Find the top of the triangle

Let A , B be the vertices of the base, C be the unknown vertex. If the sum s given, the lengths of the sides, then each side is equal to half of this sum. So, AC = BC = s/2 .

Let M be the middle of AB (its coordinates are equal to the half-sum of coordinates A and B ). Then CM is the height, from the right triangle AMC we have:

 CM = sqrt(AC^2 - AM^2) = sqrt(s^2/4 - AB^2/4) 

(If there is a negative number under the root, the task obviously has no solutions.)

So, we have the length of the vector MC , its direction is easy to find, given that it is perpendicular to the vector AB : if (p, q) is the vector AB , then the vector (-q, p) perpendicular to it, the vector (-q/l, p/l) (where l = sqrt(p^2 + q^2) ) is perpendicular to AB and has a length of 1 and vector (-q/l*L, p/l*L) (where L is the previously calculated length CM ) is perpendicular to AB and has a length equal to the length of MC .

So we have the vector MC . Adding its coordinates to the coordinates of point M , we get point C

Note that we have 2 possible solutions, different in the sign of the MC vector: to get the second solution, change the sign of the MC from the first solution.

This and other similar tasks will be coded very easily if there are classes in your arsenal representing a point, a vector, and operations on them are defined. For example, in my code, the solution usually looks like this (C #):

 var AB = B - A; var M = A + AB * 0.5; var L = Math.Sqrt(s * s - AB.Length * AB.Length) / 2; var MC = AB.Rotate(Angle.FromDegrees(90)).GetUnitVector() * L; var C1 = M + MC; var C2 = M + (-MC);