HSC 2013 MX2 Marathon (archive) (1 Viewer)

Sy123 · Jun 13, 2013

Re: HSC 2013 4U Marathon

RealiseNothing said:
My solution is I can choose the point P in $n^2$ ways as it can take any point on the square. Then Q and R can both be chosen in (n-1) ways as they can take any point in the row or column that P was chosen in. Multiplying together gives the result.

Alright, now the challenge is to find the average area of the triangles.
It can be done easily if we fix P somewhere, but I can't seem to find a good way of finding the averages at a point in general. (I have a method in mind, but it feels very brute force-y)

If no counting errors, at point (m,k), the sum of triangle areas:

$S = \frac{1}{2} \left(\frac{n-m}{2} (n-m+1) + \frac{m}{2} (m-1) \right ) \left(\frac{k}{2} (k-1) + \frac{n-m}{2} (n-m+1) \right )$

RealiseNothing · Jun 13, 2013

Re: HSC 2013 4U Marathon

Sy123 said:
Alright, now the challenge is to find the average area of the triangles.
It can be done easily if we fix P somewhere, but I can't seem to find a good way of finding the averages at a point in general. (I have a method in mind, but it feels very brute force-y)

The area of the triangles will be $\frac{1}{2} PQ.PR$

So we need to find the average length of PQ and PR. Now if (n) is odd, then we can fix the point P to be the centre of the square. The average length of PQ and PR will be $\frac{n-2}{4}$ (I think).

So from a fixed point P in the centre of the square the average area of the triangles will be $\frac{1}{2}(\frac{n-2}{4})^2$ .

Now since the rest of the points we can fix P at in the square are symmetrical then the averages will all cancel out, and so the average area of the triangles if (n) is odd will be:

$\frac{1}{2}(\frac{n-2}{4})^2$

Is the logic behind this right?

Sy123 · Jun 13, 2013

Re: HSC 2013 4U Marathon

RealiseNothing said:
The area of the triangles will be $\frac{1}{2} PQ.PR$

So we need to find the average length of PQ and PR. Now if (n) is odd, then we can fix the point P to be the centre of the square. The average length of PQ and PR will be $\frac{n-2}{4}$ (I think).

So from a fixed point P in the centre of the square the average area of the triangles will be $\frac{1}{2}(\frac{n-2}{4})^2$ .

Now since the rest of the points we can fix P at in the square are symmetrical then the averages will all cancel out, and so the average area of the triangles if (n) is odd will be:

$\frac{1}{2}(\frac{n-2}{4})^2$

Is the logic behind this right?

Can you please explain this part?

GoldyOrNugget · Jun 13, 2013

Re: HSC 2013 4U Marathon

RealiseNothing said:
From UNSW competition sample question:

A set of fifty pairwise distinct integers between 1 and 100 is chosen such that:

a) the sum of the numbers is equal to 2525; and
b) the sum of any two numbers never equals 101

Find the sum of squares of these integers.

Observation: if we pick some number n, the number 101-n is excluded. n and 101-n are always of opposite parity.

The sum of integers 1..100 is 5050. Take all the 50 odd numbers to be in our set, since they will add up to approximately 5050/2, and by the observation above, no pair will be mutually exclusive. Indeed they add up to 2500 -- we need to add 25 more.

Now we have to substitute some of the odd numbers for their respective even versions to increase the sum: i.e. we need to remove some odd integer k and replace it with the even 101-k so that -k + (101-k) = 25. The only solution for k is 38, which is even, so we need to swap out more than one number. If we swap out two numbers we'll get an even sum, so we have to swap out at least 3. So the task is to find distinct odd numbers a,b,c in our set so that -a+(101-a) -b+(101-b) -c+(101-c) = 25 --> 303 - 2(a+b+c)=25 --> a+b+c=139. Now simply pick three distinct odd numbers in [1, 100] that make this sum: say, 1, 99 and 39. Substitute them with their counterparts 100, 2 and 62 respectively.

