MHB Solving a Challenge: Finding Real x to Satisfy an Equation

[anemone]

Hi MHB,

Problem:

Find all real $x$ which satisfy $\dfrac{x^3+a^3}{(x+a)^3}+ \dfrac{x^3+b^3}{(x+b)^3}+\dfrac{x^3+c^3}{(x+c)^3} + \dfrac{3(x-a)(x-b)(x-c)}{2(x+a)(x+b)(x+c)}=\dfrac{3}{2}$.

I tried my very best to solve this intriguing problem, but failed. Now I'm even clueless than I was before, as it seems to me those inequality theorems don't play a part to solve this problem, and also, the first instinct of letting $x=a=b=c$ led to a contradiction, and second attempt to let $a=b=c$ gave no useful result.

Can someone teach me how to approach this problem, please? Thank you very much in advance.:)

 

[anemone]

Oh my goodness, I realized only now that I've completely forgotten the first three fraction could be simplified further so that each of which takes the form $\dfrac{x^3+a^3}{(x+a)^3}=\dfrac{(x+a)(x^2-ax+a^2)}{(x+a)^3}=\dfrac{x^2-ax+a^2}{(x+a)^2}=\dfrac{(x-a)^2+ax}{(x+a)^2}=dfrac{(x+a)^2-3ax}{(x+a)^2}$...I'll see if I can finish the problem and will post back if there is any progress...(Sun)

 

[kaliprasad]

Oh my goodness, I realized only now that I've completely forgotten the first three fraction could be simplified further so that each of which takes the form $\dfrac{x^3+a^3}{(x+a)^3}=\dfrac{(x+a)(x^2-ax+a^2)}{(x+a)^3}=\dfrac{x^2-ax+a^2}{(x+a)^2}=\dfrac{(x-a)^2+ax}{(x+a)^2}=dfrac{(x+a)^2-3ax}{(x+a)^2}$...I'll see if I can finish the problem and will post back if there is any progress...(Sun)

may be u can try

$\frac{x^3+ a^3}{(x+a)^3}$
= $\frac{(x+ a)^3-3ax(x+a)}{(x+a)^3}$
= $1- \frac{3ax}{(x+a)^2}$

 

[anemone]

My attempt so far...

Since $\dfrac{x^3+a^3}{(x+a)^3}=\dfrac{(x+a)(x^2-ax+a^2)}{(x+a)^3}=\dfrac{x^2-ax+a^2}{(x+a)^2}=\dfrac{(x-a)^2+ax}{(x+a)^2}=\dfrac{(x+a)^2-3ax}{(x+a)^2}$, we need to think as we don't know if $\dfrac{x^3+a^3}{(x+a)^3}=\dfrac{(x-a)^2+ax}{(x+a)^2}$ is more useful or we should instead use this $\dfrac{x^3+a^3}{(x+a)^3}=\dfrac{(x+a)^2-3ax}{(x+a)^2}$...

But we're asked to solve for $x$ of $\dfrac{x^3+a^3}{(x+a)^3}+ \dfrac{x^3+b^3}{(x+b)^3}+\dfrac{x^3+c^3}{(x+c)^3} + \dfrac{3(x-a)(x-b)(x-c)}{2(x+a)(x+b)(x+c)}=\dfrac{3}{2}$.

The fourth term from the LHS might give us a hint and we actually want to keep the $x-a$ at the top and $x+a$ at the bottom, so we must get rid of $ax$.

From $(x+a)^2=x^2+a^2+2ax$ and $(x-a)^2=x^2+a^2-2ax$, subtracting the second from the first we get $(x+a)^2-(x-a)^2=4ax$ and this gives $ax=\dfrac{(x+a)^2-(x-a)^2}{4}$. In other words, we can write $\dfrac{x^3+a^3}{(x+a)^3}=\dfrac{(x-a)^2+ax}{(x+a)^2}=\dfrac{(x-a)^2+\dfrac{(x+a)^2-(x-a)^2}{4}}{(x+a)^2}=\dfrac{3(x-a)^2}{4(x+a)^2}+\dfrac{1}{4}$ and so what we have now is:

$\dfrac{x^3+a^3}{(x+a)^3}+ \dfrac{x^3+b^3}{(x+b)^3}+\dfrac{x^3+c^3}{(x+c)^3} + \dfrac{3(x-a)(x-b)(x-c)}{2(x+a)(x+b)(x+c)}=\dfrac{3}{2}$

