2022 AMC 10A Problem 23

Use the trapezoid's symmetry: place its axis on the $y$-axis with
$$A = (-a, 0),\quad D = (a, 0),\quad B = (-b, h),\quad C = (b, h),$$
so $AD = 2a$, $BC = 2b$, and $b < a$. Let $P = (x, y)$.

Compare the distances to the two ends of each base. On the lower base,
$$PD^2 - PA^2 = \big[(x-a)^2 + y^2\big] - \big[(x+a)^2 + y^2\big] = -4ax,$$
and the given lengths make this $16 - 1 = 15$. On the upper base,
$$PC^2 - PB^2 = \big[(x-b)^2 + (y-h)^2\big] - \big[(x+b)^2 + (y-h)^2\big] = -4bx,$$
which equals $9 - 4 = 5$.

Dividing the two equations, the unknown $x$ cancels:
$$\frac{BC}{AD} = \frac{2b}{2a} = \frac{-4bx}{-4ax} = \frac{5}{15} = \frac13.$$

The answer is $\boxed{\textbf{(B)}\ \frac{1}{3}}$.

For two points symmetric about a line, the difference of squared distances from any point $P$ is linear: it equals $4 \cdot (\text{half-distance between them}) \cdot (P\text{'s signed distance to the axis})$. Both bases share the same axis, so the same factor $x$ appears twice and cancels, leaving the ratio of the bases directly. Look for differences of squares whenever several distances from one point are given.

Synthetic version: drop perpendiculars from $P$ to lines $AD$ and $BC$ with feet $F$ and $G$, and let $M, N$ be the midpoints of the bases. Then $PD^2 - PA^2 = FD^2 - FA^2 = (FD - FA)(FD + FA) = 2\,MF \cdot AD$, and likewise $PC^2 - PB^2 = 2\,NG \cdot BC$. Since $MF = NG$ (both equal $P$'s distance from the axis), $\frac{BC}{AD} = \frac{5}{15}$.

• Trying to find the trapezoid's dimensions or P's location; only the ratio is determined, and the height h cancels immediately in the differences of squares.
• Using differences of lengths ($4 - 1 = 3$ and $3 - 2 = 1$) instead of differences of squared lengths, which gives $\frac13$ by luck here but is not a valid method.