AMC 10 · 2022 · #20
Grade 8 geometry-2dPick an answer.
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 BC/AD = 2b/2a = 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.
Set symmetric coordinates
Place coordinates respecting the symmetry.
Grade 6 — use the line of symmetry as the y-axis, so symmetric vertices have opposite x-coordinates.
6.NS.C.8Draw A DiagramWrite the four distances
Write all four as equations.
Grade 8 — distance squared in the coordinate plane is (Δ x)² + (Δ y)², no square roots needed.
8.G.B.8Convert To AlgebraSubtract in pairs
Subtracting makes every square vanish.
Grade 8 — subtract paired squares so the unknown vertical coordinates cancel; only a, b, x remain.
Subtracting the paired squared distances makes the unknown coordinates cancel.
▸ Why?
Both expressions carry the identical squared block, so subtracting removes it entirely.
▸ Why?
Each squared distance is the sum of the squared coordinate gaps, which is where that block comes from.
Divide the two results
Dividing cancels the unknown.
Grade 6 — divide one equation by another to get a clean ratio.
6.RP.A.3Convert To AlgebraRead the ratio
The ratio is one third.
Grade 6 ratio — the answer is exactly the b/a we just found.
6.RP.A.3Convert To AlgebraThis AMC 12 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 BC/AD = 1/3.