AMC 10 · 2022 · #13

Grade 11 geometry-2d
Archetype Compound Area by Decomposition / Difference └ Sub-archetype Circle-Polygon Hybrid
complex-numbersminkowski-sumarea-circlesarea-rectanglespythagorean-theorem spatial-visualizationidentify-subproblemsestimation ↑ Prerequisites: complex-numbersarea-circles
📏 Medium solution 💡 2 insights
Problem
Work in the complex plane. The first point may be anywhere on the straight segment joining two fixed numbers. The second may be any complex number of magnitude at most one. The region collects every possible sum. Find its area and report the closest integer.

Pick an answer.

(A)
13
(B)
14
(C)
15
(D)
16
(E)
17
How to solve
Strategy Visualize Spatial Relationships

Nothing here needs heavy complex-number machinery. Once each complex number is read as a point, z₁ is a segment and z₂ is a unit disk, and the sum z₁ + z₂ means sliding that disk along the segment. Picturing the swept shape is the whole problem: it turns an abstract set of sums into one concrete outline. After that the shape splits into a rectangle and two round caps, so the area is two easy pieces added together, and a decimal estimate of π picks the nearest integer.

1STEP 1

Read both conditions as shapes

Read both conditions as shapes.

3 = 3+0i ⇔ (3,0), 4i = 0+4i ⇔ (0,4), |z₂| ≤ 1 ⇔ x²+y² ≤ 1
2STEP 2

The segment has length 5

Measure the segment's length.

√((3-0)²+(0-4)²)=√(9+16)=√(25)=5
3STEP 3

Adding z₂ slides a unit disk along the segment

A unit disk slides along the segment.

R = { z : dist(z, segment) ≤ 1 }
4STEP 4

Split it into a rectangle and one circle

The end caps make one full circle.

[R] = 5 · 2_rectangle + 2 · 1/2π(1)²_two caps = 10 + π
5STEP 5

Round 10 + π to the nearest integer

Rounding gives 13.

10 + π ≈ 13.14159 → closest integer = 13
Answer
13
Two crude bounds trap the area. R contains the 5 by 2 rectangle, so its area is more than 10. Turning the picture so the segment lies flat, R fits inside a rectangle of length 5 + 1 + 1 = 7 and width 2, so its area is less than 14. That leaves only 13 among the five listed integers, matching 10 + π ≈ 13.14. A second check: the caps must add exactly one unit circle to the rectangle, and 13.14 - 10 = 3.14 = π, as expected.
💡Key takeaway

Adding every vector of length at most 1 to a shape just pads that shape by 1 in all directions, so a padded segment is a rectangle with two round ends whose area is the length times 2, plus one full circle.

  • Read both conditions as shapes
  • The segment has length 5
  • Adding z₂ slides a unit disk along the segment
  • Split it into a rectangle and one circle
  • Round 10 + π to the nearest integer

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