ParthKohli
  • ParthKohli
Hey, this problem has a really elegant solution.
Mathematics
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SOLVED
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katieb
  • katieb
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ParthKohli
  • ParthKohli
If \(m^2 + n^2 = 1\), \(p^2 + q^2 = 1\) and \(mp + nq = 0\) then what is \(mn + pq\)?
ParthKohli
  • 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.
AbdullahM
  • AbdullahM
I assume we will need to use trignometry to solve it?

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ParthKohli
  • ParthKohli
That sounds like a good way to do it... in the system of real numbers.
rational
  • rational
nvm was working assuming mp+nq=1
ParthKohli
  • ParthKohli
Haha, I'm really sorry about that.
AbdullahM
  • 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
ParthKohli
  • ParthKohli
Yes, of course. But don't forget that there's a huge world of things that aren't real numbers.
rational
  • rational
Nice! here is another alternative that looks kinda neat \[mn+pq=mn(p^2+q^2)+(m^2+n^2)pq =(mp+nq)(np+mq)=0\]
ParthKohli
  • ParthKohli
That's exactly the solution I was about to post. Great job.
ParthKohli
  • ParthKohli
Where did you find this?
rational
  • rational
copied it from quora, there are plenty of solutions there but i liked that one the most
ParthKohli
  • ParthKohli
Haha, Anders Kaseorg. Same.
ParthKohli
  • ParthKohli
He's just brilliant. I've been stalking him for a few days.
AbdullahM
  • AbdullahM
Oooh, interesting :o
mathmath333
  • mathmath333
looks somewhat similar to this https://en.wikipedia.org/wiki/Brahmagupta's_identity
ParthKohli
  • ParthKohli
Haha, yes. :)

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