HSC 2015 MX2 Marathon (archive) (1 Viewer)

kawaiipotato · Oct 17, 2015

Re: HSC 2015 4U Marathon

mreditor16 said:
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!

____________________________________________
$New question$
$$ let $ f(x) = x^4 + ax^3 + bx^2 + ax + 1 = 0$

$If f(x) has at least one real root for real numbers a,b then find the minimum value of a^2 + b^2$

Drsoccerball · Oct 17, 2015

Re: HSC 2015 4U Marathon

mreditor16 said:
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!

Can't we just say that x=-1/x and it a cyclic quad forms with x then it will also form for 1/x ... Or is that what were trying to prove LOL

InteGrand · Oct 17, 2015

Re: HSC 2015 4U Marathon

Drsoccerball said:
Can't we just say that x=-1/x and it a cyclic quad forms with x then it will also form for 1/x ... Or is that what were trying to prove LOL

$Why is $x$ equal to $-\frac{1}{x}$?$

Drsoccerball · Oct 17, 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$$

Cant 12:30 be straight time? how about 1:35 ?

InteGrand · Oct 17, 2015

Re: HSC 2015 4U Marathon

Drsoccerball said:
Cant 12:30 be straight time? how about 1:35 ?

No, since the hour hand moves continuously as well. For example, at 12.30, the hour hand is halfway between 12 and 1.

Drsoccerball · Oct 17, 2015

Re: HSC 2015 4U Marathon

InteGrand said:
No, since the hour hand moves continuously as well. For example, at 12.30, the hour hand is halfway between 12 and 1.

are both hands moving at the same speed?

InteGrand · Oct 17, 2015

Re: HSC 2015 4U Marathon

Drsoccerball said:
are both hands moving at the same speed?

Of course not haha, they're moving like our clocks.

Librah · Oct 17, 2015

Re: HSC 2015 4U Marathon

Drsoccerball said:
are both hands moving at the same speed?

lmao

InteGrand · Oct 17, 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$$

Are you sure it's p possible straight times for part α? I'm getting (p – 1) straight times. (Might have screwed up somewhere.)

Sy123 · Oct 17, 2015

Re: HSC 2015 4U Marathon

InteGrand said:
Are you sure it's p possible straight times for part α? I'm getting (p – 1) straight times. (Might have screwed up somewhere.)

I got a straight time for every hour, so p straight times

PM me your working out

InteGrand · Oct 17, 2015

Re: HSC 2015 4U Marathon

Sy123 said:
I got a straight time for every hour, so p straight times

PM me your working out

It can't be a straight time for 0 o'clock (i.e. p o'clock) though can it? Since the minute hand and hour hand should be both pointing up then, rather than in opposite directions. Edit: wait I get what you mean now. I'll have another look.

Drsoccerball · Oct 17, 2015

Re: HSC 2015 4U Marathon

Ill do i

$Assuming that the clock is perfectly spherical and the minutes are perfectly separated we can construct 12 triangle with sides two of the hour marks eg: One at 1 and 2, One at 5 and 6 NOT 5 and 8...$

$$ Each angle is $ \frac{\pi}{6}$

$\frac{\pi n}{6} = \pi$

$n=6 $ is the only value of 'straight time' well at least for you guys...$$

InteGrand · Oct 17, 2015

Re: HSC 2015 4U Marathon

Drsoccerball said:
Ill do i

$Assuming that the clock is perfectly spherical and the minutes are perfectly separated we can construct 12 triangle with sides two of the hour marks eg: One at 1 and 2, One at 5 and 6 NOT 5 and 8...$

$$ Each angle is $ \frac{\pi}{6}$

$\frac{\pi n}{6} = \pi$

$n=6 $ is the only value of 'straight time' well at least for you guys...$$

$The clock hands can be at minute marks too, which have angular separation from each other of $\frac{2\pi}{60}$ rather than $\frac{\pi}{6}$. Also not sure I understand your working. It's not possible for the hour hand to be at an hour mark and the minute hand to be at a minute mark other than the 12 o'clock mark. (The clock works like how ordinary clocks work.)$

Drsoccerball · Oct 17, 2015

Re: HSC 2015 4U Marathon

InteGrand said:
$The clock hands can be at minute marks two, which have angular separation from each other of $\frac{2\pi}{60}$ rather than $\frac{\pi}{6}$. Also not sure I understand your working. It's not possible for the hour hand to be at an hour mark and the minute hand to be at a minute mark other than the 12 o'clock mark. (The clock works like how ordinary clocks work.)$

I tried to show that theres only one value for n such that the angles of the triangles added up equal to pi but i guess i excluded the minutes as you pointed out.

InteGrand · Oct 17, 2015

Re: HSC 2015 4U Marathon

Drsoccerball said:
I tried to show that theres only one value for n such that the angles of the triangles added up equal to pi but i guess i excluded the minutes as you pointed out.

