Nice complex numbers question. (1 Viewer)

Comeeatmebro · Jul 8, 2011

Wow those are some amazing marks that you achieved! And no I've got no idea for the question

Hermes1 · Jul 8, 2011

vector PR = PQ cis 60
vector RQ = PQ cis -60

therefore,
r-p = (q-p)cis 60
q-r = (q-p)cis -60

cis 60 = r-p/q-p
cis -60 = q-r/q-p

cis 60 x cis -60 = r-p/q-p x q-r/q-p
therefore,
(r-p)(q-r)/(q-p)^2 = 1

simplify from here to get above answer

nice question, this is my favourite part of complex numbers the vectors.

seanieg89 · Jul 8, 2011

Thankyou comeeatmebro and good work arj, although you have only proven the 'if' statement, not the 'only if' statement. The latter is slightly harder.

gurmies · Jul 8, 2011

arj1 said:
vector PR = PQ cis 60
vector RQ = PQ cis -60

therefore,
r-p = (q-p)cis 60
q-r = (q-p)cis -60

cis 60 = r-p/q-p
cis -60 = q-r/q-p

cis 60 x cis -60 = r-p/q-p x q-r/q-p
therefore,
(r-p)(q-r)/(q-p)^2 = 1

simplify from here to get above answer

nice question, this is my favourite part of complex numbers the vectors.

I don't think that's sufficient for an iff proof.

[EDIT: seanieg89 has addressed this already]

Hermes1 · Jul 8, 2011

seanieg89 said:
Thankyou comeeatmebro and good work arj, although you have only proven the 'if' statement, not the 'only if' statement. The latter is slightly harder.

how would you do the only if

gurmies · Jul 8, 2011

arj1 said:
how would you do the only if

Think about proving the other direction - i.e. don't assume that p, q, r are vertices of an equilateral triangle.

seanieg89 · Jul 8, 2011

Assume that the given equation holds and try to deduce that this forces the three points to be vertices of an equilateral triangle.
It isn't easy, but I will wait for more people to attempt it before i give any big hints.

Trebla · Jul 9, 2011

There's probably a shorter way but here's what I did

$Consider $(p-q)^3+(r-q)^3\\\\=(p-2q+r)[(p-q)^2-(p-q)(r-q)+(r-q)^2]\\\\=(p-2q+r)(p^2+q^2-2pq-pr+pq+qr-q^2+r^2-2qr+q^2)\\\\=(p-2q+r)(p^2+q^2+r^2-pq-pr-qr)\\\\=0 \quad $ according to the result $\\\\$Hence $\\\\-(p-q)^3=(r-q)^3\\\\\Rightarrow (p-q)^3(\cos\pi +i\sin\pi)=(r-q)^3 \\\\\Rightarrow \left[(p-q)\left(\cos\frac{\pi}{3}+i\sin\frac{\pi}{3}\right)\right]^3=(r-q)^3\\\\\Rightarrow (p-q)\left(\cos\frac{\pi}{3}+i\sin\frac{\pi}{3}\right)=(r-q)\\\\$By vector interpretation this means that vector $QP$ rotated $\frac{\pi}{3}$ anticlockwise leads to vector $QR$ with no change in length. Thus triangle $PQR$ is an isosceles triangle with one interior angle as $\frac{\pi}{3}$. This immediately implies that triangle $PQR$ must also be an equilateral triangle as well.$

cutemouse · Jul 9, 2011

Triangle PQR is equilateral iff |PQ|=|QR|=|RP| and arg (PR) - arg (PQ) = arg(QP) - arg(QR)

iff |q-p| = |r-q| = |p-r| and arg(r-p)-arg(q-p)=arg(p-q)-arg(r-q)

iff |q-p| = |r-q| = |p-r| and arg[(r-p)/(q-p)] = arg[(p-q)/(r-q)]

iff (r-p)/(q-p)=(p-q)/(r-q)

iff (r-p)(r-q)= -(p+q)^2 , q=/=p and r=/=q

iff r^2 - rq - rp +pq = - (p^2 - 2pq + q^2)

iff p^2 + q^2 + r^2 = pq + rq + rp QED

Nice complex numbers question. (1 Viewer)

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