1 problem found
The distinct points \(L,M,P\) and \(Q\) of the Argand diagram lie on a circle \(S\) centred on the origin and the corresponding complex numbers are \(l,m,p\) and \(q\). By considering the perpendicular bisectors of the chords, or otherwise, prove that the chord \(LM\) is perpendicular to the chord \(PQ\) if and only if \(lm+pq=0.\) Let \(A_{1},A_{2}\) and \(A_{3}\) be three distinct points on \(S\). For any given point \(A_{1}'\) on \(S\), the points \(A_{2}',A_{3}'\) and \(A_{1}''\) are chosen on \(S\) such that \(A_{1}'A_{2}',A_{2}'A_{3}'\) and \(A_{3}'A_{1}''\) are perpendicular to \(A_{1}A_{2},A_{2}A_{3}\) and \(A_{3}A_{1},\) respectively. Show that for exactly two positions of \(A_{1}',\) the points \(A_{1}'\) and \(A_{1}''\) coincide. If, instead, \(A_{1},A_{2},A_{3}\) and \(A_{4}\) are four given distinct points on \(S\) and, for any given point \(A_{1}',\) the points \(A_{2}',A_{3}',A_{4}'\) and \(A_{1}''\) are chosen on \(S\) such that \(A_{1}'A_{2}',A_{2}'A_{3}',A_{3}'A_{4}'\) and \(A_{4}'A_{1}''\) are respectively perpendicular to \(A_{1}A_{2},A_{2}A_{3},A_{3}A_{4}\) and \(A_{4}A_{1},\) show that \(A_{1}'\) coincides with \(A_{1}''.\) Give the corresponding result for \(n\) distinct points on \(S\).
Solution: The perpendicular bisector of the chords runs through the origin, therefore \(LM\) is perpendicular to \(PQ\) if and only if \(\frac{l+m}{2}\) is perpendicular to \(\frac{p+q}{2}\), ie \begin{align*} && (l+m) &= it (p+q) \\ \Leftrightarrow && \frac{l+m}{p+q} & \in i \mathbb{R} \\ \Leftrightarrow && 0 &= \frac{l+m}{p+q} + \frac{l^*+m^*}{p^*+q^*} \\ &&&= \frac{l+m}{p+q} + \frac{\frac{r^2}{l}+\frac{r^2}{m}}{\frac{r^2}{p}+\frac{r^2}{q}} \\ &&&=\frac{l+m}{p+q} + \frac{l+m}{p+q} \frac{pq}{lm} \\ &&&= \frac{l+m}{p+q} \left ( \frac{lm+pq}{lm} \right) \end{align*} Therefore as long as \(l+m, p+q \neq 0\) \(lm+pq = 0\) is equivalent to the chords being perpendicular. In the case where (say) \(l,m\) is a diameter, then the condition for the chords to be perpendicular is that \(p,q\) is also a diameter and at right angles, but clearly this is also equivalent to our condition. Suppose \(A_1, A_2, A_3\) are distinct points on \(S\), and \(A_1'\) is given and suppose \(a_i, a_i'\) are the corresponding complex numbers, then the conditions are: \begin{align*} A_1'A_2' \perp A_1A_2: && 0 &= a_1'a_2' + a_1a_2 \\ A_2'A_3' \perp A_2A_3: && 0 &= a_2'a_3' + a_2a_3 \\ A_3'A_1'' \perp A_3A_1: && 0 &= a_3'a_1'' + a_3a_1 \\ \\ \Rightarrow && a_2' &= -\frac{a_1a_2}{a_1'} \\ && a_3' &= -\frac{a_2a_3}{a_2'} \\ &&&= \frac{a_1'a_2a_3}{a_1a_2} \\ &&&= \frac{a_1'a_3}{a_1} \\ && a_1'' &= - \frac{a_3a_1}{a_3'} \\ &&&= \frac{a_3a_1a_1}{a_1'a_3} \\ &&&= \frac{a_1^2}{a_1'} \\ \Rightarrow && a_1'a_1'' &= a_1^2 \end{align*} Therefore \(a_1' = a_1''\) if \(a_1' = \pm a_1\) Suppose we have \(4\) points, then \begin{align*} A_1'A_2' \perp A_1A_2: && 0 &= a_1'a_2' + a_1a_2 \\ A_2'A_3' \perp A_2A_3: && 0 &= a_2'a_3' + a_2a_3 \\ A_3'A_4' \perp A_3A_4: && 0 &= a_3'a_4' + a_3a_4 \\ A_4'A_1'' \perp A_4A_1: && 0 &= a_4'a_1'' + a_4a_1 \\ \\ \Rightarrow && a_4' &= -\frac{a_3a_4}{a_3'} \\ &&&= -\frac{a_1a_3a_4}{a_1'a_3} \\ &&&= -\frac{a_1a_4}{a_1'} \\ \Rightarrow && a_1'' &= -\frac{a_4a_1}{a_4'} \\ &&&= \frac{a_4a_1a_1'}{a_1a_4} \\ &&&= a_1' \end{align*} So they coincide. For \(n\) points if there are an even number of points they coincide, an odd number and there are two points when they coincide.