Suppose the 25 integers are arranged a,b,c,...,y around the wheel,
where "a" is adjacent to "y".
For contradiction suppose every three adjacent numbers has sum < 39
Then we have this set of 25 inequalities. We add the left and right
sides of them all. Notice there each letter appears in 3 inequalities,
there is one "a" in the 1st inequality and 2 "a"'s in the last two.
Also there are 2 "b"'s in the first two inequality and 1 "b" in the
last inequality.
a+b+c < 39
b+c+d < 39
c+d+e < 39
d+e+f < 39
.....
.....
v+w+x < 39
w+x+y < 39
x+y+a < 39
y+a+b < 39
----------------------------------
3a+3b+3c+3d+...+3v+3w+3x+3y < 39·25
3(a+b+c+d+...+v+w+x+y) < 975
Divide both sides by 3
a+b+c+d+...+v+w+x+y < 975/3
a+b+c+d+...+v+w+x+y < 325
However the sum of the integers from 1 through 25, inclusive, is given
by the formula:
Sn = n(n+1)/2 where n = 25
S25 = 25(25+1)/2 = 325
So the sum a+b+c+d+...+v+w+x+y cannot be both less than 325 and also
equal to 325. So the assumption that all 25 of those inequalities hold
is false. So there must be at least one of the inequalities which is
incorrect. Therefore there are three adjacent numbers whose sum is at
least 39.
Edwin