$\dfrac{3(x-a)^2}{4(x+a)^2}+\dfrac{1}{4}+\dfrac{3(x-b)^2}{4(x+b)^2}+\dfrac{1}{4}+\dfrac{3(x-c)^2}{4(x+c)^2}+\dfrac{1}{4}+\dfrac{3(x-a)(x-b)(x-c)}{2(x+a)(x+b)(x+c)}=\dfrac{3}{2}$

$\dfrac{3}{4}\left( \dfrac{x-a}{x+a} \right)^2+\dfrac{3}{4}\left( \dfrac{x-b}{x+b} \right)^2+\dfrac{3}{4}\left( \dfrac{x-c}{x+c} \right)^2+\dfrac{3(x-a)(x-b)(x-c)}{2(x+a)(x+b)(x+c)}-\dfrac{3}{4}=0$

but I really don't know how to proceed any further...(Angry)

 

[kaliprasad]

My attempt so far...

Since $\dfrac{x^3+a^3}{(x+a)^3}=\dfrac{(x+a)(x^2-ax+a^2)}{(x+a)^3}=\dfrac{x^2-ax+a^2}{(x+a)^2}=\dfrac{(x-a)^2+ax}{(x+a)^2}=\dfrac{(x+a)^2-3ax}{(x+a)^2}$, we need to think as we don't know if $\dfrac{x^3+a^3}{(x+a)^3}=\dfrac{(x-a)^2+ax}{(x+a)^2}$ is more useful or we should instead use this $\dfrac{x^3+a^3}{(x+a)^3}=\dfrac{(x+a)^2-3ax}{(x+a)^2}$...

But we're asked to solve for $x$ of $\dfrac{x^3+a^3}{(x+a)^3}+ \dfrac{x^3+b^3}{(x+b)^3}+\dfrac{x^3+c^3}{(x+c)^3} + \dfrac{3(x-a)(x-b)(x-c)}{2(x+a)(x+b)(x+c)}=\dfrac{3}{2}$.

The fourth term from the LHS might give us a hint and we actually want to keep the $x-a$ at the top and $x+a$ at the bottom, so we must get rid of $ax$.

From $(x+a)^2=x^2+a^2+2ax$ and $(x-a)^2=x^2+a^2-2ax$, subtracting the second from the first we get $(x+a)^2-(x-a)^2=4ax$ and this gives $ax=\dfrac{(x+a)^2-(x-a)^2}{4}$. In other words, we can write $\dfrac{x^3+a^3}{(x+a)^3}=\dfrac{(x-a)^2+ax}{(x+a)^2}=\dfrac{(x-a)^2+\dfrac{(x+a)^2-(x-a)^2}{4}}{(x+a)^2}=\dfrac{3(x-a)^2}{4(x+a)^2}+\dfrac{1}{4}$ and so what we have now is:

$\dfrac{x^3+a^3}{(x+a)^3}+ \dfrac{x^3+b^3}{(x+b)^3}+\dfrac{x^3+c^3}{(x+c)^3} + \dfrac{3(x-a)(x-b)(x-c)}{2(x+a)(x+b)(x+c)}=\dfrac{3}{2}$

$\dfrac{3(x-a)^2}{4(x+a)^2}+\dfrac{1}{4}+\dfrac{3(x-b)^2}{4(x+b)^2}+\dfrac{1}{4}+\dfrac{3(x-c)^2}{4(x+c)^2}+\dfrac{1}{4}+\dfrac{3(x-a)(x-b)(x-c)}{2(x+a)(x+b)(x+c)}=\dfrac{3}{2}$

$\dfrac{3}{4}\left( \dfrac{x-a}{x+a} \right)^2+\dfrac{3}{4}\left( \dfrac{x-b}{x+b} \right)^2+\dfrac{3}{4}\left( \dfrac{x-c}{x+c} \right)^2+\dfrac{3(x-a)(x-b)(x-c)}{2(x+a)(x+b)(x+c)}-\dfrac{3}{4}=0$

but I really don't know how to proceed any further...(Angry)

taking $\frac{x-a}{x+a}= A$ ;$\frac{x-b}{x+b}= B$; $\frac{x-c}{x+c}= c$

we get $A^2 + B^2 + C^2 + 2ABC = 1$

which gives $A = \sin \frac{r}{2}$
$B = \sin \frac{s}{2}$
$C = \sin \frac{t}{2}$

where r, s t are angles of triangle

(if a proof is required I can provide the same)

