HSC 2015 MX2 Marathon (archive) (1 Viewer)

glittergal96 · Oct 14, 2015

Re: HSC 2015 4U Marathon

kawaiipotato said:
There's only one solution for ab + cd

I really do not think this is the case.

Let

$(a,b)=(\cos(\theta),\sin(\theta))$

and

$(c,d)=(\cos(\phi),\sin(\phi))$

The assumption is then that

$ac+bd=\cos(\theta-\phi)=0.$

One way to attain this is by taking

$\theta=\phi+\pi/2.$

Then

$ab+cd=\sin(\theta+\phi)=\sin(2\phi+\pi/2)$

which can take any value in the interval [-1,1].

Paradoxica · Oct 14, 2015

Re: HSC 2015 4U Marathon

l

glittergal96 said:
I really do not think this is the case.

Let

$(a,b)=(\cos(\theta),\sin(\theta))$

and

$(c,d)=(\cos(\phi),\sin(\phi))$

The assumption is then that

$ac+bd=\cos(\theta-\phi)=0.$

One way to attain this is by taking

$\theta=\phi+\pi/2.$

Then

$ab+cd=\sin(\theta+\phi)=\sin(2\phi+\pi/2)$

which can take any value in the interval [-1,1].

Matches the solution interval given by W|A. Perhaps, kawaiipotato, you have mistyped the question.

kawaiipotato · Oct 14, 2015

Re: HSC 2015 4U Marathon

glittergal96 said:
$ab+cd=\sin(\theta+\phi)=\sin(2\phi+\pi/2)$

Hmm isn't ab = sin(theta)cos(theta) and similarly for cd?
so ab + cd = sin(theta)cos(theta) + sin(phi)cos(phi)? not sin(theta)cos(phi) + sin(phi)cos(theta)

glittergal96 · Oct 14, 2015

Re: HSC 2015 4U Marathon

kawaiipotato said:
Hmm isn't ab = sin(theta)cos(theta) and similarly for cd?
si ab + cd = sin(theta)cos(theta) + sin(phi)cos(phi)? not sin(theta)cos(phi) + sin(phi)cos(theta)

Oh, my bad. (Sorry)

This quantity can then be simplified.

$\sin(\theta)\cos(\theta)+\sin(\phi)\cos(\phi)\\ =\frac{1}{2}(\sin(2\theta)+\sin(2\phi))\\ =\sin(\theta+\phi)\cos(\theta-\phi)\\=0.$

kawaiipotato · Oct 14, 2015

Re: HSC 2015 4U Marathon

glittergal96 said:
Sure, my bad. (Sorry)

This quantity can still be simplified.

$\sin(\theta)\cos(\theta)-\sin(\phi)\cos(\phi)\\ =\frac{1}{2}(\sin(2\theta)-\sin(2\phi))\\ =\sin(\theta-\phi)\cos(\theta+\phi)\\=\cos(2\phi+\pi/2).$

also it should be a + right not a -?

glittergal96 · Oct 14, 2015

Re: HSC 2015 4U Marathon

kawaiipotato said:
also it should be a + right not a -?

edited before I saw your reply haha. just wanted to get it down quickly and then fix typos.

kawaiipotato · Oct 14, 2015

Re: HSC 2015 4U Marathon

Good solution glittergal96
Also another method was:

Paradoxica · Oct 14, 2015

Re: HSC 2015 4U Marathon

Does there exist a solution involving the moduli of complex numbers?

kawaiipotato · Oct 14, 2015

Re: HSC 2015 4U Marathon

Paradoxica said:
Does there exist a solution involving the moduli of complex numbers?

I'm not sure. The only ones I saw were glittergal96's and the one I posted above.

Carrotsticks · Oct 14, 2015

Re: HSC 2015 4U Marathon

Paradoxica said:
Does there exist a solution involving the moduli of complex numbers?

Glittergal's solution implicitly does when expressing the points as functions of theta and phi, as they represent points on the unit circle.

Carrotsticks · Oct 14, 2015

Re: HSC 2015 4U Marathon

I suppose this also 'uses the modulus', but it is essentially backdooring glittergal's solution.

$\\ \frac{a+bi}{c+di} = i(bc-ad) $ and $ \frac{c+di}{a+bi} = i(ad-bc) $ which in laymans terms means that $a+ib$ and $c+id$ represent all perpendicular unit vectors, where the result $ ab+cd=0 $ comes out immediately$.$

Sy123 · Oct 16, 2015

Re: HSC 2015 4U Marathon

Here is a question I thought up while staring at a clock waiting for the time to pass:

$\\ $A clock-face has a hours hand and a minutes hand, there are marks around the clock for each hour and each minute. A specific time is 'straight' if the minute-hand and the hour-hand are each pointing at one of the marks, and the angle between the minute-hand and hour-hand is $ \pi $ radians.$$

