AIME 2022 I · 第 8 题

Solution 1 (Coordinate Geometry)

We can extend $AB$ and $AC$ to $B'$ and $C'$ respectively such that circle $\omega_A$ is the incircle of $\triangle AB'C'$.

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:

$$\begin{aligned} (x+3\sqrt3)^2 + (y-3)^2 &= 12^2, \\ (x-3\sqrt3)^2 + (y-3)^2 &= 12^2. \end{aligned}$$ 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:

Note that $OX = OY = \sqrt{117} - 3$. It follows that

$$\begin{aligned} XY &= 2 \cdot \frac{OX\cdot\sqrt{3}}{2} \\ &= OX \cdot \sqrt{3} \\ &= (\sqrt{117}-3) \cdot \sqrt{3} \\ &= \sqrt{351}-\sqrt{27}. \end{aligned}$$ Finally, the answer is $351+27 = \boxed{378}$.

~KingRavi

Solution 2 (Euclidean Geometry)

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)

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

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