Up until now, the concurrencies we have exhibited have determined some significant points in a triangle.

The concurrency point of the angle bisectors of a triangle gives us the center of the inscribed circle of the triangle. The point of intersection of the perpendicular bisectors of the sides of a triangle gives us the center of the circumscribed circle of a triangle. The point of concurrency of the altitudes of a triangle gives us the orthocenter of a triangle. The point of intersection of the medians of a triangle gives us the centroid, or center of gravity, of a triangle. As we seek further significant points of a triangle, we can list the point in a triangle where the sides subtend (i.e., determined by being opposite) equal angles. For example, in figure 3-1 the point P is so situated that ∠APB = ∠BPC = ∠CPA. This, unexpectedly, is also the very same point in a triangle from which the sum of the distances to the vertices is the smallest. That is, where AP + BP + CP is smaller than the sum of the distances from any other point in the triangle to the vertices. These are just two properties of a very special point in a triangle that will provide us with some other quite-surprising results. Follow along!

As we now begin this rather interesting exploration of this particularly significant point in a triangle, we consider triangle ABC on whose sides we shall construct equilateral triangles as shown in figure 3-2. Then we will draw the lines AA′, BB′, and CC′. Using some elementary geometry, we can easily show that these three line segments are equal. This is done by proving pairs of triangles congruent. Now we say “easily,” yet most high-school students who attempt to prove this equality—usually given as an exercise with congruence proofs early in the course—have difficulty doing it, simply because they have difficulty identifying the triangles that have to be proved congruent to establish the line-segment equality. Once the triangles have been identified, the proof is quite simple. That is, to show AA′ = CC′, we merely prove that ΔAA′B ΔBC′C, as shown in figure 3-3 (by the SAS congruence relationship). Similarly, AA′ = BB′, by showing, in similar fashion, that ΔAA′C ΔB′BC.

Now having justified that the line segments AA′, BB′, and CC′ are equal, it is interesting to note that the lines are also concurrent, at a point called the Fermat point, F, named after the French mathematician Pierre de Fermat (1607–1665).

To prove this concurrency we would draw the circumscribed circles of the three equilateral triangles and show that they contain a common point F, as shown in figure 3-4.

Let's now begin by considering the circumscribed circles of the three equilateral triangles BCA′, ACB′, and ABC′, where P, Q, and R are the centers of these circles. We show this in figure 3-4. The points of intersection of the circles Q and R are points F and A. Now we want to show that the point F is also on circle P.

Since 240°, we know that the inscribed angle ∠AFB −120°.

Similarly, ∠AFC – 120°. Therefore ∠BFC = 120°, since a complete revolution is 360°.

Since arc BA′C = 240°, ∠BFC is an inscribed angle and point F must, therefore, lie on circle P. Thus, we can see that the three circles are concurrent, intersecting at point F.

Now by joining point F with points A, B, C, A′, B′, C′, we find that ∠B′FA = ∠AFC′ = ∠C′FB = 60°, and therefore B′FB is a straight line. Similarly C′FC and A′FA are also straight lines, which establishes the concurrency of the lines AA′, BB′, and CC′. In this way, we can determine the point in triangle ABC at which the three sides subtend congruent angles. The point F is also called the equiangular point of triangle ABC since ∠AFB = ∠AFC = ∠BFC = 120°.

NAPOLEON'S THEOREM

At this point we are ready to embark on a rather famous theorem in geometry, one that is attributed to Napoleon Bonaparte (1769–1821), who, aside from his fame in French history as a major figure for his military prowess, also distinguished himself as a topflight mathematics student in school and later at the Paris Military School and earned membership in the Institut de France, a prestigious scientific society. He prided himself on his talent in mathematics, and particularly in geometry. Yet the theorem that still bears his name today was first published by Dr. W. Rutherford in The Ladies’ Diary in 1825, four years after Napoleon's death. To this day it is uncertain if Napoleon ever knew of the relationship we are about to investigate.

