Maximum number of points of intersection

There is another problem:

"When you draw two circles and two straight lines on a plain, the maximum number of points of intersection that can be obtained is $11$ .

What is the maximum number of points of intersection that can be obtained when you draw $20$ circles and $13$ straight lines on a plain?"

What method can we use in the problem(pattern recognition?)? May I know the answer and solution?Thanks fellow Brilliantians!

#Combinatorics

Easy Math Editor

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Comments

Two straight lines can have at max 1 pt of intersection. Two circles can have maximum, 2 pts of intersection and 1 st line and a circle can have 2 pts of intersection.

So, taking the case where there are maximum pts of intersection. All you need to do is figure the ways, in which 2 circles, 2 st lines, and one circle and 1 st line can be selected, and multiply them with the number of pts of intersection they can have respectively. So, in that case your solution will be:

1x 13C2 + 2x 20C2 + 2x 20C1 x 13C1 = 978

Ankan Gope - 7 years, 5 months ago

For completeness, can you explain why 978 can be achieved? How do you know that we can find 20 circles which give us $20 \times 19$ intersection points?

Calvin Lin Staff - 7 years, 5 months ago

I will give u a hint based on your idea. Use the combinatorics to find the possible combinations of circle and line and then use the possible intersections to get the result

Vishnu Kumar - 5 years, 2 months ago

how brilliant question i tried to use my own solution but it doesn't work how it happen? .can you give me your solution in this problem thanks....

AhlJhoy Balisi - 7 years, 5 months ago

infinite points u will get

bonam surendra - 7 years, 5 months ago

I AM JUST REPEATING WAT Ankan Gope HAS SAID... LET "x" BE THE TOTAL NUMBER OF CIRCLES AND "y '' BE THE TOTAL NUMBER OF STRAIGHT LINES (HERE REMEMBER ONE THING....FOR GETTING MAXIMUM NUMBER OF INTERSECTION, THE CIRCLES SHOULD INTERSECT AS A STRAIGHT CHAIN WITH TWO POINT OF INTERSECTION rather than forming ring or any other pattern)

NOW TWO INTERSECTING LINES CAN HAVE 1 INTERSECTION POINTS IN MAX AND TWO INTERSECTING CIRCLES CAN HAVE 2 INTERSECTION POINTS IN MAX AND SIMILARLY A CIRCLE AND A LINE CAN HAVE 2 INTERSECTION POINTS IN MAX

NOW THE FORMULA IS ( xC2)2 + (yC2)1+x*2

just hope it works....try yourself for 2 circle and a line or 2 circle and 2 line and so on..

danish latheef - 7 years, 5 months ago

Math	Appears as
Remember to wrap math in `\(` ... `\)` or `\[` ... `\]` to ensure proper formatting.
`2 \times 3`	$2 \times 3$
`2^{34}`	$2^{34}$
`a_{i-1}`	$a_{i-1}$
`\frac{2}{3}`	$\frac{2}{3}$
`\sqrt{2}`	$\sqrt{2}$
`\sum_{i=1}^3`	$\sum_{i=1}^3$
`\sin \theta`	$\sin \theta$
`\boxed{123}`	$\boxed{123}$