Problem
GEO-B3-M04-P012 Intersection on the Nine-Point Circle
B. New Original Problem. Points \(P\) and \(Q\) are antipodal on the circumcircle of \(ABC\). Their Simson lines meet at \(X\). Prove that \(X\) lies on the nine-point circle of triangle \(ABC\).
C. Hint 1. Each Simson line passes through the midpoint of the corresponding segment with the orthocenter.
D. Hint 2. For antipodal points, these two midpoints lie on the nine-point circle.
E. Full Solution.
Let \(H\) be the orthocenter, and let \(M,N\) be the midpoints of \(PH\) and \(QH\). By the Simson line property, the Simson line of \(P\) passes through \(M\), and the Simson line of \(Q\) passes through \(N\). The points \(M,N\) lie on the nine-point circle, because this circle is the image of the circumcircle under the homothety with center \(H\) and ratio \(\frac12\).
By the previous problem, the Simson lines are perpendicular. In the standard nine-point circle configuration, the angle \(MXN\) subtends a diameter \(MN\) of this circle, so \(X\) lies on it. Equivalently, check by angles: \(\angle MXN=90^\circ\), and \(MN\) is the corresponding diameter.
The difficulty may be increased by asking students to prove that \(MN\) is the needed diameter.