Question

Hey, this problem has a really elegant solution.

Answer 1

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

Answer 2

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.

Answer 3

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

Answer 4

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

Answer 5

nvm was working assuming mp+nq=1

Answer 6

Haha, I'm really sorry about that.

Answer 7

\(\sf m^2+n^2=1\)

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

Answer 8

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

Answer 9

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

Answer 10

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

Answer 11

Where did you find this?

Answer 12

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

Answer 13

Haha, Anders Kaseorg. Same.

Answer 14

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

Answer 15

Oooh, interesting :o

Answer 16

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

Answer 17

Haha, yes. :)