Maths Olympiad Prep

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Algebra Difficulty 5.7 AIME, harder Prove it Russia

On a checkered paper, numbers xx, yy and zz are placed in 3 squares (one number in each), other squares are empty. It is permitted to perform operations of two types:

1) To choose 2 numbers aa and bb from 2 non-empty squares and to place the number a+ba + b in some empty square;

2) To choose 3 numbers aa, bb, and cc from 3 non-empty squares and to place the number ab+c2ab + c^2 in some empty square.

Prove that using these operations one can place a number (x+y+z)2(x + y + z)^2 in some square.

Solution

Пусть исходные числа x,y,zx, y, z. Получим сначала числа x+y,y+z,z+xx+y, y+z, z+x. Затем из клеток x,y+z,yx, y+z, y получим x(y+z)+y2x(y+z) + y^2; аналогично, получим числа y(z+x)+z2y(z+x) + z^2 и z(x+y)+x2z(x+y) + x^2. Сложив их, получаем требуемое: x(y+z)+y2+y(z+x)+z2+z(x+y)+x2=x2+y2+z2+2xy+2yz+2zx=(x+y+z)2x(y+z) + y^2 + y(z+x) + z^2 + z(x+y) + x^2 = x^2 + y^2 + z^2 + 2xy + 2yz + 2zx = (x+y+z)^2.

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