We now have our set! All the odd numbers from 1..100, excluding 1, 99, 39, plus 100, 2, 62.

To find the sum of squares, you can type them into a calculator, I guess

it should only take about a minute and the competition is 3h long. It's 169175.

RealiseNothing · Jun 13, 2013

Re: HSC 2013 4U Marathon

Sy123 said:
Can you please explain this part?

1am me wasn't thinking straight.

Sy123 · Jun 15, 2013

Re: HSC 2013 4U Marathon

$$i) Prove by induction or otherwise, that for positive real numbers$ \ \ a_k$

$\frac{a_1+a_2+ \dots + a_n}{n} \geq \sqrt[n]{a_1 \cdot a_2 \cdot a_3 \dots a_n}$

$And state the conditions for equality$

$ii) Hence prove for some integer$ \ \ n \ \ $ that$

$\frac{1}{2\sqrt[2]{2!}} + \frac{1}{3\sqrt[3]{3!}} + \frac{1}{4\sqrt[4]{4!}} + \dots + \frac{1}{n\sqrt[n]{n!}} > \frac{n-1}{n+1}$

Sy123 · Jun 15, 2013

Re: HSC 2013 4U Marathon

$$Solve the following equation for$ \ \ x$

$\sum_{k=0}^{n} \binom{n}{k}x^k \cos(k\theta) = 0$

braintic · Jun 16, 2013

Re: HSC 2013 4U Marathon

Sy123 said:
$$Solve the following equation for$ \ \ x$

$\sum_{k=0}^{n} \binom{n}{k}x^k \cos(k\theta) = 0$

LHS is the real part of (1+xcisθ)^n. But not sure what to do after that.

Sy123 · Jun 16, 2013

Re: HSC 2013 4U Marathon

braintic said:
LHS is the real part of (1+xcisθ)^n. But not sure what to do after that.

Maybe you should try to find 1+xcis theta in mod-arg form, so you can use De Moivere's formula.

Sy123 · Jun 17, 2013

Re: HSC 2013 4U Marathon

$Let$

$P(x) = x^6-x^5-x^3-x^2-x$

$Q(x) = x^4-x^3-x^2 -1$

$z_1, z_2, z_3, z_4 \ \ $are roots of the polynomial$ \ \ Q(x)$

$Prove that$

$P(z_1)+P(z_2)+P(z_3)+P(z_4) = 6$

doublem · Jun 18, 2013

Re: HSC 2013 4U Marathon

Can somelone plz plz do this ques for me?

P is a point on the ellipse x^2 /a^2 + y^2 / b^2 =1 with a centre O. A line is drawn from O, parallel to the tangent to the ellipse at P and meets the ellipse at Q. Prove that the area of triangle OPQ is independent of the position P.

HeroicPandas · Jun 18, 2013

Re: HSC 2013 4U Marathon

doublem said:
Can somelone plz plz do this ques for me?

P is a point on the ellipse x^2 /a^2 + y^2 / b^2 =1 with a centre O. A line is drawn from O, parallel to the tangent to the ellipse at P and meets the ellipse at Q. Prove that the area of triangle OPQ is independent of the position P.

Here, Makematics gave u a great method

Makematics said:
Just to add a bit, use (acostheta,bsintheta) as P. Also i dont think the tangent meets the ellipse again? Should be
1. Find the gradient of the tangent at P
2.Use the point-gradient formula to find the equation of the line through the origin and parallel to the tangent.
3. Solve the equations of the line and the ellipse simultaneously to find the coordinates of Q.
4. Finding the area is the challenging bit. Use the perpendicular distance formula from the origin to PQ and then use A=1/2 (PQ) (perp. distance). After heavy algebra, you should get 1/2 ab units^2 as the area iirc. Or you could use the determinant method, which is not allowed according to my teachers -.-

RealiseNothing · Jun 18, 2013

Re: HSC 2013 4U Marathon

So who is pumped for the UNSW comp tomorrow?

asianese · Jun 18, 2013

Re: HSC 2013 4U Marathon

me

Makematics · Jun 18, 2013

Re: HSC 2013 4U Marathon

RealiseNothing said:
So who is pumped for the UNSW comp tomorrow?

hellz yeah i am!

