HSC 2013 MX2 Marathon (archive) (5 Viewers)

Sy123 · Jan 20, 2014

Re: HSC 2014 4U Marathon

dunjaaa said:
Not sure about my inequality result, but I think my solution is wrong View attachment 29810

The idea is spot on, my algebra was though

y= -(2A/C) x

$Ax^2 + B \frac{4A^2x^2}{C^2} + Cx \frac{-2A}{C}x + D = 0 \\ \\ x^2(\frac{4A^2B}{C^2}- A) = -D$

and thus $\frac{4A^2B}{C^2} - A \leq 0$

dunjaaa · Jan 20, 2014

Re: HSC 2014 4U Marathon

Oh... I common denominated. Shouldn't the inequality read < since A,B,C,D >0

Sy123 · Jan 20, 2014

Re: HSC 2014 4U Marathon

dunjaaa said:
Oh... I common denominated. Shouldn't the inequality read < since A,B,C,D >0

Well we get

$x^2(4A^2B - AC^2) = -DC^2$

$x^2 = \frac{-DC^2}{4A^2B - AC^2} = \frac{DC^2}{AC^2 - 4A^2B}$

In order for the RHS to be positive, the numerator and denominator either both need to be positive ort both be negative, since hte numerator is always positive. Then the denominator must also always be positive.

$AC^2 \geq 4A^2B \ \ \ \ C^2 \geq 4AB$

seanieg89 · Jan 21, 2014

Re: HSC 2014 4U Marathon

Sy123 said:
This can fall under graphs

$\\ $An implicit curve is sketched in the$ \ xy \ $plane with the equation$ \\ \\ Ax^2 + By^2 + Cxy + D = 0 \ \ \ \ \ $With$ \ \ A,B,C,D > 0 \\ \\ $Show that in order for the curve to be an ellipse then$ \ \ \ \ C^2 \geq 4AB$

If you enforce A,B,C,D > 0, then this statement is only (technically) true because the implicitly defined curve is NEVER an ellipse. Your inequality in fact just gives a necessary (but not sufficient) condition for at least one point in the plane to satisfy your equation.

If you let A,B,C,D be real numbers (not necessarily positive) such that the curve implicitly defined by your equation is non-degenerate, then the OPPOSITE inequality to the one you posted is equivalent to being an ellipse. That is: the implicit equation Ax^2 + By^2 + Cxy + D = 0 is an ellipse if and only if the curve defined is non-degenerate and C^2-4AB < 0.

Degeneracy vs non-degeneracy is slightly more subtle and accounts for equations like x^2-y^2=0 whose graphs are in some sense exceptions to how second order polynomial curves usually look.

bottleofyarn · Jan 22, 2014

Re: HSC 2014 4U Marathon

Here's that inequality:
$$Let a, b, c and d be non-negative numbers such that $ a+b+c+d=4. $ Prove that $\frac{4}{abcd}\ge\frac{a}{b}+\frac{b}{c}+\frac{c}{d}+\frac{d}{a}.$

And one for 2014'ers (I think I've given enough to make it reasonable).
$$Given that $ x^2+y^2\ge2xy $ and $ x+y+z\ge3\sqrt[3]{xyz},$
$$ prove that $ a^2+b^2+c^2\ge a+b+c.$

seanieg89 · Jan 23, 2014

Re: HSC 2014 4U Marathon

$Prove that not all roots of the polynomial:\\ \\ $P(x):=x^n-5x^{n-1}+12x^{n-2}-15x^{n-3}+a_{n-4}x^{n-4}+\ldots+a_0\\ \\$are positive real numbers.$

seanieg89 · Jan 23, 2014

Re: HSC 2014 4U Marathon

$The BoS problem-solving seminar consists of n students sitting at desks. Let $u_k$ be the number of ways of rearranging the seating so that exactly $k$ students stay in the same spot.\\ \\ Prove that: \\ \\ $\sum_{k=0}^n ku_k=n!.$$

dunjaaa · Jan 24, 2014

Re: HSC 2014 4U Marathon

Screen shot 2014-01-24 at 7.12.16 PM.png

I think my U(k) expression is wrong, but I couldn't get the RHS result

The expression I obtained looked like taylor's series expansion of e

Sy123 · Jan 24, 2014

Re: HSC 2014 4U Marathon

dunjaaa said:
View attachment 29844 I think my U(k) expression is wrong, but I couldn't get the RHS result The expression I obtained looked like taylor's series expansion of e

Yea the expression is incorrect, here is a hint:

$u_k = \binom{n}{k} w_{n-k}$

$Let$ \ \ w_m \ \ $be the number of ways to arrange$ \ m \ $objects such that each object is NOT in its original spot$

