HSC 2013 MX2 Marathon (archive) (1 Viewer)

Sy123 · Apr 28, 2013

Re: HSC 2013 4U Marathon

RealiseNothing said:
153846

Yep haha, nice work.

Sy123 · Apr 28, 2013

Re: HSC 2013 4U Marathon

RealiseNothing said:
It's just backwards multiplication of two numbers. The original number ends in 6, and when multiplied by 4 gets the same number but with a 6 out the front instead of the end:

$6 \times 4 = 24$

Multiply by the 4 and carry the 2 (since you multiply by the units and carry the tens)

$4 \times 4 + 2 = 18$

Multiply by the 8 and carry the 1:

$8 \times 4 + 1 = 33$

Multiply by the 3 and carry the 3:

$3 \times 4 + 3 = 15$

Multiply by the 5 and carry the 1:

$5 \times 4 + 1 = 21$

Multiply by the 1 and carry the 2:

$1 \times 4 + 2 = 6$

Since we have just 6, we stop here, and so our number is all the numbers we multiplied - 648351 - but we reverse it as we did the multiplcation in the reverse order to get 153846.

Wow......
Very intelligent solution.

Mine was simplifying the number where it must satisfy:

13n = 2 x 10^{m} - 2 for some integer m, and then I just computed a couple of values until I arrived at what I needed.
Though your method completely trumps mine haha

Sy123 · Apr 28, 2013

Re: HSC 2013 4U Marathon

(Another early IMO question).

$Solve the equation$

$\cos^2 x + \cos^2(2x) + \cos^2(3x) = 1$

RealiseNothing · Apr 28, 2013

Re: HSC 2013 4U Marathon

I was saving this question in case I felt like writing a trial paper, but that doesn't seem likely so here it is. Although it could go in the 3U thread, I thought the solution was quite nice and deserved to be here under "harder 3U" instead.

Consider the following diagram, where PQ is a tangent to the circle:

$$i) Prove that$~PM \geq PQ~$

$ii) Hence show that from any external point P, the two tangents that can be drawn to the circle are equal in length$

An alternative proof for the tangents from an external point theorem.

Sy123 · Apr 28, 2013

Re: HSC 2013 4U Marathon

RealiseNothing said:
I was saving this question in case I felt like writing a trial paper, but that doesn't seem likely so here it is. Although it could go in the 3U thread, I thought the solution was quite nice and deserved to be here under "harder 3U" instead.

Consider the following diagram, where PQ is a tangent to the circle:

$$i) Prove that$~PM \geq PQ~$

$ii) Hence show that from any external point P, the two tangents that can be drawn to the circle are equal in length$

An alternative proof for the tangents from an external point theorem.

I'll give someone else a go for this one, good question =)

Trebla · Apr 28, 2013

Re: HSC 2013 4U Marathon

$$Suppose that $ x = \cot\theta $ for some real $\theta$.$\\\\$(i) Show that $\left(\dfrac{x+i}{x-i}\right)^n=\cos 2n\theta + i\sin 2n\theta\\\\$(ii) Hence show that $\displaystyle\sum_{k=1}^{2n}\cot^2\left(\dfrac{[2k-1]\pi}{4n}\right)=2n(2n-1)$

RealiseNothing · Apr 28, 2013

Re: HSC 2013 4U Marathon

Trebla said:
$$Suppose that $ x = \cot\theta $ for some real $\theta$.$\\\\$(i) Show that $\left(\dfrac{x+i}{x-i}\right)^n=\cos 2n\theta + i\sin 2n\theta\\\\$(ii) Hence show that $\displaystyle\sum_{k=1}^{2n}\cot^2\left(\dfrac{[2k-1]\pi}{4n}\right)=2n(2n-1)$

Part i):

$(\frac{cot\theta + i}{cot\theta - i})^n$

$(\frac{\frac{cos\theta+isin\theta}{sin\theta}}{ \frac{cos - isin\theta}{sin\theta} })^ n$

$(\frac{cos\theta+isin\theta}{cos\theta-isin\theta})^n$

$\frac{(cos\theta+isin\theta)^n}{(cos\theta -isin\theta)^n}$

The numerator is just $cis^n\theta$

Now consider the denominator:

$(cos\theta - isin\theta)^ n$

$(cos(-\theta) + isin(- \theta))^n$ due to even and odd functions.

