2021 OIM Problems/Problem 4

Problem

Let $a,b,c,x,y,z$ be real numbers such that

$a^2+x^2=b^2+y^2=c^2+z^2=(a+b)^2+(x+y)^2=(b+c)^2+(y+z)^2=(c+a)^2+(z+x)^2$

Show that $a^2+b^2+c^2=x^2+y^2+z^2$

Solution

Consider the middle equality $c^2+z^2=(a+b)^2+(x+y)^2$ . By difference of squares, this is equivalent to $(a+b+c)(a+b-c)+(x+y+z)(x+y-z)=0$ If we do this for the other two arrangements, we get $(a+b+c)(b+c-a)+(x+y+z)(y+z-x)=0$ $(a+b+c)(c+a-b)+(x+y+z)(z+x-y)=0$ Adding up all of the equations: $(a+b+c)^2+(x+y+z)^2=0$ which directly implies $a+b+c=0$ and $x+y+z=0$ . Next, adding some of the given equations: $(a^2+x^2)+(b^2+y^2)+(c^2+z^2)=((a+b)^2+(x+y)^2)+((b+c)^2+(y+z)^2)+((c+a)^2+(z+x)^2)$ which, after simplifications, yields $ab+bc+ca+xy+yz+zx=0$ Going back to $a+b+c=0$ and $x+y+z=0$ , if we square the first equation and expand: $a^2+b^2+c^2+2ab+2bc+2ca=0$ But notice that $a^2+b^2+c^2$ is always nonnegative; thus $ab+bc+ca$ is never positive. This is also true for $xy+yz+zx$ . But we know from earlier that $ab+bc+ca+xy+yz+zx=0$ and since both are never positive, they must both be equal to $0$ .

Finally, $a+b+c=0$ and $x+y+z=0$ imply that $a+b+c=x+y+z$ , so squaring and expanding: $a^2+b^2+c^2+2ab+2bc+2ca=x^2+y^2+z^2+2xy+2yz+2zx$ which, when considering $ab+bc+ca=xy+yz+zx=0$ , yields the result.

Remark: The only solution to this system is $a=b=c=x=y=z=0$ .

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2021 OIM Problems/Problem 4

Problem

Solution

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