1983 AMC 12 Problema 13

Question

Si

xy=a,

xz=b,

y

yz=c,

y ninguna de estas cantidades es

0,

entonces

x^2+y^2+z^2

es igual a

If

xy=a,

xz=b,

and

yz=c,

and none of these quantities is

0,

then

x^2+y^2+z^2

equals

\frac{ab+ac+bc}{abc}

\frac{a^2+b^2+c^2}{abc}

\frac{(a+b+c)^2}{abc}

\frac{(ab+ac+bc)^2}{abc}

\frac{(ab)^2+(ac)^2+(bc)^2}{abc}

Multiplica dos de

xy=a,

xz=b,

y

yz=c,

y divide entre la tercera

Multiply two of

xy=a,

xz=b,

and

yz=c,

then divide by the third

Usa

x^2=\frac{ab}{c},

y^2=\frac{ac}{b},

y

z^2=\frac{bc}{a}

Use

x^2=\frac{ab}{c},

y^2=\frac{ac}{b},

and

z^2=\frac{bc}{a}

De los tres productos,

\begin{aligned} x^2&=\frac{ab}{c},\\ y^2&=\frac{ac}{b},\\ z^2&=\frac{bc}{a}. \end{aligned}

Al expresar la suma con denominador

abc

, obtenemos

\begin{aligned} x^2+y^2+z^2 &=\frac{(ab)^2+(ac)^2}{abc}\\ &\quad+\frac{(bc)^2}{abc}. \end{aligned}

Por lo tanto, la respuesta correcta es E.

From the three products,

\begin{aligned} x^2&=\frac{ab}{c},\\ y^2&=\frac{ac}{b},\\ z^2&=\frac{bc}{a}. \end{aligned}

Putting their sum over denominator

abc

gives

\begin{aligned} x^2+y^2+z^2 &=\frac{(ab)^2+(ac)^2}{abc}\\ &\quad+\frac{(bc)^2}{abc}. \end{aligned}

Therefore, the correct answer is E.

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