ASSIGNMENT 6

by Soo Jin Lee

11. Consider any triangle ABC. Find a construction for a point P such that the sum of the distances from P to each of the three vertices is a minimum.

To do this problem, I will first display the triangle where the point is inside of it.

Draw a triangle ABC and mark with an arbitrary dot D in the triangle.

Make segments from the point D to the each vertex of the triangle ABC.

After we make segments, rotate the inner triangle ABD 60 degree counter-clockwise.

In similar way, rotate the inner triangle ACD 60 degree clockwise.

Now draw segments from A' to the vertex C and A" to the vertex B.

As you can see from the above picture,

two yellow-green segments A'C and A"B produce one intersection which will be the point P.

(Picture 1)

Finally I get the point P such that the sum of the distances from P to each of the three vertices is a minimum.

Let's further our investigation to find the point P when I put my point outside of the triangle ABC.

Do similar process as we did when the point D is in the inside of the triangle ABC.

Draw each segment form the arbitrary point D' to the each vertex of the triangle ABC.

Then rotate the inner triangle ABD' to the 60 degree counter-clockwise and triangle ACD' 60 degree clockwise.

To get the point P, connect the point A' to the vertex C, and A" to the point B.

(Picture 2)

Since the triangle I used for both cases are same, I have to get the point P at the same location because there is only one minimum point such that the sum of the distances from P to each of the three vertices is a minimum.

To see whether it is true or not, I will overlap the Picture 1 and the Picture 2,

As you can see from the above picture, it is really interesting to see the fact that the point P is exactly at the same location!!!!

We call Point P "Fermat point ".