Maths Olympiad Prep

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Algebra Difficulty 5.9 AIME, harder Prove it Japan

For positive real numbers x,yx, y, the positive real number xyx \star y is defined as xy=xxy+1x \star y = \frac{x}{xy+1}.
Calculate the following expression:
((((((10099)98)97))3)2)1. (((\cdots (((100 \star 99) \star 98) \star 97) \star \cdots) \star 3) \star 2) \star 1.

Solution

100495001\frac{100}{495001}

For any positive real numbers x,yx, y, and zz, the following equation holds:
(xy)z=xy(xy)z+1=xxy+1xxy+1z+1=xxy+xz+1=xx(y+z)+1=x(y+z). (x \star y) \star z = \frac{x \star y}{(x \star y)z + 1} = \frac{\frac{x}{xy+1}}{\frac{x}{xy+1}z + 1} = \frac{x}{xy + xz + 1} = \frac{x}{x(y+z) + 1} = x \star (y+z).
Using this repeatedly, we have:
((((((10099)98)97)96)3)2)1=((((((100(99+98))97)96)3)2)1=((((100(99+98+97))96)3)2)1==100(99+98+97++3+2+1)=100991002. \begin{aligned} & ((\cdots((((100 \star 99) \star 98) \star 97) \star 96) \star \cdots \star 3) \star 2) \star 1 \\ &= ((\cdots((((100 \star (99 + 98)) \star 97) \star 96) \star \cdots \star 3) \star 2) \star 1 \\ &= ((\cdots((100 \star (99 + 98 + 97)) \star 96) \star \cdots \star 3) \star 2) \star 1 \\ &= \cdots \\ &= 100 \star (99 + 98 + 97 + \cdots + 3 + 2 + 1) \\ &= 100 \star \frac{99 \cdot 100}{2}. \end{aligned}
Therefore, the answer is 1004950=100495001. \text{Therefore, the answer is } 100 \star 4950 = \frac{100}{495001}.

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