Note that $w_m \neq m! \ $since this doesn't guarentee each object is out of its original position$$

$$Now try to find a recurrence relationship between$ \ w_m \ $and$ \ w_{m-1} \ $and$ \ w_{m-2}$

Sy123 · Jan 24, 2014

Re: HSC 2014 4U Marathon

$\\ $Let$ \ \ f(x) \ \ $be a continuous concave up function where$ \ \ a \leq x \leq b \\ \\ $i) Show that$ \\ \\ \frac{f(a) + f(b)}{2} \geq f \left(\frac{a+b}{2} \right) \\ \\ $ii) Hence show that$ \\ \\ \left(\sin^2 \alpha + \csc^2 \alpha \right)^2 + (\cos^2 \alpha + \sec^2 \alpha)^2 \geq \frac{25}{2}$

seanieg89 · Jan 24, 2014

Re: HSC 2014 4U Marathon

dunjaaa said:
View attachment 29844 I think my U(k) expression is wrong, but I couldn't get the RHS result The expression I obtained looked like taylor's series expansion of e

Your expression for u_k is incorrect. Some of the (n-k)! permutations of the "non-fixed" points will fix points, so that method of counting doesn't work.

dunjaaa · Jan 25, 2014

Re: HSC 2014 4U Marathon

sigh........, hopefully i got the u_k expression correct. Any tips on the next part?

Screen shot 2014-01-25 at 4.34.20 PM.png

seanieg89 · Jan 27, 2014

Re: HSC 2014 4U Marathon

Polys/Complex. Difficulty 3/5.

$Find all real polynomials $p(x)$ such that:\\ \\ $p(x+1)p(x-1)=p(x^2+1)$\\ \\ for all real $x$.$

seanieg89 · Jan 27, 2014

Re: HSC 2014 4U Marathon

dunjaaa said:
sigh........, hopefully i got the u_k expression correct. Any tips on the next part? View attachment 29849

This does seem roughly right, you definitely have the right idea for counting permutations that don't fix things (these are called derangements). Something seems to have gone wrong at some point because you have n-k as an upper limit in a sum indexed by k. This is probably just a slip in your notation earlier which would be easy to fix up.

My way was different (in fact my proof had very few equations), but your idea might still work if you can think of any clever ways to simplify the end summation. I will try to do it your way a little later to see if it works / how difficult it is.

dunjaaa · Jan 29, 2014

Re: HSC 2014 4U Marathon

Screen shot 2014-01-29 at 10.48.34 PM.png

Forgot about this question (forgot an 'up' after concave). I didn't include the other scenario when f(b)≤f(a) but either way it should still yield the same result

seanieg89 · Jan 30, 2014

Re: HSC 2014 4U Marathon

General pattern recognition rather than a specific 4U topic. Difficulty 2/5.

Sum sec(k)sec(k+1) from k=0 to 88.

(Where sec is defined in degrees.)

seanieg89 · Jan 31, 2014

Re: HSC 2014 4U Marathon

seanieg89 said:
$The BoS problem-solving seminar consists of n students sitting at desks. Let $u_k$ be the number of ways of rearranging the seating so that exactly $k$ students stay in the same spot.\\ \\ Prove that: \\ \\ $\sum_{k=0}^n ku_k=n!.$$

Imagine a rectangular grid with n! rows (one for each permutation of the students), and n columns (one for each student.)

Put a counter on each square in the grid such that the corresponding student is fixed by the corresponding permutation.

Then the LHS is clearly equal to the number of counters on our grid, since each row corresponding to a permutation with k fixed points contains k counters.

On the other hand, each of the n columns contains (n-1)! counters (one for each permutation of the other (n-1)! students.

Hence LHS=n.(n-1)!=n!.

(My actual proof was essentially the above idea written up formally using sums and not needing the counters analogy, but the idea of counting the same thing in two different ways to prove identities is probably more valuable to mx2 students.)

Sy123 · Jan 31, 2014

Re: HSC 2014 4U Marathon

$\\ $Determine the number of integer pairs$ \ (a,b) \ $such that$ \\ \\ \log_a b + 6\log_b a = 5 \\ \\ $for$ \ \ 2 \leq a \leq 2014 \ , \ 2\leq b \leq 2014$

seanieg89 · Feb 3, 2014

Re: HSC 2014 4U Marathon

Let f be a real-valued function defined on R. For which values of the real constant p does the inequality below imply that f is twice differentiable?

|f(x)-f(y)| =< C|x-y|^p for all x,y. C a constant.

Justify your answer with proof.

dunjaaa · Feb 6, 2014

Re: HSC 2014 4U Marathon

Screen shot 2014-02-06 at 2.23.22 PM.png

Felt so dumb after I realised how to do this

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