HSC 2016 MX2 Marathon (archive) (5 Viewers)

leehuan · Apr 3, 2016

Re: HSC 2016 4U Marathon

Drsoccerball said:
I didn't use a diagram you don't need to

PM me briefly the method without the diagram cause I didn't think much into it.

But the question recommended the use of a diagram so I proceeded with one

seanieg89 · Apr 3, 2016

Re: HSC 2016 4U Marathon

$\frac{z-w}{z+w}=ki\Rightarrow z=w\frac{1+ki}{1-ki}\Rightarrow |z|=|w|\frac{|1+ki|}{|1-ki|}=|w|.$

KingOfActing · Apr 3, 2016

Re: HSC 2016 4U Marathon

leehuan said:
I had to help a student do this maths question just now. I feel it would be good to leave it on the 4U marathon as an instructive exercise for any E3 or above aiming student to attempt.

$Suppose $\frac{z-1+i}{z+1-i}=ki$ where $k$ is a real number. \\ \\ By drawing a diagram or otherwise, find the value of $|z|$. Show all reasoning.$

Algebraically because I hate geometrical anything:

$\begin{align*}\frac{z-1+i}{z+1-i} &= ki\\z-1+i &= ki(z+1-i)\\z &= \frac{ki + k + 1 - i}{1-ki}\\z &= \frac{(ki + k + 1 - i)(1+ki)}{1+k^2} = \frac{- k^2 + 2k + 1 + i(k^2 + 2k - 1)}{1+k^2}\\|z| &= \frac{\sqrt{(- k^2 + 2k + 1)^2 + (k^2 + 2k - 1)^2}}{1+k^2} \\|z| &= \frac{\sqrt{k^4+4k^2+1-4k^3-2k^2+4k + k^4 + 4k^2 + 1 + 4k^3 - 2k^2 - 4k}}{1+k^2}\\|z| &= \frac{\sqrt{2k^4+4k^2+2}}{1+k^2} = \frac{\sqrt{2}\sqrt{(k^2+1)^2}}{1+k^2} \\ |z| &= \sqrt{2}\end{align*}$

Drsoccerball · Apr 3, 2016

Re: HSC 2016 4U Marathon

KingOfActing said:
Algebraically because I hate geometrical anything:

$\begin{align*}\frac{z-1+i}{z+1-i} &= ki\\z-1+i &= ki(z+1-i)\\z &= \frac{ki + k + 1 - i}{1-ki}\\z &= \frac{(ki + k + 1 - i)(1+ki)}{1+k^2} = \frac{- k^2 + 2k + 1 + i(k^2 + 2k - 1)}{1+k^2}\\|z| &= \frac{\sqrt{(- k^2 + 2k + 1)^2 + (k^2 + 2k - 1)^2}}{1+k^2} \\|z| &= \frac{\sqrt{k^4+4k^2+1-4k^3-2k^2+4k + k^4 + 4k^2 + 1 + 4k^3 - 2k^2 - 4k}}{1+k^2}\\|z| &= \frac{\sqrt{2k^4+4k^2+2}}{1+k^2} = \frac{\sqrt{2}\sqrt{(k^2+1)^2}}{1+k^2} \\ |z| &= \sqrt{2}\end{align*}$

The working out would be simplified significantly if you went from :

$z= \frac{(k+1)+i(k-1)}{1-ki}$

$z \cdot \bar{z} = |z|^2= \frac{(k+1)+i(k-1)}{1-ki} \cdot \frac{(k+1)-i(k-1)}{1+ki}$

porcupinetree · Apr 3, 2016

Re: HSC 2016 4U Marathon

$In typical BoS meet up style, KingOfActing and Ekman decide to demolish a HSP together. However, for some variety, instead of demolishing it in their usual fashion (i.e. by consumption), they decide to literally demolish it by projecting it vertically upwards through a medium with a resistance of $2v + 3v^2$, where $v$ is the velocity of the HSP (note that gravity also affects the motion of the HSP). If the speed of projection is $u$, find the maximum height of the HSP.$

leehuan · Apr 3, 2016

Re: HSC 2016 4U Marathon

^The answer is C.

Drsoccerball · Apr 3, 2016

Re: HSC 2016 4U Marathon

leehuan said:
^The answer is C.

