Problems

The four distinct points \(P_i\) (\(i=1\), \(2\), \(3\), \(4\)) are the vertices, labelled anticlockwise, of a cyclic quadrilateral. The lines \(P_1P_3\) and \(P_2P_4\) intersect at \(Q\).

1. By considering the triangles \(P_1QP_4\) and \(P_2QP_3\) show that \((P_1Q)( QP_3) = (P_2Q) (QP_4)\,\).
2. Let \(\+p_i\) be the position vector of the point \(P_i\) (\(i=1\), \(2\), \(3\), \(4\)). Show that there exist numbers \(a_i\), not all zero, such that \begin{equation} \sum\limits_{i=1}^4 a_i =0 \qquad\text{and}\qquad \sum\limits_{i=1}^4 a_i \+p_i ={\bf 0} \,. \tag{\(*\)} \end{equation}
3. Let \(a_i\) (\(i=1\),~\(2\), \(3\),~\(4\)) be any numbers, not all zero, that satisfy~\((*)\). Show that \(a_1+a_3\ne 0\) and that the lines \(P_1P_3\) and \(P_2P_4\) intersect at the point with position vector \[ \frac{a_1 \+p_1 + a_3 \+p_3}{a_1+a_3} \,. \] Deduce that \(a_1a_3 (P_1P_3)^2 = a_2a_4 (P_2P_4)^2\,\).

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*}

+ - ↺

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.