- Fermat point
geometry, the first Fermat point, or simply the Fermat point, also called Torricelli point, is the solution to the problem of finding a point F inside a triangleABC such that the total distance from the three vertices to point F is the minimum possible. It is so named because this problem is first raised by Fermat in a private letter.
To locate the Fermat point of a triangle with largest angle at most 120°:
#Construct three regular triangles out of the three sides of the given triangle.
#For each new vertex of the regular triangle, draw a line from it to the opposite triangle's vertex.
#These three lines intersect at the Fermat point.
The reader should be careful not to confuse the Fermat point with the first isogonic center also known as "X"(13).Entry X(13) in the [http://faculty.evansville.edu/ck6/encyclopedia/ETC.html Encyclopedia of Triangle Centers] ] Whilst the two points are one and the same provided no angle of the triangle exceeds 120° the coincidence ends there. When a triangle has an angle greater than 120° the first isogonic center always lies outside the triangle whereas the Fermat point is sited at the obtuse angled vertex. This means that in such a triangle the two points are never the same.
Given any Euclidean triangle ABC and an arbitrary point P let d(P) = PA+PB+PC. The aim of this section is to identify a point P0 such that d(P0) < d(P) for all P ≠ P0. If such a point exists then it will be the Fermat point. In what follows Δ will denote the points inside the triangle and will be taken to include its boundary Ω.
A key result that will be used is the dogleg rule which asserts that if a triangle and a polygon have one side in common and the rest of the triangle lies inside the polygon then the triangle has a shorter perimeter than the polygon.
[ If AB is the common side extend AC to cut the polygon at X. Then by the triangle inequality the polygon perimeter > AB+AX+XB = AB+AC+CX+XB ≥ AB+AC+BC. ]
Let P be any point outside Δ. Associate each vertex with its remote zone; that is, the half-plane beyond the (extended) opposite side. These 3 zones cover the entire plane except for Δ itself and P clearly lies in either one or two of them. If P is in two (say the B and C zones) then setting P' = A implies d(P') = d(A) < d(P) by the dogleg rule. Alternatively if P is in only one zone, say the A-zone, then d(P') < d(P) where P' is the intersection of AP and BC. So for every point P outside Δ there exists a point P' in Ω such that d(P') < d(P).
Case 1. The triangle has an angle ≥ 120o.
Without loss of generality suppose that the angle at A is ≥ 120o. Construct the equilateral triangle AFB and for any point P in Δ (except A itself) construct Q so that the triangle AQP is equilateral and has the orientation shown. Then the triangle ABP is a 60o rotation of the triangle AFQ about A so these two triangles are congruent and it follows that d(P) = CP+PQ+QF which is simply the length of the path CPQF. As P is constrained to lie within ABC, by the dogleg rule the length of this path exceeds AC+AF = d(A). Therefore d(A) < d(P) for all P є Δ, P ≠ A. Now allow P to range outside Δ. From above a point P' є Ω exists such that d(P') < d(P) and as d(A) ≤ d (P') it follows that d(A) < d(P) for all P outside Δ. Thus d(A) < d(P) for all P ≠ A which means that A is the Fermat point of Δ. In other words the Fermat point lies at the obtuse angled vertex.
Case 2. The triangle has no angle ≥ 120o.
Let P be any point inside Δ and construct the equilateral triangle CPQ. Then CQD is a 60o rotation of CPB about C so d(P) = PA+PB+PC = AP+PQ+QD which is simply the length of the path APQD. Let P0 be the point where AD and CF intersect. This point is commonly called the first isogonic center. By the angular restriction P0 lies inside Δ moreover BCF is a 60o rotation of BDA about B so Q0 must lie somewhere on AD. Since CDB = 60o it follows that Q0 lies between P0 and D which means AP0Q0D is a straight line so d(P0) = AD. Moreover if P ≠ P0 then either P0 or Q0 won't lie on AD which means d(P0) = AD < d(P). Now allow P to range outside Δ. From above a point P' є Ω exists such that d(P') < d(P) and as d(P0) ≤ d(P') it follows that d(P0) < d(P) for all P outside Δ. That means P0 is the Fermat point of Δ. In other words the Fermat point is coincident with the first isogonic center.
Since the time the problem first appeared, many methods to arrive at the solution have been developed. One method is to simply rotate BEC, where E is an arbitrary point, 60º counter-clockwise. Now the distance to minimize is the same as the path AEE'C'. Obviously the solution is when it is a straight line, from which the construction method can be derived.
This proof will show that the three lines are concurrent. One proof, using properties of
concyclic points, is as follows:
Suppose RC and BQ intersect at F, and two lines, AF and AP, are drawn. We aim to prove that AFP is a straight line.
