AMC 10 · 2015 · #12

Grade 8 arithmetic
Archetype Coordinate Plane Figure Computation └ Sub-archetype Figure Constraint Equations for Unknown Quantity
quadratic-equationscompleting-the-squareabsolute-value formula-substitution ↑ Prerequisites: quadratic-equations
📏 Medium solution 💡 3 insights
Problem
Two different points (√π, a) and (√π, b) both sit on the curve y² + x⁴ = 2x²y + 1. They share the same x-value √π, so a and b are the two y-values the curve allows at that x. Find how far apart those two y-values are, |a−b|.

Pick an answer.

(A)
1
(B)
$\frac{\pi}{2}$
(C)
2
(D)
$\sqrt{1+\pi}$
(E)
$1 + \sqrt{\pi}$

AMC 10 2015 problem © Mathematical Association of America (MAA AMC). Reproduced for educational use.

How to solve
Strategy Introduce a Variable

The two points are just the two y-values that satisfy the equation at the fixed x=√(π), so Tool #4 (Introduce a Variable) treats y as the single unknown and turns the curve into one equation in y. The expression y² + x⁴ - 2x² y looks tangled, so Tool #15 (Organize Information in More Ways) re-arranges it: grouping the terms reveals a perfect square, which collapses the whole problem. Plugging in x²=π first (Tool #9, Solve an Easier Related Problem) replaces the abstract x⁴ and x² with the concrete numbers π² and π, so there is nothing left to juggle but y.

1STEP 1

Plug in the shared x-value

Since (√π)²=π, substitute x²=π and x⁴=π² into the curve's equation.

y² + π² = 2π y + 1
2STEP 2

Gather every term on one side

Rearranging gives y²−2πy+π²=1, the shape of a squared binomial.

y² - 2π y + π² = 1
3STEP 3

Recognize the perfect square

That expression is exactly (y−π)²=1.

(y-π)² = 1
4STEP 4

Take square roots to find both y-values

So y−π=±1, giving y = π+1 and y = π−1.

y-π = ± 1 → y = π+1 or y = π-1
5STEP 5

Subtract to get the gap

Subtracting gives |a−b| = 2 — choice (C).

|a-b| = |(π+1)-(π-1)| = 2 → (C)
Answer
2
(y−π)² = 1 → y = π±1 → |a−b| = 2
Notice that π vanished from the final answer, which makes sense: the curve (y−x²)²=1 always puts its two y-values one unit above and one unit below x², so they are 2 apart no matter what x is. That instantly rules out the π-flavored choices (B) π/2, (D) √(1+π), and (E) 1+√π. A quick check confirms a and b are distinct (π+1 is not π−1), as the problem requires, and a gap of 1 (choice A) would need the square to equal 1/4, not 1. So 2 is the only consistent value.
💡Key takeaway

Plug in the x-value, tidy the equation into (y−π)²=1, and the two heights are just π±1 — exactly 2 apart.

  • Plug in the shared x-value
  • Gather every term on one side
  • Recognize the perfect square
  • Take square roots to find both y-values
  • Subtract to get the gap

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