HSC 2015 MX1 Marathon (archive) (1 Viewer)

InteGrand · Oct 7, 2015

Re: HSC 2015 3U Marathon

There is a famous paradox based on this stuff called Bertrand's paradox: https://en.wikipedia.org/wiki/Bertrand_paradox_(probability)

InteGrand · Oct 7, 2015

Re: HSC 2015 3U Marathon

Also a much more famous probability problem is the Monty Hall Problem; apparently it fooled 1000 PhD's.

rand_althor · Oct 8, 2015

Re: HSC 2015 3U Marathon

Triangle ABC is inscribed within a semi-circle with diameter AB as shown. The area of triangle ABC is equal to 2/9 of the area of the semi-circle. Find the size of the smallest angle in triangle ABC to the nearest degree.

rand_althor · Oct 8, 2015

Re: HSC 2015 3U Marathon

$$Evaluate $ \frac{100}{\log_2{100!}}+\frac{100}{\log_3{100!}}+\frac{100}{\log_4{100!}}+...+\frac{100}{\log_{100}{100!}}.$

rand_althor · Oct 8, 2015

Re: HSC 2015 3U Marathon

Several people started with $400 each and played a game with the following unusual rules. Each player pays $10 to the house at the beginning of each round. During each round, one active player is declared the loser, and he distributes all of his money in equal amounts to the remaining players. The loser must then leave, but all of the other players go on to the next round. The game is over as soon as only one player remains. At the end of the game, the surviving player was surprised to discover that he had exactly $400, equalling his starting amount. How many players were there at the beginning?

kawaiipotato · Oct 8, 2015

Re: HSC 2015 3U Marathon

rand_althor said:
$$Evaluate $ \frac{100}{\log_2{100!}}+\frac{100}{\log_3{100!}}+\frac{100}{\log_4{100!}}+...+\frac{100}{\log_{100}{100!}}.$

$= 100 \sum_{k=2}^{100} \frac{1}{\log_{k}100!}$

$= 100 \sum_{k=2}^{100} \frac{ \ln k }{\ln 100! }$

$= \frac{100}{\ln 100!} \sum_{k=2}^{100} \ln k$

$= \frac{100}{\ln 100!} ( \ln 2 + \ln 3 + \ln 4 + ... + \ln 100 )$

$= \frac{100}{\ln 100!} \left( \ln (2*3*...*100) \right)$

$= \frac{100}{\ln 100!} * \ln(100!)$

$= 100$

turnerloos · Oct 8, 2015

Re: HSC 2015 3U Marathon

kawaiipotato said:
$= 100 \sum_{k=2}^{100} \frac{1}{\log_{k}100!}$

$= 100 \sum_{k=2}^{100} \frac{ \ln k }{\ln 100! }$

$= \frac{100}{\ln 100!} \sum_{k=2}^{100} \ln k$

$= \frac{100}{\ln 100!} ( \ln 2 + \ln 3 + \ln 4 + ... + \ln 100 )$

$= \frac{100}{\ln 100!} \left( \ln (2*3*...*100) \right)$

$= \frac{100}{\ln 100!} * \ln(100!)$

$= 100$

Teach me potato senpai.

davidgoes4wce · Oct 8, 2015

Re: HSC 2015 3U Marathon

A trough 'l' metres long has a cross-section in the shape of an isosceles triangle with base length 2a metres and height b metres. Water leaks from the trough at a constant rate of c m^3/min. Find the rate at which the water level is falling when the depth of the water is b/2 metres.

I am having trouble visualizing this as a diagram. Anyone mind drawing up?

davidgoes4wce · Oct 8, 2015

Re: HSC 2015 3U Marathon

For your current HSCers, the Mathematical Induction questions over the past 17 years: 6 have been Divisibility, 9 have been Sums/Products and 2 have been inequalities.

leehuan · Oct 8, 2015

Re: HSC 2015 3U Marathon

NEXT QUESTION:

$\\ $1. Use the definition of the binomial coefficient to prove Pascal's identity, that is, $ \\ \left( \begin{matrix} a \\ b \end{matrix} \right) =\left( \begin{matrix} a-1 \\ b-1 \end{matrix} \right) +\left( \begin{matrix} a-1 \\ b \end{matrix} \right) \\ $2. Prove by mathematical induction$ \\ { \left( 1+x \right) }^{ n }=\left( \begin{matrix} n \\ 0 \end{matrix} \right) +\left( \begin{matrix} n \\ 1 \end{matrix} \right) x+\left( \begin{matrix} n \\ 2 \end{matrix} \right) { x }^{ 2 }+\dots +\left( \begin{matrix} n \\ n \end{matrix} \right) { x }^{ n }$

rand_althor · Oct 8, 2015

Re: HSC 2015 3U Marathon

leehuan said:
NEXT QUESTION:

$\\ $1. Use the definition of the binomial coefficient to prove Pascal's identity, that is, $ \\ \left( \begin{matrix} a \\ b \end{matrix} \right) =\left( \begin{matrix} a-1 \\ b-1 \end{matrix} \right) +\left( \begin{matrix} a-1 \\ b \end{matrix} \right) \\ $2. Prove by mathematical induction$ \\ { \left( 1+x \right) }^{ n }=\left( \begin{matrix} n \\ 0 \end{matrix} \right) +\left( \begin{matrix} n \\ 1 \end{matrix} \right) x+\left( \begin{matrix} n \\ 2 \end{matrix} \right) { x }^{ 2 }+\dots +\left( \begin{matrix} n \\ n \end{matrix} \right) { x }^{ n }$

