## Solution to the Russian exam problem

Recall this Russian entrance exam problem… 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 $-1 < r < 1$.

Solution. I would have wanted to see what the prospective MGU students were supposed to do with it, as it seems to me there is some deep structure here I am missing!

We need first to think and then to compute, always a good idea. Define

$f(x,y)=x^2 -(9 -y)x + y^2- 9y + 15 =x^2+yx+y^2-9y-9x+15.$

Written in this second way it is clear that $f$ is symmetric: $f(x,y)=f(y,x)$. Now let us start substituting the conditions of the problem: we must have that either $f(s,r)=0$ or $f(t,r)=0$ (or both…).

Let us assume that $f(t,r)=0$ and that $f(s,r) \neq 0$ and see what follows. Then since $f(r,s)=f(s,r) \neq 0$, we must have that $f(t,s)=0$. So if we assume that $f(s,r) \neq 0$, we have that $f(t,r)=f(r,t)=0$ and $f(s,t)=f(t,s)=0$. However, it is easy to show that if $f(r,t)=0$ and $f(s,t)=0$, then $f(s,r)=0$, a contradiction to our assumption. [This is how you do it: write the two equations $f(s,t)=0$ and $f(r,t)=0$; subtract them and solve for $t$ in terms of $r$ and $s$ and then substitute this solution into, say, $f(r,t)=0$.]

So the conclusion is that if $f(r,t)=0$, we must have $f(r,s)=0$. Similarly, if we assumed that $f(s,r)=0$ but $f(t,r) \neq 0$, we would get a contradiction. Hence the only possibility is that $f(s,r)=0$ and $f(t,r)=0$ at the same time. In other words, $s$ and $t$ are just the two solutions of the quadratic equation $f(x,r)=0$.

So far so good. Now we need to remember that $r,s,t$ are real and that $r. Let us solve the quadratic $f(x,r)=0$. We have that

$s= -\dfrac{r}{2}+ \dfrac{9}{2}- \dfrac{\sqrt{21+18r-3r^2}}{2} \hbox{ and } t= -\dfrac{r}{2}+ \dfrac{9}{2} +\dfrac{\sqrt{21+18r-3r^2}}{2}$

since $t>s$. As both $s$ and $t$ must be real, the expression under the square root must be nonnegative. Note that if the expression under the square root is zero, $s=t$ which is not allowed, so this expression must be positive, which happens if $r \in (-1, 7)$. At the same time we must have that $r < s$, solving which inequality [Do this!], we get $r<1$ (if $r=1$ then $r=s$). Combining the two conditions, we get that $-1 as required.

I am puzzled by the following point: we had to manipulate $f(r,t)=0$ and $f(s,t)=0$ algebraically to show that these two equalities imply $f(r,s)=0$. Is there a way to see that this has to be the case without manipulation? In other words, what is the structure of the function $f(x,y)$ that forces this implication?

(MG)