AMC 10 · 2020 · #15

Grade 11 algebra

Archetype Coordinate Plane Figure Computation └ Sub-archetype Pythagorean / Distance Formula for a Length

complex-numbersdifference-of-cubespolynomial-factoringquadratic-equationscoordinate-geometry identify-subproblemsphysical-representationextreme-principle ↑ Prerequisites: complex-numberspolynomial-factoring

📏 Long solution 💡 3 insights

Problem

In the complex plane, one set is the solutions of a first cubic equation and another set is the solutions of a second cubic. Treating each solution as a point, find the largest possible distance from a point of the first set to a point of the second.

Pick an answer.

(A)

$2\sqrt{3}$

(B)

(C)

(D)

$2\sqrt{21}$

(E)

$9+\sqrt{3}$

How to solve

Strategy Identify Subproblems

Tool #7 (Identify Subproblems): the two equations have nothing to do with each other, so solve them one at a time and only then compare. Tool #13 (Convert to Algebra): each cubic factors — one by the difference of cubes, the other by grouping — turning a root hunt into two easy factors. Tool #1 (Draw a Diagram): six points plotted in the plane make the longest link almost visible, which is far faster than grinding all nine distances. Tool #14 (Extreme Principle) and Tool #3 (Eliminate): all of A sits on the circle of radius 2, so the triangle inequality kills every pair that does not use the far-right point 8, leaving one short computation.

1STEP 1

Split into two root hunts

Solve the two cubics separately.

|A| = 3, |B| = 3, pairs ≤ 9

2STEP 2

Factor the first cubic

The difference of cubes works immediately.

z³ - 8 = (z - 2)(z² + 2z + 4)

3STEP 3

Finish with the quadratic formula

Solve the leftover quadratic.

A = { 2, -1 + i√(3), -1 - i√(3) }

4STEP 4

Factor the second by grouping

Grouping the terms makes it factor.

z³ - 8z² - 8z + 64 = (z-8)(z²-8), B = { 8, 2√(2), -2√(2) }

5STEP 5

Plot all six points

Plot the six points.

A: (2,0), (-1,√(3)), (-1,-√(3)); B: (8,0), (2√(2),0), (-2√(2),0)

6STEP 6

Cut the candidates down

Only the farthest point can win.

|a - b| ≤ 2 + |b| → b = ± 2√(2) gives ≤ 2 + 2√(2) ＜ 6

7STEP 7

Measure the winner

Measuring gives two root twenty-one.

√((8-(-1))² + (0-√(3))²) = √(81+3) = √(84) = 2√(21) = (D)

Answer

2√(21)

The estimate lines up: 2√(21) = √(84) ≈ 9.17, and the two points are 9 apart horizontally with only √(3) ≈ 1.73 of vertical offset, so a value just above 9 is exactly what the picture predicts. The distractors also make sense as traps. Choice (B) 6 is the distance from 2 to 8 — the answer if you forget the non-real roots. Choice (C) 9 is the horizontal gap alone — the answer if you drop the imaginary part. Choice (E) 9 + √(3) adds the two legs instead of using the Pythagorean theorem, and it must be too big since a hypotenuse is always shorter than the sum of the legs. An upper bound confirms the size: every point of A has modulus 2 and the farthest point of B is 8, so no distance can exceed 2 + 8 = 10, and 9.17 ＜ 10.

💡Key takeaway

Both equations factor, so no polar form is needed: z³ - 8 = (z-2)(z²+2z+4) gives A = {2, -1 ± i√(3)}, and grouping gives z³-8z²-8z+64 = (z-8)(z²-8), so B = {8, ± 2√(2)}. Plot the six points: A is a small triangle on the circle of radius 2, B sits on the real axis with 8 far out to the right. The farthest link runs from (-1, √(3)) to (8, 0), a right triangle with legs 9 and √(3), so the distance is √(84) = 2√(21). Answer (D).

Split into two root hunts
Factor z³ - 8
Finish set A with the quadratic formula
Factor the second cubic by grouping
Plot all six points
Cut nine pairs down to two
Measure the winning distance