AMC 10 · 2021 · #8
Grade 8 geometry-2dPick an answer.
Tool #1 (Diagram) — draw the circle, mark its center O, and drop a perpendicular from O to each of the three parallel lines. Since the two 38-chords have equal length, they are equidistant from O — so O sits on the perpendicular bisector of the band between them, and the two outer chords are the 38-chords at distance d/2 each, while the 34-chord lies at distance 3d/2 on the other side. Tool #7 (Subproblems) gives two right triangles (radius / half-chord / distance) — one per chord length. Tool #13 (Algebra) writes the two Pythagorean equations, eliminates r², and solves for d. Tool #3 (Eliminate) confirms d = 6 against the choices.
Locate the centre
The centre sits exactly midway between them.
Grade 7 circle facts: equal chords sit at equal distances from the center.
Equal chords sit at equal distances from the centre.
▸ Why?
Every point of the circle is one radius from the centre, so equal chords cut equal right triangles.
▸ Why?
The perpendicular from the centre lands on the chord's midpoint, so that distance is well defined.
Write each distance
Write all three via one gap.
Grade 5 number-line placement: the three lines sit at heights -d/2, d/2, 3d/2 from O.
5.G.A.2Draw A DiagramWrite the chord relation
Pythagoras holds for each chord.
Grade 8 Pythagorean theorem: radius, half-chord, distance form a right triangle.
8.G.B.7Identify SubproblemsWrite the two equations
Two equations come out.
Grade 8 Pythagorean theorem applied twice, once per chord length.
8.G.B.7Convert To AlgebraSubtract to kill the radius
Subtracting removes the radius.
Grade 8 square-root: distance is positive, so take the positive root of d² = 36.
8.EE.A.2Convert To AlgebraMatch the choice
The gap is 6.
Grade 8 approximation of irrationals: √(370) ≈ 19.24, so 38 fits inside a diameter ≈ 38.5.
8.NS.A.2Eliminate PossibilitiesThis AMC 12 problem only needs Grade 8 Pythagorean theorem you already know! Drop perpendiculars from the center to each chord: the two length-38 chords sit at distance d/2, the length-34 chord at distance 3d/2. Pythagoras on each: 19² + (d/2)² = 17² + (3d/2)². Simplify to 2d² = 72, so d = 6, answer (B).