# HSC 2018-2019 MX2 Marathon (1 Viewer)

#### HeroWise

##### Active Member
@blyatman once i was REALLy into programming and was doing programming comp qtns for fun. And I also remember writing code for the probability of getting a chord less than a diameter in a unit circle too. GOod old days

#### jskeza

##### Member
@blyatman once i was REALLy into programming and was doing programming comp qtns for fun. And I also remember writing code for the probability of getting a chord less than a diameter in a unit circle too. GOod old days
What course you planning on doing (if you want to say)?

#### stupid_girl

##### Active Member
Shall we have geometry again? It would be interesting if geometry can be given more weight in its final MX2 year.

AB is a chord of a circle with centre O. Extend AB to C and extend OB to D such that AC=OC, AD||OC and ∠ACD=15°. Find BC:BD.

#### TheOnePheeph

##### Active Member
Shall we have geometry again? It would be interesting if geometry can be given more weight in its final MX2 year.

AB is a chord of a circle with centre O. Extend AB to C and extend OB to D such that AC=OC, AD||OC and ∠ACD=15°. Find BC:BD.
I sure hope that geometry doesn't have more of a weight in this exam, convoluted geo questions are the absolute worst.

Is the answer to that problem $\bg_white \sqrt{6}+\sqrt{2}$ though?

#### stupid_girl

##### Active Member
Is the answer to that problem $\bg_white \sqrt{6}+\sqrt{2}$ though?
Yes

#### stupid_girl

##### Active Member
geometry again

In the figure, ABC, ADE and AFG are equilateral triangles. H, I and J are the mid-points of CD, EF and GB respectively. Find HI:IJ.

#### HeroWise

##### Active Member
TheOnePheeph can u work out the previous one if possible

#### TheOnePheeph

##### Active Member
TheOnePheeph can u work out the previous one if possible
Wait the one just posted or the one before that which I said was $\bg_white \sqrt{6}+\sqrt{2}$? If you mean the former I haven't attempted it yet as I've been studying for english lol. If you mean the latter I can write up a solution on paper a bit later and scan it. Its pretty much the exact same process as the other one I gave a solution to - finding two expressions for the ratio using sine rule, then equating them to get a value for an angle. I just didn't want to write the latex up, as it takes yonks and there is a lot of prior geometrical business in that one before getting into the sine rule algebra.

#### HeroWise

##### Active Member
The latter. Take your time man

#### TheOnePheeph

##### Active Member
The latter. Take your time man
Here you go, just wrote it up now. Know that there is likely an easier way of doing this, but my solution is the first way I saw, and does yield the correct answer.

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#### HeroWise

##### Active Member
OK so my way was not way off. Mine went off in tangents b4 i came to the required result. @stupid_girl is there a faster way?

#### stupid_girl

##### Active Member
OK so my way was not way off. Mine went off in tangents b4 i came to the required result. @stupid_girl is there a faster way?
A solution based on elementary geometry will just be more convoluted.

If you have attempted the hardest easy geometry problem (google it if you haven't), then you should see that sine rule can save you from constructing many additional lines.

#### sharky564

##### Member
geometry again

In the figure, ABC, ADE and AFG are equilateral triangles. H, I and J are the mid-points of CD, EF and GB respectively. Find HI:IJ.
View attachment 27263
We construct $\bg_white K$ such that $\bg_white CFK$ is an equilateral triangle with those points going around the triangle anti-clockwise. Then, we note that $\bg_white CB=CA$, $\bg_white CK=CF$ and $\bg_white \angle BCK = \angle BCA - \angle KCA = 60^{\circ} - \angle KCA = \angle KCF - \angle KCA = \angle ACF$ (we're using directed angles here). Therefore, $\bg_white \triangle ACF \cong \triangle BCK$ by the SAS criterion, so $\bg_white BK = AF = AG$. Similarly, $\bg_white KG = BC = AB$, so $\bg_white ABKG$ is a parallelogram, and the midpoint of $\bg_white AK$ is the midpoint of $\bg_white BG$, namely $\bg_white J$.

