Fermat point

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In geometry, 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 vertex to point F is the minimum possible. It is so named because this problem is first raised by Fermat in a private letter.

Construction

To locate the Fermat point:

1. Construct three regular triangles out of the three sides of the given triangle.
2. For each new vertex of the regular triangle, draw a line from it to the opposite triangle's vertex.
3. These three lines intersect at the Fermat points.

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

Derivation

Since the time the problem first appeared, many methods to arrive at the solution has been developed. One of the method is to simply rotate BEC, where E is an arbitrary points, anti-clockwise for 60º. 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.

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.

Properties

• In case the largest angle of the triangle is not larger than 120º, the point minimize the total distance from the three vertex to this point.
• The internal angle brought about by this point, that is, [\angle AFB], [\angle BFC], and [\angle CFA], are all equals to 120º.
• The circumcircles of the three regular triangles in the construction interset at this point.
• The triangle fromed by joining the centers of the three regular triangles in the construction is also a regular triangle(Napolean's theorem), and the circumcenter of this triangle is the fermat point of the original triangle.

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 triangle instead. His pupil, Viviani, published the solution in 1659.

See also

• Napoleon's theorem

External links

• http://mathworld.wolfram.com/FermatPoints.html
• http://www.cut-the-knot.org/Generalization/fermat_point.shtml

 

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