Because AR = AB and AC = AQ by construction,
Since and equal 60º, which are interior angles of an equilateral triangle, . This implies that triangles RAC and BAQ are congruent. Hence and . By converse of angle in the same segment, ARBF and AFCQ are both concyclic.
Thus º. Because and add up to 180º, BPCF is also concyclic. Hence º. Because º, AFP is a straight line.
Another approach to find a point within the triangle, from where sum of the distances to the
verticesof triangle is minimum, is to use one of the optimization (mathematics)methods. In particular, method of the lagrange multipliersand the law of cosines.
We draw lines from the point within the triangle to its vertices and call them X, Y and Z. Also, let the lengths of these lines be x, y, and z, respectively. Let the acute angle between X and Y be α, Y and Z be β. Then the angle between X and Z is (2π − α − β). Using the method of lagrange multipiers we have to find the minimum of the lagrangian, which is expressed as:
: "x" + "y" + "z" + : "λ"1 ("x"2 + "y"2 − 2"xy" cos("α") − "a"2) + : "λ"2 ("y"2 + "z"2 − 2"yz" cos(β) − "b"2) + : "λ"3 ("z"2 + "x"2 − 2"zx" cos("α" + "β") − "c"2) .
where "a", "b" and "c" are the lengths of the sides of the triangle.
Calculating the partial derivatives δ/δx, δ/δy, δ/δz, δ/δα, δ/δβ gives a system of 5 equations:
: δ/δx: 1 + λ1(2x − 2y cos(α)) + λ3(2x − 2z cos(α + β)) = 0
: δ/δy: 1 + λ1(2y − 2x cos(α)) + λ2(2y − 2z cos(β)) = 0
: δ/δz: 1 + λ2(2z − 2y cos(β)) + λ3(2z − 2x cos(α + β)) = 0
: δ/δα: λ1y sin(α) + λ3z sin(α + β) = 0
: δ/δβ: λ2y sin(β) + λ3x sin(α + β) = 0
After some algebraic manipulations equations for α and β separate from the rest of the parameters, giving:
: sin(α) = sin(β)
: sin(α + β) = −sin(β)
that gives: α = β = 120o
Note: if one of the vertices of triangle has angle not less than 120o, than Fermat point is at that vertex.
* In case the largest angle of the triangle is not larger than 120º, the point minimizes the total distance from the three vertices to this point.
* The internal angles brought about by this point, that is, , , and , are all equal to 120º.
circumcircles of the three regular triangles in the construction intersect at this point.
Trilinear coordinatesfor the 1st isogonic center, "X"(13)::csc("A" + π/3) : csc("B" + π/3) : csc("C" + π/3), or, equivalently,:sec("A" − π/6) : sec("B" − π/6) : sec("C" − π/6).
Trilinear coordinatesfor the 2nd isogonic center, "X"(14)::csc("A" − π/3) : csc("B" − π/3) : csc("C" − π/3), or, equivalently,:sec("A" + π/6) : sec("B" + π/6) : sec("C" + π/6). [Entry X(14) in the [http://faculty.evansville.edu/ck6/encyclopedia/ETC.html Encyclopedia of Triangle Centers] ]
* The isogonal conjugate of the 1st isogonic center is the 1st
isodynamic point, "X"(15)::sin("A" + π/3) : sin("B" + π/3) : sin("C" + π/3). [Entry X(15) in the [http://faculty.evansville.edu/ck6/encyclopedia/ETC.html Encyclopedia of Triangle Centers] ]
* The isogonal conjugate of the 2nd isogonic center is the 2nd
isodynamic point, "X"(16)::sin("A" − π/3) : sin("B" − π/3) : sin("C" − π/3). [Entry X(16) in the [http://faculty.evansville.edu/ck6/encyclopedia/ETC.html Encyclopedia of Triangle Centers] ]
* The following triangles are equilateral::
antipedal triangleof "X"(13):antipedal triangle of "X"(14): pedal triangleof the "X"(15):pedal triangle of the "X"(16):circumcevian triangle of "X"(15):circumcevian triangle of "X"(16)
* The lines "X"(13)"X"(15) and "X"(14)"X"(16) are parallel to the
Euler line. The three lines meet at the Euler infinity point, "X"(30).
* The 1st isogonic center, 2nd isogonic center, circumcenter, nine-point center lie on a Lester circle.
This question was proposed by Fermat, as a challenge to
Evangelista Torricelli. He solved the problem in a similar way to Fermat's, albeit using intersection of the circumcircles of the three regular triangles instead. His pupil, Viviani, published the solution in 1659. [MathWorld|urlname=FermatPoints |title=Fermat Points]
Geometric medianor Fermat–Weber point, the point minimizing the sum of distances to more than three given points.
*" [http://demonstrations.wolfram.com/FermatPoint/ Fermat Point] " by Chris Boucher,
The Wolfram Demonstrations Project.
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