To show Napoleon's theorem, we consider the centers of the three circumscribed circles of the three equilateral triangles drawn on the sides of triangle ABC. (See figure 3-5.) These centers determine an equilateral triangle PQR, known as the Napoleon triangle. The Napoleon triangle can be shown to be equilateral by demonstrating that the sides of triangle PQR are proportional to the equal line segments AA′, BB′, and CC′, thus making the three sides of triangle PQR equal and establishing it as an equilateral triangle. The proof of this relationship can be found in the appendix.

There are many unusual relationships tying the Napoleon triangle to the original triangle. For one, the Napoleon triangle and the original triangle share a common centroid. As we establish this interesting relationship, we will encounter some other curiosities along the way. To begin this quest, in figure 3-6 we have point G as the centroid of triangle ABC, and point P the centroid of triangle BCA′. We shall denote point M_a as the midpoint of BC. Because the centroid of a triangle trisects each of the medians, we have AM_a = 3GM_a, and A′M_a = 3PM_a. Since GP partitions AM_a and A′M_a proportionally, we can conclude that ΔM_aGP ~ ΔM_aAA′, and AA′ = 3GP, or the distance between the centroids is one-third the length of the line segment joining a vertex of the original triangle with the remote vertex of the relevant equilateral triangle.

In a similar fashion, we can show that CC′ = 3GR, and BB′ = 3GQ (see figure 3-7). We have shown earlier (pp. 66, 67) that AA′ = BB′ = CC′. Therefore, GP = GQ = GR. Since triangle PQR is equilateral and the distances from point G to its vertices are equal, we can conclude that G is also the centroid of triangle PQR. We have shown, therefore, that point G is the centroid of both the outer Napoleon triangle PQR and the original triangle ABC.

All that we have said so far about the three equilateral triangles drawn on the sides of a randomly selected triangle was based on their being drawn externally to the original triangle. Yet, as you might have expected by now, we can make all the same arguments about a configuration that has the three equilateral triangles drawn internally to the given triangle, or, shall we say, overlapping the original triangle, as shown in figure 3-8. We have triangle UVW equilateral and sharing its centroid point, G, with the centroids of triangle PQR and triangle ABC.

Referring to any of the previous few figures, say figure 3-7, suppose we now leave triangle BCA′ fixed and move point A to various positions (even on the other side of BC). As long as the point A does not land on points B or C, where triangle ABC would then have a zero area, all that we have established above will still hold true. This is truly an amazing relationship! It can be nicely shown with dynamic geometry software, such as the Geometer's Sketchpad^® or GeoGebra^®.

As if the configuration shown in figure 3-2 did not already produce enough unexpected equilateral triangles, we can find yet another one. All we need to do is to construct a parallelogram AC′CD as shown in figure 3-9, and we can identify an equilateral triangle, namely, AA′D. The same size equilateral triangle can be produced at all sides of this configuration, since each of these equilateral triangles will use for a side one of the equal lengths AA′, BB′, and CC′. To justify that triangle AA′D is, in fact, equilateral, we can show that AD = AA′, since both are equal to CC′, and ∠DAA′ = 60°, since it is alternate-interior to ∠AFC′ = 60°, where F is the Fermat point. That is, AA′D is an isosceles triangle with a 60° vertex angle, which makes it equilateral.

By connecting a vertex of each of the three equilateral triangles to the nearest vertex of the Napoleon triangle PQR, we find the lines concurrent at the center O of the circumscribed circle of triangle ABC, as shown in figure 3-10.

We are not yet finished with this rich equilateral-triangle configuration. We need to focus again on the Fermat point F. Not only is point F the equiangular point, but, as we indicated earlier, it will also turn out to be the minimum-distance point from the three vertices of triangle ABC—that is, the sum of the distances from that point to the three vertices of the triangle is less than the sum of the distances from any other point in the triangle to the vertices. In other words, this point now has two important properties: the minimum-distance point and the equiangular point of the triangle.

Let's investigate how we can justify this last claim. We begin by considering triangle ABC with no angle measuring greater than 120°, as shown in figure 3-11.