Skipping the first step:
$\begin{align*}\textup{2. Assume true for }n&=k \\(1+x)^k&=\binom{k}{0} + \binom{k}{1}x + \binom{k}{2}x^2 +...+ \binom{k}{k}x^k \\\end{align*}$

$\begin{align*}\textup{3. Prove true for }n&=k+1 \\\textup{RTP: }(1+x)^{k+1}&=\binom{k+1}{0} + \binom{k+1}{1}x + \binom{k+1}{2}x^2 +...+ \binom{k+1}{k+1}x^{k+1} \\\textup{LHS: }(1+x)^{k+1}&=(1+x)(1+x)^k \\&=(1+x) \left [ \binom{k}{0} + \binom{k}{1}x + \binom{k}{2}x^2 +...+ \binom{k}{k}x^k \right ] \\&= \binom{k}{0} + \binom{k}{1}x + \binom{k}{2}x^2 +...+ \binom{k}{k}x^k + \binom{k}{0}x + \binom{k}{1}x^2 + \binom{k}{2}x^3 +...+ \binom{k}{k-1}x^k + \binom{k}{k}x^{k+1} \\&=\binom{k}{0}+x \left [\binom{k}{0}+\binom{k}{1} \right ]+x^2 \left [\binom{k}{1}+\binom{k}{2} \right ]+...+x^{k} \left [\binom{k}{k-1}+\binom{k}{k} \right ] + \binom{k}{k}x^{k+1} \\&=\binom{k+1}{0} + \binom{k+1}{1}x + \binom{k+1}{2}x^2 +...+ \binom{k+1}{k+1}x^{k+1} \\&=\textup{RHS} \\&\textup{Using the result in part 1, and since } \binom{k}{0}=\binom{k}{k}=\binom{k+1} {0}=\binom{k+1}{k+1}=1 \end{align*}$

rand_althor · Oct 8, 2015

Re: HSC 2015 3U Marathon

$$Evaluate $\int^{\infty}_{e^{e^e}}~\frac{\mathrm{d}x}{x(\ln{x})(\ln{(\ln{x})})(\ln{(\ln{(\ln{x})})})^{\frac{4}{3}}} $ using the substitution $u=\ln{(\ln{(\ln{x})})}$.$

InteGrand · Oct 8, 2015

Re: HSC 2015 3U Marathon

rand_althor said:
$$Evaluate $\int^{\infty}_{e^{e^e}}~\frac{\mathrm{d}x}{x(\ln{x})(\ln{(\ln{x})})(\ln{(\ln{(\ln{x})})})^{\frac{4}{3}}} $ using the substitution $u=\ln{(\ln{(\ln{x})})}$.$

lol

rand_althor · Oct 8, 2015

Re: HSC 2015 3U Marathon

InteGrand said:
lol

It's actually not that bad once you do the substitution. I found it in a Stanford Math Tournament, but they gave it without the substitution.

Paradoxica · Oct 8, 2015

Re: HSC 2015 3U Marathon

$$D lies on the interval AB such that AD = BD. C is an external point such that CD is perpendicular to AB. Let the length AD = BD = y and CD = x. Suppose $ x \leq y.$ A circle is constructed such that it passes through A, B, and C. Find the radius of the circle in terms of x and y.$

kawaiipotato · Oct 8, 2015

Re: HSC 2015 3U Marathon

Paradoxica said:
$$D lies on the interval AB such that AD = BD. C is an external point such that CD is perpendicular to AB. Let the length AD = BD = y and CD = x. Suppose $ x \leq y.$ A circle is constructed such that it passes through A, B, and C. Find the radius of the circle in terms of x and y.$

Is the answer y^2 /x?

Paradoxica · Oct 8, 2015

Re: HSC 2015 3U Marathon

kawaiipotato said:
Is the answer y^2 /x?

My answers say otherwise, unless you canmoffer a valid justification...

kawaiipotato · Oct 8, 2015

Re: HSC 2015 3U Marathon

I got a different one this time. radius = (x^2 + y^2)/2x
http://imgur.com/OERRHXT

InteGrand · Oct 8, 2015

Re: HSC 2015 3U Marathon

rand_althor said:
It's actually not that bad once you do the substitution. I found it in a Stanford Math Tournament, but they gave it without the substitution.

$Yeah haha. Just found it funny seeing so many logs (and $e^{e^{e}}$).$

Drongoski · Oct 8, 2015

Re: HSC 2015 3U Marathon

Is the answer: $\frac {x^2 + y^2}{2x}$ ?

Oh kawaii already obtained this answer above.

HSC 2015 MX1 Marathon (archive) (1 Viewer)

Well-Known Member

Well-Known Member

Active Member

Active Member

Active Member

Well-Known Member

TAFE Executive Officer

Well-Known Member

Well-Known Member

Well-Known Member

Active Member

Active Member

Well-Known Member

Active Member

-insert title here-

Well-Known Member

-insert title here-

Well-Known Member

Well-Known Member

Well-Known Member

Users Who Are Viewing This Thread (Users: 0, Guests: 1)