The Fermat Point

 

By Pei-Chun Shih

 

 

In this write-up, I am going to find a point inside a triangle ABC such that the sum of the distances from this point to each of the three vertices is a minimum. This point is called the Fermat point because the problem of finding a point minimizes the sum of distances from vertices of a triangle was first raised by Fermat.

 

In order to locate the Fermat point, three equilateral triangles out of the three sides of the given triangle ABC need to be constructed first. Let A, B, and C be the apex of these three equilateral triangles. Then draw lines from A to A, B to B, and C to C. The point where these three lines intersect is the Fermat point.

 

 

There are several interesting properties about the Fermat point:

 

1.    The circumcircles of these three equilateral triangles (triangle ABC, ABC, and ABC) intersect at the Fermat point.

 

 

2.    Let the Fermat point be denoted as F. Then it can be proved easily that angle AFB, AFC, and BFC are all equal to 120 degrees.

 

[Proof]: From the diagram above, A, F, B, C are cyclic quadrilateral. So opposite angles are supplementary which add up to 180 degrees. Since triangle ABC is an equilateral triangle, each interior angle equals 60 degrees. Therefore, angle AFB, the opposite angle of angle ACB, equals 120 degrees which is 180 degrees minus 60 degree. The same reasoning can be hold for angle AFC and BFC.

 

3.    All the segments AA, BB, and CC have the same length.

 

 

[Proof]: In order to prove that AA equals CC, we need to prove that triangle BCC is congruent to triangle BAA. Since triangle ABC and BCA are equilateral triangles, then CB = AB, BA = BC, and angle CBA = 60 degrees = angle ABC. So angle CBC = angle CBA + angle ABC = angle CBC + angle ABC = angle ABA. Therefore, triangle BCC and BAA are congruent by the Side-Angle-Side congruence theorem. Thus, segment AA equals segment CC. We can prove that BB = CC by the same reasoning.

 


 

Now, its the time to prove that point F is really the one we are looking for. That is to say, point F is the point minimizes the sum of distances from vertices of a triangle.

 

 

 

Click HERE if you want to move the point P to F by yourself.

 

In the triangle ABC, select an arbitrary point P and connect it with vertices A, B, and C. Use point A as a center to rotate the triangle BAP 60 degree into position BAP. Then triangle APP is an equilateral triangle since PA = PA and angle PAP is 60 degree. Hence, PA = PP. Also, PB = PB. Therefore, PA + PB + BC can be written as PP + PB + PC which is a path from B to C. Since the shortest distance of two points is a straight line, then the path from B to C will be minimized if it is a straight line. In other words, when P and P lie on the segment BC the distance will be minimized (click on the animation bottoms above to see the demonstrations). We can observe from the animation that when P and F, the Fermat point we constructed earlier, are concurrent, both P and P lie on the segment BC. Therefore, the Fermat point, F, is the point such that the sum of its distances from the vertices of a triangle is a minimum. Furthermore, since point P and F are concurrent, angle BPC = angle BFC = 180 degrees. So angle APC = angle AFC = 180 – angle AFP = 180 – 60 = 120 degrees which is another way to prove that property 2 mentioned above is true.

 

 

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