To show that the sum of the distances from point F to each of the three vertices of triangle ABC is less than that from any other point to the vertices, we need to take any other randomly selected point D and show that the sum of the distances from this point is greater than the sum of the distances from point F to the vertices of the triangle. The justification—or proof—of this relationship is quite interesting and a bit different from other geometric proofs. Follow along and you will find it rewarding!

Let F be the equiangular point in the interior of triangle ABC, that is, where ∠AFB = ∠BFC = ∠AFC = 120°.

Draw lines through A, B, and C, which are perpendicular to AF, BF, and CF, respectively. These lines meet to form yet another equilateral triangle, A′B′C′. (To prove triangle A′B′C′ is equilateral, notice that each angle has measure 60°. This can be shown by considering, for example, quadrilateral AFBC′. Since ∠C′AF = ∠C′BF = 90°, and ∠AFB = 120°, it follows that ∠AC′B = 60°.) Let D be any other point in the interior of triangle ABC. We must then show that the sum of the distances from F to the vertices of triangle ABC is less than the sum of the distances from the randomly selected point D to the vertices of triangle ABC.

We can easily show that the sum of the distances from any point in the interior of an equilateral triangle to the sides is a constant, namely, the length of the altitude. Let's take a moment to review this, as we already encountered this as Viviani's theorem in chapter 1.

Consider equilateral triangle ABC, where PQ AB, PR BC, PS AC, and CD AB. Draw the line segments PA, PB, and PC (see figure 3-12).

Area ΔABC = Area ΔAPB + Area ΔBPC + Area ΔCPA

Since AB = BC = AC, the Area ΔABC = AB ⋅ (PQ + PR + PS). However, the Area ΔABC = AB ⋅ CD. Therefore, the sum PQ + PR + PS = CD is a constant for the given triangle.

Now using this constant relationship, we have in figure 3-11, FA + FB + FC = DK + DL + DM (where DK, DL, and DM are the perpendiculars to B′KC′, A′LC′, and A′MB′, respectively).

But DK + DL + DM < DA + DB + DC. (The shortest distance from an external point to a line is the length of the perpendicular segment from that point to the line.) By substitution: FA + FB + FC < DA + DB + DC.

You may wonder why we chose to restrict our discussion to triangles with angles of measure less than 120°. If you try to construct the point F in a triangle with one angle of measure of 150°, the reason for our restriction will become obvious. In figure 3-13, we have ∠BAC > 120°, and we find that the supposed minimum-distance point is outside of triangle ABC. When ∠BAC = 120°, as shown in figure 3-14, the minimum-distance point is on vertex A. Therefore, the minimum-distance point in a triangle (with no angle of measure greater than 120°) is the equiangular point (i.e., the point at which the sides of the triangle subtend congruent angles).

SQUARES ON THE SIDES OF A TRIANGLE

While there are other discoveries possible when equilateral triangles are placed on the sides of a randomly drawn triangle, as we have done in figure 3-2, there are also some startling results when one constructs a square on the sides of a randomly drawn triangle, as shown in figure 3-15.

Our first observation involves the lines joining the centers of the squares in figure 3-15 to the remote vertices of the triangle ABC. As you might have expected by now, they are concurrent. Then we also find that each of these concurrent lines is perpendicular to one of the sides of the triangle formed by joining the centers of the three squares. You may also describe these three concurrent lines, AF, BE, and CD, as overlapping the altitudes of triangle DEF. Moreover, each of the concurrent lines, AF, BE, and CD, is the same length as the side of triangle DEF, to which it is perpendicular. Namely, AF = DE, BE = DF, and DC = EF. You may wish to search for more such gems in this configuration. Here are a few more to entice you as you begin your search:

 

AC² + AB″² = AB² + AC′²,

A′A″² + BC² = 2 ⋅ (AB² + AC²),

AB″² + CA′² + BC″² = AC′² + CB′² + BA″².

