Fermat point

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Construction for the fermat point.

In geometry, the first Fermat point, or simply the Fermat point, also called Torricelli point, and first isogonic center, is the solution to the problem of finding a point F inside a triangle ABC 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.

1 Construction
2 Derivation
3 Proof
4 Another proof
5 Properties
6 History
7 See also
8 References
9 External links

[edit] Construction

To locate the Fermat point:

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.

For the case that the largest angle of the triangle exceeds 120°, the solution is a point on the vertex of that angle.

[edit] Derivation

One of the solution to find fermat point.

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.

[edit] Proof

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,

$\angle RAC = \angle RAB + \angle BAC$

$\angle BAQ = \angle BAC + \angle CAQ$

Since $\angle RAB$ and $\angle CAQ$ equal 60º, which are interior angles of an equilateral triangle, $\angle RAC = \angle BAQ$ . This implies that triangles RAC and BAQ are congruent. Hence $\angle ARF = \angle ABF$ and $\angle AQF = \angle ACF$ . By converse of angle in the same segment, ARBF and AFCQ are both concyclic.

Thus $\angle AFB = \angle AFC = \angle BFC = 120$ º. Because $\angle BFC$ and $\angle BPC$ add up to 180º, BPCF is also concyclic. Hence $\angle BFP = \angle BCP = 60$ º. Because $\angle BFP + \angle BFA = 180$ º, AFP is a straight line.

Q.E.D.

[edit] Another proof

Another approach to find a point within the triangle, from where sum of the distances to the vertices of triangle is minimum, is to use one of the optimization (mathematics) methods. In particular, method of the lagrange multipliers and the law of cosines.

Thus, let lines from the point within the triangle to its vertices are x, y and z. Let angle between x and y is α, y and z is β, then angle between x and z is (2π - α - β). Using method of lagrange multipiers we have to find minimum of the function:

x + y + z + λ₁ (x² + y² - 2xy cos(α) - a²) + λ₂ (y² + z² - 2yz cos(β) - b²) + λ₃ (z² + x² - 2zx cos(α+β) - c²)

where a, b and c - lengths of the triangular line segments.

Calculation of partial derivatives: δ/δx, δ/δy, δ/δz, δ/δα, δ/δβ gives the 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 computation equations for α and β separate from the rest of the parameters:

sin(α) = sin(β)

sin(α + β) = -sin(β)

that gives: α = β = 120^o

Q.E.D.

Note: if one of the vertices of triangle has angle not less than 120^o, than Fermat point is in that vertex.

[edit] Properties

Proof that the lines are concurrent and their properties.

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 angle brought about by this point, that is, $\angle AFB$ , $\angle BFC$ , and $\angle CFA$ , are all equal to 120º.
The circumcircles of the three regular triangles in the construction intersect at this point.
Trilinear coordinates for the 1st Fermat point, 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).^[1]

Trilinear coordinates for the second Fermat point, 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).^[2]

The isogonal conjugate of the 1st Fermat point is the 1st isodynamic point, X(15):

sin(A + π/3) : sin(B + π/3) : sin(C + π/3).^[3]

The isogonal conjugate of the 2nd Fermat point is the 2nd isodynamic point, X(16):

sin(A − π/3) : sin(B − π/3) : sin(C − π/3).^[4]

The following triangles are equilateral:

antipedal triangle of X(13)

antipedal triangle of X(14)

pedal triangle of 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 Fermat point, 2nd Fermat point, circumcenter, nine-point center lie on a Lester circle.

[edit] History

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.^[5]