AMC 10 · 2008 · #19
Grade 11 algebraPick an answer.
Tool #4 (Introduce a Variable) does the setup: write α = a + bi and γ = c + di, so four real numbers carry the whole problem and both moduli become square roots of sums of squares. Tool #15 (Organize Information in More Ways) is the pivot — rewrite f(1) and f(i) as (real part) + (imaginary part) i, because "is real" only speaks about the imaginary part, and once the values are sorted that way the hypothesis becomes two plain linear equations. Tool #14 (Extreme Principle) is the primary tool because the question is a minimum: after eliminating γ's influence, |α|² becomes a quadratic in one variable whose completed-square form exposes both its floor and the single point where the floor is touched. Tool #6 (Guess and Check) supplies the half that a lower bound cannot supply on its own: exhibit one concrete pair and verify f(1) and f(i) really are real, proving the bound is reached rather than merely respected. Tool #1 (Draw a Diagram) gives the independent second route in the review — the constraint on α is a straight line, and the minimum is the distance from the origin to it.
Name the four real coordinates
Splitting each coefficient gives four real unknowns.
Splitting each complex number into its two real coordinates turns "smallest modulus" into ordinary distance in the plane.
11.N-CN.A.1Introduce A VariableExpand f(1) and f(i)
Expanding the two values separates real from imaginary.
Multiplying α by i is a quarter turn that swaps its coordinates, so a and b trade places between the two equations.
Multiplying by i is a quarter turn that swaps the two coordinates, so they trade places between the equations.
▸ Why?
A complex number is a point with a horizontal and a vertical coordinate, and multiplying turns it.
▸ Why?
A quarter turn sends a direction to one at right angles, which shows up as the coordinates swapping with one sign flipped.
Read off exactly two equations
Only the imaginary parts give conditions, so there are just two.
"Is real" costs one equation per value, so two given values cost exactly two equations and leave two coordinates free.
11.N-CN.A.1Introduce A VariableEliminate d to isolate α
Eliminating one unknown isolates the first coefficient.
Subtracting the two conditions deletes γ's contribution and leaves a single straight-line condition on α.
9.A-REI.C.6Introduce A VariableComplete the square to floor |α|
Completing the square floors its size at √(2).
Rewriting as (something)² + 2 shows the floor and the one place where the floor is touched, in the same line.
9.A-SSE.B.3Extreme PrincipleCheck the two floors are compatible
The two floors are compatible, so the other size can be zero.
The value of b that shrinks α is the very value that frees γ to be zero, so both floors are reachable at the same time.
11.N-CN.A.3Extreme PrincipleBuild the pair and test it
An explicit pair reaches it, so the answer is √(2), choice (A).
A bound plus one working example is a finished proof: the bound says you cannot go lower, the example says you can get there.
11.N-CN.A.2Guess And CheckTurn "both values are real" into "both imaginary parts are zero", and the whole problem shrinks to finding the point on one straight line that sits closest to the origin.
- Name the four real coordinates
- Expand f(1) and f(i)
- Read off exactly two equations
- Eliminate d to isolate α
- Complete the square to floor |α|
- Check the two floors are compatible
- Build the pair and test it