## A Russian exam problem for the winter break

This problem is taken from Tanya Khovanova’s blog. It is a problem (the last of five; I recommend all of them) from the 1976 entrance exam to the Mathematics Department of the premier USSR University, MGU (Moscow State University). It is one of those problems that you need to stare at for a while, then it surrenders.

Let $r < s < t$ be real numbers. If you set $y$ equal to any of the numbers $r$, $s$ or $t$ in the equation $x^2 - (9 - y)x + y^2 - 9y + 15 = 0$, then at least one of the other two numbers will be a root of the resulting quadratic equation. Prove that $-1 < r < 1$.

(MG)

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