2018 AMC 10A Problem 21

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

Intersecting a parabola $y = x^2 - a$ with a circle is best done by eliminating $x^2$, not $y$: the result is a quadratic in $y$, and each $y$ must then be converted back to points using the sign of $x^2$. Symmetry explains why the answer hinges on a single axis point, and the fact that the vertex $(0,-a)$ automatically lies on the circle of radius $a$ is the hidden structure that makes the problem tractable.

Test $a = 1$: the circle $x^2 + y^2 = 1$ and parabola $y = x^2 - 1$ meet at $(0,-1)$ and at $y = 0$, $x = \pm 1$: three points, so (A), (B), (D) are out. Test $a = \tfrac13$: the quadratic $y^2 + y + \tfrac29 = 0$ has roots $-\tfrac13$ and $-\tfrac23$; only $y = -\tfrac13$ gives a point (the vertex), so one intersection, eliminating (C).

• Counting solutions y of the quadratic instead of points (x, y): each y with y + a > 0 gives two points, y = -a gives one, so 'exactly 3 points' is not 'three roots'.
• Setting the discriminant of $y^2 + y + a - a^2$ to zero (which gives $a = \tfrac12$, choice (D)) as if tangency produced three points; tangency at the vertex here produces exactly one.