AMC 8 · 2024 · #6

Grade 2 arithmetic

Archetype Path Length Comparison & Shortest Path └ Sub-archetype Shortest Route in Weighted Network

perimeterspatial-visualization path-length-comparison ↑ Prerequisites: perimetermental-arithmetic

📏 Medium solution 💡 3 insights 📊 Diagram

Problem

An ice rink (a long shape made of a rectangle with two semicircular ends) is shown with four skating paths drawn on it. Path P traces the outer boundary. Path R is a rectangle inscribed inside the rink, using straight vertical chords where P uses semicircular arcs. Path S is made of two long diagonals that cross in the middle. Path Q is a zig-zag path that crosses the rink with extra horizontal detours. We must order P, Q, R, S from shortest to longest.

Pick an answer.

(A)

P,Q,R,S

(B)

P,R,S,Q

(C)

Q,S,P,R

(D)

R,P,S,Q

(E)

R,S,P,Q

AMC 8 2024 problem © Mathematical Association of America (MAA AMC). Reproduced for educational use.

How to solve

Strategy Draw a Diagram

The whole problem IS a picture, so Tool #1 (Diagram) is unavoidable — we work directly on the figure, looking for which segments of one path match segments of another. Comparing four paths at once is hard, so we use Tool #7 (Subproblems) to break the task into three easy pairwise comparisons: (R vs P), (S vs Q), and (P vs S). Each pairwise comparison is a one-line observation about straight-vs-curved or straight-vs-zigzag. Finally, we use Tool #3 (Eliminate Possibilities) on the five answer choices: each inequality we prove rules out the choices that violate it, and only one option survives. No algebra, no measurements, no formulas — just looking at the picture carefully.

1STEP 1

Paths R and P share the two long sides; only the ends differ — P uses arcs, R uses straight chords. A chord beats an arc, so R ＜ P.

straight chord ＜ arc with same endpoints → R ＜ P

2STEP 2

Path S is a straight corner-to-corner diagonal; Q detours sideways before finishing. A straight line beats a zig-zag, so S ＜ Q.

straight diagonal ＜ zig-zag with same endpoints → S ＜ Q

3STEP 3

Only P vs S is left. P loops the perimeter once, but S crosses the long rink twice — two diagonals outrun the loop, so P ＜ S.

P ＜ S

4STEP 4

Chain the three comparisons: from R ＜ P, P ＜ S, and S ＜ Q, indirect ordering gives R ＜ P ＜ S ＜ Q.

R ＜ P, P ＜ S, S ＜ Q → R ＜ P ＜ S ＜ Q

5STEP 5

Match R ＜ P ＜ S ＜ Q to the choices: every other option misorders a pair, so only (D) R, P, S, Q survives.

Order = R, P, S, Q → (D)

Answer

R,P,S,Q

Each pairwise step relied on a single solid rule: the straight path between two points is the shortest. R beats P at the ends (straight chord vs arc); S beats Q in the middle (straight diagonal vs zig-zag); and P beats S overall because one trip around the perimeter is shorter than two full diagonal crossings of a long rink. Combining R ＜ P ＜ S ＜ Q gives exactly one of the listed orderings — choice (D) — which matches the official answer.

💡Key takeaway

This AMC 8 problem only needs Grade 2 shape recognition and length comparison you already know!

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