2018 AMC 12A Problem #16

Algebra Level 2

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?

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a = \dfrac{1}{4}

a > \dfrac{1}{2}

\dfrac{1}{4} < a < \dfrac{1}{2}

a = \dfrac{1}{2}

a > \dfrac{1}{4}

1 solution

Jerry McKenzie
Feb 9, 2018

On these two graphs, it is gaurunteed that the bottom of the parabola is at the bottom of the circle.

Noting $a$ must be posiive or the porabola will open up away from the circle.

For 2 more points of intersection, the curvature of the circle must be smaller (not as tight) as the curvature of the parabola. The curvature of the parabola is always 2 times the leading coefficient, and the curvature of a circle (in this case of radius $a$ ) is the riciprocol of the radius (namely $\frac{1}{a}$ ). Thus :

$\frac{1}{a}<2\Rightarrow\frac{1}{2}<a$

2018 AMC 12A Problem #16

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