Can x, y, and z be the side lengths of a triangle?

[anemone]

Let ABC be a triangle. Prove that $sin^2\frac{A}{2}+sin^2\frac{B}{2}+sin^2\frac{C}{2}+2sin\frac{A}{2}sin\frac{B}{2}sin\frac{C}{2}=1$.
Conversely, prove that if x, y and z are positive real numbers such that $x^2y^2+z^2+2xyz=1$, then there is a triangle ABC such that $x=sin\frac{A}{2}, y=sin\frac{B}{2}, z=sin\frac{C}{2}$.

I can solve the first part of the question.
My problem is I can't see a way to proceed with the second half of the question.

They are closely-related, I know, but I just couldn't see the right way to go.(Thinking)

Any help/hint would be deeply appreciated.

Thanks.

 

[alexmahone]

The first step would be to prove that $x,\ y,\ z\le 1$.

$x^2+y^2+z^2+2xyz=1$

Viewing this as a quadratic equation in $x$, we get

$\displaystyle x= \frac{-2yz\pm\sqrt{4y^2z^2-4(y^2+z^2-1)}}{2}$

Since $x$ is real, we must have

$4y^2z^2-4(y^2+z^2-1)\ge 0$

$y^2z^2-y^2-z^2+1\ge 0$

$y^2(z^2-1)-(z^2-1)\ge 0$

$(y^2-1)(z^2-1)\ge 0$

Similarly,

$(x^2-1)(y^2-1)\ge 0$

$(x^2-1)(z^2-1)\ge 0$

Either $x,\ y,\ z\le 1$ or $x,\ y,\ z\ge 1$.

$x,\ y,\ z\ge 1\implies x^2+y^2+z^2+2xyz\ge 5$ (contradiction)

So, $x,\ y,\ z\le 1$

We can let $x=\sin X,\ y=\sin Y,\ z=\sin Z$.

$\sin ^2X+\sin ^2Y+\sin^2Z+2\sin X\sin Y\sin Z=1$.

We now need to show that $\displaystyle X+Y+Z=\frac{\pi}{2}$.

---------- Post added at 12:14 PM ---------- Previous post was at 11:19 AM ----------

I can solve the first part of the question.

Could you post the solution? It may help me complete the proof.

 

[anemone]

Thanks for showing the great way to prove that $ x, y, z \leq 1.$, Alexmahone.

BTW, I don't think we need to let $x=sinX, y=sinY, z=sinZ $ so that proving $X+Y+Z=\frac{\pi}{2}$ will show the desired result.

I think we need to let $x=sin(\frac{A}{2}), y=sin(\frac{B}{2}), z=sin(\frac{C}{2}) $ because with these substitutions and from the given equation where $x^2y^2+z^2+2xyz=1$, we'll obtain the following equation:
$sin^2\frac{A}{2}+sin^2\frac{B}{2}+sin^2\frac{C}{2}+2sin\frac{A}{2}sin\frac{B}{2}sin\frac{C}{2}=1$.

And the equation above is true iff $A+B+C=\pi$, i.e. there exists a triangle ABC under the given conditions.

(P.S. Do you still want my working on the first part of the question? I'll show it if you want to have a look on it.)

 

[alexmahone]

$sin^2\frac{A}{2}+sin^2\frac{B}{2}+sin^2\frac{C}{2}+2sin\frac{A}{2}sin\frac{B}{2}sin\frac{C}{2}=1$.

And the equation above is true iff $A+B+C=\pi$, i.e. there exists a triangle ABC under the given conditions.

I believe you have proved that the equation above is true if $A+B+C=\pi$. But have you also proved that the equation is true only if $A+B+C=\pi$?

 

[anemone]

Hmm... I don't understand. Won't it suffice to show that the $sin^2\frac{A}{2}+sin^2\frac{B}{2}+sin^2\frac{C}{2}+2sin\frac{A}{2}sin\frac{B}{2}sin\frac{C}{2}=1$ is true if $A+B+C=\pi$ so that the prove is established?

Do you mean something like what will happen if $A+B+C=n\pi$?

 

[alexmahone]

Hmm... I don't understand. Won't it suffice to show that the $sin^2\frac{A}{2}+sin^2\frac{B}{2}+sin^2\frac{C}{2}+2sin\frac{A}{2}sin\frac{B}{2}sin\frac{C}{2}=1$ is true if $A+B+C=\pi$ so that the prove is established?

Do you mean something like what will happen if $A+B+C=n\pi$?

To complete the proof of the second half, we need to show that $\displaystyle \sin ^2X+\sin ^2Y+\sin^2Z+2\sin X\sin Y\sin Z=1\implies\displaystyle X+Y+Z=\frac{\pi}{2}$.

