Problem
GEO-B3-M05-P017 Pedal Triangle of the Lemoine Point
Let \(A_1,B_1,C_1\) be the projections of the Lemoine point \(K\) of triangle \(ABC\) onto \(BC,CA,AB\). Prove that \(K\) is the centroid of triangle \(A_1B_1C_1\).
C. Hint 1. Use the distances from \(K\) to the sides: they are proportional to \(a:b:c\).
D. Hint 2. Decompose \(\overrightarrow{KA_1},\overrightarrow{KB_1},\overrightarrow{KC_1}\) along normals to the sides.
Let \(\vec n_a,\vec n_b,\vec n_c\) be unit outward normals to \(BC,CA,AB\). Since \(K\) is the Lemoine point, its distances to the sides are proportional to \(a:b:c\). Hence \(\overrightarrow{KA_1}=-\lambda a\vec n_a\), \(\overrightarrow{KB_1}=-\lambda b\vec n_b\), \(\overrightarrow{KC_1}=-\lambda c\vec n_c\) for some \(\lambda\).
For any triangle, \(a\vec n_a+b\vec n_b+c\vec n_c=\vec0\): the sum of outward normal vectors weighted by side lengths is zero. Therefore \(\overrightarrow{KA_1}+\overrightarrow{KB_1}+\overrightarrow{KC_1}=\vec0\). This is exactly the condition that \(K\) is the centroid of triangle \(A_1B_1C_1\).
A strong problem connecting Lemoine with pedal geometry from the previous module.