I have always enjoyed geometry results that are simple to state and elegant in nature. Napoleon’s theorem certainly qualifies.

If equilateral triangles are erected externally on the sides of any triangle, their centres form an equilateral triangle.

Interestingly, this result is also true for three points on a straight line.

Here is a similar problem, this time with different triangles on two of the sides.

Let be three points in the plane. Construct an equilateral triangle and a 120-30-30 triangle (with ) both external to . Let be the midpoint of . Prove is a 90-30-60 triangle.

Both this and the previous result can be proved using transformation geometry. We use the beautiful result that a rotation is the composition of two reflections, and this result allows us to compose two rotations about two different points.

A rotation of angle about a point is a composition of reflections in two lines, both through , which have angle between them. The orientation of the two lines does not matter, only the relative angle between them. In the figure below the rotation is equivalent to a reflection in the line followed by the line , both through .

Now consider the problem of a rotation (anticlockwise) of angle about point followed by a second rotation of angle about point . We replace each rotations as a pair of reflections through and respectively. Since only the angle between the lines matters ( and respectively), we may assume one of the lines for each rotation to pass through both and . Hence we have a total of three lines, say shown below. Assume passes through .

Now the composition of two rotations can be expressed as the composition of four reflections: firstly in lines and (corresponding to the first rotation) followed by lines (again) and . Since two successive reflections in the same line are equivalent to the identity transformation (i.e. back to the original configuration), the net effect of these four reflections is a reflection in the line followed by a reflection in line . This is equivalent to a rotation in the angle twice that between the two lines and , about the point where they intersect. By a simple angle chase, this rotation angle is found to be .

Applying this result to our 90-30-60 problem above, we consider the following two rotations:

- about with rotation angle 60 degrees
- about with rotation angle 120 degrees

Under the first rotation the point maps to . Under the second the point maps to . Hence the net effect is a rotation of (60 + 120) degrees that takes to . This rotation can only occur about the midpoint of , in other words about . The centre of rotation is also found by joining and , forming lines that have angle 60/2 and 120/2 with , then finding their point of intersection. Overall a 30-60-90 triangle is formed and the centre of rotation is where the 90 degree angle occurs, as required.

One can try to prove Napoleon’s theorem in a similar way. The general result is as follows.

Given three points in the plane construct isosceles triangles , , so that , , where and . Then triangle has angles independent of the positions of and in fact .

The above results are special cases since 120 + 120 + 120 = 360 = 60 + 120 + 180. We can also allow for the case when any of the isosceles triangles are constructed internally, in which case the corresponding angles or are between and . (An angle of anticlockwise is simply equivalent to an angle of anticlockwise or clockwise.) Then the above result almost holds when or though in these cases one or more of the angles may need to be adjusted by to ensure they sum to .

In the following figure we chose , leading to as shown.

#### An example from the IMO

Now let us look at an even more general case when the triangles erected on the sides of are not necessarily isosceles. Consider the following example from the 1977 IMO. In the figure the triangle ABC is arbitrary and angles are as shown. We are required to show that and .

This time we can proceed by considering spiral similarities (dilations composed with rotations) about and . Here we consider the composition of the following two spiral similarities.

- about , scale factor mapping to
- about , scale factor mapping to

The composition is a new spiral similarity mapping to . We use the fact now that the composition of two spiral similarities is a spiral similarity with dilation factor given by the product of the two scale factors and rotation angle the sum of the two rotation angles. We know that the dilation factor of the composition factor will be (by the similarity of triangles and ) and the rotation angle will be . If we can prove the centre of this composition spiral similarity is we are done since that would mean we now have a rotation mapping to .

To do this we show that is fixed under the composite transformation. Let be the point making equilateral as shown below. Then by symmetry and . Under the first spiral similarity will map to (since and under the second will map back to (since ). This proves that is fixed under the composite transformation as required.

#### Generalisation

Let us seek now to generalise this result. What is the relationship between the angles of the three triangles external on the side of an arbitrary triangle so that the three new vertices form a triangle independent of the angles of our original triangle? It is convenient to express our spiral similarities in terms of complex numbers.

Referring to the diagram above let the points be represented by complex numbers .

- Let be the spiral similarity that maps to .
- Let be the spiral similarity that maps to .
- Let be the spiral similarity that maps to .

We may then write from which where for convenience we let for . Similarly and .

We would like the shape of to be independent of . In complex numbers one way of expressing this is to say the ratio is independent of . In other words there is some complex number independent of such that

Assuming non-zero denominators this leads to

Equality of the first and third expressions implies

(Note that the middle expression follows from the other two by addendo and so need not be used.) The above can be rearranged as

Recalling that , this implies , so the above may be written in the attractive form

This form is expected since it corresponds to the product of the three spiral similarities about the point in the above figure:

- Spiral similarity maps to
- Spiral similarity maps to
- Spiral similarity maps to

Another way of interpreting the equality is that the sum of the apex angles is an integer multiple of and

A consequence of this relationship is that the three triangles external to the original triangle may be scaled so that they “fit together” as seen above when forming the triangle . The triangles fit together in a second way to form the shape of , shown below.

The following shows how the triangles fit together in the -60,-60,120 case we considered earlier.

Note that in this case the triangles fit together to form a quadrilateral.

In general the angles of the new triangle are related to the original three triangles via the following figure (here the case of triangles erected externally is shown).

If we start with a configuration of any four points in the plane we may construct our triangles in the following way, rephrasing the result.

Let be points in the plane and three other points. Construct points so that , and . Then has angles independent of .

Another statement of this is the following found in [1].

If similar triangles and are erected outwardly on the sides of any triangle , and any three points , and are chosen so that they respectively lie in the same relative positions to these triangles, then , and form a triangle similar to the three triangles.

For me the beauty of these results comes from the fact that the angles of the triangles are not related through any obvious chasing of angles – additional insights are necessary.

#### Reference

[1] Curious triangle fact – posted in geometry.pre-college forum in Nov 1995.

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