Three circles of equal radii touch (but not cross) each other externally. Two other circles, X and Y, are drawn such that both touch (but not cross) each of the three previous circles. If the radius of X is more than that of Y, the ratio of the radii of X and Y is

Imagine three identical coins lying flat on a table, each touching the other two. Now we want to find two more circles:

• Circle $X$: A big circle that wraps around all three coins (touching them from outside)
• Circle $Y$: A small circle that fits in the middle gap (touching all three from inside)

Let's call the radius of each original circle $r$.

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When three equal circles touch each other externally, their centers form a perfect equilateral triangle!

Why? Because if two circles of radius $r$ touch externally, the distance between their centers $= r + r = 2r$

So our equilateral triangle has side length $= 2r$

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Both circles $X$ and $Y$ must be centered at point $O$ (the exact middle of the equilateral triangle) due to symmetry.

In an equilateral triangle with side length $s$, the distance from the center to any corner is:

$\text{Distance} = \dfrac{s}{\sqrt{3}}$

For our triangle with side length $2r$:

$\text{Distance from O to any original circle center} = \dfrac{2r}{\sqrt{3}} = \dfrac{2r\sqrt{3}}{3}$

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Circle $Y$ sits inside, touching all three circles.

Starting from center $O$:

• Distance to any original circle's center $= \dfrac{2r\sqrt{3}}{3}$
• But we need to stop $r$ units before reaching the center (since the original circle has radius $r$)

$\text{Radius of Y} = \dfrac{2r\sqrt{3}}{3} - r = \dfrac{2r\sqrt{3} - 3r}{3} = \dfrac{r(2\sqrt{3} - 3)}{3}$

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Circle $X$ sits outside, touching all three circles.

Starting from center $O$:

• Distance to any original circle's center $= \dfrac{2r\sqrt{3}}{3}$
• Continue $r$ more units beyond the center to touch the far edge

$\text{Radius of X} = \dfrac{2r\sqrt{3}}{3} + r = \dfrac{2r\sqrt{3} + 3r}{3} = \dfrac{r(2\sqrt{3} + 3)}{3}$

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$\dfrac{\text{Radius of X}}{\text{Radius of Y}} = \dfrac{\dfrac{r(2\sqrt{3} + 3)}{3}}{\dfrac{r(2\sqrt{3} - 3)}{3}}$

The $r$'s and $3$'s cancel:

$= \dfrac{2\sqrt{3} + 3}{2\sqrt{3} - 3}$

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Multiply top and bottom by $(2\sqrt{3} + 3)$:

Numerator: $(2\sqrt{3} + 3)^2 = (2\sqrt{3})^2 + 2(2\sqrt{3})(3) + 3^2 = 12 + 12\sqrt{3} + 9 = 21 + 12\sqrt{3}$

Denominator: $(2\sqrt{3} - 3)(2\sqrt{3} + 3) = (2\sqrt{3})^2 - 3^2 = 12 - 9 = 3$

$\dfrac{\text{Radius of X}}{\text{Radius of Y}} = \dfrac{21 + 12\sqrt{3}}{3} = \dfrac{3(7 + 4\sqrt{3})}{3} = 7 + 4\sqrt{3}$