Which is just $cis^ {-n}\theta$

So we end up with:

$\frac{cis^ n\theta}{cis^{-n}\theta}$

$cis^{2n}\theta$

$cos2n\theta + isin2n\theta$

HeroicPandas · Apr 28, 2013

Re: HSC 2013 4U Marathon

hayabusaboston said:
Do you even know how much u confused me with ur polynomials question, where u made it a^2 not a, that had me going for hours and hours trying to figure out how u got a^2 and it turns out it was a all along lol. Btw If U have a cool conics question can u plz give to me.

LOL! do the other polynomial question i posted on the 3U marathon thread

Sy123 · Apr 28, 2013

Re: HSC 2013 4U Marathon

RealiseNothing said:
I was saving this question in case I felt like writing a trial paper, but that doesn't seem likely so here it is. Although it could go in the 3U thread, I thought the solution was quite nice and deserved to be here under "harder 3U" instead.

Consider the following diagram, where PQ is a tangent to the circle:

$$i) Prove that$~PM \geq PQ~$

$ii) Hence show that from any external point P, the two tangents that can be drawn to the circle are equal in length$

An alternative proof for the tangents from an external point theorem.

From the tangent-secant theorem:

$PQ^2 = PA \cdot PB$

$PQ^2 = (PM - AM)(PM+BM)$

$$Since$ \ \ AM=BM$

$PQ^2 = PM^2 -AM^2$

$AM^2 \geq 0$

$\therefore PM^2 \geq PQ^2 \ \ \Rightarrow PM \geq PQ$

ii)

$$In order for the inequality to establish equality (i.e. when PM = PQ), from i), AM^2 = 0 If AM =0, BM = 0. If this is the case then M lies on the circle. And therefore if M lies on the circle PM = PQ. Therefore from an external point P two tangents drawn to the circle are of equal length. ====================$

RealiseNothing · Apr 28, 2013

Re: HSC 2013 4U Marathon

Sy123 said:
From the tangent-secant theorem:

$PQ^2 = PA \cdot PB$

$PQ^2 = (PM - AM)(PM+BM)$

$$Since$ \ \ AM=BM$

$PQ^2 = PM^2 -AM^2$

$AM^2 \geq 0$

$\therefore PM^2 \geq PQ^2 \ \ \Rightarrow PM \geq PQ$

ii)

$$In order for the inequality to establish equality (i.e. when PM = PQ), from i), AM^2 = 0 If AM =0, BM = 0. If this is the case then M lies on the circle. And therefore if M lies on the circle PM = PQ. Therefore from an external point P two tangents drawn to the circle are of equal length. ====================[/QUOTE] Very nice solution, it's different to what I had in mind, so I'll post mine up to for variety.$

Capt Rifle · Apr 28, 2013

Re: HSC 2013 4U Marathon

<a href="http://www.codecogs.com/eqnedit.php?latex=\int \frac{1-x}{1-\sqrt{x}}\: dx" target="_blank"><img src="http://latex.codecogs.com/gif.latex?\int \frac{1-x}{1-\sqrt{x}}\: dx" title="\int \frac{1-x}{1-\sqrt{x}}\: dx" /></a>

RealiseNothing · Apr 28, 2013

Re: HSC 2013 4U Marathon

RealiseNothing said:
I was saving this question in case I felt like writing a trial paper, but that doesn't seem likely so here it is. Although it could go in the 3U thread, I thought the solution was quite nice and deserved to be here under "harder 3U" instead.

Consider the following diagram, where PQ is a tangent to the circle:

$$i) Prove that$~PM \geq PQ~$

$ii) Hence show that from any external point P, the two tangents that can be drawn to the circle are equal in length$

An alternative proof for the tangents from an external point theorem.

From tangent-secant theorem:

$PQ^2 = PA.PB$

$PQ = \sqrt{PA.PB}$

Also since M is the midpoint of A and B:

$\frac{PA+PB}{2} = PM$

Note the AM-GM inequality:

$\frac{a+b}{2} \geq \sqrt{ab}$

Let $a=PA$ and $b=PB$

$\frac{PA+PB}{2}\geq \sqrt{PA.PB}$

$PM \geq PQ$

Also for the second part, equality holds in the AM-GM inequality when $a=b$ and so $PA=PB$ . This only occurs when M coincides with Q, and so $PM=PQ$ . Now construct the second tangent and let it touch the circle at point J. Equality holds again when $PM=PJ$ . Hence $PQ=PJ$ as both are arbitrary tangents and give the minimum possible value of $PM$ .

