HSC 2014 MX2 Marathon ADVANCED (archive) (3 Viewers)

abecina · Oct 18, 2014

Re: HSC 2014 4U Marathon - Advanced Level

Nice. I just left the t business till the end and Found the necessary distance between the two hands first i.e. \sqrt(7) Otherwise, pretty much exactly the same as yours.

Sy123 · Oct 26, 2014

Re: HSC 2014 4U Marathon - Advanced Level

$\\ a,b \ $are integers, if$ \ b \ $divides$ \ a $, show that$ \ 2^b - 1 \ $divides$ \ 2^a - 1$

glittergal96 · Oct 26, 2014

Re: HSC 2014 4U Marathon - Advanced Level

Sy123 said:
$\\ a,b \ $are integers, if$ \ b \ $divides$ \ a $, show that$ \ 2^b - 1 \ $divides$ \ 2^a - 1$

I've finished MX2 now, but this thread is too fun lol.

$$If $a=bk$ where $k$ is a positive integer, then $\\ \\ 2^a-1=2^{bk}-1=\sum_{j=1}^k 2^{bj}-2^{b(j-1)}=(2^b-1)\sum_{j=1}^k 2^{b(j-1)}.$

Axio · Oct 27, 2014

Re: HSC 2014 4U Marathon - Advanced Level

glittergal96 said:
I've finished MX2 now, but this thread is too fun lol.

$$If $a=bk$ where $k$ is a positive integer, then $\\ \\ 2^a-1=2^{bk}-1=\sum_{j=1}^k 2^{bj}-2^{b(j-1)}=(2^b-1)\sum_{j=1}^k 2^{b(j-1)}.$

Can I ask how you know to turn that statement into that sum?

Sy123 · Oct 27, 2014

Re: HSC 2014 4U Marathon - Advanced Level

Axio said:
Can I ask how you know to turn that statement into that sum?

$1 + z^k + z^{2k} + z^{3k} + \dots + z^{(n-1)k} = \frac{z^{nk}-1}{z^k-1}$

RealiseNothing · Oct 28, 2014

Re: HSC 2014 4U Marathon - Advanced Level

Sy123 said:
$\\ a,b \ $are integers, if$ \ b \ $divides$ \ a $, show that$ \ 2^b - 1 \ $divides$ \ 2^a - 1$

It's just in the form of:

$\frac{2^{ak}-1}{2^a-1} = 1+2^a+2^{2a}+...+2^{a(k-1)}$

Chlee1998 · Oct 30, 2014

Re: HSC 2014 4U Marathon - Advanced Level

Prove that the tangents are opposite vertices of a cyclic quadrilateral intersect on the secant through the other two verticies IF AND ONLY IF the two products of opposite sides of the cyclic quadrilateral are equal (so prove both ways)

Axio · Oct 30, 2014

Re: HSC 2014 4U Marathon - Advanced Level

Chlee1998 said:
Prove that the tangents are opposite vertices of a cyclic quadrilateral intersect on the secant through the other two verticies IF AND ONLY IF the two products of opposite sides of the cyclic quadrilateral are equal (so prove both ways)

Does this work?

Let A, B, C and D be the sides of a cyclic quadrilateral and E be an external point and let two equal tangents from the external point E meet the circle at A and C.

Any secant that passes through E, B and D is a diagonal to the cyclic quadrilateral ABCD.

Hence AB/AD=BC/DC, making (AB)(DC)=(AD)(BC).

Going backwards: assume (AB)(DC)=/=(AD)(BC).

Therefore triangles are not similar (unless (AB)(BC)=(AD)(DC)).

Therefore ABCD is not a cyclic quadrilateral, which is a contradiction.

Chlee1998 · Oct 30, 2014

Re: HSC 2014 4U Marathon - Advanced Level

Axio said:
Does this work?

Let A, B, C and D be the sides of a cyclic quadrilateral and E be an external point and let two equal tangents from the external point E meet the circle at A and C.

Any secant that passes through E, B and D is a diagonal to the cyclic quadrilateral ABCD.

Therefore triangle ABC is similar to triangle ADC (opposite pairs of triangles in a cyclic quadrilateral are similar).

Hence AB/AD=BC/DC, making (AB)(DC)=(AD)(BC).

Going backwards: assume (AB)(DC)=/=(AD)(BC).

Therefore triangles are not similar (unless (AB)(BC)=(AD)(DC)).

