9
$\begingroup$

Here is a diagram created by overlapping four circles. The overlapping circles create eight regions. Put a number in each of the eight regions, using each of the numbers 1 to 8 exactly once, so that the sum of the numbers in each circle is the same.

4 overlapping circles creating 8 regions


This puzzle is from http://www.earlyfamilymath.org/

$\endgroup$

3 Answers 3

8
$\begingroup$

With an aid from a computer program, there are

6

essentially distinct solutions. Here "essentially distinct" means not equal up to reflection/rotation. Each solution can be rotated/reflected into 8 different solutions.

The solutions are:

One solution with sum 12:
8 |3| 7
--+ +--
1  X  2
--+ +--
5 |6| 4

Two solutions with sum 13:
7 |5| 6
--+ +--
1  X  2
--+ +--
4 |8| 3

8 |4| 3
--+ +--
1  X  6
--+ +--
7 |5| 2

Two solutions with sum 14:
7 |4| 2
--+ +--
3  X  8
--+ +--
6 |5| 1

6 |7| 3
--+ +--
1  X  4
--+ +--
5 |8| 2

One solution with sum 15:
5 |7| 2
--+ +--
3  X  6
--+ +--
4 |8| 1

Florian F notes that there is one more symmetry involving $n$ and $9-n$. The solutions above can be shown to form such pairs, reducing the number of "essentially different" solutions to

3.

The program I used is here. (Using a constraint solver here is an absolute overkill since there are only $8! = 40320$ cases to check at most, but it does simplify coding a bit.)

An afterthought: One of the solutions can be derived from

the 3x3 magic square, as follows:

Take the following 3x3 magic square in its "standard orientation".

8 1 6
3 5 7
4 9 2

Now swap the second and third rows.

8 1 6
4 9 2
3 5 7

Note that the row sums and column sums are preserved, which are all equal to 15. Now delete the 9 at the center and fill in the diagram so that the numbers at the corners go to the inner regions.

$\endgroup$
1
  • $\begingroup$ This answer is great for several reasons; I especially like the observation about magic squares. $\endgroup$ Mar 29 at 18:24
8
$\begingroup$

After a bit of trial and error I have this:

Solution

Line of thinking:

Probably the larger numbers need to be on the outside but not all. So trying for some numbers to get the same total worked out to this. Probably there is some real math to be found.

$\endgroup$
3
  • $\begingroup$ It is easy to prove that the only possibilities for the 1 circle sum can be between 12 and 15., which have a limited possibilites of partitions coonstrained to 8 max. ... seems 6 for any of them. $\endgroup$
    – z100
    Mar 26 at 21:28
  • 3
    $\begingroup$ You can reverse the solution by replacing each number n with (9-n), for a total of 14. $\endgroup$
    – Florian F
    Mar 26 at 21:57
  • $\begingroup$ 2 solutions so far, just to check other 208 possibilities, 4 (inner, double counted) out of 8 equal 70 and there are 3 different arrangements for each other. Outer 4 fixed for each one. $\endgroup$
    – z100
    Mar 26 at 22:09
1
$\begingroup$

My line of thinking started with

parity.

As an initial consideration, in order for all the sums to be the same,

the odd numbers

must be arranged either such that

A.

each circle contains the same number of them

or

B.

some circles contain none of them and some two. More specifically, exactly one circle contains zero odd digits, and the other three each contain two, for if more than one circle were without any odd digits then there would not be enough places for the four odd digits.

or

C.

some circles contain one of them and some three. More specifically, exactly one circle contains three odd digits and the others have one each. This is the converse of (B).


(A) can be ruled out as follows

Lemma: for the circles to all have the same number of odd digits, either all the odd digits must appear in the intersection regions, or all must appear in the non-intersection regions.

Proof: There are four odd digits and four even. If all four intersection regions contain odd digits then the four circles each contain two odd digits, which satisfies the criterion. If one of the odd digits is then swapped with any of the even ones, two regions lose an odd digit and one region (not necessarily different) gains one. Since two have lost but only one has gained, the circles cannot then all have the same number of odd digits. This rules out cases with exactly three of the intersection regions containing odd digits. If another of the odd digits is then swapped for an even digit (other than the one previously swapped), then again two circles lose an odd digit but only one gains. The aggregate net loss of two odd digits cannot be evenly distributed among the four circles, so this rules out the case of exactly two intersections containing odd digits. We can rule out the case of exactly one intersection containing an odd digit either by another iteration of the same procedure, or by observing that we can make an argument analogous to the above but starting with all intersections containing even digits. Either way, this completes the proof.

