55rd Ukrainian National Mathematical Olympiad - Fourth Round

Answer. $x=y=a,z=b$ and $x=y=b,z=a$.

Let $(x,y,z)$ be a solution of the system. Consider polynomials $$\begin{array}{rl}P(t)& =(t-x{)}^3(t-z)=t^4+p_1{t}^3+q_1{t}^2+r_{1}t+s_1\text{ and}\\ Q(t)& =(t-y{)}^2(t-a)(t-b)=t^4+p_2{t}^3+q_2{t}^2+r_{2}t+s_2,\end{array}$$ and let: $$f(t)=P(t)-Q(t)=(p_1-p_2)t^3+(q_1-q_2)t^2+(r_1-r_2)t+(s_1-s_2).$$ Numbers $x,x,x,z$ are roots of $P(t)$ and $y,y,a,b$ are roots of $Q(t)$. Using Vieta's formulas, we get: $$\begin{array}{rl}{p}_1& =-(3x+z)=-(2y+(a+b))=p_2\\ q_1& =3x^2+3xz=y^2+2(a+b)y+ab=q_2\\ r_1& =-(x^3+3x^{2}z)=-(y^2(a+b)+2yab)=r_2\end{array}$$

$$P^{\prime }(t)=(t-x{)}^2(4t-3z-x)=Q^{\prime }(t)=(t-y)(4t^2-(3a+3b+2y)t+2ab+y(a+b)).$$ Consider $R(t)=4t^2-(3a+3b+2y)t+2ab+y(a+b)$ and take $t_0=\frac{1}{2}(a+b)$. We obtain $$R(t_0)=(a+b{)}^2-3(a+b)\frac{a+b}{2}-2y\frac{a+b}{2}+2ab+y(a+b)=-\frac{(a+b{)}^2}{2}+2ab=-\frac{(a-b{)}^2}{2}<0.$$ Thus $R(t)$ has two different roots and thus the discriminant of $R(t)$ is positive. But we have $P^{\prime }(t)\equiv Q^{\prime }(t)$ and $P^{\prime }(t)\ne (t-x{)}^2$. Then $Q^{\prime }(t)\ne (t-x{)}^2$. Since $R(t)$ has two different roots, we get that exactly one of them equals to $x$ and $y=x$. Moreover, $R(y)=0$, that is $$R(y)=4y^2-(3a+3b+2y)y+2ab+y(a+b)=2(y^2-(a+b)y+ab)=0.$$

But this is possible only if $y=a$ or $y=b$. Finally, we obtain $$f(t)=P(t)-Q(t)=(t-x{)}^3(z-c)=C,$$ where $c=a$ or $c=b$. But this is possible only if $z=c$. Therefore $P$ equals to $Q$ and we have two solutions $$x=y=a,z=b,\text{ or }x=y=b,z=a.$$ The second part also could be done with application of Rolle's theorem. Consider polynomial $Q(t)=(t-y{)}^2(t-a)(t-b)$. Suppose that $y,a,b$ are pairwise distinct. Then derivative $Q^{\prime }(t)$ has three distinct roots. One of them $t=y$ and two other belong to the intervals between $y,a,b$. But $P^{\prime }(t)$ equals to $Q^{\prime }(t)$ and thus $P^{\prime }(t)$ also has three distinct roots. But $P^{\prime }(t):(t-x{)}^2$. Contradiction. And thus some of the numbers $y,a,b$ are equal. Since $a\ne b$, it follows that $y=a$ or $y=b$. Consider the case $y=a$. Then $Q(t)=(t-y{)}^3(t-b)$ and $Q^{\prime }(t):(t-x{)}^2$. Using that $Q^{\prime }(t)=P^{\prime }(t):(t-x{)}^2$, we get $x=y$, and thus: $$C=(t-x{)}^3(t-z)-(t-x{)}^3(t-b)=(t-x{)}^3(z-b),$$ that is only possible if $z=b$. And we obtain the solution $x=y=a,z=b$. Similarly, in the second case we get the solution: $x=y=b,z=a$.

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Alternative solution.

Alternative solution. From the first and second equations we get: $$\{\begin{array}{l}a+b=3x+z-2y\\ ab=3x^2+3y^2-6xy+3xz-2yz.\end{array}(1)$$