Sy123 · Apr 28, 2013

Re: HSC 2013 4U Marathon

Capt Rifle said:
<a href="http://www.codecogs.com/eqnedit.php?latex=\int \frac{1-x}{1-\sqrt{x}}\: dx" target="_blank"><img src="http://latex.codecogs.com/gif.latex?\int \frac{1-x}{1-\sqrt{x}}\: dx" title="\int \frac{1-x}{1-\sqrt{x}}\: dx" /></a>

$\int \frac{1-x}{1-\sqrt{x}} \ dx = \int \frac{(1-\sqrt{x})(1+\sqrt{x})}{1-\sqrt{x}} = \int 1+\sqrt{x} \ dx = x + \frac{2}{3}x^{\frac{3}{2}} + c$

==========

$\int \frac{dx}{e^x + e^{-x}}$

Trebla · Apr 28, 2013

Re: HSC 2013 4U Marathon

Hmm perhaps I should give hints for the second part of my question?

Sy123 · Apr 28, 2013

Re: HSC 2013 4U Marathon

Trebla said:
Hmm perhaps I should give hints for the second part of my question?

Give it a bit of time, I'm working on it right now

Trebla · Apr 28, 2013

Re: HSC 2013 4U Marathon

Sy123 said:
Give it a bit of time, I'm working on it right now

Well, no hint for you then

jyu · Apr 28, 2013

Re: HSC 2013 4U Marathon

RealiseNothing said:
I was saving this question in case I felt like writing a trial paper, but that doesn't seem likely so here it is. Although it could go in the 3U thread, I thought the solution was quite nice and deserved to be here under "harder 3U" instead.

Consider the following diagram, where PQ is a tangent to the circle:

$$i) Prove that$~PM \geq PQ~$

$ii) Hence show that from any external point P, the two tangents that can be drawn to the circle are equal in length$

An alternative proof for the tangents from an external point theorem.

PQ is a tangent, so PQM is a right angle. Without stating the obvious, PM > PQ. PM = PQ when the radius = 0.

Sy123 · Apr 28, 2013

Re: HSC 2013 4U Marathon

Trebla said:
$$Suppose that $ x = \cot\theta $ for some real $\theta$.$\\\\$(i) Show that $\left(\dfrac{x+i}{x-i}\right)^n=\cos 2n\theta + i\sin 2n\theta\\\\$(ii) Hence show that $\displaystyle\sum_{k=1}^{2n}\cot^2\left(\dfrac{[2k-1]\pi}{4n}\right)=2n(2n-1)$

Moving on from part (i)

$LHS = \frac{(x+i)^{2n}}{(\cot^2 \theta + 1)^n} = \sin^{2n} \theta (x+i)^{2n} = \cos 2n\theta + i\sin 2n\theta$

Equate real parts of both sides, also make the substitution:

$y=x^2$

$\frac{\cos(2n\theta)}{\sin^{2n} \theta} = y^n - \binom{2n}{2} y^{n-1} + \dots$

$Find the theta such that the LHS is zero$

$\frac{\cos(2n\theta)}{\sin^{2n}\theta} = 0$

$\theta \neq m\pi \ \ \ m\in \mathbb{Z}$

$2n\theta = \frac{(2k-1)\pi}{2} \ \ \ \ k\in \matbb{Z}$

$\therefore \theta_k = \frac{(2k-1)\pi}{4n}$

Makes the LHS side zero, on the RHS there is a polynomial of degree n in cot^2 theta, and the theta that makes cot^2 theta a root is above.
Now taking the sum of all roots.

$\sum_{k=1}^{n} \cot^2 (\theta_k) = \binom{2n}{2} = n(2n-1)$

Now, cotangent is periodic by pi, therefore the following applies:

$\sum_{k=1}^{n} \cot^2(\theta_k) = \sum_{k=n+1}^{2n} \cot^2(\theta_k)$

$\therefore \displaystyle\sum_{k=1}^{2n}\cot^2\left(\dfrac{[2k-1]\pi}{4n}\right)=2n(2n-1)$

Sy123 · Apr 28, 2013

Re: HSC 2013 4U Marathon

jyu said:
PQ is a tangent, so PQM is a right angle. Without stating the obvious, PM > PQ. PM = PQ when the radius = 0.

We do not know that M is the centre.

RealiseNothing · Apr 28, 2013

Re: HSC 2013 4U Marathon

jyu said:
PQ is a tangent, so PQM is a right angle. Without stating the obvious, PM > PQ. PM = PQ when the radius = 0.

Sy123 said:
We do not know that M is the centre.

^^^

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