Maths Olympiad Prep

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Problem 930

AMC 12 late, AIME early
Algebra Difficulty 4.7 Prove it Selected Problems from Open Contests · Estonia

Let aa, bb, cc be fixed real numbers, where 0≤a,b,c≤40 \le a, b, c \le 4. Prove that the system of equations
{p2−aq=−3q2−br=−4r2−cp=−5 \begin{cases} p^2 - a q = -3 \\ q^2 - b r = -4 \\ r^2 - c p = -5 \end{cases}
has no real solutions (pp, qq, rr).

This one wants a proof. Work it on paper, then read the official solution and mark yourself. Be honest about it: the record is only any use to you if it is.

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Official solution

Adding up all equations gives p2−cp+q2−aq+r2−br=−12p^2 - c p + q^2 - a q + r^2 - b r = -12. From the inequality (p−c2)2≥0(p - \frac{c}{2})^2 \ge 0 we have p2−cp≥−c24≥−4p^2 - c p \ge -\frac{c^2}{4} \ge -4 and similarly, q2−aq≥−4q^2 - a q \ge -4 and r2−br≥−4r^2 - b r \ge -4. Adding up these inequalities, we see that to avoid a contradiction with the equality derived first, all three inequalities must actually be equalities, i.e. a=b=c=4a = b = c = 4 and p=q=r=2p = q = r = 2. But this does not satisfy the initial equations.

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