Permutations of Intersections of Parabola and Circles

Question

What is the maximum number of points of intersection for four distinct parabolas and three distinct circles drawn on a sheet of paper?

Anyone good enough to solve this.?

Well considering the two different parabolas can intersect at 4 different points. WHen you bring a third one in, its even more.

Two circles intersecting bring 2 points. A circle and a parabola brings 4 I guess. So how many in total?

score 5 · Answer 1 · answered Oct 06 '17 at 09:07

5

6 times 4 parabola-parabola intersections, 12 times 4 parabola-circle intersections, 3 times 2 circle-circle intersections.

answered Oct 06 '17 at 09:07

Nice!${}{}{}{}{}{}$ – quasi Oct 06 '17 at 09:11
How did you make it? – jonsno Oct 06 '17 at 09:13
@samjoe Er, by thinking. First looking for ways to intersect the parabolas. – Oct 06 '17 at 09:15
@YvesDaoust I meant which application :| – jonsno Oct 06 '17 at 09:18
@samjoe: MS Word (the parabolas are actually Beziers). – Oct 06 '17 at 09:20

quasi · Answer 2 · 2017-10-06T08:58:03.847

1

All pairs can potentially intersect in $4$ points, except for the pairs of circles, which can only have two intersection points.

It follows that an upper bound on the number of intersection points is $$4{\binom{4}{2}}+4(4\cdot 3)+2\binom{3}{2}=78$$ Explanation:

${\large{\binom{4}{2}}}$ pairs of parabolas, each pair with at most $4$ intersection points.
$4{\,\cdot\,}3$ parabola-circle pairs, each pair with at most $4$ intersection points.
$\binom{3}{2}$ pairs of circles, each pair with at most $2$ intersection points.

I see no reason why the upper bound is not achievable.

edited Oct 06 '17 at 08:58

answered Oct 06 '17 at 08:45

quasi

58,772

My mistake!!, Sorry – Satish Ramanathan Oct 06 '17 at 08:57
Your answer is correct, but some work is needed in the explanation part! +1 – jonsno Oct 06 '17 at 08:57
@samjoe: Thanks. I fixed the typos in the explanation. – quasi Oct 06 '17 at 09:12

score 1 · Answer 3 · answered Oct 06 '17 at 08:46

Total number of intersections of parabolas is 24. For any two parabolas there is 4 intersections. There is 4C2 ways to choose 2 parabolas out of 4. 4C2 x 4 = 24.

For a circle and a parabola, the maximum number of intersections is 4. Total number of intersections of circles and parabolas is 4x4x3=48

For any two circles, the maximum number of intersections is 2.

Total intersections of circles is 3C2 x 2 =6

Maximum total is 24+48+6=78

Why did you answer your own question? – quasi Oct 06 '17 at 09:00 — quasi, Oct 06 '17 at 09:00

Satish Ramanathan · Answer 4 · 2017-10-06T08:57:01.637

-1

Four parabolas can have a maximum of 24 points of intersection. A circle can have four points of intersection with a parabola and hence 16 with four parabolas. Three circles can have have 48 points of intersection.

Each circle can intersect other circle at 2 points and three circles can intersect with each other at 6 points and hence I revise it to be 78.

edited Oct 06 '17 at 08:57

answered Oct 06 '17 at 08:41

Satish Ramanathan

11,984

Very close :) :) – D.Ronald Oct 06 '17 at 08:42
Two parabolas can intersect in at most 4 points, and there are 6 pairs of parabolas. How can you achieve 32 ?? – Oct 06 '17 at 08:53
Fixed it, my apologies!! – Satish Ramanathan Oct 06 '17 at 08:58

Permutations of Intersections of Parabola and Circles

4 Answers4