62nd Czech and Slovak Mathematical Olympiad

Let $a$, $b$, $c$ be positive numbers. We search a solution of the system of equations in the set of positive reals. Due to $x+y+z=1$ the numbers $x$, $y$, $z$ are in the interval $(0,1)$. Substituting $z=1-x-y$ we obtain $$a(y-xy-y^2+x)=c(xy+1-x-y),b(x-x^2-xy+y)=c(xy+1-x-y),$$ these equations can be rewritten as $$ay(1-y)+ax(1-y)=c(1-x)(1-y),bx(1-x)+by(1-x)=c(1-x)(1-y).$$ Since $x<1$, $y<1$, we have $$ay+ax=c-cx,bx+by=c-cy.$$ From the previous two equations we obtain $$x+y=\frac{2c}{a+b+c}\text{and}x-y=\frac{(b-a)(x+y)}{c}=\frac{2(b-a)}{a+b+c};$$ then we get formulas for $x$, $y$ and finally by $z=1-x-y$ we find a formula for $z$ too: $$x=\frac{b+c-a}{a+b+c},y=\frac{c+a-b}{a+b+c},z=\frac{a+b-c}{a+b+c}(1)$$ The system has a solution in positive reals if and only if $b+c>a$, $c+a>b$, $a+b>c$ holds, which is equivalent to the existence of a triangle with sides $a$, $b$, $c$.

We need not check the values (1) due to equivalence of all rearrangements.

Other Solution:

Here is an easier way to obtain (1). Since $x+y+z=1$ we can rewrite the first part of the system as $$a(1-y)(1-z)=b(1-z)(1-x)=c(1-x)(1-y).(2)$$ Dividing $(1-x)(1-y)(1-z)$ (which is nonzero, even positive) we obtain the equivalent system $$\frac{a}{1-x}=\frac{b}{1-y}=\frac{c}{1-z}$$ If $s$ is the common (positive) value of the three previous fractions, we can easily get $$x=1-\frac{a}{s},y=1-\frac{b}{s},z=1-\frac{c}{s},(3)$$ which, substituting into the equation $x+y+z=1$, gives $$s=\frac{a+b+c}{2}$$ It follows from (3) that this $s$ yields the formulae (1), and then the proof of the problem statement.