Sarcasm?

porcupinetree said:
$In typical BoS meet up style, KingOfActing and Ekman decide to demolish a HSP together. However, for some variety, instead of demolishing it in their usual fashion (i.e. by consumption), they decide to literally demolish it by projecting it vertically upwards through a medium with a resistance of $2v + 3v^2$, where $v$ is the velocity of the HSP (note that gravity also affects the motion of the HSP). If the speed of projection is $u$, find the maximum height of the HSP.$

$Drsoccerball who said he wouldn't come to the BoS meet secretly observed Porcupinetree and Ekman. Upon the realisation that Porcupinetree had tomato sauce in his HSP, Drsoccerball stood up on the table with height H +S, where S is the distance between the floor and Porcupinetree's mouth, and at a horizontal distance of D. He launched a Cards Against Humanity card at an angle of $ \theta $ to the horizontal such that he intercepted the tomato sauce covered meat right at his mouth.$

$For simplicity you lazy people, assume no air resistance(soz not soz). If you didn't know g is acceleration due to gravity,$

$(ii) Prove that the velocity of the card required to do this is :$

$v= \sqrt{ \frac{H^2+D^2}{\frac{2 \cdot \ln{(1+\frac{3u}{2})}}{3g}} -gH +\frac{g \cdot \frac{2 \cdot \ln{(1+\frac{3u}{2})}}{3g}}{4}}$

leehuan · Apr 3, 2016

Re: HSC 2016 4U Marathon

Drsoccerball said:
Sarcasm?

Um yeah lol.

seanieg89 · Apr 4, 2016

Re: HSC 2016 4U Marathon

Find all functions f:N->N such that f(n)+f(f(n))=2n.

(N is the set of positive integers).

You might find induction useful to justify your answer.

parad0xica · Apr 4, 2016

Re: HSC 2016 4U Marathon

seanieg89 said:
Find all functions f:N->N such that f(n)+f(f(n))=2n.

(N is the set of positive integers).

You might find induction useful to justify your answer.

How would you do this using HSC techniques...

KingOfActing · Apr 4, 2016

Re: HSC 2016 4U Marathon

seanieg89 said:
Find all functions f:N->N such that f(n)+f(f(n))=2n.

(N is the set of positive integers).

You might find induction useful to justify your answer.

I have no idea if this is right

PS Am I using the term mapping right? It flows better than "renaming the variable".

$We may begin by defining the recurrence relation $ f(a_n) = a_{n+1}$. Mapping $ n \to a_n $ we obtain the recurrence relation $ a_{n+2} + a_{n+1} = 2a_n $ with characteristic equation $x^2 + x = 2$. Thus the recurrence relation has solution $ a_n = A\alpha^n + B\beta^n $ where $A$ and $B$ are constants, and $\alpha$ and $\beta$ are the roots of the characteristic equation. Hence: $\\a_n = A(-2)^n + B(1)^n = A(-2)^n + B\\$Substituting this into our functional recurrence equation, we obtain $\\ f(A(-2)^n + B) = -2A(-2)^n + B\\$First consider the case $A = 0$: $ f(B) = B \Rightarrow f(n) = n\\$In the case $A \neq 0$, map $ A(-2)^n \to n$:$\\f(n + B) = -2n + B\\n \to n - B\\f(n) = -2n + 3B\\$Recall that $B$ is a constant, hence the other possible solution is $ f(n) = -2n + C$ However, this maps to the integers, rather than the natural numbers, and thus the only solution to the original functional equation is $ f(n) = n$.$

Alternate method I just thought of - is this valid?

f(n) = n is a solution (clearly, by inspection, whatever).

Assume there is another solution such that for some n f(n) > n and the functional equation holds:

f(f(n))+ f(n) > f(n)+ n > 2n which is a contradiction

Assume there is a solution such that for some n f(n) < n and the functional equation holds:

f(f(n)) + f(n) < f(n) + n < 2n which is a contradiction

Hence for every n, n <= f(n) <= n meaning f(n) = n is the only solution

seanieg89 · Apr 4, 2016

Re: HSC 2016 4U Marathon

Your first method needs to be tidied up a bit to work. What is a_1? For any particular a_1 you can choose A and B in lots of different ways. You need to show that for ANY such choice, a_2 must be equal to a_1, or the sequence must violate the properties that iterating f must have. You have not quite done this from what I can see.

For your second attempt,how are you deducing your first inequality: f(f(n))+f(n)>f(n)+n?

You are on the right track though (to the solution I came up with, which is the one I intended 4U students to find), thinking about such inequalities.

seanieg89 · Apr 4, 2016

Re: HSC 2016 4U Marathon

parad0xica said:
How would you do this using HSC techniques...

Pretty much just using properties of inequality and induction. Out-of-HSC techniques don't come into it at all unless you approach it in a really weird way.

