## ParthKohli one year ago Hey, this problem has a really elegant solution.

1. ParthKohli

If $$m^2 + n^2 = 1$$, $$p^2 + q^2 = 1$$ and $$mp + nq = 0$$ then what is $$mn + pq$$?

2. ParthKohli

Meh - there's actually nothing elegant about it. The guy just made it look like that. But still, it looks really clever and let's see if you guys can come up with it.

3. AbdullahM

I assume we will need to use trignometry to solve it?

4. ParthKohli

That sounds like a good way to do it... in the system of real numbers.

5. rational

nvm was working assuming mp+nq=1

6. ParthKohli

Haha, I'm really sorry about that.

7. AbdullahM

$$\sf m^2+n^2=1$$ There is a trig identity: $$\sf cos^2(x)+sin^2(x)=1$$ So then: $$\sf m= cos~(\alpha)$$ $$\sf n=sin~(\alpha)$$ Do the same for: $$\sf p^2+q^2=1$$ $$\sf p= cos~(\beta)$$ $$\sf q=sin~(\beta)$$ Now, they gave us mp+nq=0 Substitute the values we got above into this second equation: $$\sf mp+nq=0$$ $$\sf cos(\alpha)cos(\beta)+sin(\alpha)sin(\beta)=0$$ And there is a trig identity: $$\sf cos(\alpha)cos(\beta)+sin(\alpha)sin(\beta)=cos(\alpha-\beta)$$ so we can simplify that further down to: $$\sf cos(\alpha-\beta)=0$$ I hope I'm going in the correct direction so far xD

8. ParthKohli

Yes, of course. But don't forget that there's a huge world of things that aren't real numbers.

9. rational

Nice! here is another alternative that looks kinda neat $mn+pq=mn(p^2+q^2)+(m^2+n^2)pq =(mp+nq)(np+mq)=0$

10. ParthKohli

That's exactly the solution I was about to post. Great job.

11. ParthKohli

Where did you find this?

12. rational

copied it from quora, there are plenty of solutions there but i liked that one the most

13. ParthKohli

Haha, Anders Kaseorg. Same.

14. ParthKohli

He's just brilliant. I've been stalking him for a few days.

15. AbdullahM

Oooh, interesting :o

16. mathmath333

looks somewhat similar to this https://en.wikipedia.org/wiki/Brahmagupta's_identity

17. ParthKohli

Haha, yes. :)