$\\ $i) Show that on a regular clock, the only straight time is$ \ 6:00$

$\\ $ii) On the planet Veldin in the Solana galaxy, there are$ \ np \ $minutes to every hour (for some positive integers$ \ n,p $). The clocks display hours$ \ 1, 2, \dots, p $. Just like normal clocks, the minutes-hand on a Veldin clock completes one revolution in an hour, and the hour-hand moves appropriately.$$

$\\ $Show that:$ \\ \\ \alpha) $ If$ \ n \ $is divisible by$ \ (p-1) \ $and$ \ p \ $is even, or$ \ n \ $is divisible by$ \ 2p - 2 \ $then there are$ \ p-1 \ $possible straight times that a clock can show$ \\ \\ \beta) $ If the conditions in part $ \alpha $ are not satisfied, then if$ \ p \ $is even, there is only 1 straight time that the clock can show, and if not, then there is no straight time that the clock can show$$

kawaiipotato · Oct 16, 2015

Re: HSC 2015 4U Marathon

Sy123 said:
Here is a question I thought up while staring at a clock waiting for the time to pass:

$\\ $A clock-face has a hours hand and a minutes hand, there are marks around the clock for each hour and each minute. A specific time is 'straight' if the minute-hand and the hour-hand are each pointing at one of the marks, and the angle between the minute-hand and hour-hand is $ \pi $ radians.$$

$\\ $i) Show that on a regular clock, the only straight time is$ \ 6:00$

$\\ $ii) On the planet Veldin in the Solana galaxy, there are$ \ np \ $minutes to every hour (for some positive integers$ \ n,p $). The clocks display hours$ \ 1, 2, \dots, p $. Just like normal clocks, the minutes-hand on a Veldin clock completes one revolution in an hour, and the hour-hand moves appropriately.$$

$\\ $Show that:$ \\ \\ \alpha) $ If$ \ n \ $is divisible by$ \ (p-1) \ $and$ \ p \ $is even, or$ \ n \ $is divisible by$ \ 2p - 2 \ $then there are$ \ p \ $possible straight times that a clock can show$ \\ \\ \beta) $ If the conditions in part $ \alpha $ are not satisfied, then if$ \ p \ $is even, there is only 1 straight time that the clock can show, and if not, then there is no straight time that the clock can show$$

When you say there are 'np' minutes, does that mean there are n multiplied by p minutes or {n}{p} minutes

InteGrand · Oct 16, 2015

Re: HSC 2015 4U Marathon

kawaiipotato said:
When you say there are 'np' minutes, does that mean there are n multiplied by p minutes or {n}{p} minutes

$Means $n$ times $p$, for example, in our clocks, $p=12$ and $n=5$.$

Sy123 · Oct 16, 2015

Re: HSC 2015 4U Marathon

kawaiipotato said:
When you say there are 'np' minutes, does that mean there are n multiplied by p minutes or {n}{p} minutes

n*p minutes

InteGrand said:
$Means $n$ times $p$, for example, in our clocks, $p=12$ and $n=5$.$

^^

Drsoccerball · Oct 16, 2015

Re: HSC 2015 4U Marathon

I'll try this when im more awake

atargainz · Oct 17, 2015

Re: HSC 2015 4U Marathon

Can someone explain/do iv for me please? Thanks. no idea how to use the img tag btw

InteGrand · Oct 17, 2015

Re: HSC 2015 4U Marathon

atargainz said:
Can someone explain/do iv for me please? Thanks. no idea how to use the img tag btw

View attachment 32554

$Call the polynomial in question $P$. We know that $P$ cannot have four real roots. Since $P$ has all real coefficients, by the conjugate root theorem (and the fact it is degree 4, so has four roots in total), it must have either four non-real complex roots and 0 real roots, or two non-real complex roots and two real roots (it can't have 0 non-real complex roots and 4 real roots, as proved in (iii)). Evaluating $P(0)$ and $P(1)$, we see that $P(x)$ changes sign between $0$ and $1$, so it must have a real root (as it is a continuous function). Hence we can rule out the possibility of it having four non-real complex roots and 0 real roots, which means it has two real roots (and two non-real complex roots). $\blacksquare$$

atargainz · Oct 17, 2015

Re: HSC 2015 4U Marathon

InteGrand said:
$Call the polynomial in question $P$. We know that $P$ cannot have four real roots. Since $P$ has all real coefficients, by the conjugate root theorem (and the fact it is degree 4, so has four roots in total), it must have either four non-real complex roots and 0 real roots, or two non-real complex roots and two real roots (it can't have 0 non-real complex roots and 4 real roots, as proved in (iii). Evaluating $P(0)$ and $P(1)$, we see that $P(x)$ changes sign between $0$ and $1$, so it must have a real root (as it is a continuous function). Hence we can rule out the possibility of it having four non-real complex roots and 0 real roots, which means it has two real roots (and two non-real complex roots). $\blacksquare$$

omg thanks, can't believe I didn't see that fml

mreditor16 · Oct 17, 2015

Re: HSC 2015 4U Marathon

A relatively nice complex numbers Q for you all -

x, y and z are complex numbers. If 0, x, y and z are concyclic, prove that 1/x, 1/y and 1/z are collinear!

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