Overview of Section 5.3 More on Extremal Problems

This section deals with the use of isometries in exploring and proving results about the solution of maximum or minimum problems. This is a vast area of geometry with many classic and rich problem environments. Three of those areas are examined in this section.

Inscribed Triangle with Minimum Perimeter (Fagnano's Problem) See [1] [2]

Given an acute angled triangle, inscribe in it a triangle whose perimeter is as small as possible.

Suppose you have a triangle ABC. Inscribe a triangle DEF so that D is on AC, E is on BC, and F is on AB.

Open GSP file

Now, if FF' and FF" are drawn, we have isosceles triangles FBF' and FBF".

So as F is moved on AC, the triangle F'BF" is isosceles and the angle at B (angle F'BF") is constant, twice the size of angle ABC.

Since triangle F'BF" is isosceles with slant sides equal to BF, then the shortest length for F'F" can be found when BF is the shortest. That will occur when BF is perpendicular to AC.

So if we locate F so that BF is perpendicular to AC, then the intersection points E' and D' with F'F" would form a triangle of minimum length and equal in length to F'F".

First, one construction:

Given triangle ABC.

Locate foot of altitude F from vertex B.

Reflect F in BC and BA.

Construct F'F''

Locate intersections D and E with F'F'' and sides BC and BA.

Triangle DEF is the minimal triangle.


Second, an alternative construction:

Construct the three altitudes. Locate the feet of the altitudes D, E, and F.

Construct triangle DEF with minimal perimeter.


Pedal Triangle. Libeskind defines the triangle with its vertices at the feet of the perpendiculars as the Pedal Triangle of the triangle. There is not agreement in the mathematics literature about this definition. The usual name for this triangle is the Orthic Triangle.

The usual definition of a Pedal Triangle is that the Pedal Triangle relative to a point P is formed by the feet of the perpendiculars to the sides (possibly extended) of the triangle.

See the Pedal Triangle lessons in EMAT 6680

When the pedal point P is the orthocenter of the triangle, then the Pedal Triangle (usual definition) is the orthic triangle.

There are some places in the mathematics literature where the Pedal Triangle is the feet of the three cevians having a point of concurrence. This is rare but quite different.

I prefer to use Orthic Triangle as the name of the triangle with vertices formed by the feet of the perpendiculars from the three vertices.

If P is located inside the triangle we have a picture like this:


This exploration leads to the following Theorem:

Corollary 5.1:

At each vertex of an Orthic Triangle, the two sides of the orthic triangle make equal angles with a side of the original triangle.


In an acute angled triangle, the altitudes are the angle bisectors of the Orthic Triangle. (Prob 5.3.1a)


The orthocenter of an acute angled triangle is the incenter of its Orthic Triangle.


In an acute angled triangle ABC with its Orthic Triangle DEF, Triangle DEF divides triangle ABC into four smaller triangles. The other three are all similar to triangle ABC.


The perimeter of the Orthic Triangle is less that twice the altitude of the original triangle. (Prob 5.3.1b)





The Shortest Network Problem For Triangles

Given a triangle in which each angle measures less than 120 degrees, construct a point in the interior of the triangle (or on the triangle) such that the sum of the distances from that point to the vertices is a minimum.

This point is called the Fermat point. We will want to explore how to find the Fermat point, examine characteristics it may have, and perhaps open up additional investigations.

The set of segments collectively is a NETWORK for the three points of the triangle. We are asking for a minimal network for the three points. Along the way we will want to understand the importance of the restriction of all angles having measure of less than 120 degrees.


Corollary: The Three lines are concurrent at P.



PROBLEM: How would you construct a figure like this one?









A comprehensive picture















PROBLEM: What is the locus of P as side AC is fixed and vertex B is moved along a straight line parallel to the base AC?

PROBLEM: What is the locus of P as side AC is fixed and vertex B is moved along a straight line NOT Parallel to the base AC?