Sy123 · Jun 20, 2013

Re: HSC 2013 4U Marathon

Can someone please post a challenging inequality to prove? (preferably the ones involving a, b, c, d etc)

asianese · Jun 20, 2013

Re: HSC 2013 4U Marathon

$For positive $ m $ and $ n$, show that$

$\dfrac{(m+n)!}{(m+n)^{m+n}}<\dfrac{m!}{m^m}\dfrac{n!}{n^n}.$

braintic · Jun 20, 2013

Re: HSC 2013 4U Marathon

asianese said:
$For positive $ m $ and $ n$, show that$

$\dfrac{(m+n)!}{(m+n)^{m+n}}<\dfrac{m!}{m^m}\dfrac{n!}{n^n}.$

$\begin{align*} \dfrac {(m+n)!}{m!.n!} . m^m . n^n &= {}^{m+n}C_m . m^m . n^n \\&= \text{term in } m^m . n^n \text{ in expansion of } (n+m)^{m+n}\\&< (n+m)^{m+n} \end{align*}$

Rearrangement gives required result.

asianese · Jun 20, 2013

Re: HSC 2013 4U Marathon

braintic said:
$\begin{align*} \dfrac {(m+n)!}{m!.n!} . m^m . n^n &= {}^{m+n}C_m . m^m . n^n \\&= \text{term in } m^m . n^n \text{ in expansion of } (n+m)^{m+n}\\&< (n+m)^{m+n} \end{align*}$

Rearrangement gives required result.

[Where is that BR coming from?]

Don't put a physical space between lines - keep it all in 1 line.

Sy123 · Jun 20, 2013

Re: HSC 2013 4U Marathon

asianese said:
$For positive $ m $ and $ n$, show that$

$\dfrac{(m+n)!}{(m+n)^{m+n}}<\dfrac{m!}{m^m}\dfrac{n!}{n^n}.$

Great question!

Here is my solution:

On the cartesian plane, sketch the curve y=ln(x). Then draw upper rectangles from x=k to x=1.

We establish the inequality:

$\ln(k) + \ln(k-1) + \dots + \ln(1) > \int_1^k \ln x \dx = k\ln k - k + 1$

$\ln \frac{k!}{k^k} > 1- k$

Now consider, the lower rectangles, this time from x=1 to x=k+1, we can establish the inequality:

$\int_1^{k+1} \ln x \ dx > \ln(k+1) + \ln(k!)$

Then

$(k+1) \ln(k+1) - k > \ln(k+1) + \ln(k!)$

Then some manipulations to get:

$\ln \left(1+ \frac{1}{k} \right )^k - k > \ln \frac{k!}{k^k}$

Now, since

$e= \lim_{n\to \infty} \left(1+ \frac{1}{n} \right )^n$

And as n increases, the RHS increases, so that means for some k:

$\left(1+\frac{1}{k} \right)^k < e$

(I'm sure there is a more rigourous way to prove this inequality, I think it appeared in a HSC question before)

This means:

$2-k >\ln \left(1+ \frac{1}{k} \right )^k - k > \ln \frac{k!}{k^k}$

Combining this inequality and the other one established, we arrive at:

$2-k > \ln \frac{k!}{k^k} > 1-k$

Substitute k=n, k=m, k=n+m

$\ln \frac{n!}{n^n} + \ln \frac{m!}{m^m} > 2-(n+m)$

$\ln \frac{(n+m)!}{(n+m)^{n+m}} < 2-(n+m)$

$\therefore \ \ln\frac{n!}{n^n} + \ln \frac{m!}{m^m} > \ln \frac{(m+n)!}{(m+n)^{m+n}}$

$\therefore \ \frac{n!}{n^n} \frac{m!}{m^m} > \frac{(m+n)!}{(m+n)^{m+n}}$

$\square$

Another one please

, I enjoyed solving that.

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