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The real numbers $x,y$ and $z$ are such that $x-7y+8z=4$ and $8x+4y-z=7$. What is the maximum value of $x^2-y^2+z^2?$

From those equations I got:

$12z-5x=13y$

$12x+5z=13$

$12y+5=13z$

$12-5y=13x$

I know that $5,12,13$ is a pythag triplet but I don’t know what to do next. I think lagrange multipliers could be used but there should be a solution that doesn’t require calculus

Hints, suggestions and solutions would be appreciated.

Taken from the 2014 KIMC https://chiuchang.org/imc/wp-content/uploads/sites/2/2018/01/2014-IWYMIC-Individual.x17381.pdf

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  • $\begingroup$ In spite of the appearance of the Pythagorean triple I don't think this has much at all to do with number theory. Anyway, this looks a bit easy to be a contest problem. The two linear equations describe the line of intersection of two planes in $\Bbb{R}^3$. You can parametrize that line, and plug in $(x,y,z)$, and you are left with the task of finding the maximum of a quadratic polynomial in a single variable. It turns out that the polynomial is a constant, but while surprising that doesn't in my opinion make this contest-worthy (may be junior high school level?) $\endgroup$ Sep 15 '19 at 9:58
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Solving the system $$x-7y+8z=4$$ $$8x+4y-z=-7$$ we get $$y=\frac{12}{5}-\frac{13}{5}x$$ $$z=\frac{13}{5}-\frac{12}{5}x$$ and we get $$x^2-y^2+z^2=1$$ Very NICE!

Below please find an image of the surface $x^2-y^2+z^2=1$ together with the (thick red) line of intersection of the two given planes.

enter image description here

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  • $\begingroup$ Thank you for making an image of the problem! $\endgroup$ Sep 15 '19 at 10:27
  • $\begingroup$ And greetings to Finland with all your brave poeple! $\endgroup$ Sep 15 '19 at 10:33
  • $\begingroup$ Im sorry but I don’t understand how you got from $$y=\frac{12}{5}-\frac{13}{5}x$$ $$z=\frac{13}{5}-\frac{12}{5}x$$ To $$x^2-y^2+z^2=1$$ $\endgroup$
    – Tyrone
    Sep 15 '19 at 10:34
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    $\begingroup$ Expand $$x^2-\left(\frac{15}{5}-\frac{13}{5}x\right)^2+\left(\frac{13}{5}-\frac{12}{5}x\right)^2=1$$ $\endgroup$ Sep 15 '19 at 10:37
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Solving for $x,y$ in terms of $z$

$$12x=13-5z$$

$$12y=13z-12$$

$$12^2(x^2-y^2+z^2)=(13-5z)^2-(13z-12)^2+144z^2$$

$$=-50z^2-(130+312)z+25$$

$$=-50\left(z+\dfrac{442}{100}\right)^2+25+50\left(\dfrac{442}{100}\right)^2$$

$$\le25+50\left(\dfrac{442}{100}\right)^2$$

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