# Fermat Point

Allyson Faircloth

This activity asks us to find a point where the sum of the distances from that point to the three vertices is minimal.  It turns out that this point is called the Fermat Point.  One way to construct this point is to make an equilateral triangle to each side of the triangle.  Then connect the vertex which does not lie on the side of the original triangle to the opposite vertex of the original triangle with a line.  Repeat this step for the other two equilateral triangles.  The point of concurrency of those three lines is the Fermat Point.  (In fact, we actually only need to construct the equilateral triangles to two sides of the triangle and then find the intersection of the lines connecting the vertices of those equilateral triangles with the opposite vertex of the original triangle.)

We now need to prove that the sum of the distances from point P to the three vertices of the original triangle is minimal.

Proof:

1. We can begin by rotating triangle APB 60 degrees around point A.

2. Now connect C' and B with a segment forming the equilateral triangle ABC'.

3. If we construct a line through points C' and C, we can not say that the shortest distance between two points is a straight line. Therefore, our Fermat point must lie on this line.

4. If we then perform the same constructions for triangle APC, we can say that the shortest distance between B' and B is a straight line. Therfore, our Fermat point must lie on the line B'B as well.

5. Since the Fermat point must lie on both line C'C and B'B, then the Fermat point is the intersection of these lines.

6. We now need to prove that triangle C'AC is congruent to triangle B'AB.

7. First of all, sides C'A and AB are congruent since they are both sides of an equilateral triangle.

8. By the same reasoning, sides AC and B'C are congruent.

9. Both equilateral triangles were formed by rotating 60 degrees. Therefore angle BAC' and angle ACB' are both 60 degrees and thus congruent.

10. Therefore, triangle C'AC and triangle B'AB are congruent by the Side-Angle-Side Theorem.

An extension of this proof would be proving that angle APB, angle APC, and angle CPB are all equal to 120 degrees.

Proof:

1. We know that line segment AP is congruent to line segment AQ since AQ is the corresponding side of the reflected triangle.

2. Therefore, triangle APQ is equilateral; and by the properties of equilateral triangles, angle PAQ is 60 degrees.

3. By similar reasoning, we can show that angle CPR is also 60 degrees.

4. Since angle CPR and QPB are vertical angles, they are congruent.

5. Angle APQ + angle QPB = angle APB

60 degrees + 60 degrees = angle APB

120 degrees = angle APB

6. The same reasoning can be used to prove angle APC and CPB are 120 degrees as well.