Problem:
Let S be the set of positive integers k such that the two parabolas
y=x2βk and x=2(yβ20)2βk
intersect in four distinct points, and these four points lie on a circle with radius at most 21. Find the sum of the least element of S and the greatest element of S.
Solution:
Note that y=x2βk has its vertex at (0,βk), which is below the line y=20, and opens upwards. Parabola x=2(yβ20)2βk has its vertex at (βk,20) and opens to the right.
If 0<kβ€4, then for βkβ€xβ€0, the maximum value of y on the first parabola is k2βkβ€12. However, for xβ€0, the minimum value of y for the second parabola is 20β2kββ>18. Thus if 0<kβ€4, the second parabola does not intersect the left half of the first parabola.
If k>5, then at x=βk the first parabola has y value (βk)2βk>20, and hence the vertex of the second parabola lies to the left of the first parabola. The lower half of the second parabola intersects the y-axis at 20β2kββ, which is above the vertex of the first parabola. Hence if k>5, the two parabolas intersect at four points.
If k=5, then the first parabola passes through the vertex (β5,20) of the second parabola. If x=β5+Ο΅, then the y-coordinate of the first parabola is 20β10Ο΅+Ο΅2, while the y-coordinate of the lower half of the second parabola is 20β2Ο΅ββ. Because 2Ο΅ββ>10Ο΅βΟ΅2 for small positive values of Ο΅, the lower half of the second parabola lies below and to the left of the left half of the first parabola. Similarly to the previous case, the lower half of the second parabola intersects the y-axis above the vertex of the first parabola. Thus for k=5, the two parabolas intersect at four points.
Adding the first equation given in the problem to half of the second equation yields
y+2xβ=(yβ20)2+x2β23βk,
which, upon completing the square, gives
(yβ241β)2+(xβ41β)2=16325β+23βk.
All four intersection points satisfy this equation, which is an equation of a circle. Hence as long as the two parabolas intersect in four distinct points, these four points are concyclic. Moreover, the square of the radius of this circle is 16325β+23βk.
Thus the desired condition is that
16325β+23βkβ€441,
which holds when kβ€280. The requested sum is 5+280=285.