Difference between revisions of "2018 AMC 10A Problems/Problem 21"
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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 intersect at exactly <math>3</math> points? | 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 intersect at exactly <math>3</math> points? | ||
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<math>QED \blacksquare</math> | <math>QED \blacksquare</math> | ||
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| + | == See Also == | ||
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| + | ==See Also== | ||
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| + | {{AMC10 box|year=2018|ab=A|num-b=20|num-a=22}} | ||
| + | {{AMC12 box|year=2018|ab=A|num-b=15|num-a=17}} | ||
| + | {{MAA Notice}} | ||
Revision as of 17:34, 8 February 2018
Problem
Which of the following describes the set of values of 
for which the curves 
and 
in the real 
-plane intersect at exactly 
points?
Solution 1
Substituting into , we get
![\[x^2+(x^2-a)^2=a^2 \implies x^2+x^4-2ax^2=0 \implies x^2(x^2-(2a-1))=0\]](http://latex.artofproblemsolving.com/8/6/7/8678d33385ba1d0b6061a2c249363a8d942e17a1.png)
Since this is a quartic, there are 4 total roots (counting multiplicity). We see that 
always at least one intersection at 
(and is in fact a double root).
The other two intersection points have 
coordinates 
. We must have 
otherwise we are in the case where the parabola lies entirely above the circle (tangent to it at the point 
). This only results in a single intersection point in the real coordinate plane. Thus, we see 
.
(projecteulerlover)
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 
. Squaring both sides and solving yields 
. Since x = 0 is already accounted for, we only need to find 1 solution for 
, where the right hand side portion is obviously increasing. Since a = 1/2 begets x = 0 (an overcount), we have 
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 
. 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, 
or 
is correct.
See Also
See Also
| 2018 AMC 10A (Problems • Answer Key • Resources) | ||
| Preceded by Problem 20 |
Followed by Problem 22 | |
| 1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 • 16 • 17 • 18 • 19 • 20 • 21 • 22 • 23 • 24 • 25 | ||
| All AMC 10 Problems and Solutions | ||
| 2018 AMC 12A (Problems • Answer Key • Resources) | |
| Preceded by Problem 15 |
Followed by Problem 17 |
| 1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 • 16 • 17 • 18 • 19 • 20 • 21 • 22 • 23 • 24 • 25 | |
| All AMC 12 Problems and Solutions | |
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