Problem
GEO-B3-M05-P007 Lemoine Point in a Right Triangle
In right triangle \(ABC\) with right angle at \(C\), let \(H\) be the foot of the altitude from \(C\) to \(AB\). Prove that the Lemoine point \(K\) is the midpoint of \(CH\).
C. Hint 1. Put \(C\) at the origin and the legs on the axes.
D. Hint 2. Use barycentric coordinates \(K=(a^2:b^2:c^2)\).
Let \(A=(u,0)\), \(B=(0,v)\), \(C=(0,0)\). Then \(a=BC=v\), \(b=CA=u\), \(c^2=AB^2=u^2+v^2\). The Lemoine point has barycentric coordinates \((a^2:b^2:c^2)\), hence its Cartesian coordinates are \(\left(\frac{uv^2}{2(u^2+v^2)},\frac{u^2v}{2(u^2+v^2)}\right)\).
The foot \(H\) of the altitude from \(C\) to \(AB\) has coordinates \(\left(\frac{uv^2}{u^2+v^2},\frac{u^2v}{u^2+v^2}\right)\). Thus the coordinates of \(K\) are half those of \(H\), so \(K\) is the midpoint of \(CH\).
A good problem for translating symmedian geometry into coordinates.