Paradoxica · Apr 6, 2016

Re: HSC 2016 4U Marathon

seanieg89 said:
Find all functions f:N->N such that f(n)+f(f(n))=2n.

(N is the set of positive integers).

You might find induction useful to justify your answer.

Since f(n) is a natural valued function, we know f(n) is natural for all n in N

f(1) + f(f(1)) = 2

The left hand side consists of two naturals. The right hand side is 2. The only way to express 2 as a sum of 2 naturals is 1 + 1. Therefore, f(1) = f(f(1)) = 1.

f(2) + f(f(2)) = 4

If f(2) = 3 then we have f(3) = 1

But then that would mean f(3) + f(f(3)) = 1 + 1 = 2 =\= 6.

Contradiction. Thus, f(2) =\= 3

If f(2) = 1, then f(f(2)) = 3 <=> f(1) = 3.

Contradiction. Thus, f(2) = f(f(2)) = 2.

P(n): f(n) = n for all n in N

P(1), P(2) are the available true base cases.

Suppose P(k) is true for all 0 < k < n + 1, and that P(n + 1) is not true.

Assumed: f(1) = 1, ... f(n) = n, f(n + 1) =\= n + 1

f(n+1) + f(f(n + 1)) = 2n + 2

Suppose f(n + 1) = m where m < n + 1

But then f(n) + f(f(n) = m + f(m) = 2m < 2n + 2

Thus, f(n + 1) cannot be less than n + 1

Suppose f(n+1) = m where m > n + 1

Then f(n + 1) + f(f(n+1)) = m + f(m) = 2n + 2

Then f(m) = 2n + 2 - m < n + 1 since m > n + 1

Let 2n + 2 - m = m* < n+1

But then f(m) + f(f(m)) = m* + m* = 2m* < 2n + 2 < 2m

Which is a contradiction. Thus, f(n+1) cannot be greater than n+1

So now we have n < f(n+1) < n+2

f(n+1) is an integer. Thus, f(n+1) can only be equal to n+1, since the inequality is necessarily true.

So if P(1), ..., P(n) is true, then P(n+1) cannot be untrue, i.e. it must also be true.

By the Axiom of Mathematical Induction, P(n) is true for all n in N.

So the only solution to the functional equation is f(n) = n.

■

seanieg89 · Apr 6, 2016

Re: HSC 2016 4U Marathon

Bingo

.

Paradoxica · Apr 6, 2016

Re: HSC 2016 4U Marathon

seanieg89 said:
Bingo .

You were quite vague on the induction hint. So I just guessed I had to use strong induction.

Next question:

In an equilateral triangle ABC, a segment is projected from A such that it intersects the side BC at D and the circumcircle again at Q

Prove the following statement:

$\frac{1}{\left|QB\right|}+\frac{1}{\left|QC\right|} = \frac{1}{\left|QD\right|}$

Here is a diagram to help with visualisation:

Paradoxica · Apr 8, 2016

Re: HSC 2016 4U Marathon

$$\noindent $ABC$ is a triangle with angles $A, B, C$, and opposite side lengths $a, b ,c$. It is known that $b$ is definitely less than $c$. Sum the following series to infinity: $\sin{A} +\frac{b}{c}\sin{2A} + \frac{b^2}{c^2}\sin{3A} + \frac{b^3}{c^3}\sin{4A} + \cdots$

porcupinetree · Apr 8, 2016

Re: HSC 2016 4U Marathon

Paradoxica said:
You were quite vague on the induction hint. So I just guessed I had to use strong induction.

Next question:

In an equilateral triangle ABC, a segment is projected from A such that it intersects the side BC at D and the circumcircle again at Q

Prove the following statement:

$\frac{1}{\left|QB\right|}+\frac{1}{\left|QC\right|} = \frac{1}{\left|QD\right|}$

Here is a diagram to help with visualisation:

Paradoxica · Apr 8, 2016

Re: HSC 2016 4U Marathon

porcupinetree said:

Would have been faster to directly use the sine area formula.

i.e.

(sin120)(QB)(QC)/2 = (sin60)(QB)(QD)/2 + (sin60)(QC)(QD)/2

This immediately re-arranges to the desired result by inspection.

InteGrand · Apr 8, 2016

Re: HSC 2016 4U Marathon

Paradoxica said:
Would have been faster to directly use the sine area formula.

i.e.

(sin120)(QB)(QC)/2 = (sin60)(QB)(QD)/2 + (sin60)(QC)(QD)/2

This immediately re-arranges to the desired result by inspection.

The fastest method of all: "By inspection, Q.E.D."

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