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If 10 circles, all with different radii, are positioned in the same [#permalink]
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Bunuel wrote:
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


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
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Re: If 10 circles, all with different radii, are positioned in the same [#permalink]
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samagra21 wrote:
\(2\) circles intersect at max \( 2\) points.
For \( n \) circles:
max point of intersection =\( nC2 * 2= 10C2*2=90\)



Can you please point out which "2" is in nC2?
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Re: If 10 circles, all with different radii, are positioned in the same [#permalink]
RitwikAgrawal wrote:
samagra21 wrote:
\(2\) circles intersect at max \( 2\) points.
For \( n \) circles:
max point of intersection =\( nC2 * 2= 10C2*2=90\)



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


refer red rings (intersection points). For any 2 circles selected, you get 2 points.
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Re: If 10 circles, all with different radii, are positioned in the same [#permalink]
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Bunuel wrote:
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


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.
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Re: If 10 circles, all with different radii, are positioned in the same [#permalink]
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bb wrote:
Did they say that there was no Geometry on the GMAT Focus? hos is this possible? Is this not a geometry question? 🤯


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
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Re: If 10 circles, all with different radii, are positioned in the same [#permalink]
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bb wrote:
Did they say that there was no Geometry on the GMAT Focus? hos is this possible? Is this not a geometry question? 🤯

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.
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Re: If 10 circles, all with different radii, are positioned in the same [#permalink]
Isn't there not meant to be geometry on FE?
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Re: If 10 circles, all with different radii, are positioned in the same [#permalink]
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sanyashah wrote:
Isn't there not meant to be geometry on FE?

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.
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If 10 circles, all with different radii, are positioned in the same [#permalink]
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Expert Reply
Bunuel wrote:
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


Every pair of circles may intersect each other in at most two points, as in the following figure:
Attachment:
Screen Shot 2024-01-14 at 1.36.01 PM.png
Screen Shot 2024-01-14 at 1.36.01 PM.png [ 11.95 KiB | Viewed 3210 times ]


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

­­­­­­­
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Re: If 10 circles, all with different radii, are positioned in the same [#permalink]
GMATGuruNY , Thank you for providing the easiest solution...and quick too.
GMATGuruNY wrote:
Bunuel wrote:
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

Every pair of circles may intersect each other in at most two points, as in the following figure:
Attachment:
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

­

­
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Re: If 10 circles, all with different radii, are positioned in the same [#permalink]
MartyMurray , Can you please tell me why you stopped at tenth only ?
MartyMurray wrote:
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.



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.



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



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).

­
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Re: If 10 circles, all with different radii, are positioned in the same [#permalink]
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sayan640 wrote:
MartyMurray , Can you please tell me why you stopped at tenth only ?
MartyMurray wrote:
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).
­

­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.
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