+129852 Distance from P to A = √ [(4-0)^2 + (2-0)^2 ] = √[16 + 4] = √20 = 2√5
Distance from P to B = √ [ (8- 4)^2 + (-1 - 2)^2 ] = √[ (4)^2 + (-3)^2] = √[16 + 9 ] = √25 = 5
Distance from P to C = √ [ (5 - 4)^2 + (4 - 2)^2 ] = √[ (1)^2 + (2)^2 ] = √ [ 1 + 4 ] = √5
So....the sum of the distances = 5 + 2√5 + √5
5 + 3√5
m = 5 n = 3
And their sum is 8
[Of much more interest would be to actually find the Fermat point.... I wonder if Wanda's "guess" is correct ??]
+129852 BTW....here's the way to determine the Fermat point :
1) Construct equlateral triangles on two sides of the original triangle
2) Connect the new vertices to the opposite vertices of the original triangle
3) The intersection of these connected segments is the Fermat point
The image shows that the Fermat point in this problem is ≈ ( 4.63, 1.75)
