The Fermat point (also called the Torricelli point) of a triangle (with no angle more than is a point which has the minimum total distance to three vertices (i.e., ).
A method to find the point is to construct three equilateral triangles out of the three sides from , then connect each new vertex to each opposite vertex, as these three lines will concur at first Fermat point.
We shall present a standard triangle inequality proof as well as a less-known vector proof:
First, we shall note that must lie inside the triangle . Otherwise, we suppose that WLOG, and are on opposite sides of . Then, consider the reflection of about . Note , , and , so thus is not the Fermat Point.
Suppose that was acute. Consider the rotation about . For any point , let the image of this point be . Then, we see that and . so is equilateral. Now, consider the point inside the triangle. Then, and , so . Thus, we get that as .
Now, WLOG let . We have that . We note that with equality if and only if .
This means that (as then ). Thus, we see that is cyclic, so thus as lies on , we see that is the intersection of the circumcircle of and (not ). Thus, note that as , . Similarly, . Thus, we have found the Fermat Point.
Vector Proof (Due to Titu Andreescu and Oleg Mushkarov)
We will let our origin be the point with .
Consider the point in the plane . Let and the unit vectors along . Then, . Similarly, and . Noting that and adding, we see that , or . Thus, the origin or point is the desired point.
There are two main generalizations:
The problem goes as following: which point minimizes , where are positive reals?
The problem goes as following: for the polygon , which point minimizes ?
Using the second solution, it is easy to see the point is the point where the unit vectors to the vertices sum to . For a quadrilateral, it is the intersection of the diagonals.
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