Az Eszterházy Károly Tanárképző Főiskola Tudományos Közleményei. 2004. Sectio Mathematicae. (Acta Academiae Paedagogicae Agriensis : Nova series ; Tom. 31)
SASHALMI, É. and HOFFMANN, M., Generalizations of Bottema's theorem on pedal points
Generalizations of Bot tenia's theorem on pedal points 27 Corollary. Let A\B\C\ be the triangle bounded by the lines containing the sides of the rectangles opposite to A Pi , Li and CP. 3. Similarly let A2B2C2 be the triangle bounded by the lines containing the sides of the rectangles opposite to PiB, P2C and P3A. These two triangles are each homothetic to ABC and the ratio of homothety is A = 1 + pcotuj. Back to the original situation, building the squares to the inner side of the segments of the side of the triangle, Theorem 1 naturally remains valid (see Fig. 1). The ratio of the homotethy, however will be changed as follows. Figure 1. Theorem 4. Consider the triangle ABC and one of its inner points P. Let the pedals of P on the sides AB, BC,CA be P 1, Pi and P3, respectively. If we build squares onto the inner side of the segments of the sides defined by the pedals, as in Tig.l., then the ratio of the homothety between the triangle ABC and AiBiCj as well as between ABC and A2B2C2 is X = cot, oj — 1. Proof. Denote the center of homothety between ABC and A\B\C\ by 0\ and the segments BP2, CP3, AP\ by a/, 6; and q. Let the distances of the sides BC,CA, AB from 0 1 be f,g,h , respectively. Obviously the distances of the sides B\C\, C\A\, A1B1 from 0 1 are (at — /), (6/ — g) and (c/ — //). Due to the homothety / : g : h = (o./ — /) : (6/ — g) : (c; — h) holds. From equation (2) af + bf 4- cf = (a - a t) 2 + (b - b t) 2 + (c - a) 2. Applying equation (1) this can be written as a 2 + b 2 + c 2 aai + bbi -f cc/ — = 'It cotu), where t is the area of the triangle ABC. Summarizing the area of the subtriangles OyBC, ()t AC and O xAB we find af + bg + ch = 21,
