$x^2 +y^2 + xy = 25$
$y^2 + z^2 + yz = 49$
$z^2 + x^2 + zx = 64$
Find $(x + y + z)^2 -100$
Here's My Approach :
$x^2 + y^2 -2xycos120 = 25$. This Equation looked too similar to the subtraction of the vectors Formula to me. So I Imagined x, y, z as 3 vectors in the x-y plane separated by the Angle of 120 degrees where vectors $|x' - y'| = 5$ and so on, We have to find the mod of the sum of the resultant of these vectors but I couldn't go any further as breaking them down into components pushes me back into equation 1 and I cannot think of any approach after this.
IF anyone has a solution to this problem using pure algebra or Vectors(Basic) please Assist(Preferably Vectors because I thought of it myself).
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$\begingroup$ Are there any restrictions on x,y,z? $\endgroup$– MathStackexchangeIsMarvellousCommented May 16 at 20:34
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$\begingroup$ Yes It does, But I would also like to see a solution based on vectors $\endgroup$– memeguyCommented May 17 at 6:34
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2 Answers
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Taking the resultant of the polynomials, with $r=(x+y+z)^2-100$, one obtains \begin{align*} r&=-91,\quad x=0,y=5, z=-8,\\ r & = 29,\quad 8y=5x,5z=8x, 129x^2 = 1600. \end{align*}
answered May 16 at 20:25
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1$\begingroup$ Can this solution be considered "basic"?? $\endgroup$ Commented May 16 at 20:30
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$\begingroup$ @MathStackexchangeIsNotSoBad The title says "advanced" algebra problem, not "basic" algebra problem. The resultant is just a determinant, so for me this is not so advanced. But this is opinion-based, I suppose. $\endgroup$ Commented May 16 at 20:31
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$\begingroup$ @DietrichBurde could you please explain how you took the resultant of the vectors? $\endgroup$– memeguyCommented May 16 at 20:48
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$\begingroup$ $13x=8y+5z$ can easily be found by hand. $\endgroup$ Commented May 16 at 20:50
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$\begingroup$ @memeguy The wikipedia page explains, how to take the resultant of, say, $x^2+y^2+xy-25$ and $y^2+z^2+yz-49$ with respect to $y$. This way one ends up finally with a polynomial $f(x)=0$ in one variable, which leads to the solutions. But this method is more computational than the other solutions here (so it has been downvoted). $\endgroup$ Commented May 17 at 9:26
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The triangle you formed by substruction of pairs of vectors has sides $5,7,8$. By Brahmagupta's formula its area is $10\sqrt3$. On the other hand, the area of the triangle is the sum of the areas of the triangles formed by pairs of the vectors. Hence, $$\frac12\sin(120^\circ)(xy+yz+zx)=10\sqrt3$$ and $xy+yz+zx=40$ which is enough the solve the question. The answer is $29$.
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1$\begingroup$ No, take $x=0$, $y=5$ and $z=-8$. Then all equations are satisfied, but $(x+y+z)^2=(-3)^2-100$ is not $29$. $\endgroup$ Commented May 16 at 20:29
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$\begingroup$ I understood what he did...it should be given that all three of them are non zero otherwise triangle isn't possible $\endgroup$ Commented May 16 at 20:32
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$\begingroup$ @DietrichBurde There may be other solutions. This is geometric one. $34$ was miscalculation. $\endgroup$ Commented May 16 at 20:33
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$\begingroup$ @BobDobbs The triangle must be formed using the subtraction of the vectors, not the vectors themselves $\endgroup$– memeguyCommented May 18 at 5:08