MHB Proving System of Equations for Algebra Challenge

[anemone]

Given the system of equation below:

$a = zb + yc$

$b = xc + za$

$c = you + xb$

Prove that $\dfrac{a^2}{1-x^2}=\dfrac{b^2}{1-y^2}=\dfrac{c^2}{1-z^2}$.

 

[Albert1]

Given the system of equation below:

$a = zb + yc---(1)$

$b = xc + za---(2)$

$c = you + xb---(3)$
Prove that $\dfrac{a^2}{1-x^2}=\dfrac{b^2}{1-y^2}=\dfrac{c^2}{1-z^2}---(4)$.

Spoiler

if (4) is true , then (5)(6)(7) also hold:
$\dfrac {a^2-b^2}{y^2-x^2}=\dfrac{c^2}{1-z^2}---(5)$
$\dfrac {b^2-c^2}{z^2-y^2}=\dfrac{a^2}{1-x^2}---(6)$
$\dfrac {a^2-c^2}{z^2-x^2}=\dfrac{b^2}{1-y^2}---(7)$
proof of (5)
(1)+(2) :$a+b=\dfrac {c(x+y)}{1-z}---(8)$
(1)-(2) :$a-b=\dfrac {-c(x-y)}{1+z}---(9)$
$(8)\times (9) :a^2-b^2=\dfrac {-c^2(x^2-y^2)}{1-z^2}$
and (5) holds
the proof of (6) and (7) are the same

 

[kaliprasad]

Spoiler

if (4) is true , then (5)(6)(7) also hold:
$\dfrac {a^2-b^2}{y^2-x^2}=\dfrac{c^2}{1-z^2}---(5)$
$\dfrac {b^2-c^2}{z^2-y^2}=\dfrac{a^2}{1-x^2}---(6)$
$\dfrac {a^2-c^2}{z^2-x^2}=\dfrac{b^2}{1-y^2}---(7)$
proof of (5)
(1)+(2) :$a+b=\dfrac {c(x+y)}{1-z}---(8)$
(1)-(2) :$a-b=\dfrac {-c(x-y)}{1+z}---(9)$
$(8)\times (9) :a^2-b^2=\dfrac {-c^2(x^2-y^2)}{1-z^2}$
and (5) holds
the proof of (6) and (7) are the same

Spoiler

$(4) => (5),(6),(7)$
but that does not mean that
$(5),(6),(7) =>(4)$

 

[Albert1]

Spoiler

$(4) => (5),(6),(7)$
but that does not mean that
$(5),(6),(7) =>(4)$

Spoiler

if $\dfrac {a_1}{b_1}=\dfrac {a_2}{b_2}=\dfrac {a_3}{b_3}=k$
then $\dfrac {a_1+a_2}{b_1+b_2}=\dfrac {a_1-a_2}{b_1-b_2}=------=k$
and if $\dfrac {a_1+a_2}{b_1+b_2}=\dfrac {a_1-a_2}{b_1-b_2}=------=k$
then $\dfrac {a_1}{b_1}=\dfrac {a_2}{b_2}=\dfrac {a_3}{b_3}=k$
of course all the denominators $\neq 0$
and more I got (5)(6)(7) from(1)(2)(3) not from (4)

 

[kaliprasad]

Spoiler

if $\dfrac {a_1}{b_1}=\dfrac {a_2}{b_2}=\dfrac {a_3}{b_3}=k$
then $\dfrac {a_1+a_2}{b_1+b_2}=\dfrac {a_1-a_2}{b_1-b_2}=------=k$
and if $\dfrac {a_1+a_2}{b_1+b_2}=\dfrac {a_1-a_2}{b_1-b_2}=------=k$
then $\dfrac {a_1}{b_1}=\dfrac {a_2}{b_2}=\dfrac {a_3}{b_3}=k$
of course all the denominators $\neq 0$
and more I got (5)(6)(7) from(1)(2)(3) not from (4)

Spoiler

I understand you proved (5),(6),(7) but that does not mean that (4) is true,

 

[Albert1]

Spoiler

if (4) is true , then (5)(6)(7) also hold:
$\dfrac {a^2-b^2}{y^2-x^2}=\dfrac{c^2}{1-z^2}---(5)$
$\dfrac {b^2-c^2}{z^2-y^2}=\dfrac{a^2}{1-x^2}---(6)$
$\dfrac {a^2-c^2}{z^2-x^2}=\dfrac{b^2}{1-y^2}---(7)$
proof of (5)
(1)+(2) :$a+b=\dfrac {c(x+y)}{1-z}---(8)$
(1)-(2) :$a-b=\dfrac {-c(x-y)}{1+z}---(9)$
$(8)\times (9) :a^2-b^2=\dfrac {-c^2(x^2-y^2)}{1-z^2}$
and (5) holds
the proof of (6) and (7) are the same

Spoiler

from (5)
proportion by addition and substraction
$\dfrac {a^2-b^2}{y^2-x^2}=\dfrac{c^2}{1-z^2}=\dfrac {a^2}{p-x^2}=\dfrac{b^2}{p-y^2}=\dfrac{c^2}{1-z^2}$
compare(6) and (7) it is easy to see p=1

 

[anemone]

Hi Albert,

Thanks for participating but I don't think your proof is complete.:( It's easier to use what we're asked to prove in our approach and then prove everything holds but more often than not, it's easy to lose sight of what we need to prove as we might have lost ourselves in our argument.

