AMC 10 · 2023 · #16

Grade 11 algebra
complex-numbersquadratic-equationscompleting-the-squareoptimization extreme-principleconvert-to-algebraidentify-subproblems ↑ Prerequisites: complex-numbersquadratic-equations
📏 Long solution 💡 4 insights
Problem
A complex number is allowed only when the absolute value of one plus itself plus its square is exactly four. That one condition traps it on a curve. Write the largest possible imaginary part in the requested square-root form and find the sum of the two whole numbers.

Pick an answer.

(A)
20
(B)
21
(C)
22
(D)
23
(E)
24
How to solve
Strategy Extreme Principle

The word "maximum" hands the finish to Tool #14 (Extreme Principle): once the height is written as a function of one quantity, the winner is whichever allowed value of that quantity pushes the function to its edge. Getting there starts with Tool #4 (Introduce a Variable) — an absolute value of a complex expression says nothing until z=x+yi turns it into ordinary real numbers. Tool #13 (Convert to Algebra) then squares the condition into a single polynomial equation in x and y, which looks hopeless at first sight. Tool #15 (Organize Information in More Ways) is what rescues it: the same block x²+x+1 hides inside both halves of that equation, and naming it turns a messy two-variable mess into a tidy quadratic in y². Tool #7 (Identify Subproblems) then splits the work in two that do not interfere — "for a fixed value of that block, how big can y² be" and "how small can the block itself be." Tool #3 (Eliminate Possibilities) closes the problem: √(m)/n could in principle be written many ways, and the coprime condition kills all but one pair.

1STEP 1

Write it in parts

Write it in real and imaginary parts.

z=x+yi ⟹ 1+z+z² = (x²+x+1-y²) + y(2x+1) i
2STEP 2

Turn the size into an equation

The size becomes one equation.

(x²+x+1-y²)² + y²(2x+1)² = 16
3STEP 3

Spot the repeated block

The same block appears twice.

A=x²+x+1, t=y², (2x+1)²=4A-3 ⟹ (A-t)²+t(4A-3)=16 ⟺ t²+(2A-3) t+(A²-16)=0
4STEP 4

Solve for the squared height

A quadratic gives the squared height.

t=(3-2A)±√((2A-3)²-4(A²-16))/2=(3-2A)+√(73-12A)/2
5STEP 5

Push the block to its minimum

The height peaks when the block is smallest.

A=(x+1/2)²+3/4 ≥ 3/4 ⟹ t ≤ ((3-3/2)+√(73-9))/2=(3/2+8)/2=19/4
6STEP 6

Match the form and add

Adding the two numbers gives 21.

y_max=√(19/4)=√(19)/2=√(m)/n ⟹ m=19, n=2, gcd(19,2)=1 ⟹ m+n=21 (B)
Answer
21
Check the winning point by hand. At x=-1/2 and y=√(19)/2, take z=-1/2+√(19)/2i. Then z²=1/4-19/4-2·1/2·√(19)/2i=-9/2-√(19)/2i, so 1+z+z²=1-1/2-9/2+(√(19)/2-√(19)/2)i=-4, whose absolute value is 4. The point really is on the curve. The size is believable too: |z|=√(1/4+19/4)=√(5)≈ 2.24, and a number of size a little over 2 has z² of size about 5, which is the right ballpark for |1+z+z²|=4. Sweeping x across the whole curve in fine steps and taking the top y each time gives a maximum of 2.17944947…, matching √(19)/2=2.17944947… to twelve decimal places, with the peak at x=-1/2. Two traps sit in the choices. Stopping at t=19/4 and reading off the numerator and denominator as m=19, n=4 without taking the square root of the bottom gives (D) 23. Dropping the halving altogether and reporting √(19), that is m=19 and n=1, gives (A) 20.
💡Key takeaway

Give z coordinates, square the size condition, and the same block x²+x+1 shows up everywhere — name it, and the curve turns into a quadratic in y² whose top sits exactly where that block is smallest.

  • Write z in real and imaginary parts
  • Turn the size condition into an equation
  • Spot the hidden repeated block
  • Solve for y squared in terms of A
  • Push A down to its smallest value
  • Match the form and read off m + n