2015 AMC 10A Problem 12

P11-15~3 minPrint

Algebraic Manipulation Quadratics Apply an identity: SFFT, sum of squares, difference of cubes, Vieta

Points $(\sqrt{\pi}, a)$ and $(\sqrt{\pi}, b)$ are distinct points on the graph of $y^2+x^4=2x^2y+1$ . What is $|a-b|$ ?

Show solution without answering

Key insight

The equation rearranges to (y minus x squared) squared equals 1. So y is x squared plus 1 or minus 1, and the two heights always differ by 2.

Solution

The equation looks scary because of $x^4$ and $\pi$ . Rearranging it shows a pattern you already know.

1. Move $2x^2y$ to the left side:
$y^2 - 2x^2y + x^4 = 1.$
2. Compare this with the pattern $A^2 - 2AB + B^2 = (A - B)^2$ . Here $A = y$ and $B = x^2$ (note $x^4 = (x^2)^2$ ). So the left side is a perfect square:
$(y - x^2)^2 = 1.$
3. A number whose square is $1$ is either $1$ or $-1$ . So
$y - x^2 = 1 \quad\text{or}\quad y - x^2 = -1,$
which means $y = x^2 + 1$ or $y = x^2 - 1$ .
4. At $x = \sqrt{\pi}$ we have $x^2 = \pi$ . The two $y$ -values are $\pi + 1$ and $\pi - 1$ . The two points are different, so $a$ and $b$ are these two numbers (in some order).
5. Their difference is
$(\pi + 1) - (\pi - 1) = 2.$

The answer is $\boxed{\textbf{(C)}\ 2}$ .

Why this works

The graph is really two parabolas, $y = x^2 + 1$ and $y = x^2 - 1$ , one stacked $2$ units above the other. So at every $x$ , not just $\sqrt{\pi}$ , the two points are $2$ apart; the $\pi$ is only there to distract you. When you see three terms like $y^2$ , $2x^2y$ and $x^4$ together, check for a perfect square.

Alternative approach

Put $x^2 = \pi$ in first: $y^2 - 2\pi y + \pi^2 - 1 = 0$ . The two solutions $a$ and $b$ add to $2\pi$ and multiply to $\pi^2 - 1$ (sum and product of the roots). Then
$(a - b)^2 = (a + b)^2 - 4ab = 4\pi^2 - 4\pi^2 + 4 = 4,$
so $|a - b| = 2$ .

Concise solution

A shorter write-up, the way contest solutions are usually written. It may take a different route from the solution above.

Group the terms involving $y$ and $x^2$ together:
$y^2 - 2x^2 y + x^4 = 1 \quad\Longrightarrow\quad (y - x^2)^2 = 1.$
So every point of the graph satisfies $y - x^2 = \pm 1$ , i.e. the graph is the union of the two parabolas $y = x^2 + 1$ and $y = x^2 - 1$ .

For a fixed $x$ (here $x = \sqrt{\pi}$ , so $x^2 = \pi$ ) the two points on the graph have $y$ -coordinates $\pi + 1$ and $\pi - 1$ . Since the points are distinct, $\{a, b\} = \{\pi + 1, \pi - 1\}$ and
$|a - b| = (\pi + 1) - (\pi - 1) = 2.$

The answer is $\boxed{\textbf{(C)}\ 2}$ .

Review in the AoPS books

Introduction to Algebra, Chapter 11: Special Factorizations (§11.2, §11.3, §11.5)
Introduction to Algebra, Chapter 10: Quadratic Equations – Part 1 (§10.2, §10.3, §10.4)
Introduction to Algebra, Chapter 22: Special Manipulations
Introduction to Algebra, Chapter 13: Quadratic Equations – Part 2 (§13.2, §13.3)