$",C,1.5*dir(C),linewidth(4)); dot("$\omega$",W,1.5*dir(270),linewidth(4)); dot("$\omega_A$",WA,1.5*dir(-WA),linewidth(4)); dot("$\omega_B$",WB,1.5*dir(-WB),linewidth(4)); dot("$\omega_C$",WC,1.5*dir(-WC),linewidth(4)); dot("$X$",X,1.5*dir(X),linewidth(4)); dot("$Y$",Y,1.5*dir(Y),linewidth(4)); dot("$Z$",Z,1.5*dir(Z),linewidth(4)); dot("$E$",E,1.5*dir(E),linewidth(4)); dot("$D$",D,1.5*dir(D),linewidth(4)); draw(WC--WB^^WC--X^^WC--E^^WA--D^^A--X); [/asy] For equilateral triangle with side length $l$, height $h$, and circumradius $r$, there are relationships: $h = \frac{\sqrt{3}}{2} l$, $r = \frac{2}{3} h = \frac{\sqrt{3}}{3} l$, and $l = \sqrt{3}r$.

There is a lot of symmetry in the figure. The radius of the big circle $\odot \omega$ is $R = 18$, let the radius of the small circles $\odot \omega_A$, $\odot \omega_B$, $\odot \omega_C$ be $r$.

We are going to solve this problem in $3$ steps:

$\textbf{Step 1:}$

We have $\triangle A \omega_A D$ is a $30-60-90$ triangle, and $A \omega_A = 2 \cdot \omega_A D$, $A \omega_A = 2R-r$ ($\odot \omega$ and $\odot \omega_A$ are tangent), and $\omega_A D = r$. So, we get $2R-r = 2r$ and $r = \frac{2}{3} \cdot R = 12$.

Since $\odot \omega$ and $\odot \omega_A$ are tangent, we get $\omega \omega_A = R - r = \frac{1}{3} \cdot R = 6$.

Note that $\triangle \omega_A \omega_B \omega_C$ is an equilateral triangle, and $\omega$ is its center, so $\omega_B \omega_C = \sqrt{3} \cdot \omega \omega_A = 6 \sqrt{3}$.

$\textbf{Step 2:}$

Note that $\triangle \omega_C E X$ is an isosceles triangle, so \[EX = 2 \sqrt{(\omega_C E)^2 - \left(\frac{\omega_B \omega_C}{2}\right)^2} = 2 \sqrt{r^2 - \left(\frac{\omega_B \omega_C}{2}\right)^2} = 2 \sqrt{12^2 - (3 \sqrt{3})^2} = 2 \sqrt{117}.\]

$\textbf{Step 3:}$

In $\odot \omega_C$, Power of a Point gives $\omega X \cdot \omega E = r^2 - (\omega_C \omega)^2$ and $\omega E = EX - \omega X = 2\sqrt{117} - \omega X$.

It follows that $\omega X \cdot (2\sqrt{117} - \omega X) = 12^2 - 6^2$. We solve this quadratic equation: $\omega X = \sqrt{117} - 3$.

Since $\omega X$ is the circumradius of equilateral $\triangle XYZ$, we have $XY = \sqrt{3} \cdot \omega X = \sqrt{3} \cdot (\sqrt{117} - 3) = \sqrt{351}-\sqrt{27}$.

Therefore, the answer is $351+27 = \boxed{378}$.

~isabelchen

Solution 3 (Simple Geometry)

AIME 2022 I 7.png

Let $O$ be the center, $R = 18$ be the radius, and $CC'$ be the diameter of $\omega.$ Let $r$ be the radius, $E,D,F$ are the centers of $\omega_A, \omega_B,\omega_C.$ Let $KGH$ be the desired triangle with side $x.$ We find $r$ using \[CC' = 2R = C'K + KC = r + \frac{r}{\sin 30^\circ} = 3r.\] \[r = \frac{2R}{3} = 12.\] \[OE = R – r = 6.\] Triangles $\triangle DEF$ and $\triangle KGH$ – are equilateral triangles with a common center $O,$ therefore in the triangle $OEH$ $OE = 6, \angle EOH = 120^\circ, OH = \frac{x}{\sqrt3}.$

We apply the Law of Cosines to $\triangle OEH$ and get \[OE^2 + OH^2 + OE \cdot OH = EH^2.\] \[6^2 + \frac{x^2}{3} + \frac{6x}{\sqrt3} = 12^2.\] \[x^2 + 6x \sqrt{3} = 324\] \[x= \sqrt{351} - \sqrt{27} \implies 351 + 27 = \boxed {378}\]

vladimir.shelomovskii@gmail.com, vvsss

Solution 4 (Mixtilinear Incircles)

Let $O$ be the center of $\omega$, $X$ be the intersection of $\omega_B,\omega_C$ further from $A$, and $O_A$ be the center of $\omega_A$. Define $Y, Z, O_B, O_C$ similarly. It is well-known that the $A$-mixtilinear inradius $R_A$ is $\tfrac{r}{\cos^2\left(\frac{\angle A}{2}\right)} = \tfrac{9}{\cos^2\left(30^{\circ}\right)} = 12$, so in particular this means that $OO_B = 18 - R_B = 6 = OO_C$. Since $\angle O_BOO_C = \angle BOC = 120^\circ$, it follows by Law of Cosines on $\triangle OO_BO_C$ that $O_BO_C = 6\sqrt{3}$. Then the Pythagorean theorem gives that the altitude of $O_BO_CX$ is $\sqrt{117}$, so $OY = OX = \text{dist}(X, YZ) - \text{dist}(O, YZ) = \sqrt{117} - 3$ and $YZ = \tfrac{O_BO_C\cdot OY}{OO_B} = \tfrac{6\sqrt{3}(\sqrt{117} - 3)}{6}=\sqrt{351} - \sqrt{27}$ so the answer is $351 + 27 = \boxed{378}$.

~Kagebaka

Video Solution

https://youtu.be/q6_LslAfFpI

~MathProblemSolvingSkills.com

Video Solution

https://youtu.be/NTbdG4IiCRY

~AMC & AIME Training

See Also

2022 AIME I (Problems • Answer Key • Resources)
Preceded by
Problem 7 		Followed by
Problem 9
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=2022_AIME_I_Problems/Problem_8&oldid=213992"