Now you should be able to proceed

 

[anemone]

taking $\frac{x-a}{x+a}= A$ ;$\frac{x-b}{x+b}= B$; $\frac{x-c}{x+c}= c$

we get $A^2 + B^2 + C^2 + 2ABC = 1$

which gives $A = \sin \frac{r}{2}$
$B = \sin \frac{s}{2}$
$C = \sin \frac{t}{2}$

where r, s t are angles of triangle

(if a proof is required I can provide the same)

Now you should be able to proceed

Now, I can even eyeball how to proceed!(Sun) Hence your help is greatly appreciated, kali! You're a good friend and an excellent challenge problems solver! But despite my saying so, I'm fairly, enormously disappointed in myself, because I should have thought of the identity $\sin^2\dfrac{A}{2}+ \sin^2\dfrac{B}{2}+\sin^2\dfrac{C}{2}+2\sin\dfrac{A}{2}\sin\dfrac{B}{2}\sin\dfrac{C}{2}=1$ if $A,\,B,\,C$ are angles in a triangle. Shame on me...(Doh)(Sadface)(Crying)(Angry)(Punch)(Tmi)(Headbang) You know, I feel really, really embarrassed and really, really disappointed...(Crying)

Okay, it would not be right if I don't finish the problem here:

If we let, like what kali has suggested that $\sin \dfrac{A}{2}=\dfrac{x-a}{x+a}$, $\sin \dfrac{B}{2}=\dfrac{x-b}{x+b}$ and $\sin \dfrac{C}{2}=\dfrac{x-c}{x+c}$, we can see the very well-known trigonometric identity that applies to the three angles inside any triangle exists in our last equation, i.e.

$\dfrac{3}{4}\left( \dfrac{x-a}{x+a} \right)^2+\dfrac{3}{4}\left( \dfrac{x-b}{x+b} \right)^2+\dfrac{3}{4}\left( \dfrac{x-c}{x+c} \right)^2+\dfrac{3(x-a)(x-b)(x-c)}{2(x+a)(x+b)(x+c)}-\dfrac{3}{4}=0$ becomes

$\left( \dfrac{x-a}{x+a} \right)^2+\left( \dfrac{x-b}{x+b} \right)^2+\left( \dfrac{x-c}{x+c} \right)^2+\dfrac{2(x-a)(x-b)(x-c)}{(x+a)(x+b)(x+c)}-1=0$

$\sin^2 \dfrac{A}{2}+\sin^2 \dfrac{B}{2}+\sin^2 \dfrac{C}{2}+2\sin \dfrac{A}{2}\sin \dfrac{B}{2}\sin \dfrac{C}{2}-1=0$

and hence we have

$\dfrac{A+B}{2}=\dfrac{180^{\circ}-C}{2}$

$\sin\left(\dfrac{A+B}{2} \right)=\sin\left(\dfrac{180^{\circ}-C}{2} \right)$

$\sin \dfrac{A}{2} \cos\dfrac{B}{2}+\cos\dfrac{A}{2}\sin \dfrac{B}{2}=\cos \dfrac{C}{2}$

Since $\sin \dfrac{A}{2}=\dfrac{x-a}{x+a}$, we have $\cos \dfrac{A}{2}=\dfrac{2\sqrt{a}}{x+a}$ and so on.

$\therefore \dfrac{x-a}{x+a} \cdot \dfrac{2\sqrt{bx}}{x+b}+\dfrac{2\sqrt{ax}}{x+a} \cdot \dfrac{x-b}{x+b}=\dfrac{2\sqrt{cx}}{x+c}$

One immediate result that we can see here is $x=0$, and after dividing the both sides of the equation by $\sqrt{x}$, all that is left is a quadratic equation and we can solve it by using the safest method, i.e. by using the quadratic formula and we're done.

Thank you again kali, for your help, I really appreciate that! (Sun)

 

[Saitama]

taking $\frac{x-a}{x+a}= A$ ;$\frac{x-b}{x+b}= B$; $\frac{x-c}{x+c}= c$

we get $A^2 + B^2 + C^2 + 2ABC = 1$

which gives $A = \sin \frac{r}{2}$
$B = \sin \frac{s}{2}$
$C = \sin \frac{t}{2}$

where r, s t are angles of triangle

(if a proof is required I can provide the same)

Now you should be able to proceed

Hi kaliprasad!

