Can you solve this matchstick triangle problem?

I think you are referring to the number of integer triangles that can be formed with a given perimeter. I was taught the following solution for that

The answer seem to be 12 for 24 matches. The formula is as follows:
* for an even amount of matches (M): the nearest integer to [M^2] /48
* for an uneven amount of matches (M): the nearest integer to [(M+3)^2] /48

See wikipedia for more on integer triangles with all kinds of special characteristics.

I did not know that this was a classical problem. Thanks for the reference. I worked out the formula for an even number of sides. It is exact when the number of sides is divisible by 12, but it seems to be accurate for any even number if you take the integer part of n^2/48.

To solve the problem for 24, you can reason as follows.

Consider the length of the longest side. By the triangle inequality, it must be less than the sum of the two other sides. If it were exactly equal then the longest side and the sum of the two other sides would both be 12, but since the longest side must be less than the sum of the two other sides, the longest side can be at most 11.

If all sides were the same length then they would all be 8. That is the shortest that the longest side can be, since if it were any less then it would no longer be the longest side.

Therefore the longest side ranges from 8 to 11.

Next look at each of the possible lengths for the longest side and consider the possibilities for the next to longest side.

If the longest side is 11, the sum of the other two sides is 13. The next to the longest side must be at least 7 and at most 11, for 5 possibilities: (11,7,6), (11,8,5), (11,9,4), (11,10,3) and (11,11,2)

Doing similar reasoning for the other possible lengths of the longest side, you would get 4 cases for length 10, 2 for length 9 and 1 for length 8. Adding these together gives 5 + 4 + 2 + 1 = 12.

I know I am probably just talking to myself, but on the off chance that anyone else is paying attention, there is a simple way of showing that if the n^2/48 formula works for even numbers then the (n+3)^2 formula works for odd numbers. The way to do it is to show that if n is odd then the number of corresponding triangles is the same as that for n+3. Since n+3 would be even, it follows that the number of triangles for n must be (n+3)^2/48.

A way of showing equality is to show that there is a one-to-one correspondence between the triangles for each of the two numbers. For each triangle (i,j,k) for n, associate the triangle (i+1, j+1, k+1) for n+3. The association from the smaller number to the larger is fairly obvious. Going in the reverse direction takes just a little bit of work. First you have to show that the triangle (i -1, j -1, k -1) is defined, which would be true if none of the sides equals 1. For an even number, this will be the case. You then have to show that if the triangle inequality applies for (i,j,k) then it applies for (i-1,j-1,k-1). This follows from the fact that (i+j) – k has the same parity as i+j+k, meaning that it is even, so i+j – k is at least equal to 2, and therefore (i-1) + (j-1) – (k-1) >= 1.