2026 AMC 10A Problem 19

Look down at the table from above. The problem becomes a flat question about a circle drawn on the table, plus one use of Pythagoras for the height of the cubes.

1. Draw the footprint. Put coordinates on the table.
- The two joined cubes stand side by side and together cover a $4 \times 2$ rectangle: $x$ from $-2$ to $2$ and $y$ from $-1$ to $1$. Their shared face stands above the segment $x = 0$.
- That shared face has two vertical edges, standing above the points $(0, 1)$ and $(0, -1)$. The third cube's face is centered on the edge above $(0, 1)$. The face is $2$ wide, so it runs from $x = -1$ to $x = 1$, and the third cube sticks out beyond the pair: $x$ from $-1$ to $1$ and $y$ from $1$ to $3$.

From above, the cubes form a T shape: a $4 \times 2$ rectangle with a $2 \times 2$ square attached to the middle of one long side.

1. What "under the cover" means. The cover is half of a ball with radius $R$. Its rim rests on the table, so its center $P$ is a point on the table. A point is under the cover exactly when its distance from $P$ is at most $R$.

3. Only the top corners matter. The point of a box farthest from $P$ is always one of its corners. A corner on top of a cube, at height $2$, stands directly above a corner $Q$ of the footprint. The flat distance $PQ$ and the vertical edge of length $2$ meet at a right angle, so by Pythagoras that top corner is at distance $\sqrt{PQ^2 + 2^2}$ from $P$. The bottom corner at $Q$ is closer. So we need
$$R^2 \ge PQ^2 + 4 \quad \text{for every corner } Q \text{ of the footprint.}$$
In other words: find the smallest circle on the table that surrounds the whole T (call its radius $r$), and then $R^2 = r^2 + 4$.

1. Use the symmetry. The T is symmetric about the $y$-axis, so try a center on that line, $P = (0, c)$. (Step 8 checks that moving off the line cannot help.)

5. The corners that pull hardest. The far corners of the T are the bottom corners $(\pm 2, -1)$ and the top corners $(\pm 1, 3)$. Their squared distances from $(0, c)$ are
$$\text{bottom: } 4 + (c + 1)^2, \qquad \text{top: } 1 + (3 - c)^2.$$
Moving the center up brings the top corners closer but pushes the bottom corners away, and moving it down does the opposite. The best spot is where the two are equal:
$$4 + (c + 1)^2 = 1 + (3 - c)^2$$
$$4 + c^2 + 2c + 1 = 1 + 9 - 6c + c^2$$
$$8c = 5, \qquad c = \frac58.$$

6. The flat radius. With $c = \frac58$:
$$r^2 = 4 + \left(\frac{13}{8}\right)^2 = \frac{256}{64} + \frac{169}{64} = \frac{425}{64}.$$
Check with a top corner: $1 + \left(\frac{19}{8}\right)^2 = \frac{64}{64} + \frac{361}{64} = \frac{425}{64}$. The same, as planned.

1. The other corners are inside. The corners $(\pm 2, 1)$ are at squared distance $4 + \left(\frac38\right)^2 = 4 + \frac{9}{64}$, and the inner corners $(\pm 1, 1)$ are at $1 + \frac{9}{64}$. Both are less than $\frac{425}{64}$.

8. Moving off the line does not help. Suppose the center moves to the right of the $y$-axis (or stays on it). Then it is at least as far sideways from the two left corners $(-2, -1)$ and $(-1, 3)$ as before.
- If it also moves up, it gets farther from $(-2, -1)$ vertically too, so that corner ends up beyond $r$.
- If it moves down, it gets farther from $(-1, 3)$, so that corner ends up beyond $r$.
- If it only moves sideways, both left corners get farther.

Moving to the left fails in the same way, using the right corners $(2, -1)$ and $(1, 3)$. So $\left(0, \frac58\right)$ is the best center, and $r^2 = \frac{425}{64}$ is the smallest possible.

9. Add the height.
$$R^2 = r^2 + 2^2 = \frac{425}{64} + \frac{256}{64} = \frac{681}{64}, \qquad R = \frac{\sqrt{681}}{8} \approx 3.26.$$

The answer is $\boxed{\textbf{(C)}\ \frac{\sqrt{681}}{8}}$.

A hemisphere resting on a table has its center on the table, so Pythagoras splits every distance into a flat part and a height part. Since every top corner has the same height, the 3D question becomes "find the smallest circle around the footprint". The smallest circle is decided by the points that are farthest out; here a balance between the bottom corners and the top corners. Symmetry tells you which line to search along, and setting two distances equal gives the exact spot.