Finally, if we let $\bg_white EF$ and $\bg_white CD$ intersect at $\bg_white P$, then there exists a dilation at $\bg_white P$ sending $\bg_white CFK$ to $\bg_white DEA$. As this transformation occurs, these points pass through the midpoints of the segments $\bg_white DC, EF, AK$, which are just $\bg_white H,I,J$ respectively so we must have $\bg_white HIJ$ equilateral.

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#### stupid_girl

##### Active Member
$\bg_white \noindent Let T_n be a sequence such that T_1=1, T_2=2 and T_{n+2}=T_{n+1}+T_n. Show by mathematical induction that T_nT_{n+2}-(T_{n+1})^2=\left(-1\right)^n.$

If you like to play with induction, you may also try
$\bg_white \noindent Let a and b be two relatively prime integers. Show by mathematical induction that (a+b)^{2n+1}-a^{2n+1}-b^{2n+1} is divisible by ab(a+b).$

Having said that, it is in fact easier to do it without induction.

#### blyatman

##### Well-Known Member
Can anyone help me on this one?

Consider the Riemann zeta function $\bg_white \zeta$, defined as
$\bg_white \zeta(s)=\sum_{n=1}^\infty\frac{1}{n^s}$
where $\bg_white s$ is complex with a real part greater than 1.
Prove that the real part of every non-trivial zero of the Riemann zeta function is $\bg_white \frac{1}{2}$.

If you work it out, please don't post it here but DM me.

No, DM me - I'll pay $100 for it! #### Drdusk ##### π Moderator Can anyone help me on this one? Consider the Riemann zeta function $\bg_white \zeta$, defined as $\bg_white \zeta(s)=\sum_{n=1}^\infty\frac{1}{n^s}$ where $\bg_white s$ is complex with a real part greater than 1. Prove that the real part of every non-trivial zero of the Riemann zeta function is $\bg_white \frac{1}{2}$. If you work it out, please don't post it here but DM me. I'll pay you$10, and I'm already taking a risk here. Also, after you send it to me, you must destroy your copy (this part is VERY IMPORTANT and non-negotiable).
I'll pay \$1000, how about that DM ME GUYS

#### sharky564

##### Member
$\bg_white \noindent Let T_n be a sequence such that T_1=1, T_2=2 and T_{n+2}=T_{n+1}+T_n. Show by mathematical induction that T_nT_{n+2}-(T_{n+1})^2=\left(-1\right)^n.$

If you like to play with induction, you may also try
$\bg_white \noindent Let a and b be two relatively prime integers. Show by mathematical induction that (a+b)^{2n+1}-a^{2n+1}-b^{2n+1} is divisible by ab(a+b).$

Having said that, it is in fact easier to do it without induction.
First q: Note that
$\bg_white T_n T_{n+2} - T_{n+1}^2 = T_n^2 + T_n T_{n+1} - T_n^2 - 2T_n T_{n-1} - T_{n-1}^2$ $\bg_white = T_n (T_{n+1} - 2T_{n-1}) - T_{n-1}^2$ $\bg_white = T_n T_{n-2} - T_{n-1}^2,$
and then check base cases to get the result.

Second q: Note that
$\bg_white (a+b)^{2n+1} - a^{2n+1} - b^{2n+1} = (a+b)^2[(a+b)^{2n-1} - a^{2n-1} - b^{2n-1}]$ $\bg_white + ab(a+b)\left [\frac{2(a^{2n-1} + b^{2n-1})}{a+b} + \frac{ab(a^{2n-3} + b^{2n-3})}{a+b}\right ],$
where we note the fractions are integers as $\bg_white a+b$ divides $\bg_white a^{2k+1} + b^{2k+1}$, and then just do the base cases.