2022 AMC 10A Problem 23

Step 1: set up coordinates using the symmetry. An isosceles trapezoid has a mirror line through the middles of both bases. Put that line on the $y$-axis:
$$A = (-a, 0), \quad D = (a, 0), \quad B = (-b, h), \quad C = (b, h).$$
Then $AD = 2a$ and $BC = 2b$. Let $P = (x, y)$.

Step 2: why subtract? Each squared distance has many unknowns, but $A$ and $D$ are mirror images, so the difference $PD^2 - PA^2$ should lose most of them.

Step 3: the bottom base.
$$PA^2 = (x + a)^2 + y^2, \qquad PD^2 = (x - a)^2 + y^2.$$
The $y^2$ parts cancel when we subtract, and
$$(x - a)^2 - (x + a)^2 = (x^2 - 2ax + a^2) - (x^2 + 2ax + a^2) = -4ax.$$
The given lengths make $PD^2 - PA^2 = 16 - 1 = 15$. So $-4ax = 15$.

Step 4: the top base, the same way. The $(y - h)^2$ parts cancel, and
$$PC^2 - PB^2 = (x - b)^2 - (x + b)^2 = -4bx.$$
The given lengths make this $9 - 4 = 5$. So $-4bx = 5$.

Step 5: divide. 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 that are mirror images across a line, the difference of the squared distances from any point $P$ is simple: it only depends on how far apart the two points are and how far $P$ is from the mirror line. Both bases share the same mirror line, so the same unknown appears twice and cancels. When several distances from one point are given, try subtracting their squares.