Triangles: The Fermat-Torricelli Point

A lesson by Chris J.Koeppen


 

Lesson 1: The Fermat-Torricelli Point
-adapted from Rethinking Proof by Michael D. de Villiers
 
During Lesson 1, we will be investigating and proving conjectures about a right triangle.

(All questions should be answered on separate paper in your journal in a complete and organized manner.)

 

 


1. Open file: fermat1.gsp

What kind of triangles are constructed on the sides of triangle ABC? question 1

2. Construct lines DC, EA, and FB.

What do you notice about these lines? Drag any vertex to test your observations. question 2

3. Measure the distances DC, EA, and FB.

What do you notice about these distances? Carefully check your observations by further dragging. question 3

4. Drag C past B.

What happens to the triangles? question4

 

When you drag C past B, do your observations from Questions 2 and 3 still hold? question 5


 

Now you have noticed that if equilateral triangles DBA, ECB, and FAC are constructed on the sides of any right triangle ABC,

  • the the lines DC, EA, and FB are concurrent

  • Segments DC, EA, and FB are equal in length.

Further, the result appears to be true even if the triangles lie inwardly.

 

This point of concurrency is known as the Fermat-Torricelli point.

 

In order to confirm our conjecture, we must be careful, and fully investigate the problem further to come up with ideas for a thorough proof.

 


 

 

5. Shade in the interior of triangle DBC.

6. Rotate triangle DBC around point B by -60.

What do you notice about the rotated triangle? Find other pairs of triangles that act similarly. question 6

7. Construct a point at the point of concurrency and label it O.

8. Next measure the six angles formed around point O.

What do you notice about the six angles around O? Drag a vertex of triangle ABC to check your observations. question 7

9. Show the circumcircle of triangle ADB. (Remember that the circumcircle is centered at the intersection of the perpendicular bisectors of an equilateral triangle.) Repeat for the other two equilateral triangles.

What do you notice about the relationship between angle AOB and angle ADB? What can you conclude from that? (Hint: Look at quadrilateral AOBD) question 8

After constructing the other circumcircles for the equilateral triangles, what do you notice when looking at the other two triangles? question 9
 


 

 

A. Now, lets prove segments equal.......

 

The conjecture is the following: 

If equilateral triangles DBA, ECB, and FAC are constructed on the sides of any right triangle ABC, then the lengths DC, EA, and FB are equal.
 


1. Hide the circumcircles and the circumcenters. Keep everything as is, and just remember we want to focus on the initial construction.

(questions a-h for proof 1)

a. In triangles DBC and ABE, what can you say about the corresponding sides DB and AB? Why?

b. What can you say about corresponding sides BC and BE? Why?

c. What can you say about corresponding angles DBC and ABE?

d. From question c, what can you conclude about corresponding sides DC and AE?

e. Repeat the above for triangles EAC and BFC to complete the proof.

g. Did your answers to a-e use the fact that angle ABC measures 90 degrees? What does that imply about the conjecture you just proved?


 

 

 

 

 

Consider the following quotation below in relation to your conclusion in Question g:

A good proof should convey an insight into exactly why the proposition is true. Such insight sometimes reveals the pleasant, unanticipated surprise that the proposition is merely a special case of a more general one, thus allowing for its immediate generalization.               

                                                     -- M. De Villiers 1998

 

In what way has your proof provided you with insight that led to an immediate generalization? question h


 

 

B. Now lets prove lines concurrent.............

The conjecture is the following:

If equilateral triangles DBA, ECB, and FAC are constructed on the sides of any right triangle ABC, the lines DC, EA, and FB are concurrent.


1. Open file: fermat2.gsp

2. Pictured is the original construction (A) as well as the modified construction (B) from the last proof. Circumcircles ADB and BEC are shown where they should intersect at point B.

3. Now, construct the other point of intersection of these two circles. This will be your new point O.

4. Use the segment tool to construct the six segments OA, OB, OC, OD, OE and OF.

 

First, we will prove that AOE and DOC are straight lines, and then that the circumcircle AFC also passes through O. Using this fact, we will then show that BOF is also a straight line, which implies that lines AE, DC, and BF are concurrent at O.

 

Answer the questions on a separate piece of paper accordingly to hopefully prove what we need.

(questions a-j for proof 2)

a. What can you say about the size of angle BCE ? Why?

b. From a, what can you now say about the size of angle BOE? Why?

c. What can you say about the size of angle BOA? Why?

d. From b and c, what can you now conclude about angle AOE?

e. Repeat the same argument to show that DOC is a straight line.

f. From the angles determined above, calculate the size of angle AOC.

g. From f, what can you now conclude about quadrilateral CFAO? Why?

5. Now, we need to construct circumcircle of triangle AFC. Remember that the center of the circumcircle is the intersection of the perpendicular bisectors of an equilateral triangle. Be sure to only show the circle and the center, hiding the construction lines.

h. Repeat the same argument as in a- e

i. Would the preceding argument still be valid if angle ABC were not equal to 90 degrees? What can you conclude from that?


 

Consider the quotation below from a Russian mathematician, in relation to your conclusion in i.

A good proof is one that makes us wiser.           - Yu Manin (1981: 107)

 

 j. In what way has the proof made you "wiser"?

 


 


 

 

 

 

So what did you learn???

  • If equilateral triangles DBA, ECB, and FAC are constructed on the sides of any right triangle ABC,

    • the the lines DC, EA, and FB are concurrent

    • Segments DC, EA, and FB are equal in length.

  • Using Geometer's Sketchpad, we provided a complete and organized construction and proof.

  • This point of concurrency is known as the Fermat-Torricelli point.

 

Now that we know what the Fermat-Torricelli point is, lets see what the big fuss is about.........see if we can we apply it somewhere not on a piece of paper or on a computer screen.

On To- Lesson Two: Lets Build a Hospital --->

 

 

 

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