Here is the brilliant and neat solution of other that I wanted to share with the folks of MHB:

Spoiler

Express $x,\,y,\,z$ in terms of $a,\,b,\,c$, we see that we've:

$a = zb + yc \implies z=\dfrac{a-yc}{b}$

$b = xc + za\implies z=\dfrac{b-xc}{a}$

So equating the two we get:

$\dfrac{a-yc}{b}=\dfrac{b-xc}{a}$---(1)

But from $c = you + xb$, we can make $y$ the subject and then replace it into (1):

$y=\dfrac{c-xb}{a}$

$\therefore \dfrac{a-\left(\dfrac{c-xb}{a}\right)c}{b}=\dfrac{b-xc}{a}$

$x=\dfrac{b^2+c^2-a^2}{2bc}$

By the same token, we also get:

$y=\dfrac{a^2+c^2-b^2}{2ac}$; $z=\dfrac{a^2+b^2-c^2}{2ab}$

Now, we can find, separately, the expressions for $\dfrac{a^2}{1-x^2},\,\dfrac{b^2}{1-y^2},\,\dfrac{c^2}{1-z^2}$ all in terms of $a,\,b,\,c$:

$\begin{align*}\dfrac{a^2}{1-x^2}&=\dfrac{a^2}{1-\left(\dfrac{b^2+c^2-a^2}{2bc}\right)^2}\\&=\dfrac{4a^2b^2c^2}{(2bc)^2-(b^2+c^2-a^2)^2}\\&=\dfrac{4a^2b^2c^2}{(2bc-b^2-c^2+a^2)(2bc+b^2+c^2-a^2)}\\&=\dfrac{4a^2b^2c^2}{(a^2-(b-c)^2)((b+c)^2-a^2)}\\&=\dfrac{4a^2b^2c^2}{(a-b+c)(a+b-c)(b+c-a)(b+c+a)}\end{align*}$

Similarly,

$\dfrac{b^2}{1-y^2}=\dfrac{4a^2b^2c^2}{(a-b+c)(a+b-c)(b+c-a)(b+c+a)}$

$\dfrac{c^2}{1-z^2}=\dfrac{4a^2b^2c^2}{(a-b+c)(a+b-c)(b+c-a)(b+c+a)}$

Therefore we can conclude by now that $\dfrac{a^2}{1-x^2}=\dfrac{b^2}{1-y^2}=\dfrac{c^2}{1-z^2}$.

 

[Euge]

Hi anemone,

Here is my solution:

Spoiler

Since $a = zb + yc$ and $b = xc + za$, then $a = z(xc + za) + yc = (zx + y)c + z^2 a$, and thus

$$(1 - z^2)a = (zx + y)c.\tag{1}$$

As $c = you + xb$, substituting the expression $xc + za$ for $b$ yields $c = (y + xz)a + x^2 c$, or

$$(1 - x^2)c = (y + xz)a.\tag{2}$$

Using $(1)$ and $(2)$, the following proportion holds:

$$\frac{(1 - z^2)a}{(1 - x^2)c} = \frac{c}{a}.$$

This is equivalent to the proportion

$$\frac{a^2}{1 - x^2} = \frac{c^2}{1 - z^2}.\tag{3}$$

Substituting $zb + yc$ for $a$ in the equation $b = xc + za$ results in $(1 - z^2)b = (x + zy)c$, and using the same substitution in the equation $c = you + xb$ yields $(1 - y^2)c = (x + yz)b$. So the following proportion holds:

$$\frac{(1 - z^2)b}{(1 - y^2)c} = \frac{c}{b}.$$

Equivalently,

$$\frac{b^2}{1 - y^2} = \frac{c^2}{1 - z^2}.\tag{4}$$

Equations $(3)$ and $(4)$ together yield

$$\frac{a^2}{1 - x^2} = \frac{b^2}{1 - y^2} = \frac{c^2}{1 - z^2}.$$

 

[anemone]

Hi anemone,

Here is my solution:

Spoiler

Since $a = zb + yc$ and $b = xc + za$, then $a = z(xc + za) + yc = (zx + y)c + z^2 a$, and thus

$$(1 - z^2)a = (zx + y)c.\tag{1}$$

As $c = you + xb$, substituting the expression $xc + za$ for $b$ yields $c = (y + xz)a + x^2 c$, or

$$(1 - x^2)c = (y + xz)a.\tag{2}$$

Using $(1)$ and $(2)$, the following proportion holds:

$$\frac{(1 - z^2)a}{(1 - x^2)c} = \frac{c}{a}.$$

This is equivalent to the proportion

$$\frac{a^2}{1 - x^2} = \frac{c^2}{1 - z^2}.\tag{3}$$

Substituting $zb + yc$ for $a$ in the equation $b = xc + za$ results in $(1 - z^2)b = (x + zy)c$, and using the same substitution in the equation $c = you + xb$ yields $(1 - y^2)c = (x + yz)b$. So the following proportion holds:

$$\frac{(1 - z^2)b}{(1 - y^2)c} = \frac{c}{b}.$$

Equivalently,

$$\frac{b^2}{1 - y^2} = \frac{c^2}{1 - z^2}.\tag{4}$$

Equations $(3)$ and $(4)$ together yield

$$\frac{a^2}{1 - x^2} = \frac{b^2}{1 - y^2} = \frac{c^2}{1 - z^2}.$$

Awesome, Euge! And thanks for participating!