Problem
GEO-B3-M05-P002 First Six Points of the Nine-Point Circle
In an acute triangle \(ABC\) with orthocenter \(H\), prove that the side midpoints and the midpoints of \(AH,BH,CH\) lie on one circle.
C. Hint 1. Let \(O\) be the circumcenter and \(N\) the midpoint of \(OH\).
D. Hint 2. Use the homothety centered at \(H\) with ratio \(\frac12\), then check the side midpoints.
The homothety centered at \(H\) with ratio \(\frac12\) sends the circumcircle of \(ABC\) to the circle centered at \(N\), the midpoint of \(OH\), with radius \(\frac R2\). The points \(A,B,C\) go to the midpoints of \(AH,BH,CH\), so these three points lie on the new circle.
Let \(M_a\) be the midpoint of \(BC\). In the vector model from the previous problem, \(\vec n=\frac{\vec a+\vec b+\vec c}{2}\), \(\vec m_a=\frac{\vec b+\vec c}{2}\). Thus \(|\vec m_a-\vec n|=\frac{|\vec a|}{2}=\frac R2\). The other two side midpoints are identical. Hence all six points lie on one circle.
After this, the altitude feet can be added to obtain the full nine-point circle.