2026 AMC 10A Problem 22

Each touching pair of ellipses gives one equation. The trick is knowing where two equal ellipses touch, and symmetry answers that.

1. Put the centers on coordinates. The picture has three rows: $B, A$ on top; $C, O, F$ in the middle; $D, E$ at the bottom. Put $O$ at $(0, 0)$.
- Same row, $4$ apart: $C = (-4, 0)$ and $F = (4, 0)$.
- The top and bottom rows sit halfway between the middle-row centers, so $A$ and $E$ have $x = 2$, and $B$ and $D$ have $x = -2$.
- $A$ and $E$ are in the same column and $9$ apart, but they are two rows apart. So neighboring rows are $\frac{9}{2} = 4.5$ apart, and
$$A = (2, 4.5), \quad B = (-2, 4.5), \quad D = (-2, -4.5), \quad E = (2, -4.5).$$

2. The ellipse around $O$. The ellipses all have the same size and stand upright, taller than they are wide. Write $O$'s ellipse as
$$\frac{x^2}{a^2} + \frac{y^2}{b^2} = 1,$$
where $a$ is half its width and $b$ is half its height. Every other ellipse is this one slid over to a new center.

3. Where two equal ellipses touch. Take two of the ellipses, say around $O$ and $F$, and let $M$ be the point halfway between their centers.
- Spin the whole picture half a turn around $M$. The center $O$ lands on $F$ and $F$ lands on $O$. An ellipse looks exactly the same after half a turn around its own center, so the spin swaps the two ellipses.
- Suppose the two ellipses share a point $T$. After the spin, $T$ lands on a point $T'$ on the opposite side of $M$, and $T'$ is also shared. An ellipse has no dents, so the straight segment from $T$ to $T'$ lies inside both ellipses, and that segment passes through $M$. So whenever the ellipses meet at all, $M$ is inside both of them.
- "Closely packed" means neighbors just touch without overlapping. So $M$ must sit exactly on the edge of each ellipse: the touching point is the halfway point.

(Two equal coins touch halfway between their centers for the same reason.)

1. $O$ and $F$ give the width. The halfway point of $O = (0, 0)$ and $F = (4, 0)$ is $(2, 0)$. It lies on $O$'s ellipse, so $\frac{2^2}{a^2} = 1$ and $a = 2$. (That makes sense: the ellipses are $4$ wide, and centers in a row are $4$ apart.)

5. $O$ and $A$ give the height. The halfway point of $O = (0, 0)$ and $A = (2, 4.5)$ is $\left(1, \frac94\right)$. It lies on $O$'s ellipse:
$$\frac{1^2}{4} + \frac{(9/4)^2}{b^2} = 1 \quad\Longrightarrow\quad \frac{81/16}{b^2} = \frac34 \quad\Longrightarrow\quad b^2 = \frac{81}{16} \cdot \frac43 = \frac{27}{4}.$$
So $b = \frac{\sqrt{27}}{2} = \frac{3\sqrt3}{2}$.

1. The major axis. The height $2b = 3\sqrt3 \approx 5.2$ is bigger than the width $2a = 4$, so the major (long) axis is the vertical one, of length $3\sqrt3$. Here $m = 3$ and $p = 3$, and $3$ has no square factor.

1. Check the other neighbors. Every other touching pair is a copy of one of these two: $B$ and $A$ are $4$ apart in a row like $O$ and $F$, and pairs such as $A$ and $F$ or $O$ and $E$ are $2$ across and $4.5$ up or down, like $O$ and $A$. So one ellipse size makes all the neighbors touch.

So $m + p = 3 + 3 = 6$.

The answer is $\boxed{\textbf{(A)}\ 6}$.

Two equal ellipses facing the same way are swapped by a half turn around the point halfway between their centers. That symmetry pins the touching point to the halfway point, so each touching pair gives one easy equation: "this point is on the ellipse". The same idea works for any two equal shapes that look the same after a half turn, such as circles, squares and rectangles.

Squash the picture until the ellipses become circles.

1. Shrink every height in the picture by the same factor, chosen so that each ellipse becomes a circle of radius $2$ (its half-width). Shapes that just touch still just touch after the squash.
2. Circles of radius $2$ that touch have centers $2 + 2 = 4$ apart. The row of $C, O, F$ is unchanged, and $O$, $F$ and the squashed $A$ now form an equilateral triangle of side $4$: the familiar honeycomb pattern of coins.
3. The height of an equilateral triangle with side $4$ is $2\sqrt3$. So squashing changed the row spacing from $4.5$ to $2\sqrt3$, a factor of $\frac{2\sqrt3}{4.5}$.
4. The circle's height $4$ was squashed from the ellipse's height by that same factor, so the ellipse's height is
$$4 \div \frac{2\sqrt3}{4.5} = \frac{18}{2\sqrt3} = \frac{9}{\sqrt3} = 3\sqrt3.$$

The major axis is $3\sqrt3$, and $m + p = 6$.