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I saw this formula used in quite a few places, with the following text:


If $A,B,C$ are the points $z_1,z_2,z_3$ and the angles $B$ and $C$ are each $\pi$ - $\frac{\alpha}{2}$ then

$$ (z_2-z_3)^2 = 4(z_3-z_1)(z_1-z_2) \sin^2\frac{\alpha}{2} $$


Does anyone know what this means? Does it actually mean anything?

Thank you.

PS: I searched all over google for this formula, haven't found anything interesting.

PS2: I have no idea what tags I should put, help would be appreciated.

Trying to answer

Since angles B and C are both equal to $\pi$ - $\frac{\alpha}{2}$ this means that

angle A = 2$\pi$ - 2 ($\pi - \frac{\alpha}{2}$)

A = $\alpha$

After this, I'll use the Law of sines of a triangle, and tranform $sin (\pi - \frac{\alpha}{2}) = (-1)(cos(\frac{\alpha}{2})) $ :

$\frac{|z2-z3|}{\sin \alpha} = \frac{|z1-z3|}{(cos(\frac{\alpha}{2})} = \frac{|z1-z2|}{(cos(\frac{\alpha}{2})}$

This is as far as I got. Any ideas on how to proceed ?

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Please reformat the question. –  nbubis Jan 23 '13 at 14:30
    
Reformatted. This being my first post, I don't know all the formatting rules yet, sorry. –  Vlad Preda Jan 23 '13 at 14:31
    
what's a eachpie? –  nbubis Jan 23 '13 at 14:32
    
I don't know, this is how I found the "problem" in multiple places ( here's one: wiki.answers.com/Q/… ), including the typo ant instead of and –  Vlad Preda Jan 23 '13 at 14:35
    
I think it's: each PI - alpha/2 –  Vlad Preda Jan 23 '13 at 14:39

1 Answer 1

up vote 1 down vote accepted

Here is the picture you want to draw.

The triangle

Now, the rest is not hard, but I must say that the use of $z$ variables throws you off, as I was expecting a complex variable result. In fact, as stated the problem is not right and should be interpreted in vector terms. Also note that we know that this must be the sides AB and AC are equal in length.Let us write it as: $|\bf{r}_3-\bf{r}_2|^2 =4|\bf{r}_2-\bf{r}_1||\bf{r}_3-\bf{r}_1|$ $\sin^2{\alpha/2}$. Since AB and AC are equal in length $|\bf{r}_3-\bf{r}_1|=|\bf{r}_2-\bf{r}_1|$. For each one of the two right triangles shown we have hypotenuse $|\bf{r}_3-\bf{r}_1|$, and opposite side $(1/2)|\bf{r}_3-\bf{r}_2|$ and angle $\alpha/2$. Then: $$ (1/2)|\bf{r}_3-\bf{r}_2|=|\bf{r}_3-\bf{r}_1|\sin\alpha/2$$. Square this and rearrange: $$ |\bf{r}_3-\bf{r}_2|^2=4\sin^2\alpha/2|\bf{r}_3-\bf{r}_1|^2$$ And substitute once the alternative expression for the hypotenuse: $$ |\bf{r}_3-\bf{r}_2|^2=4\sin^2\alpha/2|\bf{r}_3-\bf{r}_1||\bf{r}_2-\bf{r}_1|$$.

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