AMC 8 · 2003 · #24

Grade 8 rate-ratio

Archetype Small-Domain Constrained Search └ Sub-archetype Logic-Deduction / Elimination

graph-readingspatial-visualizationpattern-recognitionpythagorean-theorem identify-subproblemscaseworksystematic-enumeration ↑ Prerequisites: graph-readingspatial-visualization

📏 Medium solution 💡 3 insights 📊 Diagram

Problem

A ship sails from A to B along a semicircle centered at Island X, then sails in a straight line from B to C. Pick the graph that best shows the ship's distance from X over the whole trip. From the figure, X is the center of the semicircle and C is the point above X where the radius going up meets the circle (so XB = XC and angle BXC = 90°).

Pick an answer.

(A)

Graph with an irregular/incorrect distance pattern

(B)

Horizontal segment at radius r, then a smooth symmetric curve dipping below r and returning to r

(C)

Horizontal segment at radius r, then a V-shape (two straight lines) dipping to a minimum and returning to r

(D)

Graph starting with a non-horizontal initial segment (incorrect first leg)

(E)

Graph starting with a non-horizontal initial segment (incorrect first leg)

AMC 8 2003 problem © Mathematical Association of America (MAA AMC). Reproduced for educational use.

How to solve

Strategy Draw a Diagram

The problem is about positions, paths, and a graph — exactly what Tool #1 (Draw a Diagram) is for. Putting X at the origin turns the question into something we can measure. Tool #7 (Identify Subproblems) tells us to handle leg A → B and leg B → C separately, because they obey completely different rules (circle vs. straight line). Tool #9 (Solve an Easier Related Problem) lets us sidestep messy algebra on the second leg: instead of computing the distance at every moment, we only check the start, the middle, and the end of BC, which is enough to pin down the shape of the graph.

1STEP 1

Put X at the origin. Then B = (r, 0) and C = (0, r), so distance from X is just distance from the origin.

X=(0,0), A=(-r,0), B=(r,0), C=(0,r)

2STEP 2

Leg 1 (A → B): every point of a circle centered at X is the same distance from X, so the distance stays r the whole way.

distance from X = r for all of A → B

3STEP 3

So on the graph, leg 1 is a flat horizontal segment at height r — any choice whose first piece isn't flat is out.

Leg 1 graph piece: horizontal line at height r

4STEP 4

Leg 2 (B → C): sample just three points — start B, midpoint M = ( $\frac{r}{2}$ , $\frac{r}{2}$ ), end C. That's the easier sub-problem.

M = (

\frac{r}{2}

\frac{r}{2}

)

5STEP 5

By the Pythagorean theorem the endpoints are r from X, but the midpoint is closer: XM ≈ 0.71 r.

XB = r, XM = √((

\frac{r}{2}

)² + (

\frac{r}{2}

)²) =

\frac{r}{√(2)}

≈ 0.71 r, XC = r

6STEP 6

So leg 2 makes a smooth symmetric dip from r down to about 0.71 r and back — a curve, not a straight-line V.

Leg 2 graph piece: smooth U-shaped dip from r to

\frac{r}{√(2)}

back to r

7STEP 7

Stitch them: a flat segment at height r, then the smooth dip back to r. The only choice with that exact shape is (B).

Answer = (B)

Answer

Horizontal segment at radius r, then a smooth symmetric curve dipping below r and returning to r

Both end-of-trip distances should match the radius, and they do: the ship starts at A (distance r), stays at distance r along the whole semicircle, ends leg 2 at C (distance r again). The minimum distance on leg 2 is

\frac{r}{√(2)}

≈ 0.71 r, which is less than r but still positive — consistent with the smooth dip in graph (B). The V-shape in (C) would mean the distance changes at a constant rate on each side of the minimum, which is wrong because the distance from a fixed point to a sliding point on a line is not linear.

💡Key takeaway

Whenever a ship rides a circle around a point, its distance to that point cannot change. Whenever it cuts across in a straight line, the distance changes smoothly — never in a sharp V. Spotting those two shapes turns this AMC 8 problem into a Grade 8 graph-reading exercise.

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