Problem
GEO-B2-M10-P022 The Orthocenter as Radical Centre
In an acute triangle \(ABC\), circles with diameters \(AB\), \(BC\), \(CA\) are drawn. Prove that their radical centre is the orthocenter of triangle \(ABC\).
Prove that the radical axis of the circles with diameters \(AB\) and \(AC\) is the altitude from \(A\).
For a point \(X\), the power with respect to the circle with diameter \(AB\) is \(\overrightarrow{XA}\cdot\overrightarrow{XB}\), and with respect to the circle with diameter \(AC\) it is \(\overrightarrow{XA}\cdot\overrightarrow{XC}\). Equality of powers gives \(\overrightarrow{XA}\cdot(\overrightarrow{XB}-\overrightarrow{XC})=0\), that is \(XA\perp BC\). Hence the radical axis of these two circles is the altitude from \(A\). Similarly, the other two radical axes are the altitudes from \(B\) and \(C\). Their common point is the orthocenter.
The problem connects circles, dot product, and radical axes.