HSC 2015 MX1 Marathon (archive) (4 Viewers)

InteGrand · Oct 8, 2015

Re: HSC 2015 3U Marathon

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

Oh kawaii already obtained this answer above.

Yes. This question can be done quickly using the extended sine rule and a bit of Pythagoras and right-angled trig.

Drsoccerball · Oct 8, 2015

Re: HSC 2015 3U Marathon

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

Are you sure i got $\sqrt{x^2 +y^2}$

rand_althor · Oct 8, 2015

Re: HSC 2015 3U Marathon

Drsoccerball said:
Are you sure i got $\sqrt{x^2 +y^2}$

I think that's the circle passing through A and B with center C, not passing through A, B, C.

InteGrand · Oct 8, 2015

Re: HSC 2015 3U Marathon

$The normal sine rule you learn is that for any triangle, $\frac{a}{\sin A}=\frac{b}{\sin B}=\frac{c}{\sin C}$. The extended sine rule just adds to this equality that these ratios are all equal to $2R$, where $R$ is the circumradius of the triangle (i.e. the radius of the unique circle which passes through the vertices of the triangle, which is called the triangle's \textsl{circumcircle} (note that every non-degenerate triangle has one and only one circumcircle)).$

Drsoccerball · Oct 8, 2015

Re: HSC 2015 3U Marathon

rand_althor said:
I think that's the circle passing through A and B with center C, not passing through A, B, C.

I used circle geo and double angles do get it are you sure ?

Paradoxica · Oct 8, 2015

Re: HSC 2015 3U Marathon

Drsoccerball said:
I used circle geo and double angles do get it are you sure ?

Kawaiipotato and Drongoski are both correct. Not sure what you're doing lol.

rand_althor · Oct 8, 2015

Re: HSC 2015 3U Marathon

Drsoccerball said:
I used circle geo and double angles do get it are you sure ?

Well I drew the segment AC. Since CD is perpendicular to AB, angle CDA is equal to 90 degrees. One of the circle geometry theorems is "In a circle, the perpendicular bisector of a chord passes through the center of the circle." So if we have chord AB and center C, the radius is then AC, which can be found using Pythagoras to be what you got. However, that is the radius of the circle passing through A and B with center C, i.e. not what we're after.

Zlatman · Oct 8, 2015

Re: HSC 2015 3U Marathon

I just used some Pythagoras and circle geometry.

AB will be a chord of the circle, and since CD is the perpendicular bisector of AB, it passes through O. Letting the radius equal r, and using Pythagoras in triangle DOB, we get:

$r^2 = (r - x)^2 + y^2$

which simplifies to:

$r = \frac{x^2 + y^2}{2x}$

EDIT: rand describes the same thing in the post above
EDIT 2: no he doesn't, never mind, lol

Drongoski · 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 $u=\ln{(\ln{(\ln{x})})}$.$

Is the answer: 3 ??

rand_althor · Oct 8, 2015

Re: HSC 2015 3U Marathon

Drongoski said:
Is the answer: 3 ??

Yep.

InteGrand · Oct 8, 2015

Re: HSC 2015 3U Marathon

rand_althor said:
Well I drew the segment AC. Since CD is perpendicular to AB, angle CDA is equal to 90 degrees. One of the circle geometry theorems is "In a circle, the perpendicular bisector of a chord passes through the center of the circle." So if we have chord AB and center C, the radius is then AC, which can be found using Pythagoras to be what you got. However, that is the radius of the circle passing through A and B with center C, i.e. not what we're after.

$If the centre were $C$ and $A$ and $B$ lay on the circle then $AC$ and $BC$ would be radii, and they have lengths $\sqrt{x^2+y^2}$ by Pythagoras' theorem in triangle $ADC$ or $BDC$, so that'd be the radius of that circle, no need for that circle geometry.$

Edit: oh, you were describing the misunderstanding, nvm.

Drongoski · Oct 8, 2015

Re: HSC 2015 3U Marathon

Paradoxica said:
Kawaiipotato and Drongoski are both correct. Not sure what you're doing lol.

I just used: Pythag Thm and similar triangles.