Can you please show why is it ok to assume $A=\sin\frac{r}{2}$ and similarly the others? Using this substitution, you are restricting $\frac{x-a}{x+a}$ to be between 0 and 1, why?

Thanks!

 

[kaliprasad]

Hi kaliprasad!

Can you please show why is it ok to assume $A=\sin\frac{r}{2}$ and similarly the others? Using this substitution, you are restricting $\frac{x-a}{x+a}$ to be between 0 and 1, why?

Thanks!

thanks pranav,
My mistake. I presumes that each term > 0 but it is not so then the above holds. Thanks for pointing that

 

[Saitama]

thanks pranav,
My mistake. I presumes that each term > 0 but it is not so then the above holds. Thanks for pointing that

But how would you solve the problem then? I am very confused.

 

[anemone]

Hey Pranav,

Thanks for chiming in in this thread of mine to remind us that if we wanted to use trigonometric to solve for an algebra problem, we needed to be extremely careful as we've to check if the condition to do so has been met, (i.e. we are allowed to proceed with the trigonometric substitution to tackle the problem).

I'm not sure if what I've done here enough to show that we could, assume that $-1\le \dfrac{x-a}{x+a} \le 1$ (similarly to the other two terms).

From the final simplified version of the original given equation, we're at:

$\left( \dfrac{x-a}{x+a} \right)^2+\left( \dfrac{x-b}{x+b} \right)^2+\left( \dfrac{x-c}{x+c} \right)^2+\dfrac{2(x-a)(x-b)(x-c)}{(x+a)(x+b)(x+c)}-1=0$

If we rewrite it a bit by moving the terms $\left( \dfrac{x-a}{x+a} \right)^2+\left( \dfrac{x-b}{x+b} \right)^2-1$ to the RHS, we see that we have:

$\left( \dfrac{x-c}{x+c} \right)^2+2\left(\dfrac{(x-a)(x-b)}{(x+a)(x+b)} \right) \left(\dfrac{x-c}{x+c} \right)=1-\left( \dfrac{x-a}{x+a} \right)^2-\left( \dfrac{x-b}{x+b} \right)^2$

Complete the square of the LHS, we get:

$\left( \dfrac{x-c}{x+c}+\dfrac{(x-a)(x-b)}{(x+a)(x+b)} \right)^2-\left(\dfrac{(x-a)(x-b)}{(x+a)(x+b)} \right)^2=1-\left( \dfrac{x-a}{x+a} \right)^2-\left( \dfrac{x-b}{x+b} \right)^2$

$\left( \dfrac{x-c}{x+c}+\dfrac{(x-a)(x-b)}{(x+a)(x+b)} \right)^2=1-\left( \dfrac{x-a}{x+a} \right)^2-\left( \dfrac{x-b}{x+b} \right)^2+\left(\dfrac{(x-a)(x-b)}{(x+a)(x+b)} \right)^2$

$\left( \dfrac{x-c}{x+c}+\dfrac{(x-a)(x-b)}{(x+a)(x+b)} \right)^2=1-\left( \dfrac{x-a}{x+a} \right)^2-\left( \dfrac{x-b}{x+b} \right)^2\left(1-\left(\dfrac{x-a}{x+a} \right)^2 \right)$

$\left( \dfrac{x-c}{x+c}+\dfrac{(x-a)(x-b)}{(x+a)(x+b)} \right)^2=\left(1-\left( \dfrac{x-a}{x+a} \right)^2\right)\left(1-\left( \dfrac{x-b}{x+b} \right)^2\right)$

$\therefore \left( \dfrac{x-c}{x+c}+\dfrac{(x-a)(x-b)}{(x+a)(x+b)} \right)=\sqrt{\left(1-\left( \dfrac{x-a}{x+a} \right)^2\right)}\sqrt{\left(1-\left( \dfrac{x-b}{x+b} \right)^2\right)}$

Now, it's obvious that either $-1\le \dfrac{x-a}{x+a} \le 1$, or $-1\le \dfrac{x-b}{x+b} \le 1$ and this is also true for $-1\le \dfrac{x-c}{x+c} \le 1$ and therefore the trigonometric substitution of $\sin \dfrac{A}{2}=\dfrac{x-a}{x+a}$, $\sin \dfrac{B}{2}=\dfrac{x-b}{x+b}$ and $\sin \dfrac{C}{2}=\dfrac{x-c}{x+c}$ can be applied safely into the problem...

How does this sound to you all?:(