Problem
GEO-B3-M04-P024 Complex Check of Direction
B. New Original Problem. Let points \(A,B,C,P\) lie on the unit circle of the complex plane and have complex coordinates \(a,b,c,p\). Prove that the direction of the Simson line of \(P\) with respect to \(ABC\) can be expressed by a number proportional to \((p-a)(p-b)(p-c)/p\), and use this to explain why antipodal points give perpendicular Simson lines.
C. Hint 1. Write the formula for the projection of a point onto a line through two points of the unit circle.
D. Hint 2. For the antipodal point \(q=-p\), compare the direction with the original one.
E. Full Solution.
In the complex plane, for the line through two unit-circle points \(u,v\), the foot of the perpendicular from \(p\) can be written as a linear expression in \(p,u,v\) and their conjugates. Since \(\overline{u}=1/u\), \(\overline{v}=1/v\), the formula simplifies to a rational expression.
Subtracting the coordinates of two perpendicular feet, for example on \(CA\) and \(AB\), gives a direction number for the Simson line. After cancellation, it is proportional to \((p-a)(p-b)(p-c)/p\).
If \(p\) is replaced by the antipodal point \(q=-p\), the direction number is multiplied by a purely imaginary factor. This means a rotation of direction by \(90^\circ\). Hence the Simson lines of antipodal points are perpendicular.
This is for strong students: it shows how complex numbers compress a long angle-chasing technique.