AMC 10 · 2013 · #25

Grade 11 algebracounting

Archetype Small-Domain Constrained Search └ Sub-archetype Logic-Deduction / Elimination

complex-numberscompleting-the-squarequadratic-equations work-backwardsconvert-to-algebrasystematic-enumeration ↑ Prerequisites: complex-numbersquadratic-equations

📏 Long solution 💡 4 insights

Problem

A quadratic map must send an upper-half input to a small grid of outputs. Count the inputs.

Pick an answer.

(A)

399

(B)

401

(C)

413

(D)

431

(E)

441

How to solve

Strategy Work Backwards

The set of admissible z is a scatter of irrational points in a half-plane — there is nothing to list. But the set of admissible outputs is a tidy 21 × 21 grid of 441 lattice points. Tool #11 (Work Backwards) flips the search: instead of asking which z qualify, ask which of the 441 targets c are reachable from the upper half-plane, and how many times. Tool #13 (Convert to Algebra) supplies the machinery — complete the square so that inverting f is just taking a square root — and tool #4 (Introduce a Variable) renames w=z+i/2 so the finish line Im(z) ＞ 0 becomes the clean height test Im(w) ＞ 1/2. Two facts then have to be proved, not assumed: no target is hit twice (from the sum of the roots), and every target passing the test really is hit (from an explicit square root). Once that pairing is a genuine bijection, tool #2 (Make a Systematic List) counts the surviving grid pairs column by column, and tool #16 (Change Focus / Count the Complement) rechecks the same number from the other side by counting the few pairs that fail.

1STEP 1

Complete the square, then shift

Completing the square makes it a plain root.

f(z)=(z+i/2)²+5/4; w=z+i/2 ⟹ w²=c-5/4, Im(w) ＞ 1/2

2STEP 2

No target is hit twice

No output is hit twice.

z²+iz+(1-c)=0 → z₁+z₂=-i → Im(z₁)+Im(z₂)=-1

3STEP 3

How high a square root reaches

How high a root reaches is a plain formula.

w²=u=p+qi, r=|u|: x²-y²=p, x²+y²=r → y²=(r-p)/2; Im(w) ＞ 1/2 possible⇔ r-p ＞ 1/2

4STEP 4

The height test becomes b² ≥ a

The height test becomes one inequality.

p=a-5/4, q=b: r ＞ p+1/2⇔ b² ＞ a-1⇔ b² ≥ a

5STEP 5

One z per admissible pair

So counting inputs is counting pairs.

#{z:Im(z) ＞ 0, f(z)=a+bi admissible}=#{(a,b)inZ²: |a| ≤ 10, |b| ≤ 10, b² ≥ a}

6STEP 6

Count the surviving pairs

That count is 399, choice (A).

a ≤ 0: 11 · 21=231; a ≥ 1: 20+3 · 18+5 · 16+14=168; 231+168=399→(A)

Answer

399

The grid holds 441 points, which is choice (E), so the answer must be a bit less — and only 441-399=42 pairs fail, which is believable because a is capped at 10 while b² climbs to 100, so the test b² ≥ a is easy to pass. Two spot checks confirm the criterion at its two most delicate places. First c=0, i.e. a=b=0, which passes since 0 ≥ 0: solving z²+iz+1=0 gives z=-1±√(5)/2i, whose imaginary parts are (√5-1)/2≈ 0.618 ＞ 0 and (-√5-1)/2≈-1.618 ＜ 0 — exactly one admissible root, and the two imaginary parts sum to -1 as step 2 promised. Second, the boundary case c=10+3i, where b²=9 ＜ 10=a and the criterion says "missed": the two roots are z=3 and z=-3-i, since f(3)=9+3i+1=10+3i and f(-3-i)=(8+6i)+(1-3i)+1=10+3i. One sits exactly on the real axis and the other below it, so neither counts — the strict inequality Im(z) ＞ 0 is what rejects this target, and it shows the criterion is tight at a=b²+1. The wrong choices are all "stopped checking too early": the 42 failing pairs split as 10 with b=0, 18 with |b|=1, 12 with |b|=2 and 2 with |b|=3; removing only the first group gives 431 (D), the first two gives 413 (C), the first three gives 401 (B), and removing nothing gives 441 (E).

💡Key takeaway

Stop hunting for z and count outputs instead: completing the square shows a+bi is reached from above the real axis exactly once when b² ≥ a and never otherwise, so the answer is just how many of the 441 grid pairs pass that one test.

Complete the square, then shift
No target is hit twice
How high a square root reaches
The height test becomes b² ≥ a
One z per admissible pair
Count the surviving pairs