Competition · AMC preparation · step 4 of 4

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

Set up coordinates

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

Track the distance along leg 1

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

Draw leg 1 as a flat line

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

Sample three points on leg 2

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 = (r/2, r/2)

5STEP 5

Compute the three distances

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

XB = r, XM = √((r/2)² + (r/2)²) = r/√(2) ≈ 0.71 r, XC = r

6STEP 6

Shape leg 2 as a dip

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 r/√(2) back to r

Grade 8: a V-shape on a graph means a linear change with a sudden corner. Distance from a point to a moving line-point is nonlinear, so the graph curves.

On the straight leg from B to C, the ship's distance from X starts at the radius, sinks to a low point in the middle, and climbs back to the radius, tracing a smooth curved dip rather than a straight-sided V.

▸ Why?

Both ends of this leg, B and C, still sit on the original circle around X, so each is exactly one radius from X; that pins the dip to begin and end at the same height the flat part left off.

A circle's radius is constantEvery point on a circle sits exactly one radius from the center, so all radii of a circle are equal.

▸ Why?

In between, the straight segment BC is a chord that cuts across the inside of the circle, so every point on it other than B and C sits nearer to X than the radius — that is why the height drops below where it started.

▸ Why?

The circle is exactly the ring of points one radius from X, so any point resting inside that ring, like the interior of the chord, has to be less than one radius away.

A circle's radius is constantEvery point on a circle sits exactly one radius from the center, so all radii of a circle are equal.

▸ Why?

A straight-sided V would mean the distance shrinks and then grows at one steady rate with a sharp corner at the bottom, but the distance from X to the moving ship is the square root of its sideways gap squared plus its upward gap squared, and that square root bends gradually through its lowest point instead of turning a corner.

▸ Why?

With X at the origin, the ship's sideways gap and its upward gap are the two legs of a right triangle whose hypotenuse is the straight line back to X, so that distance equals √((sideways gap)² + (upward gap)²); as the ship slides along the line this quantity eases down and back up, so its graph is a curve, not two straight pieces meeting at a point.

Pythagorean theoremIn a right triangle the two legs' squares add up to the hypotenuse's square (a²+b²=c²).

8.F.B.5Identify Subproblems

7STEP 7

Stitch the pieces and pick the graph

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}{\sqrt{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.

Set up coordinates
Track the distance along leg 1
Draw leg 1 as a flat line
Sample three points on leg 2
Compute the three distances
Shape leg 2 as a dip
Stitch the pieces and pick the graph

A parent dashboard for the family lives at sensimlab.com.

Set up coordinates

Track the distance along leg 1

Draw leg 1 as a flat line

Sample three points on leg 2

Compute the three distances

Shape leg 2 as a dip

Stitch the pieces and pick the graph

More like this