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2022 AIME I Problems/Problem 8

Revision as of 15:42, 28 March 2022 by Inventivedant (talk | contribs)

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

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

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