AMC 10 · 2022 · #23

Grade 8 geometry-2d
coordinate-geometrypythagorean-theoremratio-proportion identify-subproblemsconvert-to-algebra ↑ Prerequisites: coordinate-geometry
📏 Medium solution 💡 3 insights
Problem
An isosceles trapezoid ABCD has AD ∥ BC with BC ＜ AD and legs AB = CD. There is a point P somewhere in the plane with PA = 1, PB = 2, PC = 3, PD = 4. Find the ratio BCAD\frac{BC}{AD}.

Pick an answer.

(A)
$\frac{1}{4}$
(B)
$\frac{1}{3}$
(C)
$\frac{1}{2}$
(D)
$\frac{2}{3}$
(E)
$\frac{3}{4}$

AMC 10 2022 problem © Mathematical Association of America (MAA AMC). Reproduced for educational use.

How to solve
Strategy Draw a Diagram

Tool #1 (Diagram) — place the trapezoid in coordinates so its symmetry simplifies the algebra. Let the symmetry axis be the y-axis, AD on the x-axis: A = (-a, 0), D = (a, 0), B = (-b, h), C = (b, h) with b ＜ a. The wanted ratio is BCAD\frac{BC}{AD} = 2b2a\frac{2b}{2a} = ba\frac{b}{a}. Tool #7 (Subproblems) — focus separately on the two pairs of symmetric vertices: {A, D} and {B, C}. Tool #13 (Algebra) — write each distance squared, then subtract the two equations in each pair. The a and b pop out cleanly without ever needing y or h. Tool #3 (Eliminate) — sanity check against the listed fractions.

1STEP 1

Use the symmetry: axis on the y-axis, base AD on the x-axis, so A, D = (∓a, 0) and B, C = (∓b, h); the target ratio is just ba\frac{b}{a}.

A = (-a, 0), D = (a, 0), B = (-b, h), C = (b, h)
2STEP 2

Write the four distance-squared equations; since the target is ba\frac{b}{a}, both h and y are nuisance variables to eliminate.

PA² &= (x+a)² + y² = 1 ; PD² &= (x-a)² + y² = 16 ; PB² &= (x+b)² + (y-h)² = 4 ; PC² &= (x-b)² + (y-h)² = 9
3STEP 3

Subtract within each symmetric pair so the vertical terms cancel: PA² minus PD² gives 4ax = -15, and PB² minus PC² gives 4bx = -5.

4ax = -15, 4bx = -5
4STEP 4

Since a and x are nonzero, divide the equations: 4bx4ax\frac{4bx}{4ax} = −5−15\frac{-5}{-15}, so ba\frac{b}{a} = 13\frac{1}{3}.

ba\frac{b}{a} = −5−15\frac{-5}{-15} = 13\frac{1}{3}
5STEP 5

Translate back: BCAD\frac{BC}{AD} = 2b2a\frac{2b}{2a} = ba\frac{b}{a} = 13\frac{1}{3}, choice (B).

BCAD\frac{BC}{AD} = ba\frac{b}{a} = 13\frac{1}{3} → (B)
Answer
13\frac{1}{3}
Sanity. The distances 1, 2, 3, 4 are very asymmetric — P is much closer to A than to D, so P is way off-center (large |x|), and similarly P is closer to B than to C. The pair PA, PD feels the long base AD; the pair PB, PC feels the short base BC. The differences PA² - PD² = -15 and PB² - PC² = -5 scale with the bases, and their ratio 515\frac{5}{15} = 13\frac{1}{3} directly gives BCAD\frac{BC}{AD}. Choice (B) 13\frac{1}{3} is well in the listed range (14\frac{1}{4} to 34\frac{3}{4}) and matches the gut feeling that BC is noticeably shorter than AD but not tiny.
💡Key takeaway

This AMC 10 problem only needs Grade 8 coordinate distance you already know — drop the trapezoid onto axes using its symmetry, square the four distances, subtract within each symmetric pair to kill the y's, divide 4bx = -5 by 4ax = -15, and read BCAD\frac{BC}{AD} = 13\frac{1}{3}.