Now suppose all the intersections contain odd digits. There are two cases:

Case 1. The 1 and 7 are in the same circle, for a subtotal of 8 in that circle. This cannot happen because then the opposite circle would contain 3 and 5, also with subtotal 8, such that both those circles would require the same third digit for their sums to be equal.

Case 2. The 1 and 7 are not in the same circle. In this case, each must appear in the same circle with each of the other two odd digits. But then one of the circles must have an odd-digit subtotal of 1+3=4, and another must have an odd-digit subtotal of 5+7=12. The difference between these is 8, which is larger than the largest difference between any of the even digits available, so there is no way to make these two circles have the same sum.

Thus, if all the intersections contain odd digits, the circles' digit sums cannot all be equal. But then the same applies if all the intersections contain even digits, for in that case, transforming every digit $d$ to $9-d$ produces a corresponding pattern with all intersections odd, which cannot yield a solution. As I discuss in more detail below, that transformation converts solutions to solutions and non-solutions to non-solutions.


Now consider case B,

one circle contains zero odd digits, and the others contain two odd digits each. Without loss of generality, let the top-left circle be the one without odd digits. Only two regions then remain in the top right and bottom left circles for odd digits, so those must all be filled, and that puts two odd digits in the bottom right circle as well, as needed. The overall pattern is:

e | e | o
----------
e |   | o
----------
o | o | e

Observe that whichever even digit we choose for the bottom-right circle, the sum of the top-left circle is that of the remaining even digits. We can therefore rule out 2 for the bottom-right position, because the resulting target sum, 4+6+8=18, is larger than can be formed from 2 and any two of the available odd digits to form the sum of the bottom-right circle.

We can also rule out 4 for the bottom-right even digit. If that were 4 then the target sum would be 2+6+8=16. But this would require the top-right circle and the bottom-left circle, which are disjoint, to add up to 32 altogether. The sum of all 8 digits is only 36, and if we exclude two even digits that are outside these two circles then we cannot reach a sum of 32 for the remaining digits.

Now suppose the bottom-right even digit is 6, giving a target sum of 2+4+8=14. The sum of the odd digits is 1+3+5+7=16, so to fill out the top-right and bottom-left circles with a combined total of 14+14=28, we need to use both the 4 and the 8 from the top-left circle. To make 14 with the 4, we have to use the 3 and 7, and that leaves the 1 and 5 to make 14 with the 8. We need to choose one digit from each of those pairs to group with the 6 to form 14, and we can choose either the 1 and 7 or the 3 and 5. That gives us two distinct solutions:

2 | 4 | 3     2 | 4 | 7
---------     ---------
8 |   | 7     8 |   | 3
---------     ---------
5 | 1 | 6     1 | 5 | 6

And suppose the bottom-right even digit is 8, giving a target sum of 2+4+6=12. To fill out the top-right and bottom-left circles to a total of 12+12=24, we need a contribution of 8 from the top-left circle, achievable only by using the 2 and 6. We must combine the 2 with the 7 and 3, and the 6 with the 5 and 1. for the bottom right we can choose only the 1 and 3. That gives one more distinct solution:

4 | 2 | 7
---------
6 |   | 3
---------
5 | 1 | 8


Case C could be analyzed analogously, but let us observe that

patterns of type B can be converted to patterns of type C and vice versa by replacing each digit $d$ with $9-d$. When one of the circles' sums is $S$, the sum of the transformed circle is $27-S$, so this transformation converts solutions into (different) solutions and non-solutions into non-solutions. It follows that applying this transformation to the solutions for case B produces solutions for case C. There cannot be any other solutions for case C, because if there were, they would produce additional case B solutions via the same transformation.


Thus, the

three

solutions to case B are the only distinct ones, up to geometric symmetry and

the $9-d$ transformation.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.