If O is the centre of the circumcircle, and P the mid-point of AC, then triangles OPC and ADC are similar.

rand_althor · Oct 9, 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 $u=\ln{(\ln{(\ln{x})})}$.$

$\begin{align*}u&=\ln{(\ln{(\ln{x})})} \\\frac{\mathrm{d}u}{\mathrm{d}x}&=\frac{1}{\ln{(\ln{x})}} \times\ \frac{\mathrm{d}}{\mathrm{d}x}\ln{(\ln{x})} \\&=\frac{1}{\ln{(\ln{x})}} \times\ \frac{1}{\ln{x}} \times \frac{\mathrm{d}}{\mathrm{d}x}\ln{x} \\&=\frac{1}{\ln{(\ln{x})}} \times\ \frac{1}{\ln{x}} \times \frac{1}{x} \\&=\frac{1}{x(\ln{x})(\ln{(\ln{x})})} \\\mathrm{d}x&=\mathrm{d}u\times x(\ln{x})(\ln{(\ln{x})}) \\\end{align*}$

$\begin{align*}\textup{As }x &\to \infty,~ u \to \infty \\x&=e^{e^e}\to u= \ln(\ln{(\ln{(e^{e^e})})})=1\end{align*}$

$\begin{align*}&\int^\infty_1~\frac{1}{x(\ln{x})(\ln{(\ln{x})})(u)^{\frac{4}{3}}} \times \mathrm{d}u\times x(\ln{x})(\ln{(\ln{x})}) \\&=\int^\infty_1~u^{-\frac{4}{3}}~\mathrm{d}u \\&=-3 \left [u^{-\frac{1}{3}} \right ]^\infty_1 \\&=-3 \left [\lim_{u\to\infty}\frac{1}{\sqrt[3]{u}}-1 \right ] \\&=3\end{align*}$

rand_althor · Oct 9, 2015

Re: HSC 2015 3U Marathon

Twenty guests sit at twenty seats around a round table according to the following process. One guest, who is selected uniformly at random from the unseated guests, selects an empty seat uniformly at random and sits in that seat. This process repeats until all guests are seated. Alice is one of the guests. Find the probability that when Alice sits down the both the seat to her left and the seat to her right are already occupied.

kawaiipotato · Oct 9, 2015

Re: HSC 2015 3U Marathon

In a round table, having one person sitting and two people sitting to the left and right of them means that there's just at least 3 people sitting (including the specific person, Alice). So she is the >= 3rd person to be picked.
This is equivalent to 1- P(picked 1st) - P(picked 2nd)
= 1 - [( 1/20 + 19/20 * 1/19 ) ] = 9/10
edit: nvm I assumed that there aren't fixed chairs

rand_althor · Oct 9, 2015

Re: HSC 2015 3U Marathon

kawaiipotato said:
In a round table, having one person sitting and two people sitting to the left and right of them means that there's just at least 3 people sitting (including the specific person, Alice). So she is the >= 3rd person to be picked.
This is equivalent to 1- P(picked 1st) - P(picked 2nd)
= 1 - [( 1/20 + 19/20 * 1/19 ) ] = 9/10
edit: nvm I assumed that there aren't fixed chairs

The answer is 1/3. I can't explain why your answer is wrong though.

InteGrand · Oct 9, 2015

Re: HSC 2015 3U Marathon

kawaiipotato said:
In a round table, having one person sitting and two people sitting to the left and right of them means that there's just at least 3 people sitting (including the specific person, Alice). So she is the >= 3rd person to be picked.
This is equivalent to 1- P(picked 1st) - P(picked 2nd)
= 1 - [( 1/20 + 19/20 * 1/19 ) ] = 9/10
edit: nvm I assumed that there aren't fixed chairs

Being 3rd or later to be picked is a necessary condition for having two people sitting next to her, but it is not also a sufficient condition – Alice could be picked 3rd or later and still not sit down somewhere next to two people.

braintic · Oct 10, 2015

Re: HSC 2015 3U Marathon

The answer is 1/3 regardless of the number of people.

Rewind the seating process so that you see people standing and leaving the table.
You need only focus on Alice and the two seats to her side.
There is one chance in three that Alice will be the first of those three people to stand.

braintic · Oct 10, 2015

Re: HSC 2015 3U Marathon

rand_althor said:
Twenty guests sit at twenty seats around a round table according to the following process. One guest, who is selected uniformly at random from the unseated guests, selects an empty seat uniformly at random and sits in that seat. This process repeats until all guests are seated. Alice is one of the guests. Find the probability that when Alice sits down the both the seat to her left and the seat to her right are already occupied.

Same question with a twist:

20 guests are placed by the same process around a round table.

What is the minimum number of seats required at the table so that the probability is less than 5% that when Alice sits down both seats adjacent to her are occupied?

Answer: 51

(No 3U work is required - only 2U probability)

Paradoxica · Oct 10, 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.$

Now that everyone knows the answer to this question...

$Hence, find any restrictions on the value of r$

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