AMC 10 · 2021 · #14

Grade 8 geometry-2d

Archetype Coordinate Plane Figure Computation

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 intersect a circle, creating three chords of lengths $38,38,$ and $34$ . What is the distance between two adjacent parallel lines?

$\textbf{(A) }5\frac12 \qquad \textbf{(B) }6 \qquad \textbf{(C) }6\frac12 \qquad \textbf{(D) }7 \qquad \textbf{(E) }7\frac12$

Pick an answer.

(A)

$5\frac12$

(B)

(C)

$6\frac12$

(D)

(E)

$7\frac12$

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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

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

#1 Draw a Diagram 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.

$$\text{Center } O \text{ lies midway between the two length-}38\text{ chords.}$$

💡 Grade 7 circle facts: equal chords sit at equal distances from the center.

#1 Draw a Diagram 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 $O$.

$$h_{38} = \dfrac{d}{2}, \qquad h_{34} = \dfrac{3d}{2}$$

💡 Grade 5 number-line placement: the three lines sit at heights $-\tfrac{d}{2}, \tfrac{d}{2}, \tfrac{3d}{2}$ from $O$.

#7 Identify Subproblems 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: $h^2 + \ell^2 = r^2$.

$$h^2 + \ell^2 = r^2 \;\;\text{for each chord}$$

💡 Grade 8 Pythagorean theorem: radius, half-chord, distance form a right triangle.

#13 Convert to Algebra 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.
$\tfrac{9d^2}{4} + 289 = r^2$.

$$\tfrac{d^2}{4} + 361 = r^2 \quad \text{and} \quad \tfrac{9d^2}{4} + 289 = r^2$$

💡 Grade 8 Pythagorean theorem applied twice, once per chord length.

#13 Convert to Algebra 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).

$$72 = 2d^2 \;\Rightarrow\; d^2 = 36 \;\Rightarrow\; d = 6$$

💡 Grade 8 square-root: distance is positive, so take the positive root of $d^2 = 36$.

#3 Eliminate Possibilities 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.

$$d = 6 \;\Rightarrow\; \textbf{(B)}$$

💡 Grade 8 approximation of irrationals: $\sqrt{370} \approx 19.24$, so $38$ fits inside a diameter $\approx 38.5$.

[1] #1 7.G.B.4 Sketch the circle with center $O$ and the three parallel chords. Equal-length ch

[2] #1 5.G.A.2 Mark the gap $d$ between adjacent parallel lines. The two length-$38$ chords are

[3] #7 8.G.B.7 For each chord, drop a perpendicular from $O$ to its midpoint and connect $O$ to

[4] #13 8.G.B.7 Apply Pythagoras to each chord length. For the $38$-chords ($\ell = 19$, $h = d/

[5] #13 8.EE.A.2 Both expressions equal $r^2$, so set them equal: $\tfrac{d^2}{4} + 361 = \tfrac{

[6] #3 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.2 Represent 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.4 Know the formulas for area and circumference of a circle (Using the circle fact that equal chords are equidistant from the center.)
8.G.B.7 Apply the Pythagorean theorem to determine unknown side lengths in right triangles (Forming $h^2 + \ell^2 = r^2$ for each chord.)
8.EE.A.2 Use square root and cube root symbols to represent solutions (Solving $d^2 = 36$ to get $d = 6$.)
8.NS.A.2 Use 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).

Problem

Toolkit + CCSS Solution

Understand

Plan

Execute — Answer: B

Review

CCSS standards used (min grade 8)

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