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

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- October 24th 2009, 12:26 AMvuze88Intersections Q
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

- October 24th 2009, 04:24 AMHallsofIvy
No, I can't because I don't believe that 79 is correct! What reason do you have to believe that is the correct answer? I get a maximum of 42 points of intersection.

- October 24th 2009, 07:05 AMtonio

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.

Tonio - October 24th 2009, 07:12 AMaidan
What does "distinct parabola" & "distinct circle" mean in this case?

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

c1&c3, c2&c3

Intersections of circles = 6

Total intersections: 24 + 48 + 6 = 78

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