AMC 10 · 2023 · #24

Grade 8 geometry-2d

vector-parameterizationcoordinate-geometryperimeterminkowski-sumspatial-visualization physical-representationidentify-subproblems ↑ Prerequisites: coordinate-geometryperimeter

📏 Long solution 💡 4 insights

Problem

Let R be the set of points (2u - 3w, v + 4w) as u, v, w each range over [0, 1]. Find the perimeter of the boundary of R.

Pick an answer.

(A)

$10\sqrt{3}$

(B)

(C)

(D)

(E)

AMC 10 2023 problem © Mathematical Association of America (MAA AMC). Reproduced for educational use.

How to solve

Strategy Draw a Diagram

The region is built from three direction-vectors added in scaled amounts. Tool #1 (Diagram) — plot the three vectors (2, 0), (0, 1), (-3, 4) and sketch the shape they sweep out. Tool #2 (Systematic List) enumerates the 2³ = 8 corner images by trying u, v, w ∈ {0, 1}. Tool #7 (Subproblems) breaks 'find perimeter' into 'identify the boundary vertices' then 'add up six edge lengths.' Tool #10 (Physical) is the Minkowski-sum picture: R is the unit segment in direction a plus the unit segment in direction b plus the unit segment in direction c — a hexagon whose six edges come in pairs ± a, ± b, ± c.

1STEP 1

P = u(2, 0) + v(0, 1) + w(-3, 4) is a Minkowski sum of three segments, so R is a centrally symmetric hexagon with sides in parallel pairs.

R = [0, a] + [0, b] + [0, c], a = (2, 0), b = (0, 1), c = (-3, 4)

2STEP 2

Plug (u, v, w) ∈ {0, 1}³ into P to get the 8 corner images, from (0, 0) out to (-1, 5).

corners: (0,0),(2,0),(0,1),(-3,4),(2,1),(-1,4),(-3,5),(-1,5)

3STEP 3

(0, 1) and (-1, 4) fall inside, so the boundary is the hexagon A(0,0) B(2,0) C(2,1) D(-1,5) E(-3,5) F(-3,4) (CCW).

hexagon vertices (CCW): A(0,0), B(2,0), C(2,1), D(-1,5), E(-3,5), F(-3,4)

4STEP 4

Reading lengths off a, b, c gives |a| = 2, |b| = 1, |c| = 5, with |(-3, 4)| = 5 from the 3-4-5 triangle.

|a| = 2, |b| = 1, |c| = 3² + 4² = 5

5STEP 5

Each direction contributes two equal sides, so the perimeter is 2(2 + 1 + 5) = 16, choice (E).

Perimeter = 2(2 + 1 + 5) = 16 → (E)

Answer

Cross-check by tracing ABCDEFA as a closed loop: walk a to B, b to C, c to D, -a to E, -b to F, -c back to A. Net displacement is a + b + c - a - b - c = 0, confirming the polygon closes. The two interior points (0, 1) and (-1, 4) correspond to corners where only one or two parameters are at 1 — geometrically they're tucked inside because their direction vectors don't 'stick out' compared to the convex hull. Perimeter 16 is choice (E).

💡Key takeaway

This AMC 10 problem only needs Grade 8 Pythagoras you already know — the region is built by adding three unit segments (2,0), (0,1), (-3,4), sweeping out a hexagon whose six sides come in pairs of those three lengths. Side lengths 2, 1, 5 each twice, so perimeter = 2(2+1+5) = 16, choice (E).