What do you mean angles though? We can't place the hands of the clock arbitrarily on the clock, since their movement is governed by how clocks work. E.g. it is impossible to have the hour hand at 2 and the minute hand at 5 at the same instant.

Drsoccerball · Oct 17, 2015

Re: HSC 2015 4U Marathon

InteGrand said:
What do you mean angles though? We can't place the hands of the clock arbitrarily on the clock, since their movement is governed by how clocks work. E.g. it is impossible to have the hour hand at 2 and the minute hand at 5 at the same instant.

LOOOOOOOOOOOOOOOOOOOOOOOOOL my bad.

Paradoxica · Oct 17, 2015

Re: HSC 2015 4U Marathon

11 straight times can only occur if the clock is continuous (established already). The clock, however is not continuous.
The straight times on a continuous clock occur in equal elapses of time (Provable, although I don't see how to do it yet). This means that on a 12-clock, the straight lines occur every 1440/11 minutes. Ten of these cannot be integer times, and only one can be integral. Since we have established that 6:00 is a straight line, it is the only straight line on the clock.

Sy123 · Oct 17, 2015

Re: HSC 2015 4U Marathon

InteGrand has pointed out an error in the question, for part ii.a there are (p-1) straight times, not p straight times, edited

-------------------------

$\\ $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$$

InteGrand · Oct 17, 2015

Re: HSC 2015 4U Marathon

Sy123 said:
InteGrand has pointed out an error in the question, for part ii.a there are (p-1) straight times, not p straight times, edited

-------------------------

$\\ $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$$

$Well this was my attempt for the general case.$

$(ii) Note that the clock has $np$ minute marks, which we can label $0,1,2,...,np-1$ going clockwise from the top. Also, we should have $p\geq 2$; if $p=1$, the clock doesn't really make sense lol. Also note that it is only possible to have straight times if $np$ (the no. of minute marks) is even, since if $np$ is odd, then there do not exist diametrically opposite minute marks on the clock face, so we can't get straight times at all, even if we could place the hands on the clock arbitrarily.$

$$\alpha$) Now, suppose we start at $0$ o'clock, so both hands are pointing `up' at the start. One `period' for the clock is $p$ hours (12 hours in our world). We consider the movement of the clock in one period. Consider when the hour hand is at the $x^{\text{th}}$ minute mark ($x=0,1,2,...,np-1$). In one cycle, there is one and only one time when this happens, so the position of the minute hand is unique for each $x$. The minute hand moves $p$ times as fast as the hour hand, so it is at position $px^{\text{th}}$ minute mark (take it modulo $np$). Now, we have a straight time if and only if the difference between the two hands' positions is $\frac{np}{2}$ (i.e. half a circle), i.e. we need$

$px-x \equiv \frac{np}{2}\mod (np)$

$$\Longleftrightarrow (p-1)x \equiv \frac{np}{2}\mod (np)\ \ (*)$.$

$Note that if $n$ is divisible by $(p-1)$ and $p$ is even, we have $np$ even (so it is in principle possible to have straight times), and also we have $(p-1) | np$. So, $\gcd (p-1,np)=1$, and also $(p-1)| n \cdot \frac{p}{2}$. Thus we can divide the entire congruence (*) through to get$

$x\equiv \frac{np}{2(p-1)}\mod \left(\frac{np}{p-1} \right)\ \ (**)$

$We again obtain (**) if $n$ is divisible by $2p-2=2(p-1)$, since then, $np$ is divisible by $p-1$ and so is $\frac{np}{2}$, as $\frac{np}{2(p-1)}$ is integral. So in either of the cases, we obtain the congruence (**), which has exactly one solution modulo $\frac{np}{p-1}$, and hence $p-1$ solutions modulo $np$. Hence there are $(p-1)$ straight times.$

$$\beta )$ If the conditions in $\alpha)$ do not hold, I think we can still have multiple straight times. For example, take $p=10,n=12$. Then $p-1=9,2p-2=18$ and we see that the conditions in $\alpha)$ are not satisfied. But our congruence (*) is $9x\equiv 60 \mod (120)$. Since $\gcd(9,120)=3$ and this divides 60, this congruence will have 3 solutions modulo 120. I think the conditions to have exactly one straight time should be that $np$ be even, and $(p-1)x\equiv \frac{np}{2}\mod (np)$ have a unique solution modulo $np$, i.e. $\gcd(p-1,np)=1\Longleftrightarrow \gcd(p-1,n)=1$ (as $\gcd(p-1,p)=1$ for all positive integers $p$ already, so $\gcd(p-1,np)=\gcd(p-1,p)$). The condition to have no straight times is that $\gcd(p-1,np)\nmid \frac{np}{2}\Longleftrightarrow \gcd(p-1,n)\nmid \frac{np}{2}$ (or have $np$ be odd).$

relativity1 · Oct 17, 2015

Re: HSC 2015 4U Marathon

Please dont tell me this is a typical 4u question...

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