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HSC 2013 MX2 Marathon (archive) (6 Viewers)

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Sy123

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Re: HSC 2013 3U Marathon Thread

An interesting volumes question

It took me ages to visualise everything lol
Through drawing a diagram, we can see that the side of a square, s is given by:

Hence area of one square is: 4(r-y)^2

Volume thus by integrating from r to -r, i.e. from the top to the bottom, we get 32/3 r^3

Hopefully that's right.

====



I'm still trying to do that polynomial one....

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Re: HSC 2013 3U Marathon Thread

Through drawing a diagram, we can see that the side of a square, s is given by:

Hence area of one square is: 4(r-y)^2

Volume thus by integrating from r to -r, i.e. from the top to the bottom, we get 32/3 r^3

not quite :p
 

Sy123

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Re: HSC 2013 4U Marathon



Is this what you mean?



Not quite, that might be its locus, but we don't know what alpha is (i.e. Re(z-i)/(z+1) = 0 )

===

Also twinklegal, my explanation is wrong, the variable square is above the fixed square of units y, but still using similar triangles as we wake a triangle to the top of the thing, I still get the same answer :/

EDIT: NVM, we just take from O to a side of the square directly, then pythagoras theorem, to get sidelength 2sqrt(r^2-x^2), squaring this and integrating from -r to r we get 16r^3/3?
 
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Re: HSC 2013 4U Marathon

Also twinklegal, my explanation is wrong, the variable square is above the fixed square of units y, but still using similar triangles as we wake a triangle to the top of the thing, I still get the same answer :/

EDIT: NVM, we just take from O to a side of the square directly, then pythagoras theorem, to get sidelength 2sqrt(r^2-x^2), squaring this and integrating from -r to r we get 16r^3/3?
yep :)
 

Sy123

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Re: HSC 2013 4U Marathon

Hint: Choose as the largest root(s) and consider the discriminant of

Dividing using long division, we get that:



Discriminant is greater than zero:



Expanding:





Solving for alpha:

We find that the answer directly after as an upper bound
 

study1234

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Re: HSC 2013 4U Marathon

Prove that four consecutive terms in a row of the Pascal triangle cannot form an AP.
 

Sy123

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Re: HSC 2013 4U Marathon

Prove that four consecutive terms in a row of the Pascal triangle cannot form an AP.
The numbers in pascal's triangle can be represented by binomial coefficients, the nth row and kth spot number is given by:

Thus, if there exists an AP in Pascal's triangle, for some



Is an arithmetic progression, if it is an arithmetic progression then the difference between the terms must be equal:















Take the second equality:









======================







 

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Re: HSC 2013 4U Marathon

Show that, if the four points representing the complex numbers z1, z2, z3, z4 are concyclic, the fraction (z1-z2)(z3-z4)/(z3-z2)(z1-z4) must be real.
 

Sy123

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Re: HSC 2013 4U Marathon

Show that, if the four points representing the complex numbers z1, z2, z3, z4 are concyclic, the fraction (z1-z2)(z3-z4)/(z3-z2)(z1-z4) must be real.
Let





Add them side by side, opposite angles of a cyclic quadrilateral add to pi so, and use the identity:





We get:



Since the argument of the complex number is pi, that means is real.
 
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