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There are $4$ distinct parabolas, $5$ distinct circles and $3$ distinct lines in the $x-y$ plane. Find the maximum possible value of the number of their points of intersection.

My Attempt at the Question: I first to try and figure out the maximum number of points of intersection. For that I figured that parabola and circle intersect at max. $4$ points. And straight line will intersect the circle and parabola at $4$ distinct points. But this lead me nowhere.

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First we find the kinds of intersection that take place b/w the lines, circles and parabolas, which are as follows:-

  1. Parabola with a parabola
  2. Parabola with a circle
  3. Parabola with a line
  4. Circle with a circle
  5. Circle with a line
  6. Line with a line

Now, lets count the maximum number of point of intersection in each of the above cases and consequently count the maximum number of intersections that cantake place given that there are $4$ distinct parabolas, $5$ distinct circles and $3$ distinct lines:-

CASE 1:- Intersection of two parabolas

intersection of parabola with a parabola

As we can see in the above image there are maximum $4$ possible points of intersection for two distinct parabolas. Since there are $4$ distinct parabolas hence there are maximum $4\cdot\binom{4}{2}$ maximum distinct point of intersections.

$\therefore$ Maximum number of points of intersection of $4$ distinct parabola$=4\cdot\displaystyle\binom{4}{2}=24$

CASE 2:- Intersection of parabola with a circle

Intersection of parabola with a circle

As can be seen in the above image there are at max. $4$ distinct points of intersection of a parabola and a circle.

Ways to select a parabola and a circle$= \displaystyle\binom{4}{1}\cdot\binom{5}{1}$

$\therefore$ Max. no. of distinct points of intersection$=4\cdot\displaystyle\binom{4}{1}\cdot\binom{5}{1}=80$

CASE 3:- Intersection of a parabola with a line

Intersection of parabola with a line

As can be seen in the above image there can be max. $2$ distinct points of intersection b/w a line and a parabola.

Ways to select a parabola and a line $=\displaystyle\binom{4}{1}\cdot\binom{3}{1}$

$\therefore$ Max. no. of distinct points of intersection$=2\cdot\displaystyle\binom{4}{1}\cdot\binom{3}{1}=24$

CASE 4:- Intersection of two circles

Intersection of a circle with a circle

As can be seen in the above image there can be max. $2$ distinct points of intersection b/w two circles.

Ways to select two circles $=\displaystyle\binom{5}{2}$

$\therefore$ Max. no. of distinct points of intersection$=2\cdot\displaystyle\binom{5}{2}=20$

CASE 5:- Intersection of a circle with a line

Intersection of a circle with a line

As can be seen in the above image there can be max. $2$ distinct points of intersection b/w a line and a circle.

Ways to select a circle and a line $=\displaystyle\binom{5}{1}\cdot\binom{3}{1}$

$\therefore$ Max. no. of distinct points of intersection$=2\cdot\displaystyle\binom{5}{1}\binom{3}{1}=30$

CASE 6:- Intersection of two lines

enter image description here

As can be seen in the above image there can be max. $1$ distinct points of intersection b/w two lines.

Ways to select two lines $=\displaystyle\binom{3}{2}$

$\therefore$ Max. no. of distinct points of intersection$=1\cdot\displaystyle\binom{3}{2}=3$

Since all the above cases are disjoint, and when anyone of them happens an intersection takes place, hence the total no of intersections are the sum of all the disjoint events.

Hence, total number of points of intersection is given by

$$4\cdot\displaystyle\binom{4}{2}+ 4\cdot\displaystyle\binom{4}{1}\cdot\binom{5}{1} + 2\cdot\displaystyle\binom{4}{1}\cdot\binom{3}{1} + 2\cdot\displaystyle\binom{5}{2} + 2\cdot\displaystyle\binom{5}{1}\binom{3}{1} + 1\cdot\displaystyle\binom{3}{2}\\ =24+80+24+20+30+3=\boxed{181}$$

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  • $\begingroup$ Got it.. Thanks $\endgroup$ – user36160 Dec 30 '16 at 4:31
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Use the fact that 2 parabolae intersect at 4 points.So does the circles. Generalising it, n parabolas will intersect (maximum)on $4(^nC_2)$ points. Thus total = $4\times(^4C_2) + 2\times(^5C_2) + 1\times(^3C_2) + 4\times(5\times4) + 2\times(4\times3) +2\times(5\times3) = 181$ points. Your doubts about it are most welcome!

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  • $\begingroup$ The answer in the key us given as 181 $\endgroup$ – user36160 Dec 29 '16 at 16:49
  • $\begingroup$ @user36160 , seems did some calculation mistake. $\endgroup$ – Anand Kishore Dec 29 '16 at 16:51
  • $\begingroup$ And please properly format and explain the answer $\endgroup$ – user36160 Dec 29 '16 at 16:51
  • $\begingroup$ From starting, point of intersections of P with P, C with C, L with L, P with C, P with L, C with L. $\endgroup$ – Anand Kishore Dec 29 '16 at 16:57
  • $\begingroup$ Got it.. thanks $\endgroup$ – user36160 Dec 30 '16 at 4:32

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