MHB What is the value of the expression given these equations?

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The discussion revolves around solving for the expression $\dfrac{x^2}{m^2}+\dfrac{y^2}{n^2}+\dfrac{z^2}{p^2}$ given two equations: $\dfrac{x}{m}+\dfrac{y}{n}+\dfrac{z}{p}=1$ and $\dfrac{m}{x}+\dfrac{n}{y}+\dfrac{p}{z}=0$. Participants explore whether this expression yields a fixed numerical value. The conversation also touches on the related expression $\dfrac{m^2}{x^2}+\dfrac{n^2}{y^2}+\dfrac{p^2}{z^2}$. Overall, the focus is on deriving a solution based on the provided equations.
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given :

$\dfrac {x}{m}+\dfrac{y}{n} +\dfrac {z}{p}=1$

$\dfrac {m}{x}+\dfrac{n}{y} +\dfrac {p}{z}=0$

find:

$\dfrac {x^2}{m^2}+\dfrac{y^2}{n^2} +\dfrac {z^2}{p^2}=?$
 
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Re: find value

Albert said:
given :

$\dfrac {x}{m}+\dfrac{y}{n} +\dfrac {z}{p}=1$

$\dfrac {m}{x}+\dfrac{n}{y} +\dfrac {p}{z}=0$

find:

$\dfrac {x^2}{m^2}+\dfrac{y^2}{n^2} +\dfrac {z^2}{p^2}=?$

Hello.
\dfrac {x^2}{m^2}+\dfrac{y^2}{n^2} +\dfrac {z^2}{p^2}=1

1º)

\dfrac {m}{x}+\dfrac{n}{y} +\dfrac {p}{z}=0

myz+nxz+pxy=0

(mnp)(myz+nxz+pxy)<br /> =0

mnpmyz+mnpnxz+mnppxy=0(*)

2º)

\dfrac {x}{m}+\dfrac{y}{n} +\dfrac {z}{p}=1

xnp+myp+mnz=mnp

(xnp+myp+mnz)^2=m^2n^2p^2

x^2n^2p^2+m^2y^2p^2+m^2n^2z^2+

+2xnpmyp+2xnpmnz+2mypmnz=m^2n^2p^2

For (*):

2xnpmyp+2xnpmnz+2mypmnz=0

Then:

x^2n^2p^2+m^2y^2p^2+m^2n^2z^2=m^2n^2p^2

\dfrac{x^2n^2p^2+m^2y^2p^2+m^2n^2z^2}{m^2n^2p^2}=1

\dfrac {x^2}{m^2}+\dfrac{y^2}{n^2} +\dfrac {z^2}{p^2}=1

Regards.
 
Hello, Albert!

\text{Given: }\:\dfrac {x}{m}+\dfrac{y}{n} +\dfrac {z}{p}=1

. . . . . . . \dfrac {m}{x}+\dfrac{n}{y} +\dfrac {p}{z}=0

\text{Find: }\:\dfrac {x^2}{m^2}+\dfrac{y^2}{n^2} +\dfrac {z^2}{p^2}
\text{Let: }\:a\,=\,\frac{x}{m},\;\;b \,=\, \frac{y}{n},\;\;c \,=\, \frac{z}{p}

\text{We have: }\:\begin{Bmatrix}a+b+c &amp;=&amp; 1 &amp; [1] \\ \frac{1}{a}+\frac{1}{b} + \frac{1}{c} &amp;=&amp; 0 &amp; [2] \end{Bmatrix}

\text{And we want: }\:a^2+b^2+c^2.\text{From [2]: }\:\frac{ab+bc+ac}{abc} \:=\:0 \quad\Rightarrow\quad ab + bc + ac \:=\:0

\text{Square [1]: }\: (a+b+c)^2 \:=\:1 ^2

. . a^2+2ab+2ac+b^2+2bc+c^2 \:=\:1

. . a^2+b^2+c^2 + 2\underbrace{(ab + bc + ac)}_{\text{This is }0} \:=\:1

\text{Therefore: }\:a^2+b^2+c^2 \:=\:1
 
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Albert said:
given :

$\dfrac {x}{m}+\dfrac{y}{n} +\dfrac {z}{p}=1$

$\dfrac {m}{x}+\dfrac{n}{y} +\dfrac {p}{z}=0$

find:

$\dfrac {x^2}{m^2}+\dfrac{y^2}{n^2} +\dfrac {z^2}{p^2}=?$

now the following value can be found (is it a fixed number)?
$\dfrac {m^2}{x^2}+\dfrac{n^2}{y^2} +\dfrac {p^2}{z^2}=?$
 
Here is a little puzzle from the book 100 Geometric Games by Pierre Berloquin. The side of a small square is one meter long and the side of a larger square one and a half meters long. One vertex of the large square is at the center of the small square. The side of the large square cuts two sides of the small square into one- third parts and two-thirds parts. What is the area where the squares overlap?

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