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HSC 2013 MX2 Marathon (archive) (1 Viewer)

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Sy123

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Re: HSC 2013 4U Marathon

My solution is I can choose the point P in 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:

 
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RealiseNothing

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Re: HSC 2013 4U Marathon

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

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 (I think).

So from a fixed point P in the centre of the square the average area of the triangles will be .

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:



Is the logic behind this right?
 

Sy123

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Re: HSC 2013 4U Marathon

The area of the triangles will be

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 (I think).

So from a fixed point P in the centre of the square the average area of the triangles will be .

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:



Is the logic behind this right?
Can you please explain this part?
 

GoldyOrNugget

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Re: HSC 2013 4U Marathon

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

Sy123

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Re: HSC 2013 4U Marathon

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.
 

doublem

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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

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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.
Here, Makematics gave u a great method
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 -.-
 

Sy123

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Re: HSC 2013 4U Marathon

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

Sy123

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Re: HSC 2013 4U Marathon

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:





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



Then



Then some manipulations to get:



Now, since



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



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

This means:



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



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











Another one please :p, I enjoyed solving that.
 
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