Would this make a good high school math problem?

So the solution is in the set of a group of three integers that add to 13, with none being greater than six.

But if A+B+C=13, does that mean B+C+A=13 is distinct and different? for instance, side A =1, side B = 6, side C=6. is that different from A=6, B=1, C=6?

I should have mentioned that we are only interested in the lengths of the sides. For triangles, any way of arranging the sides still gives a congruent triangle.

I am guessing that you figured that none could be greater than 6, because if one side were greater than 6, the sum of the other two sides would be less than 6, which would violate the triangle inequality.

That is a very good start that narrows down the possibilities quite a bit. One way of putting it would be to say that the size of the largest side is no greater than 6. What can we say about the size of the smallest side?

I (hater of many math problems) quite like this math problem.

A more applicable problem could be: collect cost and quantity of products of various container sizes from a trip to the store e.g. ketchup sold in various sizes of bottles. Figure out the real cost per ounce of the product vs cost of packaging and shipping. That would be a real-world use of math.

To add to what @zenvelo said, if we were to arrange the sides from smallest to largest, saying that the largest side can be no greater than 6 is equivalent to saying that the sum of the two smallest sides must be at least 7.

There is another constraint that we can use. Since the sum of the sides is 13, in every case their average value is 13/3, a little over 4. If there is a side that is larger than 4, making it above the average, there must be another side that is less than 4. We conclude that the smallest side can be no greater than 4.

It is possible to use all the constraints to design an algorithm for listing all possibilities for 13 or any other perimeter without doing any guessing. I used this as a programming exercise.

Here is how the algorithm works for the case of 13. Let x, y and z be the shortest, middle and longest sides of the triangle.
We have four separate cases for x from 1 to 4.

For each choice of x, we have three constraints:
y >= x
y <= z
x+y> z

We know that x+y+z=13. In the above inequalities, we can substitute z=13-x-y and solve for y giving constraints of:
y >= x
y <= z, y <= 13-x-y, y <= ½(13-x)
x+y > z, x+y > 13-x-y, y > ½(13–2x)
For a given value of x, we use the first and third constraints to find the minimum value of y, and use the second constraint to find the maximum value of y.

For example, for x = 3, we have y>=3 and y>7/2, combining to give y > =4 , and y <=10/2, giving solutions for y=4 or y=5: (3,4,6) and (3,5,5)