MathDB

3

Part of 1998 North Macedonia National Olympiad

Problems(1)

\sum_{k=0}^{n-1}f(x+k,y+k,z+k) = n^3f(x/n,y/n,z/n) , ratio of triangle areas

Source: North Macedonian Mathematical Olympiad 1998 p3

2/12/2020
A triangle ABCABC is given. For every positive numbers p,q,rp,q,r, let A′,B′,C′A',B',C' be the points such that BA′→=pAB→,CB′→=qBC→\overrightarrow{BA'} = p\overrightarrow{AB}, \overrightarrow{CB'}=q\overrightarrow{BC} , and AC′→=rCA→\overrightarrow{AC'}=r\overrightarrow{CA}. Define f(p,q,r)f(p,q,r) as the ratio of the area of △A′B′C′\vartriangle A'B'C' to that of △ABC\vartriangle ABC. Prove that for all positive numbers x,y,zx,y,z and every positive integer nn, ∑k=0n−1f(x+k,y+k,z+k)=n3f(xn,yn,zn)\sum_{k=0}^{n-1}f(x+k,y+k,z+k) = n^3f\left(\frac{x}{n},\frac{y}{n},\frac{z}{n}\right).
ratiotriangle areaarea