# Chaitanya's Random Pages

## January 6, 2013

### Triangles on the sides of a triangle

Filed under: mathematics — ckrao @ 5:34 am

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 $A,B,C$ be three points in the plane. Construct an equilateral triangle $EAC$ and a 120-30-30 triangle $FAB$ (with $\angle AFB = 120^{\circ}$) both external to $ABC$. Let $D$ be the midpoint of $AB$. Prove $DEF$ 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 $\theta$ about a point $P$ is a composition of reflections in two lines, both through $P$, which have angle $\theta/2$ 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 $\ell$ followed by the line $m$, both through $P$.

Now consider the problem of a rotation (anticlockwise) of angle $\theta_1$ about point $P_1$ followed by a second rotation of angle $\theta_2$ about point $P_2$. We replace each rotations as a pair of reflections through $P_1$ and $P_2$ respectively. Since only the angle between the lines matters ($\theta_1/2$ and $\theta_2/2$ respectively), we may assume one of the lines for each rotation to pass through both $P_1$ and $P_2$. Hence we have a total of three lines, say $\ell, m, n$ shown below. Assume $m$ passes through $P_1P_2$.

Now the composition of two rotations can be expressed as the composition of four reflections: firstly in lines $\ell$ and $m$ (corresponding to the first rotation) followed by lines $m$ (again) and $n$. 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 $\ell$ followed by a reflection in line $n$. This is equivalent to a rotation in the angle twice that between the two lines $\ell$ and $m$, about the point $Q$ where they intersect. By a simple angle chase, this rotation angle is found to be $\theta_1 + \theta_2$.

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

• about $E$ with rotation angle 60 degrees
• about $F$ with rotation angle 120 degrees

Under the first rotation the point $C$ maps to $A$. Under the second the point $A$ maps to $B$. Hence the net effect is a rotation of (60 + 120) degrees that takes $C$ to $B$. This rotation can only occur about the midpoint of $BC$, in other words about $D$. The centre of rotation is also found by joining $E$ and $F$, forming lines that have angle 60/2 and 120/2 with $EF$, 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 $A, B, C$ in the plane construct isosceles triangles $DBC$, $ECA$, $FAB$ so that $\angle CDB = 2\alpha$, $AEC = 2\beta$, $BFA = 2\gamma$ where $\alpha + \beta + \gamma =180^{\circ}$ and $DB=DC, EC = EA, FA = FB$. Then triangle $DEF$ has angles independent of the positions of $A,B,C$ and in fact $\angle EDF = \alpha, \angle FED = \beta, \angle DFE = \gamma$.

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 $\alpha, \beta$ or $\gamma$ are between $-180^{\circ}$ and $0^{\circ}$. (An angle of $-\theta$ anticlockwise is simply equivalent to an angle of $360-\theta$ anticlockwise or $\theta$ clockwise.) Then the above result almost holds when $\alpha + \beta + \gamma =0^{\circ}$ or $-180^{\circ}$ though in these cases one or more of the angles $\angle EDF, \angle FED, \angle DFE$ may need to be adjusted by  $180^{\circ}$ to ensure they sum to $\pm 180^{\circ}$.

In the following figure we chose $\alpha = 60^{\circ}, \beta = \gamma = -30^{\circ}$, leading to $\angle EDF = 60-180=-120^{\circ}, \angle FED = \angle DFE = -30^{\circ}$ 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 $ABC$ 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  $DE=DF$ and $DE \perp DF$.

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

• $45^{\circ}$ about $C$, scale factor $AC/EC$ mapping $E$ to $A$
• $45^{\circ}$ about $B$, scale factor $FB/AB$ mapping $A$ to $F$

The composition is a new spiral similarity mapping $E$ to $F$. 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 $AC/EC \times FB/AB = 1$ (by the similarity of triangles $AEC$ and $AFB$) and the rotation angle will be $45^{\circ} + 45^{\circ} = 90^{\circ}$. If we can prove the centre of this composition spiral similarity is $D$ we are done since that would mean we now have a $90^{\circ}$ rotation mapping $E$ to $F$.

