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I did the following: Take the vectors $A=(a,0), B=(x,y), -A=(-a,0)$.

Then multiply each vector $X$ by $\cfrac{1}{|X|}$ to obtain the unit vector. (I don't know if this step is really necessary.)

Then we will have:

$$A'=(\frac{a}{\sqrt{a^2}},0)=(1,0) \quad B'=(\frac{x}{\sqrt{x^2+y^2}},\frac{y}{\sqrt{x^2+y^2}} )$$

Then to bissect the first and second angle, I'd use:

$$C=\cfrac{1}{2}(B'-A')+A'\quad \quad \quad D=\cfrac{1}{2}(B'+A')-A' $$

And hence show that it's inner product equals zero. But for some unknown reason, It's not working. I'm computing $\langle C,D \rangle$ it with Mathematica and obtaining:

$$\frac{x^2}{4 \left(a^2+b^2\right)}-\frac{a x}{4 \sqrt{a^2+b^2}}+\frac{x}{4 \sqrt{a^2+b^2}}+\frac{y^2}{4 \left(a^2+b^2\right)}+\frac{a}{4}-\frac{1}{2}$$

I don't know what is wrong.

Red Banana
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1 Answers1

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Let see what happened. $$C=\frac{1}{2}(B'+A') = \frac{1}{2}\left(1+\frac{x}{\sqrt{x^2+y^2}},\frac{y}{\sqrt{x^2+y^2}}\right).$$

$$D=\frac{1}{2}(B'-A') = \frac{1}{2}\left(-1+\frac{x}{\sqrt{x^2+y^2}},\frac{y}{\sqrt{x^2+y^2}}\right).$$

Then, $$CD = \frac{1}{4}\left( \frac{x^2}{x^2+y^2}-1 + \frac{y^2}{x^2+y^2}\right) = 0.$$

GAVD
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  • I still don't understand it, why did you omit the sum and subtraction of $A'$ for each vector? See:

    $$C=\cfrac{1}{2}(B'-A')+A'\quad \quad \quad D=\cfrac{1}{2}(B'+A')-A'$$

     – Red Banana Sep 08 '15 at 07:45