 Construction for the fermat point. 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. Table of Geometry, from the 1728 Cyclopaedia. ...
A triangle is one of the basic shapes of geometry: a polygon with three vertices and three sides which are straight line segments. ...
Pierre de Fermat Pierre de Fermat (August 17, 1601 – January 12, 1665) was a French lawyer at the Parlement of Toulouse, southwestern France, and a mathematician who is given credit for his contribution towards the development of modern calculus. ...
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 points.
For the case that the largest angle of the triangle exceeds 120°, the solution is a point on the vertex of that angle. For alternate meanings, such as the musical instrument, see triangle (disambiguation). ...
Derivation  One of the solution to find fermat point. 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: Parallel programming (also concurrent programming), is a computer programming technique that provides for the execution of operations concurrently, either within a single computer, or across a number of systems. ...
In geometry, a set of points is said to be concyclic if they lie on a common circle. ...
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. See also: congruence relation In geometry, two shapes are called congruent if one can be transformed into the other by a series of translations, rotations and reflections. ...
Thus º. Because and add up to 180º, BPCF is also concyclic. Hence º. Because º, AFP is a straight line.
Q.E.D. Q.E.D. is an abbreviation of the Latin phrase quod erat demonstrandum (literally, which was to be demonstrated). This is a translation of the Greek (hoper edei deixai) which was used by many early mathematicians including Euclid and Archimedes. ...
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 minimize the total distance from the three vertex to this point.
- The internal angle brought about by this point, that is, , , and , 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. Evangelista Torricelli, portrait by an unknown artist. ...
See also In mathematics, Napoleons theorem is a theorem that states that if we construct equilateral triangles on the sides of any triangle (all outward or all inward), the centres of those equilateral triangles themselves form an equilateral triangle. ...
External links - http://mathworld.wolfram.com/FermatPoints.html
- http://www.cut-the-knot.org/Generalization/fermat_point.shtml
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