To do this we show that $D$ is fixed under the composite transformation. Let $G$ be the point making $\triangle BCG$ equilateral as shown below. Then by symmetry $\angle CGD = \angle GBD = 60 - 15 = 45^{\circ}$ and $\angle CGD = \angle BGD = 30^{\circ}$. Under the first spiral similarity $D$ will map to $G$ (since $\triangle DCG \sim \triangle ECA$ and under the second $G$ will map back to $D$ (since $\triangle GBD\sim \triangle ABF$). This proves that $D$ 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 $A, B, C, D, E, F$ be represented by complex numbers $a, b, c, d, e, f$.

• Let $z_1$ be the spiral similarity that maps $DC$ to $DB$.
• Let $z_2$ be the spiral similarity that maps $EA$ to $EC$.
• Let $z_3$ be the spiral similarity that maps $FB$ to $FA$.

We may then write $z_1 = (b-d)/(c-d)$ from which $d = c - (b-c)/(z_1 - 1) = c - w_1(b-c)$ where for convenience we let $w_i = 1/(z_i - 1)$ for $i = 1,2,3$. Similarly $e = a - w_2(c-a)$ and $f = b - w_3(a - b)$.

We would like the shape of $\triangle DEF$ to be independent of $ABC$. In complex numbers one way of expressing this is to say the ratio $(f-d)/(e-d)$ is independent of $a, b, c$. In other words there is some complex number $\alpha (w_1, w_2, w_3)$ independent of $a,b,c$ such that

$\displaystyle \alpha (w_1, w_2, w_3) := \frac{f-d}{e-d} = \frac{a(-w_3) + b(1 + w_3 + w_1) + c(-1 - w_1)}{a(1 + w_2) + b w_1 + c(-w_2 - 1 - w_1)}.$

Assuming non-zero denominators this leads to

$\displaystyle \alpha = \frac{-w_3}{1 + w_2} = \frac{1 + w_3 + w_1}{w_1} = \frac{1+w_1}{w_2 + 1 + w_1}.$

Equality of the first and third $w_i$ expressions implies

$\displaystyle w_3 = \frac{-(1 + w_2)(1 + w_1)}{1 + w_1 + w_2}.$

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

$\displaystyle 1 + \frac{1}{w_3} = \frac{-(1 + w_1 + w_2)}{(1 + w_1)(1 + w_2)} +1 = \frac{w_1w_2}{(1 + w_1)(1 + w_2)}.$

Recalling that $w_i = 1/(z_i - 1)$, this implies $(w_i + 1)/w_i = 1 + 1/w_i = z_i$, so the above may be written in the attractive form

$\displaystyle z_1 z_2 z_3 = 1.$

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

• Spiral similarity $z_3$ maps $DB$ to $DG$
• Spiral similarity $z_2$ maps $DG$ to $DC$
• Spiral similarity $z_1$ maps $DC$ to $DB$

Another way of interpreting the equality $z_1 z_2 z_3 = 1$ is that the sum of the apex angles $\angle CDB, \angle AEC, \angle BFA$ is an integer multiple of $360^{\circ}$ and

$\displaystyle \frac{DB}{DC} \cdot \frac{EC}{EA} \cdot \frac{FA}{FB} = 1.$

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 $BCG$. The triangles fit together in a second way to form the shape of $DEF$, 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 $z_1z_2 z_3 = 1$ result.

Let $P,Q,R,S$ be points in the plane and $A,B,C$ three other points. Construct points $D, E, F$ so that $\triangle BCD \sim \triangle SQR$, $\triangle ACE \sim \triangle SPR$ and $\triangle ABF \sim \triangle QPR$. Then $\triangle DEF$ has angles independent of $\triangle ABC$.

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

If similar triangles $ADB, CBE$ and $FAC$ are erected outwardly on the sides of any triangle $ABC$, and any three points $P$, $Q$ and $R$ are chosen so that they respectively lie in the same relative positions to these triangles, then $P$, $Q$ and $R$ 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.