Triangle ABC has an altitude drawn from C to AB, meeting the base at point P. This altitude divides the triangle into two unequal right triangles.

Triangle A'B'C' also has a point, P' on its base, with a line segment connecting it to vertex C', but chosen so that angle A'P'C' is 60°, with the resulting triangle A'P'C' though, not being equilateral.

All eight line segments are of integer length, and each triangle has a perimeter less than 50. The bases, AB and A'B', are the longest sides in each of the two respective original triangles, and they differ by 1 unit in length.

• What are the dimensions of the triangles ABC and A'B'C'?
• What are the lengths of CP and C'P'?

I had been looking for a more elegant solution than this, but didn't find one.

A brief search through the smaller pythagorean triples gave only a few choices of pairs that share a common leg length.
Of these, only two yield a perimeter of ABC less than 50 (5,12,13;9,12,15 and 6,8,10;8,15,17).
Only the latter has AB as its longest side (as per the problem), so we have the first triangle:
AB = 21
AC = 10 (or 17)
BC = 17 (or 10)
CP = 8

From the problem, A'B' must thus be 20 or 22.
Drawing a picture and applying the law of cosines:
(A'C')² = (A'P')² + (C'P')² - 2(A'P')(C'P')cos(60), or simply (A'C')² = (A'P')² + (C'P')² - (A'P')(C'P').

Similarly, from the supplementary base angle (120°):
(B'C')² = (B'P')² + (B'P')² + (B'P')(C'P').

After this I just wrote a small program to loop over values of A'P' (and derived values for B'P') to find combinations which gave integral results for A'C' and B'C'.

The loop limits in my program found three possibilities:
A'P'=6,B'P'=14,C'P'=16,A'C'=14,B'C'=26,P=60
A'P'=15,B'P'=7,C'P'=8,A'C'=13,B'C'=13,P=48
A'P'=16,B'P'=6,C'P'=10,A'C'=14,B'C'=14,P=50

Of these, only one has a total perimter less than 50, so the values for the second triangle must be:
A'B' = 22
A'C' = 13
B'C' = 13
C'P' = 8