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Algebra Difficulty 4.7 AIME Find the answer

Suppose xx and yy are real numbers such that 1<x<y<1-1<x<y<1. Let GG be the sum of the geometric series whose first term is xx and whose ratio is yy, and let GG^{\prime} be the sum of the geometric series whose first term is yy and ratio is xx. If G=GG=G^{\prime}, find x+yx+y.

A number or a short expression. Fractions can be typed as 3/2, and spacing doesn't matter.

Solution

We note that G=x/(1y)G=x /(1-y) and G=y/(1x)G^{\prime}=y /(1-x). Setting them equal gives x/(1y)=x /(1-y)= y/(1x)x2x=y2x(x+y1)(xy)=0y /(1-x) \Rightarrow x^{2}-x=y^{2}-x \Rightarrow(x+y-1)(x-y)=0, so we get that x+y1=0x+y=1x+y-1=0 \Rightarrow x+y=1.

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