AMC 10 · 2018 · #22

Grade 11 geometry-2d
complex-numberscomplex-polar-formsystems-of-equationsarea-trianglessymmetry-argument convert-to-algebracoordinate-geometryidentify-subproblems ↑ Prerequisites: complex-numberssystems-of-equations
📏 Long solution 💡 4 insights
Problem
Two squared equations are given, and each has two complex solutions. Plotting all four solutions in the complex plane gives the vertices of a parallelogram. Write that parallelogram's area as one whole number times a square root minus another whole number times a square root, with neither root divisible by a square, then report the sum of the four whole numbers.

Pick an answer.

(A)
20
(B)
21
(C)
22
(D)
23
(E)
24
How to solve
Strategy Introduce a Variable

Tool #4 (Introduce a Variable): writing z = a + bi turns each complex equation into two real equations, which is the only way to get coordinates for the vertices. Tool #15 (Organize in More Ways): the two equations from matching parts lead to a quartic, but comparing lengths supplies a third equation for free, and all three are linear in a² and b² — the same information reorganized is much easier to solve. Tool #7 (Subproblems): solve the two square-root problems separately, then treat the area as its own geometry problem. Tool #1 (Draw a Diagram): plotting the four points shows they come in ± pairs, which fixes both the correct vertex order and the fastest area formula.

1STEP 1

Split z into real and imaginary parts

Split into real and imaginary parts.

(a² - b²) + 2abi = 4 + 4√(15) i ⟹ a² - b² = 4, ab = 2√(15)
2STEP 2

Compare lengths for a free third equation

Comparing sizes hands over a third equation for free.

a² + b² = |4 + 4√(15) i| = √(16 + 240) = 16
3STEP 3

Solve for the first pair of roots

Three equations fix the first pair of roots.

a² = 10, b² = 6 ⟹ z = √(10) + √(6) i or z = -√(10) - √(6) i
4STEP 4

Run the same routine on the second equation

The second equation follows the same routine.

a² = 3, b² = 1 ⟹ z = √(3) + i or z = -√(3) - i
5STEP 5

Opposite pairs force a parallelogram

Opposite pairs become the diagonals.

diagonals v₁ ⇔ -v₁ of length 2 · 4 = 8 and v₂ ⇔ -v₂ of length 2 · 2 = 4, crossing at the origin
6STEP 6

Four triangles of equal area

The four triangles have equal areas.

Area = 4 · 1/2|x₁y₂ - x₂y₁| = 2|x₁y₂ - x₂y₁|
7STEP 7

Simplify into the required form

Tidying and adding gives 20.

Area = 2|√(10) - 3√(2)| = 6√(2) - 2√(10) ⟹ 6 + 2 + 2 + 10 = 20 → (A)
Answer
20
Numerically 6√(2) - 2√(10) ≈ 8.49 - 6.32 ≈ 2.16, a small positive number — and small is exactly right. The arm v₁ = (√(10), √(6)) points at about 37.8° while v₂ = (√(3), 1) points at exactly 30°, only about 7.8° apart, so the parallelogram is a thin sliver rather than anything close to the 16 it would enclose if the arms were perpendicular. Cross-check with the sine form: Area = 2|z₁||z₂|sinθ = 2 · 4 · 2·sin 7.8° ≈ 16 · 0.135 ≈ 2.16, matching. Two natural slips are worth naming, and the answer list quietly refuses both. Stopping at 2√(18) - 2√(10) without simplifying gives 2 + 18 + 2 + 10 = 32; the squarefree condition is what rules it out. Reporting only one pair of triangles, 3√(2) - √(10), gives 3 + 2 + 1 + 10 = 16. Neither 32 nor 16 appears among the choices, which is a signal to go back rather than to round toward the nearest option.
💡Key takeaway

To square-root a complex number, set z = a + bi and match real parts, imaginary parts, and lengths — three equations that are linear in a² and b² — and since the four roots come in ± pairs they form a parallelogram whose diagonals cross at the origin, with area 2|x₁y₂ - x₂y₁| = 6√(2) - 2√(10).

  • Split z into real and imaginary parts
  • Compare lengths for a free third equation
  • Solve for the first pair of roots
  • Run the same routine on the second equation
  • Opposite pairs force a parallelogram
  • Four triangles of equal area
  • Simplify into the required form