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What is the greatest possible number of points

Expert replies
by NandishSS » Tue Jul 18, 2017 7:52 am
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

OA: E

Source: Math Revolution
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Source: — Problem Solving |

by regor60 » Tue Jul 18, 2017 9:00 am
NandishSS wrote: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

OA: E

Source: Math Revolution
Two circles can intersect at two points at the most.

How many different pairs of two circles can be generated from 11 ?

11!/2!9! = 55 unique pairs x 2 intersections/pair = E
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by Ian Stewart » Tue Jul 18, 2017 8:47 pm
Same idea as the post above, but using more elementary principles: draw just one circle first. Then draw another - it can intersect the first at 2 points, at most. Then draw a third. It can intersect each of the first two circles at two points, so we can make 4 new intersection points. Similarly the next circle can create 6 new intersection points, and so on. So the maximum total number of intersection points will be

2 + 4 + 6 + ... + 18 + 20

which is an equally spaced sum with 10 terms. The average term in that sum is the average of the smallest and largest terms, so is 11, and since sum = avg*number of terms, the sum is thus 11*10 = 110.
For online GMAT math tutoring, or to buy my higher-level Quant books and problem sets, contact me at ianstewartgmat at gmail.com

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by Jay@ManhattanReview » Tue Jul 18, 2017 11:32 pm
NandishSS wrote: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

OA: E

Source: Math Revolution
Two circles with unequal radii can intersect each other at the most two points.

The first circle can intersect the other 10 circles in 2 x 10 = 20 points.

Thus 11 circles would intersect each other in (20/2)*11 = 110 points. We divided the number of points by '2' because the intersecting points were counted twice.

The correct answer: E

Hope this helps!

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