Fill in the missing numbers along the sides of the triangle so that it contains each of the numbers from 4 through 12 exactly once. Furthermore, each side of the triangle should contain four numbers whose sim is 32.

 A B == 9 + 11   CD == 5 +7  E F == 6 + 10

Step-by-step explanation:

(This reminds me of a "formula" I developed a few years ago when I was photographing formal groups, and had to figyre out how many people to put in each row, and the rows had to be different lengths)  Anyway--

Get the sum of the numbers at the angles, subtract that number from 32, and that's the amount needed for the row.

Once I have that number, divide by 2 and split as "n + 1 and n-1

AB side:  angles given: 8 +4=12 , 32-12 =20  20/2 = 10;  10+1 =11  10-1=9

CD side: angles given:  8 + 12 =20, 32-20=12; 12/2 =6  6+1 =7 6-1 =5

EF side: angles given:  4 + 12 =16 32-16=16  My formula would have yielded 7+9, but those numbers were already used. Conveniently, 10 + 6 are left over.   Same principle, but + or - 2.

Thanks for the opportunity to work in this!

It is a fun way for me to end my participation in for a while. My school year is over!

Good Luck, Stay Safe!