You can define your task well with an analytic approach. Orient your triangle so the vertex at the right angle is at the Origin. Did you observe it was a right triangle. Put the other two vertices at (4,0) and (0,3).
Now, it's easy to check the total distance from any vertex:
(0,0) ==> 3^2 + 4^2 = 25
(4,0) ==> 4^2 + 5^2 = 41
(0,3) ==> 3^2 + 5^2 = 34
(0,0) wins the vertex competition.
Okay, it's edge time.
Anywhere on the x-axis, say (a,0) we have 4^2 + (a^2 + 3^2) = 25 + a^2 < 25 + 16 = 41
Minimizing this, we get a = 0, which is back at the Origin.
Anywhere on the y-axis, say (0,b) we have 3^2 + (b^2 + 4^2) = 25 + b^2 < 25 + 9 = 34
Minimizing this, we get b = 0, which is back at the Origin.
Anywhere on the hypotenuse, say (r,s) we have 5^2 + (r^2 + s^2) = 25 + r^2 + s^2 < 25 + 16 + 9 = 50
Minimizing this, we get r = 0 and s = 3, which gives 34, but we already know that.
Lesson learned. If we wander along the edges, we get greater values than at the vertices!! Wow! I smell a theorem!!!
Okay, what happens if we wander into the interior??? Really big hint: In may be similar to the hypotenuse except that we are not constrained to that line.
Note: This is called EXPLORATION. One needn't know how to proceed in order to get started.
