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Algebra Difficulty 6.3 National olympiad Find the answer

Find all quadruples (x1,x2,x3,x4)\left(x_{1}, x_{2}, x_{3}, x_{4}\right) of real numbers such that the following six equalities hold:

x1+x2=x32+x42+6x3x4,x1+x3=x22+x42+6x2x4,x1+x4=x22+x32+6x2x3,x2+x3=x12+x42+6x1x4,x2+x4=x12+x32+6x1x3,x3+x4=x12+x22+6x1x2. \begin{aligned} & x_{1}+x_{2}=x_{3}^{2}+x_{4}^{2}+6 x_{3} x_{4}, \\ & x_{1}+x_{3}=x_{2}^{2}+x_{4}^{2}+6 x_{2} x_{4}, \\ & x_{1}+x_{4}=x_{2}^{2}+x_{3}^{2}+6 x_{2} x_{3}, \\ & x_{2}+x_{3}=x_{1}^{2}+x_{4}^{2}+6 x_{1} x_{4}, \\ & x_{2}+x_{4}=x_{1}^{2}+x_{3}^{2}+6 x_{1} x_{3}, \\ & x_{3}+x_{4}=x_{1}^{2}+x_{2}^{2}+6 x_{1} x_{2} . \end{aligned}

Solution

Subtracting the first two equations from each other gives x2x3=x32x22+6x4(x3x2)x_{2}-x_{3}=x_{3}^{2}-x_{2}^{2}+6 x_{4}(x_{3}-x_{2}), which we can factor as 0=(x3x2)(x3+x2+1+6x4)0=(x_{3}-x_{2})(x_{3}+x_{2}+1+6 x_{4}). We see that x2=x3x_{2}=x_{3} or x2+x3+1+6x4=0x_{2}+x_{3}+1+6 x_{4}=0. Similarly, it also holds that x2=x3x_{2}=x_{3} or x2+x3+1+6x1=0x_{2}+x_{3}+1+6 x_{1}=0. Therefore, if x2x3x_{2} \neq x_{3}, then in both cases the second equality holds; subtracting these from each other gives x1=x4x_{1}=x_{4}. We conclude that x2=x3x_{2}=x_{3} or x1=x4x_{1}=x_{4}. Similarly, for any permutation (i,j,k,l)(i, j, k, l) of (1,2,3,4)(1,2,3,4), it holds that xi=xjx_{i}=x_{j} or xk=xlx_{k}=x_{l}.

We now prove that at least three of the xix_{i} are equal. If they are all equal, this is obviously true. Otherwise, there are two that are unequal, say without loss of generality x1x2x_{1} \neq x_{2}. Then x3=x4x_{3}=x_{4}. If now x1=x3x_{1}=x_{3}, then there are three elements equal. Otherwise, x1x3x_{1} \neq x_{3}, so x2=x4x_{2}=x_{4} and again there are three elements equal. The quadruple (x1,x2,x3,x4)(x_{1}, x_{2}, x_{3}, x_{4}) is thus, up to order, of the form (x,x,x,y)(x, x, x, y), where it may be that x=yx=y.

Substituting this into the given equations gives x+y=8x2x+y=8 x^{2} and 2x=x2+y2+6xy2 x=x^{2}+y^{2}+6 x y. Adding these two equations: 3x+y=9x2+y2+6xy3 x+y=9 x^{2}+y^{2}+6 x y. The right-hand side can be factored as (3x+y)2(3 x+y)^{2}. With s=3x+ys=3 x+y we thus have s=s2s=s^{2}, from which it follows that s=0s=0 or s=1s=1. It holds that s=3x+y=2x+(x+y)=2x+8x2s=3 x+y=2 x+(x+y)=2 x+8 x^{2}. Therefore, 8x2+2x=08 x^{2}+2 x=0 or 8x2+2x=18 x^{2}+2 x=1.

In the first case, x=0x=0 or x=14x=-\frac{1}{4}. We find y=03x=0y=0-3 x=0 respectively y=03x=34y=0-3 x=\frac{3}{4}. In the second case, we can factor the equation as (4x1)(2x+1)=0(4 x-1)(2 x+1)=0, so x=14x=\frac{1}{4} or x=12x=-\frac{1}{2}. We find y=13x=14y=1-3 x=\frac{1}{4} respectively y=13x=52y=1-3 x=\frac{5}{2}.

All together, we have found the following quadruples: (0,0,0,0),(14,14,14,34)(0,0,0,0),\left(-\frac{1}{4},-\frac{1}{4},-\frac{1}{4}, \frac{3}{4}\right) and the permutations thereof, (14,14,14,14)\left(\frac{1}{4}, \frac{1}{4}, \frac{1}{4}, \frac{1}{4}\right) and (12,12,12,52)\left(-\frac{1}{2},-\frac{1}{2},-\frac{1}{2}, \frac{5}{2}\right) and the permutations thereof. Checking shows that all these quadruples satisfy the equations.

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Source: NuminaMath-1.5, licensed Apache-2.0. Statement and solution reproduced as published; topic and difficulty added by this site.