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

AI Thread Summary
The discussion centers on proving a relationship involving the sine of half-angles in a triangle, specifically that sin²(A/2) + sin²(B/2) + sin²(C/2) + 2sin(A/2)sin(B/2)sin(C/2) = 1. Participants discuss the necessity of showing that if x, y, and z are positive real numbers satisfying x²y² + z² + 2xyz = 1, then they correspond to the sine of half-angles of a triangle. A key point raised is the need to establish that x, y, z must be less than or equal to 1, which is derived from analyzing the quadratic equation formed. The conversation also emphasizes the importance of proving the converse relationship, specifically that the original sine equation implies A + B + C = π, thus confirming the existence of triangle ABC. The thread concludes with participants sharing insights and working towards completing the proof.
anemone
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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.
 
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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 ----------

anemone said:
I can solve the first part of the question.

Could you post the solution? It may help me complete the proof.
 
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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.)
 
anemone said:
$sin^2\frac{A}{2}+sin^2\frac{B}{2}+sin^2\frac{C}{2}+2sin\frac{A}{2}sin\frac{B}{2}sin\frac{C}{2}=1$.

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 $A+B+C=\pi$?
 
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 $A+B+C=n\pi$?
 
anemone said:
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: $\displaystyle\sin^2\frac{A}{2}+\sin^2\frac{B}{2}+\sin^2\frac{C}{2}+2\sin\frac{A}{2}\sin\frac{B}{2} \sin\frac{C}{2}=1\implies A+B+C=\pi$.
 
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Alexmahone said:
...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.
 
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.
 
Last edited:
Before I had time to response, you gave me the complete proof. :o

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.
 

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