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Difference between revisions of "2018 AMC 10A Problems/Problem 21"

(Created page with "Which of the following describes the set of values of <math>a</math> for which the curves <math>x^2+y^2=a^2</math> and <math>y=x^2-a</math> in the real <math>xy</math>-plane i...")
 
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\textbf{(E) }a>\frac12 \qquad
 
\textbf{(E) }a>\frac12 \qquad
 
</math>
 
</math>
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== Solution ==
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Substituting <math>y=x^2-a</math> into <math>x^2+y^2=a^2</math>, we get
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<cmath>
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x^2+(x^2-a)^2=a^2 \implies x^2+x^4-2ax^2=0 \implies x^2(x^2-(2a-1))=0
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</cmath>
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Since this is a quartic, there are 4 total roots (counting multiplicity). We see that <math>x=0</math> always at least one intersection at <math>(0,-a)</math> (and is in fact a double root).
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The other two intersection points have <math>x</math> coordinates <math>\sqrt{2a-1}</math>. We must have <math>2a-1> 0,</math> otherwise we are in the case where the parabola lies entirely above the circle (tangent to it at the point <math>(0,a)</math>). This only results in a single intersection point in the real coordinate plane. Thus, we see <math>a>\frac{1}{2}</math>.

Revision as of 17:25, 8 February 2018

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

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 $a>\frac{1}{2}$.

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