2018 AMC 10A Problems/Problem 21

Problem

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

(projecteulerlover)

Solution 2

$[asy] Label f; f.p=fontsize(6); xaxis(-2,2,Ticks(f, 0.2)); yaxis(-2,2,Ticks(f, 0.2)); real g(real x) { return x^2-1; } draw(graph(g, 1.7, -1.7)); real h(real x) { return sqrt(1-x^2); } draw(graph(h, 1, -1)); real j(real x) { return -sqrt(1-x^2); } draw(graph(j, 1, -1)); [/asy]$

Looking at a graph, it is obvious that the two curves intersect at (0, -a). We also see that if the parabola goes 'in' the circle. Then, 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 $\boxed{\textbf{(E) }a>\frac12}$ is the right answer.

Solution by JohnHankock

Solution 3

This describes a unit parabola, with a circle centered on 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, $\boxed{\textbf{(E) }a>\frac12}$ is correct.

$QED \blacksquare$

Solution 4

Notice, the equations are of that of a circle of radius a centered at the origin and a parabola translated down by a units. They always intersect at the point $(0, a)$ , and they have symmetry across the y-axis, thus, for them to intersect at exactly 3 points, it suffices to find the y solution.

First, rewrite the second equation to $y=x^2-a\implies x^2=y+a$ And substitute into the first equation: $y+a+y^2=a^2$ Since we're only interested in seeing the interval in which a can exist, we find the discriminant: $1-4a+4a^2$ . This value must not be less than 0 (It is the square root part of the quadratic formula). To find when it is 0, we find the roots: $4a^2-4a+1=0 \implies a=\frac{4\pm\sqrt{16-16}}{8}=\frac{1}{2}$ Since $\lim_{a\to \infty}(4a^2-4a+1)=\infty$ , our range is $\boxed{\textbf{(E) }a>\frac12}$ .

Solution by ktong

Solution 5 (Cheating with Answer Choices)

Simply plug in $a = 0, \frac{1}{2}, \frac{1}{4}, 1$ and solve the systems. (This shouldn't take too long.) And then realize that only $a=1$ yields three real solutions for $x$ , so we are done and the answer is $\boxed{a>\frac{1}{2}}$

~ ccx09

2018 AMC 10A (Problems • Answer Key • Resources)
Preceded by Problem 20	Followed by Problem 22
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All AMC 10 Problems and Solutions

2018 AMC 12A (Problems • Answer Key • Resources)
Preceded by Problem 15	Followed by Problem 17
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All AMC 12 Problems and Solutions

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