AMC 10 · 2008 · #23

Grade 12 algebrageometry-2d

Archetype Coordinate Plane Figure Computation └ Sub-archetype Axis-Aligned Figure Area from Coordinates

complex-numberscomplex-polar-formbinomial-theorem pattern-recognitionsymmetry-argument ↑ Prerequisites: complex-numbers

📏 Long solution 💡 3 insights

Problem

A quartic equation with complex coefficients has four solutions plotted as points. Find the area of the polygon they form.

Pick an answer.

(A)

$2^{\frac{5}{8}}$

(B)

$2^{\frac{3}{4}}$

(C)

(D)

$2^{\frac{5}{4}}$

(E)

$2^{\frac{3}{2}}$

How to solve

Strategy Organize Information in More Ways

Solving a general quartic head-on is hopeless, so the move is to rewrite the left side rather than attack it. The coefficient magnitudes 1, 4, 6, 4 are the fourth row of Pascal's triangle, which suggests the whole polynomial is a fourth power in disguise plus a constant. Renaming that inner expression as a single variable w turns the problem into the far easier equation w⁴ = c, whose roots are a familiar evenly spaced ring on a circle. Since w differs from z only by a translation, the polygon for w and the polygon for z are congruent, so a picture of the ring gives the area directly.

1STEP 1

Rewrite the quartic as a fourth power

The coefficients hide a fourth power.

(z+i)⁴ = z⁴ + 4iz³ - 6z² - 4iz + 1 ⟹ z⁴ + 4z³ i - 6z² - 4zi - i = (z+i)⁴ - (1+i)

2STEP 2

Translate the roots by +i

Translating the roots leaves the area untouched.

w = z + i ⟹ w⁴ = 1 + i, area(polygon in z) = area(polygon in w)

3STEP 3

Solve w⁴ = 1+i in polar form

Polar form places the four roots on one circle.

1+i = 2¹/2(cosπ/4 + isinπ/4), w_k = 2¹/8(cosθ_k + isinθ_k), θ_k = π/16 + kπ/2

4STEP 4

Check the polygon really has four vertices

Their angles are all distinct, so there really are four vertices.

|w_k| = R = 2¹/8 for all k, θ_k+1 - θ_k = π/2 ≠ 0 (mod 2π)

5STEP 5

Identify the quadrilateral as a square

A quarter-turn symmetry makes it a square.

w⁴ = (iw)⁴ → {w₀, w₁, w₂, w₃} is fixed by rotation through π/2

6STEP 6

Turn the circumradius into area

The circumradius gives the area 2⁵/4, choice (C).

Area = 1/2(2R)(2R) = 2R² = 2·(2¹/8)² = 2 · 2¹/4 = 2⁵/4

Answer

2⁵/4

Two independent size checks agree. Using the side instead of the diagonals: the side of a square with circumradius R is R√(2) = 2¹/8 · 2¹/2 = 2⁵/8, and squaring gives 2⁵/4, the same area. Numerically R ≈ 1.0905, so the area is about 2.378, which must sit below the area of the circle that holds the square, π R² ≈ 3.73, and above half of it; it does. The five choices are about 1.54, 1.68, 2, 2.38, 2.83, and only (D) matches. The near-miss choices are informative: 2⁵/8 is the side length rather than the area, and 2³/2 is what appears if the modulus √(2) of 1+i is given only a square root instead of a fourth root. Both traps are avoided here because the modulus was tracked as (2¹/2)¹/4 explicitly.

💡Key takeaway

When the coefficients look like 1, 4, 6, 4, the polynomial is a fourth power in disguise, and the fourth roots of any nonzero number always sit as a square on a circle.

Rewrite the quartic as a fourth power
Translate the roots by +i
Solve w⁴ = 1+i in polar form
Check the polygon really has four vertices
Identify the quadrilateral as a square
Turn the circumradius into area