Arrange 1-12 around a circle

Question

This is a very interesting word problem that I came across in an old textbook of mine. So I know its got something to do with plain old algebra, which yields the shortest, simplest proofs, but other than that, the textbook gave no hints really and I'm really not sure about how to approach it. Any guidance hints or help would be truly greatly appreciated. Thanks in advance :) So anyway, here the problem goes:

The numbers $1-12$ are to be placed around a circle, as on a clock, but in any order. Show that there are three consecutive numbers in the arrangement with a sum of at least $19$.

Answer 1

Taken directly from http://gottfriedville.net/mathprob/nt-circle12.html

Proof #1:

The sum of the $12$ numbers is $78$. If we add up all the sums of three adjacent numbers, we use each number three times, for a total of $234$. If all the sums are to be $< 19$, then their maximum total would be $12 \times 18 = 216$, which is too small.

In fact, there must be a sum of at least $20$, since $12 \times 19 = 228$, which is still too small.

Proof #2:

To avoid a sum of at least $19$, the numbers $12, 11, 10$, and $9$ must all be separated by at least two other numbers. Otherwise, two of them, plus any of the others totals at least $20$. Let the $12$ places be as follows: _ $X$ _ _ $X$ _ _ $9$ _ _ $X$ _ where $X$ indicates $10, 11$, and $12$ in some order. Since each of the dashes is part of some $3$-element sum with one of the $X$'s, none of them can accommodate $8$, without creating a sum that is at least $8 + 10 + 1 = 19$.