HMMT February

Algebra Difficulty 4.7 AIME Find the answer

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

Your answer

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

Solution

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

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