2021 AMC Fall 10A Problem 17

Because the panel is a flat plane, the height $h(P)$ above a ground point $P$ is a linear (affine) function of the position of $P$. Two consequences we will use: the height above the midpoint of a segment is the average of the heights at its endpoints, and for any parallelogram $PQRS$ (where $\vec P + \vec R = \vec Q + \vec S$) we have $h(P) + h(R) = h(Q) + h(S)$.

Let $O$ be the center of the hexagon. In a regular hexagon, $OA = AB = BC = CO$ and $OA \parallel CB$, so $OABC$ is a rhombus, a parallelogram with opposite vertices $O, B$ and $A, C$. Hence
$$h(O) = h(A) + h(C) - h(B) = 12 + 10 - 9 = 13.$$

Next, $O$ is the midpoint of the long diagonal $BE$, so $h(O) = \dfrac{h(B) + h(E)}{2}$, giving
$$h(E) = 2 \cdot 13 - 9 = 17.$$

The answer is $\boxed{\textbf{(D)}\ 17}$.

"Flat panel" means the height function is linear, and linear functions turn geometric relations among points (midpoints, parallelograms) into the same relations among heights. The center of a regular hexagon is the hub: it is the midpoint of every long diagonal and forms a rhombus with any three consecutive vertices, so three known heights determine the center and the center determines everything opposite.

Coordinates: take side length $2$ with $A = (2, 0)$, $B = (1, \sqrt3)$, $C = (-1, \sqrt3)$, $E = (-1, -\sqrt3)$, and let the panel be $z = ax + by + c$. Then $2a + c = 12$, $a + \sqrt3 b + c = 9$, $-a + \sqrt3 b + c = 10$. Subtracting the last two gives $a = -\tfrac12$, so $c = 13$ and $\sqrt3 b = -\tfrac72$. Then $z_E = -a - \sqrt3 b + c = \tfrac12 + \tfrac72 + 13 = 17$.

• Guessing that opposite pillars are related by the same difference as neighbors (for example E = B + 3 = 12 or E = 9), instead of using the plane's linearity through the center.
• Treating $ABCE$ or $ABCD$ as a parallelogram; in a regular hexagon the parallelograms through three consecutive vertices go through the center $O$.