 

However, be aware that triangle DEF in figure 3-15 is not necessarily equilateral as you might expect. In figure 3-16, we again have squares drawn on the sides of triangle ABC. Points P and Q are selected so that PB′BB″ and QC′CC″ are parallelograms. Then, surprisingly, triangle PAQ is an isosceles right triangle—that is, AP = AQ, and ∠PAQ is a right angle.

The Luxembourger mathematician Joseph (Jean Baptiste) Neuberg (1840–1926) discovered the following theorem, which truly depicts a noteworthy point in a triangle. In figure 3-17, we draw the three squares (externally) on the sides of the randomly drawn triangle ABC. The centers of these three squares, D, E, and F, form triangle DEF. We now draw the three squares on the sides of triangle DEF, yet overlapping triangle DEF. We find that the centers of these squares are the midpoints of the sides of the original triangle ABC. Truly a remarkable relationship of the points of a triangle!

In order to prove this is true, we would merely need to show that the midpoint M_a of BC is the center of the square on DE, that is, that M_aD and M_aE are equal and perpendicular. (See figure 3-18.)

Rotating 90° the triangle A′AC, it will coincide with triangle BAA″ (i.e., A′ → B, A → A, C → A″), showing that A′C and BA″ are equal and perpendicular. Since D and M_a are the midpoints of two sides of the triangle BA′C, we find that DM_a must be parallel to the third side, A′C, and half as long; similarly, EM_a is parallel to BA″ and half its length. Since A′C and BA″ are equal and perpendicular, then so are DM_a and EM_a. That means that M_a is the point of intersection of the diagonals of the (overlapping) square on side DE of triangle DEF.

We can enjoy some of the truly unexpected relationships that come from this configuration—squares on the sides of a randomly drawn triangle—by modifying the figure a bit. In figure 3-19, we have squares drawn on the sides of a randomly drawn triangle (ABC). Now here comes the tricky part: we draw a triangle on the remote side of each of the squares, so that each side is parallel to a side of the original triangle (ABC). That is,

 

for

for and

for

 

When we draw the lines connecting the remote vertices of the three triangles to those of the original triangle, you will find that these three lines are also concurrent (at Q). This is shown in figure 3-19.

Now, taking the squares on the sides of a randomly drawn triangle even further, we will consider the parallelograms drawn between the squares. As shown in figure 3-20, the parallelograms are:

 

parallelogram A₁AA₂A₃ with center point G,

parallelogram B₁B₃B₂B with center point H, and

parallelogram C₁C₃C₂C with center point I.

 

We get concurrent lines (at R) here by joining the center of each parallelogram with the opposite square's center point. Again, we have a rather unexpected result.

This configuration yields a number of additional concurrent lines, as we will see in the next several figures. In figure 3-21 we notice that the lines joining the midpoints of the sides of the original triangle with the center point of the opposite parallelogram are concurrent (at S).

In figure 3-22 we notice that the diagonal extensions of the parallelograms between the squares are concurrent (at T). We have found, therefore, a second concurrency in the figure formed by joining the center of each square with the remote vertex of the opposite parallelogram (at U).

Yes, there are even more concurrencies on this configuration, as we will see in figure 3-23. This time we will join the midpoint of the side of each square's exterior side with the remote vertex of the opposite parallelogram. The lines are concurrent at V.

There is still another concurrency (shown in figure 3-24), when we join each vertex of the original triangle with the midpoint of the remote side of the square on the opposite side (at W).

Although we have exhibited quite a few concurrencies in this configuration, there are many more such relationships to be found. We encourage you to search for some of them. Using a dynamic geometry program, such as the Geometer's Sketchpad or GeoGebra, can be helpful in this search. The ambitious reader will be enchanted to find that on his website “The Encyclopedia of Triangle Centers—ETC” (http://faculty.evansville.edu/ck6/encyclopedia/ETC.html) Clark Kimberling has located well over 3,600 triangle centers and shows how they can be found. These are the points of concurrency of three triangle-related lines and may serve as further motivation to search for other points of concurrency.