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

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ParthKohli
 one year ago
Best ResponseYou've already chosen the best response.0If \(m^2 + n^2 = 1\), \(p^2 + q^2 = 1\) and \(mp + nq = 0\) then what is \(mn + pq\)?

ParthKohli
 one year ago
Best ResponseYou've already chosen the best response.0Meh  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.

AbdullahM
 one year ago
Best ResponseYou've already chosen the best response.1I assume we will need to use trignometry to solve it?

ParthKohli
 one year ago
Best ResponseYou've already chosen the best response.0That sounds like a good way to do it... in the system of real numbers.

rational
 one year ago
Best ResponseYou've already chosen the best response.4nvm was working assuming mp+nq=1

ParthKohli
 one year ago
Best ResponseYou've already chosen the best response.0Haha, I'm really sorry about that.

AbdullahM
 one year ago
Best ResponseYou've already chosen the best response.1\(\sf m^2+n^2=1\) 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

ParthKohli
 one year ago
Best ResponseYou've already chosen the best response.0Yes, of course. But don't forget that there's a huge world of things that aren't real numbers.

rational
 one year ago
Best ResponseYou've already chosen the best response.4Nice! here is another alternative that looks kinda neat \[mn+pq=mn(p^2+q^2)+(m^2+n^2)pq =(mp+nq)(np+mq)=0\]

ParthKohli
 one year ago
Best ResponseYou've already chosen the best response.0That's exactly the solution I was about to post. Great job.

ParthKohli
 one year ago
Best ResponseYou've already chosen the best response.0Where did you find this?

rational
 one year ago
Best ResponseYou've already chosen the best response.4copied it from quora, there are plenty of solutions there but i liked that one the most

ParthKohli
 one year ago
Best ResponseYou've already chosen the best response.0Haha, Anders Kaseorg. Same.

ParthKohli
 one year ago
Best ResponseYou've already chosen the best response.0He's just brilliant. I've been stalking him for a few days.

AbdullahM
 one year ago
Best ResponseYou've already chosen the best response.1Oooh, interesting :o

mathmath333
 one year ago
Best ResponseYou've already chosen the best response.0looks somewhat similar to this https://en.wikipedia.org/wiki/Brahmagupta's_identity
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