# STEP III 2009 6 We have:

\displaystyle \left|e^{i \beta} - e^{i \alpha}\right| = \left|e^{i \left(\frac {\beta - \alpha} 2\right)}\right| \left|e^{i \left(\frac {\beta - \alpha} 2\right)} - e^{-i \left(\frac {\beta - \alpha} 2\right)}\right|

remember that:

\displaystyle e^{i \theta} - e^{-i \theta} = 2i \sin \theta

and \left|e^{i \theta}\right| = 1 when \theta is real, so:

\begin{align*}\left|e^{i \left(\frac {\beta - \alpha} 2\right)}\right| \left|e^{i \left(\frac {\beta - \alpha} 2\right)} - e^{-i \left(\frac {\beta - \alpha} 2\right)}\right| &= \left|2i \sin \frac {\beta - \alpha} 2\right| \\ & = 2 \left|\sin \frac {\beta - \alpha} 2\right|\end{align*}

note that since 0 < \alpha < \beta < 2 \pi, we have 0 < \dfrac {\beta - \alpha} 2 < \pi, so \displaystyle \sin \frac {\beta - \alpha} 2 > 0, and:

\displaystyle \left|e^{i \beta} - e^{i \alpha}\right| = 2 \left|\sin \frac {\beta - \alpha} 2\right| = 2 \sin \frac {\beta - \alpha} 2

as required.

The given equality is equivalent to:

\displaystyle 4 \sin \frac {\beta - \alpha} 2 \sin \frac {\delta - \gamma} 2 + 4 \sin \frac {\gamma - \beta} 2 \sin \frac {\delta - \alpha} 2 = 4 \sin \frac {\gamma - \alpha} 2 \sin \frac {\delta - \beta} 2

We use the product-to-sum formula:

\displaystyle 2 \sin a \sin b = \cos (a - b) - \cos (a + b)

(this was back when STEP had a formula booklet, if this question were in a modern exam, something equivalent to this would most likely be given)

To get:

\displaystyle 4 \sin \frac {\beta - \alpha} 2 \sin \frac {\delta - \gamma} 2 = 2 \left(\cos \frac {\beta - \alpha - \delta + \gamma} 2 - \cos \frac {\beta - \alpha + \delta - \gamma} 2\right)

\displaystyle 4 \sin \frac {\gamma - \beta} 2 \sin \frac {\delta - \alpha} 2 = 2 \left(\cos \frac {\gamma - \beta - \delta + \alpha} 2 - \cos \frac {\gamma - \beta + \delta - \alpha} 2\right)

\displaystyle 4 \sin \frac {\gamma - \alpha} 2 \sin \frac {\delta - \beta} 2 = 2 \left(\cos \frac {\gamma - \alpha - \delta + \beta} 2 - \cos \frac {\gamma - \alpha + \delta - \beta} 2\right)

Note now that:

\displaystyle \frac {\gamma - \beta - \delta + \alpha} 2 = -\frac {\beta - \alpha + \delta - \gamma} 2

so:

\displaystyle \cos \frac {\beta - \alpha + \delta - \gamma} 2 = \cos \frac {\gamma - \beta + \delta - \alpha} 2

so that:

\begin{align*}4 \sin \frac {\beta - \alpha} 2 \sin \frac {\delta - \gamma} 2 + 4 \sin \frac {\gamma - \beta} 2 \sin \frac {\delta - \alpha} 2 & = 2 \left(\cos \frac {\beta - \alpha - \delta + \gamma} 2 - \cancel{\cos \frac {\beta - \alpha + \delta - \gamma} 2}\right) \\ & + 2 \left(\cancel{\cos \frac {\gamma - \beta - \delta + \alpha} 2} - \cos \frac {\gamma - \beta + \delta - \alpha} 2\right) \\ & = 2 \left(\cos \frac {\beta - \alpha - \delta + \gamma} 2 - \cos \frac {\gamma - \beta + \delta - \alpha} 2\right) \\ & = 4 \sin \frac {\gamma - \alpha} 2 \sin \frac {\delta - \beta} 2\end{align*}

Hence the claim.

Note that in the complex plane, e^{i \alpha}, e^{i \beta}, e^{i \gamma}, e^{i \delta} are four distinct points in the complex plane, lying on the unit circle, so these are the vertices of some cyclic quadrilateral say ABCD. (since \alpha < \beta < \gamma < \delta, this is labelled counterclockwise)

Remember that the distance between two complex numbers z,w is given by |z - w|. So the \left|e^{i \alpha} - e^{i \beta}\right| corresponds to the distance between the points A and B, ie. |AB|, and so on.

So, this result is equivalent to the claim:

Let ABCD be a cyclic quadrilateral labelled in counterclockwise order. Then |AB| |CD| + |BC| |AD| = |AC| |BD|.

This is Ptolemy’s theorem. Or, in words:

The sums of products of opposite lengths in a cyclic quadrilateral is equal to the product of the lengths of its diagonals.

An example diagram with \alpha = 5^\circ, \beta = 78^\circ, \gamma = 123^\circ, \delta = 247^\circ: 1 Like     