Problems

Show that $\big\vert \e^{\i\beta} -\e^{\i\alpha}\big\vert = 2\sin\frac12 (\beta-\alpha)\,\( for \)0<\alpha<\beta<2\pi\,$. Hence show that \[ \big\vert \e^{\i\alpha} -\e^{\i\beta}\big\vert \; \big\vert \e^{\i\gamma} -\e^{\i\delta}\big\vert + \big\vert \e^{\i\beta} -\e^{\i\gamma}\big\vert \; \big\vert \e^{\i\alpha} -\e^{\i\delta}\big\vert = \big\vert \e^{\i\alpha} -\e^{\i\gamma}\big\vert \; \big\vert \e^{\i\beta} -\e^{\i\delta}\big\vert \,, \] where \(0<\alpha<\beta<\gamma<\delta<2\pi\). Interpret this result as a theorem about cyclic quadrilaterals.

Four complex numbers \(u_1\), \(u_2\), \(u_3\) and \(u_4\) have unit modulus, and arguments \(\theta_1\), \(\theta_2\), \(\theta_3\) and \(\theta_4\), respectively, with \(-\pi < \theta_1 < \theta_2 < \theta_3 < \theta_4 < \pi\). Show that \[ \arg \l u_1 - u_2 \r = \tfrac{1}{2} \l \theta_1 + \theta_2 -\pi \r + 2n\pi \] where \(n = 0 \hspace{4 pt} \mbox{or} \hspace{4 pt} 1\,\). Deduce that \[ \arg \l \l u_1 - u_2 \r \l u_4 - u_3 \r \r = \arg \l \l u_1 - u_4 \r \l u_3 - u_2 \r \r + 2n\pi \] for some integer \(n\). Prove that \[ | \l u_1 - u_2 \r \l u_4 - u_3 \r | + | \l u_1 - u_4 \r \l u_3 - u_2 \r | = | \l u_1 - u_3 \r \l u_4 - u_2 \r |\;. \]

Prove that if \(A+B+C+D=\pi,\) then \[ \sin\left(A+B\right)\sin\left(A+D\right)-\sin B\sin D=\sin A\sin C. \] The points \(P,Q,R\) and \(S\) lie, in that order, on a circle of centre \(O\). Prove that \[ PQ\times RS+QR\times PS=PR\times QS. \]

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Solution: \begin{align*} \sin(A+B)\sin(A+D) - \sin B \sin D &= \sin (A+B)\sin(\pi - B-C) - \sin B \sin (\pi - A - B - C) \\ &= \sin (A+B)\sin(B+C) - \sin B \sin(A+B+C) \\ &= \sin(A+B)\sin (B+C) - \sin B (\sin (A+B)\cos C +\cos(A+B) \sin C) \\ &= \sin(A+B)\cos B \sin C + \cos(A+B)\sin B \sin C \\ &= \sin A \sin C \cos^2 B + \cos A \sin B \cos B \sin C - \cos A \cos B \sin B \sin C + \sin A \sin^2 B \sin C \\ &= \sin A \sin C (\cos^2 B + \sin^2 B) \\ &= \sin A \sin C \end{align*}

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Using the extended form of the sine rule: \(\frac{a}{\sin A} = \frac{b}{\sin B} = \frac{c}{\sin C} = 2R\) where \(R\) is the circumradius, we have \begin{align*} PR \times QS &= 2R \sin (A+D) \times 2R \sin (A+B) \\ &= 4R^2 \l \sin A \sin C + \sin B \sin D \r \\ &= 2R \sin A \times 2R \sin C + 2R \sin B 2R \sin D \\ &= PS \times QR + PQ \times RS \end{align*} as required.