AMC 10 · 2021 · #14

Grade 8 geometry-2d

Archetype Coordinate Plane Figure Computation └ Sub-archetype Pythagorean / Distance Formula for a Length

chord-perpendicular-from-centerpythagorean-theoremcoordinate-geometryequal-spacing convert-to-algebraidentify-subproblems ↑ Prerequisites: pythagorean-theoremcoordinate-geometry

📏 Long solution 💡 3 insights

Problem

Three equally spaced parallel lines cut across a circle, producing chords of lengths 38, 38, and 34. The two adjacent gaps between the lines are equal — call that gap d. Find d.

Pick an answer.

(A)

$5\frac12$

(B)

(C)

$6\frac12$

(D)

(E)

$7\frac12$

AMC 10 2021 problem © Mathematical Association of America (MAA AMC). Reproduced for educational use.

How to solve

Strategy Draw a Diagram

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 $\frac{d}{2}$ each, while the 34-chord lies at distance $\frac{3d}{2}$ 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.

1STEP 1

Equal chords are equidistant from the center, so the two 38-chords are the outer lines and the 34-chord is the middle one.

Center O lies midway between the two length-38 chords.

2STEP 2

So the two 38-chords sit at distance $\frac{d}{2}$ from O and the 34-chord at distance $\frac{3d}{2}$ .

h₃8 =

\frac{d}{2}

, h₃4 =

\frac{3d}{2}

3STEP 3

For each chord the radius, half-chord ℓ, and distance h from O form a right triangle: h² + ℓ² = r².

h² + ℓ² = r² for each chord

4STEP 4

Pythagoras on each chord gives $\frac{d²}{4}$ + 361 = r² (38-chords, ℓ = 19) and $\frac{9d²}{4}$ + 289 = r² (34-chord, ℓ = 17).

\frac{d²}{4}

+ 361 = r² and

\frac{9d²}{4}

+ 289 = r²

5STEP 5

Setting the two equal cancels r²: 72 = 2d², so d² = 36 and d = 6 (positive root, since d is a distance).

72 = 2d² → d² = 36 → d = 6

6STEP 6

So d = 6 matches (B); sanity check r² = 9 + 361 = 370, r ≈ 19.24, so the 38-chord fits just inside the diameter.

d = 6 → (B)

Answer

Plug back: d = 6, h₃8 = 3, h₃4 = 9. Check the two Pythagorean equalities: 3² + 19² = 9 + 361 = 370 and 9² + 17² = 81 + 289 = 370. Both give r² = 370 — same radius, as required. The radius √(370) ≈ 19.24 is just over 19 =

\frac{38}{2}

, so the longest chord nearly hits the diameter, which lines up with h₃8 = 3 being small. Magnitudes all sensible.

💡Key takeaway

This AMC 10 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 $\frac{d}{2}$ , the length-34 chord at distance $\frac{3d}{2}$ . Pythagoras on each: 19² + ( $\frac{d}{2}$ )² = 17² + ( $\frac{3d}{2}$ )². Simplify to 2d² = 72, so d = 6, answer (B).

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