PROBLEM: What is the locus of P as side AC is fixed and vertex B is moved randomly in the plane?


INVESTIGATION: Construct a point P' in triangle ABC by constructing the equilateral triangles so that they are toward the inside of the triangle. This is called the Second Fermat point of the triangle. Which of the result we have seen for the First Fermat point hold for the Second Fermat Point?


Converse: If a point P in triangle ABC is the Fermat point with congruent angles of 120 degrees around P, then the sum AP + BP + CP is the minimum network


Proof: For an equilateral triangle, the sum of the distances from an internal point P to each of the sides is a constants equal to the altitude. Use this previous result to show that for triangle ABC, with P as the Fermat point.

Construct a triangle DEF by constructing lines through A, B, and C perpendicular to AP, BP, and CP, respectively.

Show this triangle DEF is equilateral. This is immediate because in EBPA the angles at A and B are 90 degrees and hence in is a cyclic quadrilateral with the sum of the angle at E and P = 180 degrees. But the angle at P is 120 degrees (Fermat Point) so angle E has a measure of 60 degrees. Similarly we can find the angles at D and F are each 60 degrees.

Now, from our previous result about an equilateral triangle we know that for any other P1 inside triangle ABC,

A1P1 +B1P1 + C1P1 = AP + BP + CP

and this is a constant equal to an altitude of DEF. However, in each cas A1P1 > AP1, B1P1 < BP1, and C1P1 > CP1. So,

AP1 +BP1 + CP1 > AP + BP + CP

That is, for any point P1 in ABC other than the Fermat Point, the sum of the distances will be greater.

Another twist: Try to use the result that the Fermat point in a triangle is the minimum network to prove that in an equilateral triangle, the sum of the distances to the side the triangle is a constant equal to the altitude. Hint: For an equilateral triangle ABC, any point P with perpendiculars dropped to a side will be a Pedal point and the triangle QRS formed by the feet of the perpendiculars will be a Pedal Triangle for P. At some point let P be the point at the midpoint of one of the sides of ABC.




The Shortest Network Problem for Polygons

Given n points A1, A2, . . ., An, find a connected set of line segments of shortest total length such that any two of the given points can be joined by a polygonal path consisting of some of the line segment.

These are called Steiner Networks. Note that the endpoints of the individual segments in the network are not restricted to points A1, A2, . . ., An. Not all segments in the network are need be used in a polygonal path from Ai to Aj.

Consider four points on the vertices of a quadrilateral. Not all quadrilaterals will have a minimal network. Let's start with a network for points on a square (or on a quadrilateral that is somewhat like a square).

We want the minimal network for four points.



Consider looking at two triangles


Now if we construct the minimal network for each triangle by locating the Fermat Point for each, we will have the minimal network in each triangle. If we can do this in a way that the middle point we used for forming two triangles lines on the line connecting the two Fermat points, then combined we have a minimal network for the four points.


Here is a GSP Sketch with a tool for finding Fermat points in any triangle.


The minimal network has the shape like the last picture in the sequence.







If we used the opposite sides of the quadrilateral to construct an equilateral triangle and a circumcircle for each, then a line segment between the two outer vertices of the equilateral triangles would intersect the circles at E and F. Each of E and F would be the fourth vertex of a CyQl and so each would be the Fermat point of some triangle.

Any point of the segment EF could determine the common vertex for the two subtriangles. Regardless, E and F would be the two 'extra' points to complete the minimal network from those two sides. A minimal network from using the other two sides can be determined and the minimal network for the quadrilateral is the shorter of the two.


The non-vertex point in a network each have 120 degree angles around them because they are Fermat points of the appropriate triangles. In general these points are called Steiner Points with the usual domain of graph theory. Who was Jacob Steiner? This link will take you to his biography in the St. Andrews History of Mathematics site.


Construct a minimal network for points at the vertices of a regular pentagon.


Problem Set 5.3