4 distinct parabolas and 3 distinct circles are drawn on a large sheet of paper. Find the maximum number of points of intersection of these curves. The answer is 78. Is there a better way than drawing it out :S
This is confusing and I'm far from being sure, but I get 66:
Every parabola can intersect with a circle at most at 4 points ==> we get 4*3*4 = 48 intersection points between the three circles and the 4 parabolas
Every two parabolas can intersect in 2 different ponts ==> another 2*(1+2+3) = 12 points
Every circle can intersect another circle at two points ==> 6 points more
Sum them all you get 66 points...or I'm wrong, of course.
Any parabola perpendicular (or not oriented parallel) will have 4 intersections with any other parabola.
Parabola1 & Parabola2 will have 4 intersections.
P1&P3, P1&P4, P2&P3, P2&P4, P3&P4
Intersections of all Parabolas = 24
Each Circle and Parabola can have 4 intersection points.
Curve1 & Parabola1 will have 4 intersections.
c1&p2, c1&p3, c1&p4
Intersections of all Parabolas & 1 circle = 16
Intersections of all 3 circles and all 4 parabola = 48
Intersection of circle1 & circle 2 = 2 points
Intersections of circles = 6
Total intersections: 24 + 48 + 6 = 78