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Difference between revisions of "2021 OIM Problems/Problem 4"

(Created page with "== Problem == Let <math>a,b,c,x,y,z</math> be real numbers such that <cmath>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</cmath> Show that <math>...")
 
(solution)
 
Line 8: Line 8:
  
 
== Solution ==
 
== Solution ==
−
{{solution}}
+
Consider the middle equality <math>c^2+z^2=(a+b)^2+(x+y)^2</math>. By difference of squares, this is equivalent to
 +
<cmath>(a+b+c)(a+b-c)+(x+y+z)(x+y-z)=0</cmath>
 +
If we do this for the other two arrangements, we get
 +
<cmath>(a+b+c)(b+c-a)+(x+y+z)(y+z-x)=0</cmath>
 +
<cmath>(a+b+c)(c+a-b)+(x+y+z)(z+x-y)=0</cmath>
 +
Adding up all of the equations:
 +
<cmath>(a+b+c)^2+(x+y+z)^2=0</cmath>
 +
which directly implies <math>a+b+c=0</math> and <math>x+y+z=0</math>. Next, adding some of the given equations:
 +
<cmath>(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)</cmath>
 +
which, after simplifications, yields
 +
<cmath>ab+bc+ca+xy+yz+zx=0</cmath>
 +
Going back to <math>a+b+c=0</math> and <math>x+y+z=0</math>, if we square the first equation and expand:
 +
<cmath>a^2+b^2+c^2+2ab+2bc+2ca=0</cmath>
 +
But notice that <math>a^2+b^2+c^2</math> is always nonnegative; thus <math>ab+bc+ca</math> is never positive. This is also true for <math>xy+yz+zx</math>. But we know from earlier that
 +
<cmath>ab+bc+ca+xy+yz+zx=0</cmath>
 +
and since both are never positive, they must both be equal to <math>0</math>.
 +
 
 +
Finally, <math>a+b+c=0</math> and <math>x+y+z=0</math> imply that <math>a+b+c=x+y+z</math>, so squaring and expanding:
 +
<cmath>a^2+b^2+c^2+2ab+2bc+2ca=x^2+y^2+z^2+2xy+2yz+2zx</cmath>
 +
which, when considering <math>ab+bc+ca=xy+yz+zx=0</math>, yields the result.
 +
 
 +
<i>Remark:</i> The only solution to this system is <math>a=b=c=x=y=z=0</math>.
 +
 
 +
~ [https://artofproblemsolving.com/wiki/index.php/User:Eevee9406 eevee9406]
  
 
== See also ==
 
== See also ==
 
https://olcoma.ac.cr/internacional/oim-2021/examenes
 
https://olcoma.ac.cr/internacional/oim-2021/examenes

Latest revision as of 19:57, 8 April 2025

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 $a=b=c=x=y=z=0$.

~ eevee9406

See also

https://olcoma.ac.cr/internacional/oim-2021/examenes

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