I Micromass' big October challenge

[Charles Link]

I think you've got the basic idea=the resultant of $A +B$ and $C+D$ which lies along their respective angle bisectors must be equal and opposite. One question I have for you that I'm not sure you answered completely: What if $A+B$ and $C+D$ lie in opposite directions, (along with the angle bisectors from these vectors), as is required, does it guarantee that the angle between $A$ and $B$ is the same as the angle between $C$ and $D$? Let the angle between $A$ and $B$ be $\theta_1$, and the angle between $C$ and $D$ be $\theta_2$ where $\theta_1 \neq \theta_2$. Can we still have for that case $A+B =-(C+D)$? i.e. can we have |A+B|=|C+D|? Why or why not?

 

[EpidermalOblivion]

I think you've got the basic idea=the resultant of $A +B$ and $C+D$ which lies along their respective angle bisectors must be equal and opposite. One question I have for you that I'm not sure you answered completely: What if $A+B$ and $C+D$ lie in opposite directions, (along with the angle bisectors from these vectors), as is required, does it guarantee that the angle between $A$ and $B$ is the same as the angle between $C$ and $D$? Let the angle between $A$ and $B$ be $\theta_1$, and the angle between $C$ and $D$ be $\theta_2$ where $\theta_1 \neq \theta_2$. Can we still have for that case $A+B =-(C+D)$? i.e. can we have |A+B|=|C+D|? Why or why not?

Yes, this is necessary. I mentioned that when the magnitudes of two opposite sides are altered together, the angles within each of the two pairs of numbers are increased/decreased by the same amount. Any other translation that would result in different angles is impossible. I edited my main proof to include a more in-depth explanation of this.

 

[Charles Link]

Qualitatively speaking, "increasing the angle..affects the sum..." , your logic is basically correct, but can you express it mathematically?: i.e. Given $|A|=|B|=1$ with an angle $\theta_1$ between them, please compute the length of the vector sum $|A+B |$. The computation involves just a little trigonometry. Have you taken a course in trigonometry yet?

 

[EpidermalOblivion]

Qualitatively speaking, "increasing the angle..affects the sum..." , your logic is basically correct, but can you express it mathematically?: i.e. Given $|A|=|B|=1$ with an angle $\theta_1$ between them, please compute the length of the vector sum $|A+B |$. The computation involves just a little trigonometry. Have you taken a course in trigonometry yet?

I think that would be $\sqrt{(\cos\theta+1)^2+sin^2\theta}$, which can be simplified to $\sqrt{2\cos\theta+2}$.

 

[Charles Link]

I think that would be $\sqrt{(\cos\theta+1)^2+sin^2\theta}$, which can be simplified to $\sqrt{2\cos\theta+2}$.

Yes, that is correct. And for the amplitudes $|A+B|$ and $|C+D|$ to be equal, what can you say about $\theta_1$ and $\theta_2$? ... From what I can see, I think you have successfully solved it...Hopefully @mfb will concur.

 

[mfb]

I agree.

They key element is the unique way to get a (non-zero) sum. With that, you can skip all the discussion of the rotations. Every group of 4 numbers will have pairs with opposite non-zero sums, and those opposite sums uniquely identify the elements, with the same angles for both pairs. Done.