Interlocking Circles

1. ____ ____ / \ / \ A X B / / \ \ | D E | | |_F_| | | /| |\ | \ G H I J / \ / X \ / X__K_/ \_L__X | | \ / \ / \ / --C--

   Three interlocking circles create seven distinct regions and divide the three circles into 12 segments (labels A-L on the figure). Find A-L as distinct positive integers such that each of the seven regions has the same sum (when adding the three values surrounding each region.) Try to find the numbers such that the maximum of (A..L) is as small as possible. Repeat the previous problem with the added constraint that the sum around each of the three circles should also be the same (different than that of the seven regions, of course.) If possible, for comparison purposes, please submit your diagrams or solutions such that A is your maximum value, and B is your minimum value for that A. (You should be able to reassign numbers in your solution to do this.)

2. The centers of the three circles create an equilateral triangle. If we assume the side-length of this triangle is 1, what is the radius of the circles (all identical) which makes the lengths of segments D and I the same?

Source: Original.

1. 1. ____ ____ / \ / \ A.13 X B.1 / / \ \ | 5 12 | | |-3-| | | /| |\ | \ 2 8 9 7 / \ / X \ / X_10_/ \_4__X | | \ / \ / \ / -C.6-

      ____ ____ / \ / \ A.13 X B.1 / / \ \ | 4 9 | | |-7-| | | /| |\ | \ 3 5 8 10 / \ / X \ / X_12_/ \_2__X | | \ / \ / \ / -C.6-

      My own analysis: Consider the central region and the three outer regions. Together, these four regions use all 12 numbers, so the common sum must be 1/4 of the sum of all 12 numbers. The numbers 1 thru 12 sum to 78, which is not divisible by 4, so we must change a number to 13 (the next lowest maximum value) and make the total divisible by 4. Changing 11 to 13 makes the sum 80, and each region must sum to 20. By considering only the remaining three regions, we can see that A+B+C must also be 20. So there are actually 8 sums which must equal 20. A little trial and error will find these solutions.

      One solver pointed out that this problem is exactly the same as trying to label the edges of a cube such that the sum at each vertex is the same. Thanks for pointing that out!

   2. ____ ____ / \ / \ A.13 X B.2 / / \ \ | 3 10 | | |-8-| | | /| |\ | \ 5 12 1 9 / \ / X \ / X__4_/ \_11_X | | \ / \ / \ / -C.6-

      Now we need the sum of all 12 numbers to be divisible by 4 (as above) and also divisible by 3 (for the 3 circles.) The first total sum that satisfies this condition is 84. By replacing 7 with 13 we achieve this sum and the solution here is found.

2. The radius of all circles must be 1. Then it is easy to see that all of the arcs comprise 1/6 of a circle.