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Annual Harvard-MIT Mathematics Tournament

Algebra Difficulty 5.1 AIME, harder Prove it United States

Problem:

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 $G'$ be the sum of the geometric series whose first term is $y$ and ratio is $x$ . If $G = G'$ , find $x + y$ .

Solution

Solution:

We note that $G = \dfrac{x}{1 - y}$ and $G' = \dfrac{y}{1 - x}$ . Setting them equal gives
$\frac{x}{1 - y} = \frac{y}{1 - x}$
Cross-multiplying, we get
$x(1 - x) = y(1 - y)$
Expanding both sides:
$x - x^2 = y - y^2$
Bringing all terms to one side:
$x - x^2 - y + y^2 = 0$
$(x - y) + (y^2 - x^2) = 0$
$(x - y) + (y - x)(y + x) = 0$
$(x - y) - (x - y)(x + y) = 0$
$(x - y)[1 - (x + y)] = 0$
Since $x < y$ , $x - y \neq 0$ , so $1 - (x + y) = 0$ , i.e., $x + y = 1$ .

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