2021 AMC Fall 10A Problem 22

Break the 3D picture into two flat pictures: one looking down from above, and one slice through the side.

Looking down from above.

1. Each sphere sits on the base, so each center is at height $r$.
2. Each pair of spheres touches, so their centers are $2r$ apart. The three centers form an equilateral triangle with side $2r$, and by symmetry its center is on the cone's middle line (the axis).
3. The distance from the center of an equilateral triangle to a corner is $\frac{\text{side}}{\sqrt3}$. (For side $s$, the height is $\frac{\sqrt3}{2}s$, and the center is $\frac23$ of the way down it: $\frac23 \times \frac{\sqrt3}{2}s = \frac{s}{\sqrt3}$.) So each sphere's center is
$$d = \frac{2r}{\sqrt3}$$
away from the axis.

Slicing through the axis and one center.

4. The slice of the cone is a triangle with base from $x = -5$ to $x = 5$ and top at $(0, 12)$. Its right side goes from $(5, 0)$ to $(0, 12)$ and has length $13$ (a $5$-$12$-$13$ right triangle).
5. The sphere shows up as a circle of radius $r$ with center $(d, r)$, touching the right side.
6. Find where the side is at height $r$. The side drops from $x = 5$ at height $0$ toward $x = 0$ at height $12$, moving in $\frac{5}{12}$ for each unit up. At height $r$ it is at $x = 5 - \frac{5r}{12}$. So the horizontal gap from the center to the side is
$$g = 5 - \frac{5r}{12} - d.$$
7. Turn the horizontal gap into the true distance. The circle touches the side, so the shortest distance from the center to the side is $r$. The side makes the same angles as the $5$-$12$-$13$ triangle, so the shortest distance is $\frac{12}{13}$ of the horizontal gap:
$$\frac{12}{13}\left(5 - \frac{5r}{12} - d\right) = r.$$
8. Multiply by $13$: $60 - 5r - 12d = 13r$, so $60 = 18r + 12d$.

Solve.

9. Put in $d = \frac{2r}{\sqrt3}$: $12d = \frac{24r}{\sqrt3} = 8\sqrt3\, r$. So $60 = (18 + 8\sqrt3)\, r$ and
$$r = \frac{60}{18 + 8\sqrt3} = \frac{30}{9 + 4\sqrt3}.$$
10. Remove the root from the bottom by multiplying top and bottom by $9 - 4\sqrt3$. The bottom becomes $81 - 48 = 33$:
$$r = \frac{30(9 - 4\sqrt3)}{33} = \frac{10(9 - 4\sqrt3)}{11} = \frac{90 - 40\sqrt3}{11}.$$

As a check, this is about $\frac{90 - 69.3}{11} \approx 1.9$, a sensible size.

The answer is $\boxed{\textbf{(B)}\ \frac{90-40\sqrt{3}}{11}}$.

Spheres touching each other and a cone are hard to picture all at once, so split the problem. From above, the centers make a regular shape, which tells you how far each is from the axis. In a slice through the axis, each sphere is a circle touching a straight line, and a $5$-$12$-$13$ triangle turns that into one equation. Most 3D ball-packing problems work the same way.

Use the distance from a point to a line. The right side of the slice is the line $12x + 5y = 60$. The distance from $(d, r)$ to this line is $\frac{60 - 12d - 5r}{13}$. Setting it equal to $r$ gives $60 = 18r + 12d$, the same equation as in step 8.

Locate the centers. Each sphere touches the base, so every center is at height $r$. The spheres are mutually tangent, so the centers are pairwise $2r$ apart: they form an equilateral triangle of side $2r$ in the horizontal plane at height $r$, centered on the cone's axis by symmetry. The distance from a center to the axis is the circumradius of that triangle,
$$d = \frac{2r}{\sqrt3} = \frac{2\sqrt3}{3}\,r.$$

Slice through the axis and one center. The cross-section of the cone is a triangle with apex at height $12$ and base from $-5$ to $5$; its right slant side is the line through $(5, 0)$ and $(0, 12)$, i.e. $12x + 5y = 60$ (a $5$-$12$-$13$ triangle). The sphere appears as a circle of radius $r$ centered at $(d, r)$, tangent to this line, so the distance from the center to the line equals $r$:
$$\frac{60 - 12d - 5r}{13} = r \quad\Longrightarrow\quad 60 = 18r + 12d = 18r + 8\sqrt3\, r.$$

Solve.
$$r = \frac{60}{18 + 8\sqrt3} = \frac{30}{9 + 4\sqrt3} = \frac{30(9 - 4\sqrt3)}{81 - 48} = \frac{30(9 - 4\sqrt3)}{33} = \frac{90 - 40\sqrt3}{11}.$$

The answer is $\boxed{\textbf{(B)}\ \frac{90-40\sqrt{3}}{11}}$.