Swing between Geometry and Algebra

The sum of the lengths is equal the distance of one of the vertices to the vertex of the equilateral triangle constructed on the opposite side. This can be easily found using Pythagorean theorem.

http://www.cut-the-knot.org/Generalization/fermat_point.shtml

Aim is to find O, the Fermat point of ∆ABC such that each of the angles AOB, BOC, and COA is equal to 120 degrees. Notice that with side lengths 5, 12 and 13 the area of this right-angled triangle is 30. Also the areas of three smaller triangles ∆OAB, ∆OBC and ∆OCA add up as ½{OA.OB Sin (2π/3)} + ½{OB.OC Sin (2π/3)} + ½{OC.OA Sin (2π/3)} in the light of the fact that enclosed angle is 120 degrees each. Equating the two calculated areas, OA.OB + OB.OC + OC.OA = 40√3. Coming to algebra now, three simultaneous cyclic expressions using the cosine formula are: 5^2 = OA^2 + OB^2 + OA.OB, 12^2 = OB^2 + OC^2 + OB.OC and 13^2 = OC^2 + OA^2 + OC.OA. The order is immaterial as these equations are simply added to give: 169 = OA^2 + OB^2 + OC^2 + 20√3. And then some elementary algebraic manipulation of the expression (OA + OB + OC)^2, i.e. p^2 here equals 169 + 60√3. Hence answer is 229.