Therefore ABCD is not a cyclic quadrilateral, which is a contradiction.

firstly, i don't think you know what if and only if means.
secondly, I think you're making wrong assumptions, for example opposite triangles in cyclic quads aren't necessarily similar.

in other words, you've only proven it only one way.
For backwards, you're supposed to explicitly prove that if (AB)(DC)=(AD)(BC), then the inteserction of the tangents at A and C go through the secant of B and D, which you've NOT done. What you've proven is that if tangents at A and C go through secant of B and D, then (AB)(DC)=(AD)(BC)

So your answer is incomplete. You didn't even do the hard bit lol

Axio · Oct 30, 2014

Re: HSC 2014 4U Marathon - Advanced Level

Chlee1998 said:
firstly, i don't think you know what if and only if means.
secondly, I think you're making wrong assumptions, for example opposite triangles in cyclic quads aren't necessarily similar.

in other words, you've only proven it only one way.
For backwards, you're supposed to explicitly prove that if (AB)(DC)=(AD)(BC), then the inteserction of the tangents at A and C go through the secant of B and D, which you've NOT done. What you've proven is that if tangents at A and C go through secant of B and D, then (AB)(DC)=(AD)(BC)

So your answer is incomplete. You didn't even do the hard bit lol

Lol I just realised what I was doing wrong with the similar triangles :S...

Chlee1998 · Oct 31, 2014

Re: HSC 2014 4U Marathon - Advanced Level

Can anyone do the Q?

glittergal96 · Oct 31, 2014

Re: HSC 2014 4U Marathon - Advanced Level

Chlee1998 said:
Can anyone do the Q?

Sure.

So let ABCD be the cyclic quad in anticlockwise order, and we are drawing tangents at A and C.

Assume these tangents meet at E on BD (without loss of generality we can assume E is on the side of D.)

Then the alternate segment theorem tells us triangles EAD and EBA are similar, and also ECD and EBC are similar.

So AE/BE=AD/AB and CE/BE=CD/BC.

As AE and CE are both tangents to the same circle from the same point, they are equal, and we can conclude AD/AB = CD/BC => AD.BC=AB.CD.

Conversely, suppose AD.BC=AB.CD.

Let X be the intersection of the tangent at A with BD and Y be the intersection of the tangent at C with BD.

Then we get XAD and XBA similar and YCD and YBC similar.

Hence DX/BX=(AX/BX).(DX/AX)=(AD/AB)^2=(CD/BC)^2=(CY/BY).(DY/CY)=DY/BY.

But the quantity DP/BP increases as you move P further away from the circle, so we must have X=Y as claimed.

(Note the this is only truly if and only if we interpret parallel lines as meeting at infinity, and we then allow for the tangents to meet the secant at equal distances away from the circle on either side. Eg a square obviously has products of opposite sides equal, but tangents at opposite vertices are parallel.)

Sy123 · Nov 1, 2014

Re: HSC 2014 4U Marathon - Advanced Level

$\\ $Find all integer solutions$ \ (x,y) \ $where$ \ x^y = y^x \ $and$ \ x \neq y$

glittergal96 · Nov 1, 2014

Re: HSC 2014 4U Marathon - Advanced Level

Sy123 said:
$\\ $Find all integer solutions$ \ (x,y) \ $where$ \ x^y = y^x \ $and$ \ x \neq y$

Cool question! I think I have covered cases correctly:

First note that there is no solution with either of x,y equal to 0.

Case A: x negative, y positive

Unless y=1, only one side will be an integer. A trivial check shows there is no solution with y=1 and x negative.

Case B: x, y both negative.

By factoring out the (-1)^x and (-1)^y, we get a 1-1 correspondence between negative solution pairs and positive solution pairs with both variables having the same parity. This reduces the problem to...

Case C: x,y positive integers.

Wlog assume x < y.

By inspection there is no solution with x=1.

The condition is then equivalent to x/log(x)=y/log(y).

But differentiating shows that f(t)=t/log(t) decreases from 1 to e and increases thereafter, so we must have 1 < x < e. So x=2 is the only possible source of solutions and there is exactly one real solution to x^2=2^x in the interval (e,infinity).

As it happens, this solution is an integer! It is 4.

Hence the only solution pairs are (2,4),(4,2),(-2,-4),(-4,-2).

