AMC 10 · 2010 · #13

Grade 9 algebra
Archetype Small-Domain Constrained Search └ Sub-archetype Logic-Deduction / Elimination
symmetric-polynomialssystems-of-equationssign-analysis convert-to-algebracasework ↑ Prerequisites: systems-of-equations
📏 Medium solution 💡 3 insights
Problem
A circle and a hyperbola both depend on the same whole number. Count the values that keep them apart.

Pick an answer.

(A)
0
(B)
1
(C)
2
(D)
4
(E)
8
How to solve
Strategy Organize Information in More Ways

Tool #15 (Organize Information in More Ways): chasing x and y separately leads to a messy quartic, but the system only ever mentions x²+y² and xy — the exact ingredients of (x+y)² and (x-y)² — so regrouping around the sum and the difference makes the whole system collapse into two squares. Tool #3 (Eliminate Possibilities): a real square cannot be negative, which knocks out values of k. Tool #11 (Work Backwards): eliminating only proves which k are impossible; to be sure the surviving k really do give an intersection, I run the substitution in reverse and construct the point. Tool #16 (Change Focus / Count the Complement): the question asks when the graphs miss, so I first describe exactly when they meet and then take what is left over.

1STEP 1

Regroup into two perfect squares

Adding and subtracting makes two perfect squares.

(x+y)² = k²+2k, (x-y)² = k²-2k
2STEP 2

A real square is never negative

A real square is never negative.

k²+2k ≥ 0 and k²-2k ≥ 0
3STEP 3

Those two conditions are also enough

Those conditions also build a shared point back.

x=(u+v)/2, y=(u-v)/2 ⟹ xy=(u²-v²)/4=k, x²+y²=(u²+v²)/2=k²
4STEP 4

Solve both sign conditions

Solving both leaves a narrow gap of values.

k(k+2) ≥ 0 ⇔ k ≤ -2 or k ≥ 0; k(k-2) ≥ 0 ⇔ k ≤ 0 or k ≥ 2
5STEP 5

Count the integers left out

Exactly 2 whole numbers fall in it, choice (C).

Z ∖ ({k ≤ -2}∪{0}∪{k ≥ 2})={-1, 1} → (C) 2
Answer
2
Check the boundary values by hand, since the whole answer rests on where the gap starts. For k=2: x=y=√2 gives x²+y²=4=2² and xy=2, so they meet. For k=-2: x=√2, y=-√2 gives 4 and -2, so they meet. For k=3 the construction in Step 3 gives u=√(15), v=√3, x≈ 2.803, y≈ 1.070, and indeed xy≈ 3 and x²+y²≈ 9. So the failures really do stop at ±1. A size check agrees: the circle x²+y²=k² has radius |k|, which grows in step with k, while the nearest point of xy=k only drifts out like √(2|k|), so large |k| must intersect and only the smallest sizes can fail. Choices 4 and 8 would require failures beyond k=±1, which the sign analysis rules out, and choice 0 would require k=1 to work, which the negative square forbids. So 2, choice (C), is right.
💡Key takeaway

When a system only mentions x²+y² and xy, rebuild it as (x+y)² and (x-y)² — and then show the leftover conditions not only must hold but actually build the point.

  • Regroup into two perfect squares
  • A real square is never negative
  • Those two conditions are also enough
  • Solve both sign conditions
  • Count the integers left out

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