2018 AMC 10A Problems/Problem 21

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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?

$\textbf{(A) }a=\frac14 \qquad \textbf{(B) }\frac14 < a < \frac12 \qquad \textbf{(C) }a>\frac14 \qquad \textbf{(D) }a=\frac12 \qquad \textbf{(E) }a>\frac12 \qquad$

Solution 1

Substituting $y=x^2-a$ into $x^2+y^2=a^2$, we get \[x^2+(x^2-a)^2=a^2 \implies x^2+x^4-2ax^2=0 \implies x^2(x^2-(2a-1))=0\] Since this is a quartic, there are 4 total roots (counting multiplicity). We see that $x=0$ always at least one intersection at $(0,-a)$ (and is in fact a double root).

The other two intersection points have $x$ coordinates $\sqrt{2a-1}$. We must have $2a-1> 0,$ otherwise we are in the case where the parabola lies entirely above the circle (tangent to it at the point $(0,a)$). This only results in a single intersection point in the real coordinate plane. Thus, we see $a>\frac{1}{2}$.


Solution 2

Looking at a graph, it is obvious that the two curves intersect at (0, -a). We also see that if the parabola go's 'in' the circle, than by going out of it (as it will) it will intersect five times, an impossibility. Thus we only look for cases where the parabola becomes externally tangent to the circle. We have $x^2 - a = -\sqrt(a^2 - x^2)$. Squaring both sides and solving yields $x^4 - (2a - 1)x^2 = 0$. Since x = 0 is already accounted for, we only need to find 1 solution for $x^2 = 2a - 1$, where the right hand side portion is obviously increasing. Since a = 1/2 begets x = 0 (an overcount), we have $a > 1/2 -> E$ is the right answer.

Solution by JohnHankock

Solution 3

This describes a unit parabola, with a circle centered at the axis of symmetry and tangent to the vertex. As the curvature of the unit parabola at the vertex is 2, the radius of the circle that matches it has a radius of $\frac{1}{2}$. This circle is tangent to an infinitesimally close pair of points, one on each side. Therefore, it is tangent to only 1 point. When a larger circle is used, it is tangent to 3 points because the points on either side are now separated from the vertex. Therefore, $\boxed{a > \frac{1}{2}}$ or $\boxed{E}$ is correct.

$QED \blacksquare$

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