Right side of system of equations as an arithmetic sequence

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Right side of system of equations as an arithmetic sequence

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Kezer

#1 h Apr 12, 2016, 9:29 PM • 2 Y

Y by Adventure10, Mango247

Solve the system of equations in real numbers $\begin{align*} x^2+y^2+xy &= 37 \\ x^2+z^2+zx &= 28 \\ y^2+z^2+yz &= 19 \end{align*}$

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#2 h Apr 12, 2016, 10:14 PM • 2 Y

Y by Kezer, Adventure10

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#3 h Apr 12, 2016, 10:16 PM • 3 Y

Y by Couper, saagar, Adventure10

For some reason this reminds me of a triangle with sidelengths $\sqrt{19}$ , $\sqrt{28}$ , and $\sqrt{37}$ (law of cosines, a point P with distances $x$ , $y$ , $z$ , to the vertices, and three 120 degree angles)

(Unfortunately this observation is pretty much useless.)

This post has been edited 2 times. Last edited by mathguy5041, May 4, 2016, 9:06 PM

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#4 h Apr 13, 2016, 1:12 PM • 3 Y

Y by Kezer, Adventure10, Mango247

Given the system $\left\{\begin{array}{lll} x^{2}+y^{2}+xy = 37 \quad\quad (1) \\ x^{2}+z^{2}+zx = 28 \quad\quad (2) \\ y^{2}+z^{2}+yz = 19 \quad\quad (3) \end{array}\right.$

Subtracting $(1)-(2)\ :\ (y-z)(x+y+z)=9 \quad\quad (4)$ .
Subtracting $(1)-(3)\ :\ (x-z)(x+y+z)=18 \quad\quad (5)$ .

Dividing $(5)\$ by $\ (4)\$ gives $\frac{x-z}{y-z}=2$ or $y=\frac{x+z}{2}$ .
In $(1)\ :\ x^{2}+(\frac{x+z}{2})^{2}+x \cdot \frac{x+z}{2} = 37$ or $7x^{2}+4xz+z^{2}=148 \quad\quad (6)$ .
Subtracting $(6)-(2)\ :\ 6x^{2}+3xz=120$ or $z =\frac{40-2x^{2}}{x}$ .
In $(2)\ : \ x^{2}+(\frac{40-2x^{2}}{x})^{2}+\frac{40-2x^{2}}{x} \cdot x = 28$ .

Solving $3x^{4}-148x^{2}+1600=0$ gives us 4 solutions.

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#5 h Apr 13, 2016, 1:50 PM • 2 Y

Y by Kezer, Adventure10

From the fact that $2y=x+z$ , we have that $x, y, z$ form an arithmetic progression. Let $x=y-a$ and $z=y+a$ . The system of equations turns into $\begin{align*}3y^2-3ya+a^2 &= 37 \\ 3y^2+a^2 &= 28 \\ 3y^2+3ya+a^2 &= 19,\end{align*}$ Which means that $ya=-3\implies a=-\frac{3}{y}$ . This means that $3y^4-28y^2+9=0$ , which has 4 solutions.

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#6 h Apr 18, 2016, 12:37 AM • 2 Y

Y by Adventure10, Mango247

See PP4 from here.

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#7 h Jun 13, 2018, 5:03 PM • 1 Y

Y by Adventure10

law of cosines

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#8 h Jun 15, 2018, 2:56 PM • 1 Y

Y by Adventure10

use cosine rule

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#9 h Jun 16, 2018, 12:12 AM • 1 Y

Y by Adventure10

Hmmm, I'm not sure that you can use LoC for this, but post #3 is an interesting observation.

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#10 h Jun 16, 2018, 12:29 AM • 1 Y

Y by Adventure10

vatatmaja wrote:

Hmmm, I'm not sure that you can use LoC for this, but post #3 is an interesting observation.

Maybe on triangle with sides x, y with $z$ in between the sides.

Another interesting observation is that $120*3=360$ .

This post has been edited 1 time. Last edited by IceParrot, Jun 16, 2018, 12:31 AM

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#11 h Jun 16, 2018, 2:44 AM • 1 Y

Y by Adventure10

you can definitely use law of cosines for this, we went over a very similar problem in the precalculus aops class (i think just different numbers)

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