Assignment 6

By Dorothy Evans

Problem # 11

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.

I
chose this problem because I saw a recent episode on the TV series NUMB3RS that
utilized this concept to find a fugitive and I believe this investigation is
applicable to a high school geometry course.
In the TV show the fugitive had recently been spotted at three locations
and the mathematician explained how they could narrow the search by determining
the shortest distance from one point to all three of the other points where the
fugitive had been spotted. By working on
the likely assumption that the fugitive had taken the shortest path to all
three locations they could determine the fugitive’s most likely point of origin
(where he was hiding). Texas Instruments
a sponsor of the show and educators have also created an activity from the show
to help students explore this concept and its practical application.

http://www.cbs.com/primetime/numb3rs/ti/activities/Act4_Slippery_Toxin_final.pdf.

This
model is also used in logistics problems in determining locations of
manufacturing plants, distribution centers, and other locations of interest for
companies. The mathematical proof in the
concept problem (above) utilizes an equilateral triangle and is therefore a
simplified proof of determining the shortest sum of distances from a point to 3
other points. In my example I do not use
an equilateral triangle. My first
hypothesis was that the shortest distance was one of the centers of a triangle. In my first attempt to validate (while not a
formal proof) I convinced myself the centroid was the solution. Here I recreated
the problem in GSP and evaluated the sum of the distances manually.

Can
you find a contradiction? While the
centroid is a close approximation and in an equilateral triangle the centroid would
work it’s not the shortest distance in a right triangle or an obtuse or acute
scalene triangle. See if you can figure
it out before going on. My next
hypothesis is that it’s a triangle center I have not yet studied. I believe it’s a center that exists where
each of the angles ÛCAD =ÛDAB=ÛBAC= 60 degrees. As you can see this placement would be the
centroid if the triangle were equilateral.
At this point in time I am still attempting to create a GSP diagram to
find this center but have yet to be successful.

An
update on my investigation … I have found the name of the center and it is the Fermat
point. I am still working on the
construction.

For
full access to first GSP script for your own investigation click here.