2018 AMC 10A Problem 21

P21-25~5 minPrint

Coordinate Geometry Quadratics Systems of Equations Substitute to simplify (u = x+1/x, shifting, scaling)Exploit symmetry to reduce work or pair up objects

Which of the following describes the set of values of $a$ for which the curves $x^2+y^2=a^2$ and $y=x^2-a$ in the real $xy$ -plane intersect at exactly $3$ points?

Show solution without answering

Key insight

Put y + a in place of x squared. The equation in y factors: one root is the parabola's bottom point, the other adds two points when a > 1/2.

Solution

Both curves are unchanged when $x$ is replaced by $-x$ , so any meeting point off the $y$ -axis comes with a mirror partner. To get an odd number, $3$ , we need exactly one point on the $y$ -axis and one mirror pair.

1. Get rid of $x$ . The parabola says $x^2 = y + a$ . Put that into the circle:
$(y + a) + y^2 = a^2, \qquad\text{so}\qquad y^2 + y + a - a^2 = 0.$
2. Factor. This factors as
$(y + a)(y - a + 1) = 0.$
(Multiply out to check: $y^2 - ay + y + ay - a^2 + a = y^2 + y + a - a^2$ .)
So $y = -a$ or $y = a - 1$ .
3. Turn each $y$ back into points. For a given $y$ , the points have $x^2 = y + a$ .
- If $y + a$ is positive, there are two points ( $x$ and $-x$ ).
- If $y + a = 0$ , there is one point ( $x = 0$ ).
- If $y + a$ is negative, there are none.
4. The root $y = -a$ . Here $y + a = 0$ , so it gives exactly one point, $(0, -a)$ : the bottom of the parabola.
5. The root $y = a - 1$ . Here $y + a = 2a - 1$ .
- If $2a - 1 > 0$ , that is $a > \tfrac12$ , it gives two more points. Total $3$ .
- If $a = \tfrac12$ , both roots are the same $y = -\tfrac12$ , giving just $1$ point.
- If $a < \tfrac12$ , it gives no points. Total $1$ .

So there are exactly $3$ meeting points precisely when $a > \tfrac12$ .

The answer is $\boxed{\textbf{(E)}\ a>\frac12}$ .

Why this works

When a parabola $y = x^2 - a$ meets another curve, replace $x^2$ rather than $y$ : you get an equation in $y$ alone. Each $y$ you find must then be turned back into points, and that is where the counting happens (two points, one point, or none). The mirror symmetry explains why an odd count needs a point on the axis.

Alternative approach

Test values against the choices. With $a = 1$ : the circle $x^2 + y^2 = 1$ and parabola $y = x^2 - 1$ meet at $(0, -1)$ , $(1, 0)$ and $(-1, 0)$ , which is $3$ points. That rules out every choice that leaves out $a = 1$ , leaving (C) and (E). With $a = \tfrac13$ : the roots are $y = -\tfrac13$ (one point) and $y = -\tfrac23$ , which needs $x^2 = -\tfrac13$ (no points). Only $1$ point, so (C) is out too.

Concise solution

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

Both curves are symmetric about the $y$ -axis, so intersection points off the axis come in pairs $(\pm x, y)$ . An odd total of $3$ therefore requires exactly one point on the axis plus one symmetric pair.

Substitute $x^2 = y + a$ from the parabola into the circle:
$(y + a) + y^2 = a^2 \quad\Longrightarrow\quad y^2 + y + (a - a^2) = 0.$
Each root $y$ of this quadratic yields points with $x^2 = y + a$ : two points if $y + a > 0$ , one point if $y + a = 0$ , none if $y + a < 0$ .

Notice $y = -a$ is always a root: $a^2 - a + a - a^2 = 0$ . It corresponds to the single axis point $(0, -a)$ , the parabola's vertex sitting on the circle. By Vieta the roots sum to $-1$ , so the other root is $y = a - 1$ .

For exactly $3$ points we need the second root to give two points: $y + a = 2a - 1 > 0$ , i.e. $a > \tfrac12$ . (If $a = \tfrac12$ the roots coincide and there is only $1$ point; if $a < \tfrac12$ the second root gives nothing, again $1$ point.)

The answer is $\boxed{\textbf{(E)}\ a>\frac12}$ .

Review in the AoPS books

Introduction to Geometry, Chapter 17: Analytic Geometry
Introduction to Algebra, Chapter 10: Quadratic Equations – Part 1 (§10.2, §10.3, §10.4)
Introduction to Algebra, Chapter 5: Multi-Variable Linear Equations (§5.2, §5.3, §5.6)
Intermediate Algebra, Chapter 20: Some Advanced Strategies (§20.2)