Question

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 $\triangle ABC$. If she obtains $m + n\sqrt{5}$, where $m$ and $n$ are integers, what is $m + n$?

Answer 1

Answer: 8

Step-by-step explanation:

The distance formula is used to find the distances to each of the points:

 d = √((x2-x1)² +(y2-y1)²)

 d = √((4-0)² +(2-0)²) +√((4-8)² +(2-(-1))²) +√((4-5)² +(2-4)²)

 d = √20 +√25 +√5 = 2√5 +5 +√5

 d = 5 +3√5

Comparing to the form

 d = m +n√5

we see that m=5, n=3. The the sum m+n = 5+3 = 8.

(sorry if im wrong)

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