Fermat point
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.,
).
Contents
[hide]Construction
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.
Proofs
We shall present a standard triangle inequality proof as well as a less-known vector proof:
Geometric 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.
Generalizations
There are two main generalizations:
Weighted Generalization
The problem goes as following: which point minimizes
, where
are positive reals?
Polygon Generalization
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.
See Also
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