You said that you have proved $\displaystyle A+B+C=\pi\implies \sin^2\frac{A}{2}+\sin^2\frac{B}{2}+\sin^2\frac{C}{2}+2\sin\frac{A}{2}\sin\frac{B}{2}\sin\frac{C}{2}=1$.

However, what we now need is the converse: $\displaystyle\sin^2\frac{A}{2}+\sin^2\frac{B}{2}+\sin^2\frac{C}{2}+2\sin\frac{A}{2}\sin\frac{B}{2} \sin\frac{C}{2}=1\implies A+B+C=\pi$.

 

[anemone]

...what we now need is the converse: $\displaystyle\sin^2\frac{A}{2}+\sin^2\frac{B}{2}+\sin^2\frac{C}{2}+2\sin\frac{A}{2}\sin\frac{B}{2}\sin\frac{C}{2}=1\implies A+B+C=\pi$.

Ah! You're right! Will think hard on that part and report back once I get the idea or workout the whole proof.

 

[alexmahone]

The last part of the proof:

Let $\displaystyle x=\sin\frac{A}{2},\ y=\sin\frac{B}{2},\ \sin\frac{C}{2}$.

$\displaystyle \sin^2\frac{A}{2}+\sin^2\frac{B}{2}+\sin^2\frac{C}{2}+2\sin\frac{A}{2}\sin\frac{B}{2}\sin\frac{C}{2}=1$

$\displaystyle \frac{1-\cos A}{2}+\frac{1-\cos B}{2}+\sin^2\frac{C}{2}+2\sin\frac{A}{2}\sin\frac{B}{2}\sin\frac{C}{2}=1$

$\displaystyle -\frac{\cos A}{2}-\frac{\cos B}{2}+\sin^2\frac{C}{2}+2\sin\frac{A}{2}\sin\frac{B}{2}\sin\frac{C}{2}=0$

$\displaystyle -\cos\left(\frac{A+B}{2}\right)\cos\left(\frac{A-B}{2}\right)+\sin^2\frac{C}{2}+\left[\cos\left(\frac{A-B}{2}\right)-\cos\left(\frac{A+B}{2}\right)\right]\sin\frac{C}{2}=0$

$\displaystyle \left[\sin\frac{C}{2}+\cos\left(\frac{A-B}{2}\right)\right]\left[\sin\frac{C}{2}-\cos\left(\frac{A+B}{2}\right)\right]=0$

$\displaystyle \sin\frac{C}{2}+\cos\left(\frac{A-B}{2}\right)=0$ or $\displaystyle \sin\frac{C}{2}-\cos\left(\frac{A+B}{2}\right)=0$

$\displaystyle \frac{A-B}{2}=\frac{\pi}{2}+\frac{C}{2}$ or $\displaystyle \frac{A+B}{2}=\frac{\pi}{2}-\frac{C}{2}$

$\displaystyle A-B-C=\pi$ or $\displaystyle A+B+C=\pi$

If $\displaystyle A+B+C=\pi$, we are done.

If $\displaystyle A-B-C=\pi$, put $\displaystyle \frac{A'}{2}=\pi-\frac{A}{2}$ ie $\displaystyle A'=2\pi-A$. (Note that $\displaystyle \sin\frac{A'}{2}=\sin\frac{A}{2}$.)

$\displaystyle 2\pi-A'-B-C=\pi$

$\displaystyle A'+B+C=\pi$ and we are done.

 

[anemone]

Before I had time to response, you gave me the complete proof.

BTW, here is my initial work:
$sin^2\frac{A}{2}+sin^2\frac{B}{2}+sin^2\frac{C}{2}+2sin\frac{A}{2}sin\frac{B}{2}sin\frac{C}{2}$

$=\frac{3}{2}-\frac{cosA+cosB+cosC}{2}+sin\frac{C}{2}[(-)(cos\frac{A+B}{2}-cos\frac{A-B}{2})]$.

$=\frac{3}{2}-\frac{cosA+cosB+cosC}{2}-sin\frac{C}{2}[sin\frac{C}{2}-cos\frac{A-B}{2}]$.

$=\frac{3}{2}-\frac{cosA+cosB+cosC}{2}-sin^2\frac{C}{2}+sin\frac{C}{2}cos\frac{A-B}{2}$.

$=\frac{3}{2}-\frac{cosA+cosB+cosC}{2}-(\frac{-1+cosC}{2})+\frac{1}{2}(sin\frac{A+C-B}{2}-sin\frac{A-(B+C)}{2})$.

$=\frac{3}{2}-\frac{cosA+cosB+cosC}{2}-(\frac{-1+cosC}{2})+\frac{1}{2}(cosB+cosA)$.

=1

Now that I know I need to get down to the product of two factors that equals to zero because the relationship between A, B and C can clearly be drawn from there.

Thanks.