What is the greatest possible number of points at which 11 circles wit

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Any Circle intersects another circle at maximum two intersection points.

With two circles : greatest number of intersection points = 2
With third circle introduced to the first two (already intersecting at two points), the third circle can intersect both the circles at two different intersection points, introducing 4 more intersection points ( third circle cuts each of the first two at two different poitns) . in addition to the two intersection points of first two.
i.e., with three circles , max # of intersection points = 2 + 2*2 = 6
with 4th circle introduced, it cuts the first three circles at two intersection points each i.e, 2*3 additional intersection points introduced now.
with 4 circles , max # of intersection points = (2 + 2*2) (intersection points for 3 circles) + 2*3 (additional intersection points introduced by 4th circle) = 2 + 2*2 + 2*3
Similarily , with 5th circle . total = 2 + 2*2 + 2*3 + 2*4
we see a series emerging here
greatest number of of intersection points for m circles = 2 + 2*2 + 2*3 + .....+ 2*(m-1)

Applying the series extrapolation for 11 circles, greatest number of intersection points = 2 + 2*2 + 2*3 + .....+ 2*10

= 2 * (1 + 2 + 3 + 4 + 5 + 6 + 7 + 8 + 9 + 10)
= 2 * (110)/2 = 110 ( calculation for the sum shown below)
-- Calculation for the sum.
since the sum in the paranthesis is a sum of consecutive integers , we can calculate the average by (last + first) / 2
Sum in paranthesis = average * # of terms = 10*((1 + 10)/2) = 110/2
--

Correct Answer : E

Answer is E i.e 110

Two circles can intersect at maximum 2 points.

Three circles can intersect at maximum 6 points..

Because we select 2 circles from 3 and each two circles can intersect at maximum 2 points

2\(c^3_2\) ... Three circles can intersect at 6 points

So 11 circles can intersect at maximum

\(2c^{11}_2\) ... 2 * 11 * 5 = 110

The first 2 circles intersect at 2 points.
3rd circle can intersect the first two circles at 4 points. refer to the figure.

i.e Circle 1 = no intersection
Circle 2 = 2 points
Circle 3 = 2+4 points
Circle 4 = 2+4+8 points
.
.
.
Circle 11= 2+4+8+10+12+14+16+18+20 = n(n+1)= 10*11=110 points.
Implies N circles can be drawn to intersect N(N-1) points . In this case, 11 circles can be drawn that can intersect at 11*10=110 points

Ans E

1st Circle doesn't intersect itself - 2*(1-1)=0 points
2nd Circle intersects 1st at 2*(2-1)=2 Points
3rd Circle intersect 2nd and 3rd at 2*(3-1)=4 Points
4th Circle intersects other Circles at 2*(4-1)=6 points
...
11th circle intersects other circles at 2*(11-1)=20 points

Therefore total no of points = 2 + 4 + 6 + ... + 20
\(Sum = 10/2 * (20+2) = 5*22 =110\)

Answer is E

Answer is E -> 110.

2 circle intersects at 2 points, 3 circles intersect at 4 points, 4 circles intersect at 6 points, 5 circles at 8 points, etc., Hence max # of points will be the result of addition of these points - as you can draw a circle anyway you want to ensure there are distinct points with each of these circle's intersection.

S = 2+4+6+... +20 (n=10) as there are 11 circles
S = (a1+an)/2*n = (2+20)/2*10 = 110

Given: Number of circles with different radii = 11

We can draw a few circles to obtain the below results

For 2 circles, max number of points of intersection = 2 = 2 * 1
For 3 circles, max number of points of intersection = 6 = 3 * 2
For 4 circles, max number of points of intersection = 12 = 4 * 3
We can observe a pattern here.

For n circles with different radii, max number of points of intersection can be deduced from the above results as n * (n-1).

Therefore, for 11 circles with different radii, max number of points of intersection = 11 * 10 = 110

Answer: E

The answer is E
By drawing circles and deducing
2 circles intersect at 2 points(1x2)
3 circles intersect at 6 points(2x3)
4 circles intersect at 12 points(3x4)
11 circles intersect at 10x11=110 points

2 circles can intersect at the most 2 points. ---> which is 2P2
3 circles can intersect at the most 6 points. ----> which is 3P2

Therefore it is similar to number of arrangements possible for 11 objects taken two at a time.
Therefore 11 circles with different radii will intersect at 11P2 = 11!/9! = 11*10= 110 points. Answer choice E.

1 circles have 0 intersection point ( 1 x 0)
2 circles have 2 intersection points (2 x 1)
3 circles have 6 intersection points (3 x 2)
4 circles have 12 intersection points (4 x 3)
....
....
a pattern develops!
So, 11 circles have (11 x 10) = 110 intersection points

However, an expression exists as well ie; C x (C-1) = I, where C is the number of circles and I the number of intersections.

MATH REVOLUTION OFFICIAL SOLUTION:

The greatest possible number of points of intersection between two circles is 2. The greatest possible number of points of intersection of three circles is 2+2*2 (The greatest possible number of points of intersection between two circles + The greatest possible number of points of intersection that one circle can have with other two circles). The greatest possible number of points of intersection of four circles is 2+2*2+2*3 (The greatest possible number of points of intersection of three circles + The greatest possible number of points of intersection that one circle can have with other two circles).

So, the greatest possible number of points of intersection of 11 circles is, 2+2*2+2*3+…..2*10=2(1+2+…..+10)=2[10(10+1)/2]=110. The correct answer is E.

Note: 1+2+…..+n=n(n+1)/2
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[quote="Bunuel"]

Math Revolution and GMAT Club Contest Starts!

QUESTION #14:

What is the greatest possible number of points at which 11 circles with different radii intersected one another?

A. 45
B. 60
C. 85
D. 90
E. 110
quote]


IanStewart Sir, can we expect such a question on the GMAT? I've never seen anything like it before in my long, arduous preparation. If possible, could you please explain the concept in simpler terms? Thanks a lot.

I've seen several thousand official questions, and I've never seen a question similar to this one. So it's very unlikely you'd encounter something like this on a real test. But I suppose it's not impossible. If a question doesn't require any specialized math knowledge and can be solved within 2 minutes, then it's within the scope of the GMAT, and that's true of this problem.

If you notice that two distinct circles intersect at 2 points at most, then to count the number of points where 11 circles can meet, we just need to count how many pairs of circles we have, which is 11C2 = (11)(10)/2! = 55, and multiply by 2, because each pair of circles meets in 2 places, not 1, so the answer is 2*55 = 110.
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