AMC 10 · 2021 · #14
Grade 8 geometry-2dProblem
Three equally spaced parallel lines intersect a circle, creating three chords of lengths 38,38, and 34. What is the distance between two adjacent parallel lines?
Pick an answer.
AMC 10 2021 problem © Mathematical Association of America (MAA AMC). Reproduced for educational use.
Try it yourself first — the explanation is most useful after you’ve attempted it.
Toolkit + CCSS Solution
Understand
Restated: 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 $d$.
Givens: Three parallel lines, equally spaced (consecutive gaps both $= d$); Chord lengths $38, 38, 34$ (the two length-$38$ chords must be the outer ones equidistant from the center, since equal chords are equidistant from the center); Answer choices: $5\tfrac{1}{2},\; 6,\; 6\tfrac{1}{2},\; 7,\; 7\tfrac{1}{2}$
Unknowns: The distance $d$ between two adjacent parallel lines
Understand
Restated: 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 $d$.
Givens: Three parallel lines, equally spaced (consecutive gaps both $= d$); Chord lengths $38, 38, 34$ (the two length-$38$ chords must be the outer ones equidistant from the center, since equal chords are equidistant from the center); Answer choices: $5\tfrac{1}{2},\; 6,\; 6\tfrac{1}{2},\; 7,\; 7\tfrac{1}{2}$
Plan
Primary tool: #1 Draw a Diagram
Secondary: #7 Identify Subproblems, #13 Convert to Algebra, #3 Eliminate Possibilities
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 $r^2$, and solves for $d$. Tool #3 (Eliminate) confirms $d = 6$ against the choices.
Execute — Answer: B
7.G.B.4 Step 1 - Sketch the circle with center $O$ and the three parallel chords.
- Equal-length chords are equidistant from the center, so the two length-$38$ chords are mirror images across the line through $O$ perpendicular to them.
- The middle line in the triple of parallel lines is therefore the midline between the two $38$-chords (otherwise the spacing would not be equal), so the two $38$-chords are the outer two lines and the $34$-chord is the middle line.
💡 Grade 7 circle facts: equal chords sit at equal distances from the center.
5.G.A.2 Step 2 - Mark the gap $d$ between adjacent parallel lines.
- The two length-$38$ chords are at distance $d/2$ from $O$ (one above, one below), and the length-$34$ chord is two gaps away from one of them, at distance $d/2 + d = 3d/2$ from $O$.
💡 Grade 5 number-line placement: the three lines sit at heights $-\tfrac{d}{2}, \tfrac{d}{2}, \tfrac{3d}{2}$ from $O$.
8.G.B.7 Step 3 - For each chord, drop a perpendicular from $O$ to its midpoint and connect $O$ to one endpoint.
- This makes a right triangle with legs $h$ (distance from center) and $\ell$ (half the chord), and hypotenuse $r$ (the radius).
- Pythagoras: $h^2 + \ell^2 = r^2$.
💡 Grade 8 Pythagorean theorem: radius, half-chord, distance form a right triangle.
8.G.B.7 Step 4 - Apply Pythagoras to each chord length.
- For the $38$-chords ($\ell = 19$, $h = d/2$): $\bigl(\tfrac{d}{2}\bigr)^2 + 19^2 = r^2$, i.e.
- $\tfrac{d^2}{4} + 361 = r^2$.
- For the $34$-chord ($\ell = 17$, $h = 3d/2$): $\bigl(\tfrac{3d}{2}\bigr)^2 + 17^2 = r^2$, i.e.
- $\tfrac{9d^2}{4} + 289 = r^2$.
💡 Grade 8 Pythagorean theorem applied twice, once per chord length.
8.EE.A.2 Step 5 - Both expressions equal $r^2$, so set them equal: $\tfrac{d^2}{4} + 361 = \tfrac{9d^2}{4} + 289$.
- Subtract $\tfrac{d^2}{4}$ from both sides and subtract $289$: $72 = \tfrac{8d^2}{4} = 2d^2$.
- So $d^2 = 36$ and $d = 6$ (positive root since $d$ is a distance).
💡 Grade 8 square-root: distance is positive, so take the positive root of $d^2 = 36$.
8.NS.A.2 Step 6 - Match to the choices: $d = 6$ is (B).
- Quick sanity: with $d = 6$, $r^2 = 9 + 361 = 370$, so $r = \sqrt{370} \approx 19.24$.
- The longest chord $38$ is just under the diameter $\approx 38.47$ — that's reasonable.
💡 Grade 8 approximation of irrationals: $\sqrt{370} \approx 19.24$, so $38$ fits inside a diameter $\approx 38.5$.
7.G.B.4 Sketch the circle with center $O$ and the three parallel chords. Equal-length ch 5.G.A.2 Mark the gap $d$ between adjacent parallel lines. The two length-$38$ chords are 8.G.B.7 For each chord, drop a perpendicular from $O$ to its midpoint and connect $O$ to 8.G.B.7 Apply Pythagoras to each chord length. For the $38$-chords ($\ell = 19$, $h = d/ 8.EE.A.2 Both expressions equal $r^2$, so set them equal: $\tfrac{d^2}{4} + 361 = \tfrac{ 8.NS.A.2 Match to the choices: $d = 6$ is (B). Quick sanity: with $d = 6$, $r^2 = 9 + 361 Review
Reasonableness: Plug back: $d = 6$, $h_{38} = 3$, $h_{34} = 9$. Check the two Pythagorean equalities: $3^2 + 19^2 = 9 + 361 = 370$ and $9^2 + 17^2 = 81 + 289 = 370$. Both give $r^2 = 370$ — same radius, as required. The radius $\sqrt{370} \approx 19.24$ is just over $19 = 38/2$, so the longest chord nearly hits the diameter, which lines up with $h_{38} = 3$ being small. Magnitudes all sensible.
Alternative: Tool #3 (Eliminate Possibilities) on the choices using only the Pythagorean equations. For each candidate $d \in \{5\tfrac{1}{2}, 6, 6\tfrac{1}{2}, 7, 7\tfrac{1}{2}\}$, compute $r^2$ two ways: $r^2 = (d/2)^2 + 361$ and $r^2 = (3d/2)^2 + 289$. The two values agree only when $d = 6$ (giving $370 = 370$); every other candidate gives a mismatch. So (B) by elimination.
CCSS standards used (min grade 8)
5.G.A.2Represent real-world and mathematical problems by graphing points (Placing the three parallel lines on a vertical number line at heights $-d/2, d/2, 3d/2$ from $O$.)7.G.B.4Know the formulas for area and circumference of a circle (Using the circle fact that equal chords are equidistant from the center.)8.G.B.7Apply the Pythagorean theorem to determine unknown side lengths in right triangles (Forming $h^2 + \ell^2 = r^2$ for each chord.)8.EE.A.2Use square root and cube root symbols to represent solutions (Solving $d^2 = 36$ to get $d = 6$.)8.NS.A.2Use rational approximations of irrational numbers to compare their size (Estimating $r = \sqrt{370} \approx 19.24$ for the sanity check.)
⭐ 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 $d/2$, the length-$34$ chord at distance $3d/2$. Pythagoras on each: $19^2 + (d/2)^2 = 17^2 + (3d/2)^2$. Simplify to $2d^2 = 72$, so $d = 6$, answer (B).
⭐ 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 $d/2$, the length-$34$ chord at distance $3d/2$. Pythagoras on each: $19^2 + (d/2)^2 = 17^2 + (3d/2)^2$. Simplify to $2d^2 = 72$, so $d = 6$, answer (B).
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