2023 AMC 10A Problem 18

We cannot easily picture this solid, so count instead.

1. Edges. There are $12$ faces with $4$ edges each, which is $48$. But every edge is shared by exactly $2$ faces, so each edge was counted twice:
$$E = \frac{48}{2} = 24.$$
2. Vertices. Euler's formula says that for a solid like this, vertices minus edges plus faces equals $2$: $V - E + F = 2$. (Check on a cube: $8 - 12 + 6 = 2$.) Here
$$V - 24 + 12 = 2, \quad\text{so}\quad V = 14.$$
3. Split the vertices. Say $a$ vertices have $3$ edges and $b$ vertices have $4$ edges. Then $a + b = 14$.
4. Count edge ends. Every edge has $2$ ends, so there are $2 \times 24 = 48$ edge ends in total. Each $3$-edge vertex holds $3$ ends and each $4$-edge vertex holds $4$:
$$3a + 4b = 48.$$
5. Solve. If all $14$ vertices had $3$ edges, that would be $42$ ends. We need $6$ more, and each $4$-edge vertex adds $1$ extra, so $b = 6$. Then $a = 14 - 6 = 8$.

The answer is $\boxed{\textbf{(D)}\ 8}$.

For solids you cannot picture, three counts do all the work: faces times sides, divided by $2$, gives edges; Euler's formula gives vertices; and counting edge ends gives a second equation. The same plan works for a soccer ball or any other solid with mixed kinds of corners.

Count corners of the rhombi instead. Each rhombus has $2$ sharp (acute) and $2$ wide (obtuse) angles, so there are $24$ sharp and $24$ wide angles in total. In this solid, a $4$-edge vertex is surrounded by $4$ sharp angles and a $3$-edge vertex by $3$ wide angles (four wide angles would add up to more than $360^\circ$, which cannot happen at a corner of a solid). So $3a = 24$ gives $a = 8$.