A Math Forum Project

Geometry Forum - Problem of the Week Solutions - Six Circle Intersection, Oct. 31-Nov. 4, 1994 Annie says: We had a veritable flood of answers to this problem. There were several variations on the solution and the problem-solving process. There was a group who defined intersection in one way and opted for "place all the circles on top of one another and get an infinite number of intersections." Most of the rest conducted trials with different numbers of overlapping but not coincident circles and saw a pattern to the answers which resulted in the algorithms and formulas presented below. And a few offered both approaches. (BTW, as we all know, the "puzzle guy's" name is Will Shortz. Thanks Henri.) Sean Nichols As a solution to the Problem of the Week for November 1st, I have deduced that there must be at most 30 points where 6 coplanar circles of the same size can intersect. My reasoning is thus: If there is 1 circle lying on a plane, it can have 0 points of intersection (there is nothing to intersect with). If there are 2 circles, the second circle has at most 2 points of intersection (should be pretty obvious). If there are 3 circles, the third circle has 4 points of intersection (The third circle intersects each of the others twice; 2x2=4). If there are 4 circles, the 4th has 6 points of intersection. And so on... So, the 5th has 8 points, and the 6th has 10 points. So, when one adds up all the points of intersection, one gets: 0+2+4+6+8+10 = 30 points. Yesterday (Nov. 2), I sent in a solution to the Problem of the Week, stating that the six circles could intersect at a maximum of 30 points. Since then, I have been thinking some more about the problem, and if possible, I now wish to change that to: If the circles are all drawn _exactly_ one on top of another, there are an infinite number of points of intersection between the 6 circles (All points on all of the circles intersect). I hope it is possible at this time to change my solution to the one above. Mike Veltri and Mike Woods The maximum number is 30. The reason being because each circle can cut 10 times. But one cancels the other out. So multiply 10 by the number of circles (6) and you get 60. But if one cancels the other out, then divide it by 2 and you get 30. There's the answer. We also got this interesting chart for this problem, Circles Intersections ------- ------------- 1 0 2 2 3 6 4 12 5 20 6 30 You see when multiply the first circle by the second circle you get the number of times that they touch (i.e., circle number 4 times circle number 5 equals 20 intersections). Sandra Duruisseau If you have six circles and each circle intersects with five other circles in two places each you will come to an answer of sixty intersections. (6*5*2=60) This answer seems logical but each of these points lies on two circles so we must divide 60 by 2 to get the correct number of intersections. Therefore, the final answer is 30 intersections. Anita Lau My Solution; Each circle can meet every other circle twice each. There are 6 circles. . . Circle 1 2 3 4 5 6 X represents where the intersections have already been counted for previously or is the same circle. 1 X 2 2 2 2 2 2 X X 2 2 2 2 3 X X X 2 2 2 4 X X X X 2 2 The 2 represents the number of 5 X X X X X 2 intersections between the two 6 X X X X X X circles. Therefore, there are 30 intersections possible with 6 equal circles. Kristina Almquist If 6 circles of equal size intersect as many times as possible, I believe there would be 30 intersections. At first I randomly drew 6 intersecting circles & got low numbers. Then I realized that not every circle was touching every other circle and not every circle was crossing as many lines as it possibly could. Also, some circles had common intersections & some just touched on the sides. What a waste of possible intersections! So, going through again, I made sure that the circles didn't touch sides, didn't share intersections and did intersect every other circle. I saw that circle 1 had 0, #2 had 2, #3 could intersect 4 times, #4, 6 times, #5 had 8 & #6 had 10 more to add. So, every new circle added twice as many intersections as there were already existing circles. This is because, since they were the same size, they had to intersect only as many as its radius will allow. So, in adding their intersections 0,2,4,6,8,10.. there are 30 intersections. Without even drawing the circles, it is possible to figure the number. Susanna Puntel The maximum number of times 6 circles can intersect is 30. Here's why. Start with 2 circles. They can intersect 2 times max. Add a 3rd circle. Now there are 6 pts of intersection (2+4). 4th circle adds 6 more pts for a total of 12 (2+4+6). 5th circle adds 8 more pts (2+4+6+8) for a total of 20. 6th circle adds 10 more pts (2+4+6+8+10) for a total of 30 points. You can follow this pattern for any number of circles. Susan Garges Six circles can intersect a maximum number of 5 circles each. Each circle can intersect 5 others. A formula can be used to figure this out. x= number of circles x(x-1) 6(6-1)=6(5)=30 A intersects B,C,D,E,F B intersects A,C,D,E,F C intersects A,B,D,E,F D intersects A,B,C,E,F E intersects A,B,C,D,F F intersects A,B,C,D,E Total = 30 Greg Hull I got an answer of 30 intersections between 6 congruent circles. The first thing I did was draw a set of "x" and "y" axes . Then drew The first circle with the midpoint at the origin. The size of my circles were 2''. For the second circle I moved its midpoint up the y-axis 4/16''. I followed this procedure for the next 4 circles then added up the number of intersections. Farah Khan, Brian Smith, Allen Shacklock, Braden Shipe, and Jen Thomas All wrote solutions to the effect that "you would have an infinite amount of points if you put them together--all six on top of one another." Before joining the others in this conclusion Steve Barron noted that initially he "tried several drawings and got 22, then 23, then 24, then 26, then 29, then 30 as my final number of points of intersection." Mandy Coffman, Devon Cooper, Misty Lent, and Eric Reynolds The maximum number of intersections of six circles of the same size is 30. First, we made a table: circles | intersections ________________________ 2 3 4 5 6 Then, we drew the number of circles and got the maximum number of intersections for the first three. Our table looked like this: circles | intersections _______________________ 2 2 3 6 4 12 5 6 We concluded that each number times the previous number gave us the number of intersections. We checked by drawing the last two, and our table looked like this: circles | intersections _______________________ 2 2 3 6 4 12 5 20 6 30 We also decided that the number of intersections for n circles is n*(n-1). Previous page || Next problem || Previous problem || Table of Contents || Forum Home Page