Given that - $$a=a_1 a_2$$ $$b=b_1 b_2$$ $$c=c_1 c_2$$ $$h=a_2 b_1 + b_2 a_1$$ $$g=a_1 c_2 + a_2 c_1$$ $$f=b_1 c_2 + b_2 c_1$$

Find the relationship between $a, b, c, f, g, h$

My Attempt: I could not see how I could exploit the symmetry of the equations to directly get an answer, so I tried to solve them as 6 simultaneous equations by - $$a_2=\frac{a}{a_1}$$ $$b_2=\frac{b}{b_1}$$ $$c_2=\frac{c}{c_1}$$ and put these values into the remaining 3 equations to obtain - $$a_1 b_1 h = a {b_1}^2 + b {a_1}^2$$ $$a_1 c_1 g = a {c_1}^2 + c {a_1}^2$$ $$c_1 b_1 f = c {b_1}^2 + b {c_1}^2$$ Then, I treated the last 2 equations as quadratics in $a_1$ and $b_1$; found their values using the quadratic formula and input those into the first of the last 3 equations above.

Then I found the value of $c_1$ using that and then the value of the remaining - $a_1, a_2, b_1, b_2, c_2$.

Then, when I finally input these values into any of the equations I got a tautology (which, as I now realize - too late - was doomed to happen from the very beginning, due to my approach).


So, How should I go about finding the relation between $a, b, c, f, g, h$?


Equally - How do I eliminate $a_1, a_2, b_1, b_2, c_1, c_2$ from the equations?

I just need a hint on how I could exploit the symmetry of the equations.

  • $\begingroup$ What does "Find the relationship between a,b,c,f,g,h" mean exactly? One could argue that the initial six equalities already gives a relationship among them. Are you looking for a polynomial expression in a,b,c,f,g,h (and not involving the subscripted variables) that equals 0? $\endgroup$ – Greg Martin May 19 '17 at 3:36
  • $\begingroup$ Hint: consider $af^2+bg^2+ch^2$. $\endgroup$ – Greg Martin May 19 '17 at 3:38
  • $\begingroup$ @GregMartin Yes, you're right I need a single expression in a, b, c, f, g, h without the subscripted variables. As an example of such an expression - $\frac{c^{2}}{2a}=\frac{f^{3}}{5g^{2}} +h tan(\frac{b}{a})$ The expression could include logs and trig functions (although I don't think they will be necessary) $\endgroup$ – Quantum Sphinx May 19 '17 at 3:39
  • $\begingroup$ @GregMartin Yes, that is the kind of expression I need $\endgroup$ – Quantum Sphinx May 19 '17 at 3:43

Notice that $a,b,c,h,g,f$ are represented by staggered multiplication of $a_1,a_2,b_1,b_2,c_1,c_2$, and one way to align them is to multiply them together, and then we could rearrange how they are combined together. So symmetry is the key here.

$$h\cdot g \cdot f=(a_2 b_1 + b_2 a_1)(a_1 c_2 + a_2 c_1)(b_1 c_2 + b_2 c_1)$$ $$=a_2b_1a_1c_2b_1c_2 + a_2b_1a_1c_2b_2c_1+a_2b_1a_2c_1b_1c_2 + a_2b_1a_2c_1b_2c_1$$$$+b_2a_1a_1c_2b_1c_2+b_2a_1a_1c_2b_2c_1+b_2a_1a_2c_1b_1c_2+b_2a_1a_2c_1b_2c_1$$

$$=ab_1^2c_2^2+abc+a_2^2b_1^2c+a_2^2c_1^2b+a_1^2c_2^2b+a_1^2b_2^2c+abc+b_2^2c_1^2a$$ $$=2abc + a(b_1^2c_2^2+b_2^2c_1^2)+b(a_1^2c_2^2+a_2^2c_1^2) + c(a_2^2b_1^2+a_1^2b_2^2)$$ $$=2abc+a((b_1c_2 + b_2c_1)^2-2b_1c_2b_2c_1)+b((a_1c_2+a_2c_1)^2-2a_1c_2a_2c_1)+c((a_2b_1+a_1b_2)^2-2a_2b_1a_1b_2)$$ Thus $$hgf=2abc + a(f^2-2bc)+b(g^2-2ac)+c(h^2-2ab)$$ $$hgf +4abc -af^2-bg^2-ch^2=0$$


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.