Cut the face

You have a watch-face, numbered from 1 to 12 around a circle.

Can you with two cuts divide the face so you get the numbers in groups with the same sum as the others?

How do you do?
Is there more solutions than one?

Example: Cut in a cross and you get 1, 2, 3 in one group, 4, 5, 6 in another group etc. This is not a solution because 1+2+3=6 and 4+5+6=15 has not the same sum.

Originally posted by FabianFnas
You have a watch-face, numbered from 1 to 12 around a circle.

Can you with two cuts divide the face so you get the numbers in groups with the same sum as the others?

How do you do?
Is there more solutions than one?

Example: Cut in a cross and you get 1, 2, 3 in one group, 4, 5, 6 in another group etc. This is not a solution because 1+2+3=6 and 4+5+6=15 has not the same sum.

Sum of all numbers is 78 which is not divisible by 4. We are therefore looking to divide the clockface into 3.

78/3 = 26

11,12,1,2

9,10,3,4

5,6,7,8

Looks unique to me.

10,11,12,1,2,3 and 4,5,6,7,8,9

you just use one cut to get half the way across.

Unfortunately, the problem is how to do it with exactly TWO cuts.

I am wondering if there might be a second solution which involves cutting a two digit number into 2 separate numbers.. That might technically meet the qualifications.

it is two cuts. As I said, you use one cut to get half way across.

Or have the two cuts at an angle to each other, not parallel.

I wonder if the stipulation requires the cuts to be a chord.