Sy123 · Nov 1, 2014

Re: HSC 2014 4U Marathon - Advanced Level

glittergal96 said:
Cool question! I think I have covered cases correctly:

First note that there is no solution with either of x,y equal to 0.

Case A: x negative, y positive

Unless y=1, only one side will be an integer. A trivial check shows there is no solution with y=1 and x negative.

Case B: x, y both negative.

By factoring out the (-1)^x and (-1)^y, we get a 1-1 correspondence between negative solution pairs and positive solution pairs with both variables having the same parity. This reduces the problem to...

Case C: x,y positive integers.

Wlog assume x < y.

By inspection there is no solution with x=1.

The condition is then equivalent to x/log(x)=y/log(y).

But differentiating shows that f(t)=t/log(t) decreases from 1 to e and increases thereafter, so we must have 1 < x < e. So x=2 is the only possible source of solutions and there is exactly one real solution to x^2=2^x in the interval (e,infinity).

As it happens, this solution is an integer! It is 4.

Hence the only solution pairs are (2,4),(4,2),(-2,-4),(-4,-2).

Yep pretty similar to my solution!

Quite interesting how numbers like 4 and 2 can come out of something that seems so simple

glittergal96 · Nov 1, 2014

Re: HSC 2014 4U Marathon - Advanced Level

Sy123 said:
Yep pretty similar to my solution!

Quite interesting how numbers like 4 and 2 can come out of something that seems so simple

Definitely! It feels more like a coincidence than anything actually deep, but still.

Chlee1998 · Nov 3, 2014

Re: HSC 2014 4U Marathon - Advanced Level

Prove that (2^p)-1 is coprime to (2^q)-1 if p and q are distinct primes. Use HSC methods

Sy123 · Nov 7, 2014

Re: HSC 2014 4U Marathon - Advanced Level

Here is a nice probability question

$\\ $There are$ \ n \ $coins called$ \ C_1, C_2, \dots, C_n \ $and they are all biased so that the probability of$ \ C_k \ $landing heads is$ \ \frac{1}{2k+1} \\ \\ $Find the probability in terms of$ \ n \ $of flipping these$ \ n \ $coins and getting an odd number of heads$$

glittergal96 · Nov 7, 2014

Re: HSC 2014 4U Marathon - Advanced Level

Chlee1998 said:
Prove that (2^p)-1 is coprime to (2^q)-1 if p and q are distinct primes. Use HSC methods

I'll prove the stronger result that

$a|b\Leftrightarrow 2^a-1| 2^b-1$

for positive integers a and b. Note that if this is true, then any common factor of your two quantities must also be a common factor of two distinct primes, and hence can only be 1.

The $(\Rightarrow)$ part was proven in the last page of this marathon. I'll show the other direction now.

We can write b=aq+r uniquely, with q,r non-negative integers and $0\leq r < a$

If $2^a-1|2^b-1$ , then $2^a-1|((2^b-1)-(2^{aq}-1))$ from the $(\Rightarrow)$ result.

So $2^a-1|2^{aq}(2^r-1)$ .

But since the LHS is odd, it is coprime to any power of 2. So we get

$2^a-1 | 2^r-1.$

As the RHS is strictly less than the LHS, this is only possible if the RHS is 0. Ie r=0, which means that $a|b$ as claimed.

glittergal96 · Nov 7, 2014

Re: HSC 2014 4U Marathon - Advanced Level

Sy123 said:
Here is a nice probability question

$\\ $There are$ \ n \ $coins called$ \ C_1, C_2, \dots, C_n \ $and they are all biased so that the probability of$ \ C_k \ $landing heads is$ \ \frac{1}{2k+1} \\ \\ $Find the probability in terms of$ \ n \ $of flipping these$ \ n \ $coins and getting an odd number of heads$$

Let p(n) be the probability of flipping an odd number of heads with n such coins.

Let q(n)=(2n+1)p(n).

Now if flipping (n+1) coins is to give an odd number of heads, either our final coin is a head and we have an even number of heads in our first n coins, or our final coin is a tail and we have an odd number of heads in our first n coins.

So

p(n+1)=(1-p(n))/(2n+3)+p(n)(2n+2)/(2n+3)=(1+(2n+1)p(n))/(2n+3).

But this is just saying q(n+1)=q(n)+1.

As p(1)=1/3, we have q(1)=1 and hence q(n)=n.

This lets us conclude that $p(n)=\frac{n}{2n+1}.$

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