+0

0
22
2
+248

Wanda is trying to locate the Fermat point P of triangle ABC, where A is at the origin, B is at (8,-1), and C is at (5,4) (the Fermat point is the point such that the sum of its distances from the vertices of a triangle is minimized). She guesses that the point is at P=(4,2), and computes the sum of the distances from P to the vertices of ABC. If she obtains $m + n\sqrt{5}$, where m and n are integers, what is m+n?

Jun 11, 2019

#1
+101085
+2

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  ??]

Jun 11, 2019
#2
+101085
+1

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)

Jun 11, 2019