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<s<t. 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<r<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?


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