In a given triangle angles less than
first Fermat point
(1)
This problem is called Fermat's problem or Steiner's segment problem (Courant and Robbins
1941) and was proposed by Fermat to Torricelli. Torricelli's solution was published
by his pupil Viviani in 1659 (Johnson 1929). The first Fermat point has equivalent
triangle center functions
and is Kimberling center
If all angles of the triangle are less than first
Fermat point is the interior point angle
of
(4)
The first Fermat point can be constructed by drawing equilateral triangles on the outside
of the given triangle and connecting opposite vertices .
The three diagonals in the figure then intersect
in the first Fermat point .
The second Fermat point triangle center function
(5)
and is Kimberling center
The antipedal triangle of equilateral and
has area
where Brocard angle . The antipedal
triangle of equilateral and has area
The Fermat points are also known as the isogonic centers, since they are isogonal
conjugates of the isodynamic points .
The two Fermat points are collinear with the symmedian point of midpoint of the segment triangle centroid
and orthocenter of midpoint
of the two Fermat points lies on the nine-point
circle of
Given three positive real numbers acute triangle
(10)
is a minimum (Greenberg and Robertello 1965, van de Lindt 1966, Tong and Chua 1995)
See also Brocard Angle ,
Equilateral Triangle ,
Fermat Axis ,
First
Fermat Point ,
Isodynamic Points ,
Isogonal
Conjugate ,
Lester Circle ,
Second
Fermat Point ,
Steiner's Segment Problem
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References Courant, R. and Robbins, H. What Is Mathematics?, 2nd ed. Gallatly,
W. "The Isogonic Points." §151 in The
Modern Geometry of the Triangle, 2nd ed. Greenberg,
I. and Robertello, R. A. "The Three Factory Problem." Math. Mag. 38 ,
67-72, 1965. Honsberger, R. Mathematical
Gems I. Johnson,
R. A. Modern
Geometry: An Elementary Treatise on the Geometry of the Triangle and the Circle. Kimberling, C.
"Central Points and Central Lines in the Plane of a Triangle." Math.
Mag. 67 , 163-187, 1994. Kimberling, C. "Triangle Centers
and Central Triangles." Congr. Numer. 129 , 1-295, 1998. Kimberling,
C. "Fermat Point." http://faculty.evansville.edu/ck6/tcenters/class/fermat.html . Kimberling,
C. "Encyclopedia of Triangle Centers: X(13)=1st Isogonic Center." http://faculty.evansville.edu/ck6/encyclopedia/ETC.html#X13 . Kimberling,
C. "Encyclopedia of Triangle Centers: X(14)=2nd Isogonic Center." http://faculty.evansville.edu/ck6/encyclopedia/ETC.html#X14 . Mowaffaq,
H. "An Advanced Calculus Approach to Finding the Fermat Point." Math.
Mag. 67 , 29-34, 1994. Nelson, D. "Napoleon Revisited."
Math. Gaz. 58 , 108-116, 1974. Pottage, J. Geometrical
Investigations. Spain, P. G.
"The Fermat Point of a Triangle." Math. Mag. 69 , 131-133,
1996. Tong, J. and Chua, Y. S. "The Generalized Fermat's Point."
Math. Mag. 68 , 214-215, 1995. van de Lindt, W. J.
"A Geometrical Solution of the Three Factory Problem." Math. Mag. 39 ,
162-165, 1966. Wells, D. The
Penguin Dictionary of Curious and Interesting Geometry. Referenced on Wolfram|Alpha Fermat Points
Cite this as:
Weisstein, Eric W. "Fermat Points." From
MathWorld https://mathworld.wolfram.com/FermatPoints.html
Subject classifications
with all angles less than
(
, the first Fermat point 
or 
(sometimes simply called "the Fermat point," Torricelli
point, or first isogonic center) is the point 
which minimizes the sum of distances from 
, 
,
and 
,
![bc[c^2a^2+(c^2+a^2-b^2)^2][a^2b^2-(a^2+b^2-c^2)^2][4Delta-sqrt(3)(b^2+c^2-a^2)]](/images/equations/FermatPoints/Inline15.svg)
(Kimberling 1998, p. 67).
(
), then the first
Fermat point is the interior point 
from which each side subtends an angle
of 
, i.e.,
or 
is constructed analogously using equilateral triangles pointing
inwards. It has triangle center function
(Kimberling 1998, p. 68).
is equilateral and
has area
![2Delta[cotomegacot(1/3pi)+1]](/images/equations/FermatPoints/Inline27.svg)
is the Brocard angle. The antipedal
triangle of 
is also an equilateral and has area
![2Delta[cotomegacot(1/3pi)-1]](/images/equations/FermatPoints/Inline35.svg)
,
and the midpoint of the segment 
, where 
is the triangle centroid
and 
is the orthocenter of 
(left figure). Furthermore, the midpoint
of the two Fermat points lies on the nine-point
circle of 
(right figure).
,
the "generalized" Fermat point is the point 
of a given acute triangle 
such that
