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1992 Paper 2 Q10
D: 1600.0 B: 1529.8
complex numbers locus Argand diagram argument modulus circle inscribed angle theorem geometric proof cross-ratio

Let \(\alpha\) be a fixed angle, \(0 < \alpha \leqslant\frac{1}{2}\pi.\) In each of the following cases, sketch the locus of \(z\) in the Argand diagram (the complex plane):

  1. \({\displaystyle \arg\left(\frac{z-1}{z}\right)=\alpha,}\)
  2. \({\displaystyle \arg\left(\frac{z-1}{z}\right)=\alpha-\pi,}\)
  3. \(|\dfrac{z-1}{z}|=1.\)
Let \(z_{1},z_{2},z_{3}\) and \(z_{4}\) be four points lying (in that order) on a circle in the Argand diagram. If \[ w=\frac{(z_{1}-z_{2})(z_{3}-z_{4})}{(z_{4}-z_{1})(z_{2}-z_{3})} \] show, by considering \(\arg w\), that \(w\) is real.

Solution:

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TikZ diagram
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\begin{align*} \arg w &= \arg \frac{(z_{1}-z_{2})(z_{3}-z_{4})}{(z_{4}-z_{1})(z_{2}-z_{3})} \\ &= \arg \frac{(z_{1}-z_{2})(z_{3}-z_{4})}{(z_{2}-z_{3})(z_{4}-z_{1})} \\ &= \arg \frac{(z_{1}-z_{2})}{(z_{3}-z_{2})}\frac{(z_{3}-z_{4})}{(z_{1}-z_{4})} \\ &= \arg \frac{(z_{1}-z_{2})}{(z_{3}-z_{2})} + \arg \frac{(z_{3}-z_{4})}{(z_{1}-z_{4})}\\ &= \beta + \pi - \beta = \pi \end{align*} Therefore \(w\) is real

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1990 Paper 2 Q2
D: 1600.0 B: 1500.0
trigonometry product-to-sum formulas circle theorems Ptolemy's theorem inscribed angle theorem geometric proof sine rule cyclic quadrilateral

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 \[ PQ\times RS+QR\times PS=PR\times QS. \]

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.

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