AMC 10 · 2019 · #17
Grade 11 algebraPick an answer.
Tool #13 (Convert to Algebra): "equilateral" is a statement about three distances, and in the complex plane a distance is a modulus, so the geometric condition becomes the single chain of equations |z| = |z³| = |z³ - z| — no picture-based case analysis needed. Tool #4 (Introduce a Variable): naming r = |z| turns the first equality into r = r³, a one-line equation. Tool #9 (Easier Related Problem): every surviving condition mentions only z², so replacing the unknown by w = z² solves a smaller problem first and takes square roots at the very end — that final square-root step is exactly where the count doubles. Tool #1 (Draw a Diagram): the reduced condition says w is distance 1 from 0 and distance 1 from 1, which is two circles crossing — a picture that makes "exactly two" obvious before any algebra. Tool #3 (Eliminate Possibilities): choice (E) claims infinitely many, so the argument must show both the size and the direction of z get pinned down, not just one of them.
Turn the triangle into three distances
Turn the triangle into three distances.
A side length in the complex plane is just the size of a difference, so "equilateral" is three moduli that agree.
11.N-CN.A.1Convert To AlgebraFirst equality pins the size
The first equality pins the size.
Cubing scales a length by its own square, and the only positive length that survives that unchanged is 1.
11.N-CN.A.3Introduce A VariableFactor the third side
Factor the third side.
Pulling z out front costs nothing when |z| = 1, and it drops the problem from degree three to degree two.
9.A-SSE.A.2Introduce A VariableSwitch the unknown to w = z squared
Switch the unknown to its square.
Both surviving conditions only ever mention z², so make z² the thing you solve for.
11.N-CN.A.2Solve An Easier Related ProblemTwo circles, two crossings
Two circles cross at two points.
Two equal circles whose centers are nearer than a diameter always meet in exactly two points, one above the line of centers and one below.
8.G.B.8Draw A DiagramSolve the two-circle system
Solving gives the two values.
Subtracting two circle equations destroys the squared terms and leaves one straight line to intersect.
11.A-REI.C.7Introduce A VariableUndo the squaring — the count doubles
Undoing the square doubles them.
Every nonzero complex number has exactly two square roots, opposite in sign, so passing back from z² to z doubles the tally.
Every nonzero complex number has exactly two square roots, opposite in sign, so the tally doubles.
▸ Why?
A number and its opposite have the same square, so the two roots come as a matched pair.
▸ Why?
A complex number is a point with a length and a direction, so halving the direction gives two choices.
Confirm the vertices are distinct and count
All are valid, so the answer is 4.
Nothing is left free — the length is forced to 1 and the direction to four choices — so the answer is finite, not "infinitely many".
11.N-CN.A.1Eliminate PossibilitiesEvery side of the triangle is a modulus, so equilateral means |z| = |z³| = |z³ - z|; the first equality forces |z| = 1 and the second collapses to |z² - 1| = 1, which puts z² at exactly two spots, and each spot has two square roots — 4 values of z, choice (D).
- Turn the triangle into three distances
- First equality pins the size
- Factor the third side
- Switch the unknown to w = z squared
- Two circles, two crossings
- Solve the two-circle system
- Undo the squaring — the count doubles
- Confirm the vertices are distinct and count