AMC 8 · 2004 · #23

Grade 8 rate-ratio
graph-readingspatial-visualizationpattern-recognitionpythagorean-theorem pattern-recognitionpath-length-comparison ↑ Prerequisites: spatial-visualizationgraph-reading
📏 Medium solution 💡 3 insights 📊 Diagram
Problem
Tess runs counterclockwise around the rectangular block JKLM, starting and ending at her home corner J. We must pick the graph (out of A–E) whose vertical axis — Tess's straight-line distance from J — rises, peaks at the corner opposite J, and falls back to 0 as time goes by.

Pick an answer.

(A)
Distance increases steadily for the entire trip
(B)
Distance rises and falls in stairstep jumps with flat plateaus
(C)
Distance rises to a long flat plateau, then falls (rounded arcs)
(D)
Distance rises to a single peak in the middle, then falls back to 0
(E)
Distance rises, levels off, rises again, then levels off

AMC 8 2004 problem © Mathematical Association of America (MAA AMC). Reproduced for educational use.

How to solve
Strategy Eliminate Possibilities

Five concrete graphs are offered, so Tool #3 (Eliminate Possibilities) is the natural AMC multiple-choice move: list three qualitative features the right graph must have, then strike any graph that fails one. Tool #1 (Draw a Diagram) makes those features visible — sketching the rectangle and marking distances at the four corners reveals the rise-peak-fall shape. Tool #9 (Solve an Easier Related Problem) replaces the curved Pythagorean pieces with their easier endpoint values: we only need the corner distances (0, side JK, diagonal JL, side JM, 0), not a calculus-style formula.

1STEP 1

Sketch JKLM and mark the distance from J at each corner in order: 0, side JK, diagonal JL (the farthest), side JM, then 0 again.

J → 0, K → JK, L → JL, M → JM, J → 0
2STEP 2

Turn those into three rules for the graph: start and end at 0, one single peak at the midpoint, and never flat anywhere.

shape: 0 ↗ JL ↘ 0, single peak at t = t_mid
3STEP 3

Reject (A): it only climbs and never returns to 0, but Tess runs back home.

(A) ends above 0 → rejected
4STEP 4

Reject (B): its flat stretches mean constant distance, which needs a circle around J — but she runs straight sides.

(B) has horizontal segments → rejected
5STEP 5

Reject (C): two peaks would mean the farthest point is hit twice, but only L is farthest.

(C) has two maxima → rejected
6STEP 6

Reject (E): it has flat stretches and never returns to 0 — it breaks two rules at once.

(E) has horizontals and ends above 0 → rejected
7STEP 7

Confirm (D): it rises to one midpoint peak, falls back to 0, splits into four segments, and never flattens — every rule holds.

(D) matches all three features → (D)
Answer
Distance rises to a single peak in the middle, then falls back to 0
Sanity-check with a 6 × 4 block: side JK = 4, diagonal JL = √(4²+6²) ≈ 7.21, side JM = 6. The corner distances 0, 4, 7.21, 6, 0 rise to a single peak in the middle and fall back to 0 — exactly the shape of graph (D). The two side-runs (JK and MJ) give straight-line pieces on the graph, and the two cross-runs (KL and LM) give pieces that curve gently toward the peak. None of them is ever flat, which is why graphs (B), (C), (E) had to go.
💡Key takeaway

When the answer choices are graphs, list two or three must-have features (starts at 0, one peak, ends at 0) and cross off any graph that breaks even one — Grade 8 qualitative graph reading is enough!