If 10 circles, all with different radii, are positioned in the same

Question

﻿If 10 circles, all with different radii, are positioned in the same plane, what is the maximum possible number of distinct points where 2 or more of the circles intersect?

A. 90
B. 100
C. 180
D. 200
E. 360

MartyMurray · Accepted Answer

If you're faced with a GMAT question that is different from anything you've seen before, one approach you can use to answer it is to try some different scenarios and look for a pattern.

In the case of this question, we can look for a pattern to how circles of different sizes can intersect.

Start with two large circles of similar but different sizes. They can intersect at two points, and there's no way for them to intersect at more than two points.

https://i.postimg.cc/RWZP7sY7/Capture-4.png

Then, if we add a third, slightly smaller circle, it can intersect the other two at 4 points. So, with three circles, we now have 6 intersections.

https://i.postimg.cc/5XKsn5cZ/Capture5.png

Each successive circle can intersect each of the others at two points. So the fourth circle can intersect the first three at 6 points.

https://i.postimg.cc/bZR9BQ7z/Capture7.png

So, we see the pattern.

A circle can intersect any other circle at 2 points maximum. So, at maximum, each successive circle intersects the others at two more points than the previous circle intersected the others at.

Thus, we have the following:

The first circle intersects no others at no points.

The second intersects 1 other at 2 points.

The third intersects 2 others at 4 points.

The fourth intersects 3 others at 6 points.

The fifth intersects 4 others at 8 points.

The sixth circle intersects 5 others at 10 points.

The seventh intersects 6 other at 12 points.

The eighth intersects 7 others at 14 points.

The ninth intersects 8 others at 16 points.

The tenth intersects 9 others at 18 points.

2 + 4 + 6 + 8 + 10 + 12 + 14 + 16 + 18 = 90

The correct answer is (A).

Oppenheimer1945 · Answer

2  circles intersect at max   2  points. 
For   n   circles:
max point of intersection =  nC2 * 2= 10C2*2=90

gmatophobia · Answer

We can find a pattern to solve this question. Let's assume that we have  n  circles

n = 2  → circles: A and B

A can intersect with B at two points

Total : 2*1 = 2 points

n = 3  → circles: A, B and C

A can intersect with B at two points
A can intersect with C at two points

B can intersect with C at two points

Total : (2*2) + (2) = 6 points

n = 4  → circles: A, B, C and D

A can intersect with B at two points
A can intersect with C at two points
A can intersect with D at two points

B can intersect with C at two points
B can intersect with D at two points

C can intersect with D at two points

Total : (2*3) + (2*2) + (2) = 12 points

General Pattern for  n  circles   =  2 + 2*2 + 2*3 + ..... +2(n-1)

=  2(1+2+3+4..+(n-1))

1+2+3+4..+(n-1) represents sum of n-1 terms which are in A.P.

=  2(\frac{(n-1)*n}{2})

For  n = 10

=  2(\frac{9 * 10}{2}) = 90

Option A

RitwikAgrawal · Answer

Can you please point out which "2" is in nC2?

Oppenheimer1945 · Answer

refer red rings (intersection points). For any 2 circles selected, you get 2 points.

JeffTargetTestPrep · Answer

The circles have different radii, so each one can have at most 2 points of intersection with another.

So, each of the 10 circles can have at most (9)(2) distinct points of intersection with the other 9 circles.

Therefore, the maximum possible number of distinct points of intersection is:

(10)(9)(2)/(2) = 90

Answer: A

Note:

Why did we divide by 2 ? Because we have to compensate for the fact that (10)(9)(2) includes every point of intersection twice. For example, circle A’s points of intersection with circle B are the same as circle B’s points of intersection with circle A.

Bunuel · Answer

Actually, there are two more questions that have elements of geometry in them: https://gmatclub.com/forum/a-spherical- ... 16425.html https://gmatclub.com/forum/marie-wishes ... 14546.html

MartyMurray · Answer

It's interesting how Geometry is sneaking in.

I haven't seen anything with triangles, angles, or working from one shape, such as a circle, to conclusions about another, such as a rectangle. So, maybe as long as a question doesn't have those elements, they don't consider it a Geometry question.

This circles question doesn't take much geometric knowledge, and the balloon question can be handled basically as an Algebra question, but that one about Marie's house and the fence sure seems like a Geometry question to me.

mbaaspirantss07 · Answer

Isn't there not meant to be geometry on FE?

MartyMurray · Answer

That's what they say, but there are some questions that involve geometric shapes.

At the same time, these are not hardcore Geometry questions. They don't require test-takers to have much knowledge of Geometry or to relate angles, lengths, areas, or other geometric characteristics or properties to each other in sophisticated ways.

For example, this question doesn't require knowledge of how circles relate to other shapes or advanced knowledge of properties of circles. It's basically a logic puzzle that happens to use circles.

GMATGuruNY · Answer

Every pair of circles may intersect each other in at most two points, as in the following figure: Screen Shot 2024-01-14 at 1.36.01 PM.png From 10 circles, the number of pairs that be formed = 10C2 = \frac{10*9}{2*1} = 45 Since each of these 45 pairs may intersect in at most two points, the maximum possible number of intersections = 45*2 = 90 A

sayan640 · Answer

GMATGuruNY  , Thank you for providing the easiest solution...and quick too.

sayan640 · Answer

MartyMurray  , Can you please tell me why you stopped at tenth only ?

MartyMurray · Answer

I stopped at the tenth because the question asks about ten circles. So, we don't have to go any further to find the answer.

KarishmaB · Answer

Responding to a pm: It is a question testing Combinations and not Geometry per se. The context used is Geometry but that much Geometry does appear in GMAT Focus too. Draw two intersecting circles. We see that they intersect at 2 points. Draw a third circle intersecting both previous circles at distinct points (since we need to maximize the number of points of intersection). Are you able to? Yes. Next draw a fourth circle intersection all previous 3 circles at 2 points each. Again we are able to. So there is no reason why we should not be able to draw 10 circles, each pair intersecting in 2 points. This is similar to the handshake problem in probability. "If there are n people in a group and each person shakes hands with all other people" problem. The only difference here is that each pair intersects in 2 points. So number of points = 10C2 * 2 = 90 Answer (A) Video on Combinations: https://youtu.be/tUPJhcUxllQ

GMATinsight · Answer

the maximum points of intersection between any two circles = 2

i.e. with one selection two circles there are two distinct points of intersection

Total ways to choose two circles out of 10 = 10C2 = 45

So total intersections = 10C2 x 2 = 45x2 = 90

Answer: Option A

If 10 circles, all with different radii, are positioned in the same

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If 10 circles, all with different radii, are positioned in the same

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