MX2 Marathon (1 Viewer)

HeroWise · Oct 5, 2019

@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

InteGrand · Oct 5, 2019

For the chord in a circle problem, see also https://en.wikipedia.org/wiki/Bertrand_paradox_(probability).

jskeza · Oct 5, 2019

HeroWise said:
@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)?

blyatman · Oct 5, 2019

InteGrand said:
For the chord in a circle problem, see also https://en.wikipedia.org/wiki/Bertrand_paradox_(probability).

That was an interesting read, never thought about it that way.

stupid_girl · Oct 5, 2019

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 · Oct 5, 2019

stupid_girl said:
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 $\sqrt{6}+\sqrt{2}$ though?

stupid_girl · Oct 5, 2019

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

Yes

stupid_girl · Oct 7, 2019

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 · Oct 8, 2019

TheOnePheeph can u work out the previous one if possible

TheOnePheeph · Oct 8, 2019

HeroWise said:
TheOnePheeph can u work out the previous one if possible

Wait the one just posted or the one before that which I said was $\sqrt{6}+\sqrt{2}$ ? 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 · Oct 8, 2019

The latter. Take your time man

TheOnePheeph · Oct 8, 2019

HeroWise said:
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.

HeroWise · Oct 9, 2019

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 · Oct 9, 2019

HeroWise said:
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 · Oct 10, 2019

stupid_girl said:
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 $K$ such that $CFK$ is an equilateral triangle with those points going around the triangle anti-clockwise. Then, we note that $CB=CA$ , $CK=CF$ and $\angle BCK = \angle BCA - \angle KCA = 60^{\circ} - \angle KCA = \angle KCF - \angle KCA = \angle ACF$ (we're using directed angles here). Therefore, $\triangle ACF \cong \triangle BCK$ by the SAS criterion, so $BK = AF = AG$ . Similarly, $KG = BC = AB$ , so $ABKG$ is a parallelogram, and the midpoint of $AK$ is the midpoint of $BG$ , namely $J$ .

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

stupid_girl · Oct 12, 2019

Was this one solved?
$$\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
$$\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 · Oct 12, 2019

Can anyone help me on this one?

Consider the Riemann zeta function $\zeta$ , defined as
$\zeta(s)=\sum_{n=1}^\infty\frac{1}{n^s}$
where $s$ 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 $\frac{1}{2}$ .

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).

fan96 · Oct 12, 2019

blyatman said:
Can anyone help me on this one?

Consider the Riemann zeta function $\zeta$ , defined as
$\zeta(s)=\sum_{n=1}^\infty\frac{1}{n^s}$
where $s$ 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 $\frac{1}{2}$ .

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).

No, DM me - I'll pay $100 for it!

Drdusk · Oct 12, 2019

blyatman said:
Can anyone help me on this one?

Consider the Riemann zeta function $\zeta$ , defined as
$\zeta(s)=\sum_{n=1}^\infty\frac{1}{n^s}$
where $s$ 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 $\frac{1}{2}$ .

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 · Oct 12, 2019

stupid_girl said:
Was this one solved?
$$\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
$$\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
$T_n T_{n+2} - T_{n+1}^2 = T_n^2 + T_n T_{n+1} - T_n^2 - 2T_n T_{n-1} - T_{n-1}^2$ $= T_n (T_{n+1} - 2T_{n-1}) - T_{n-1}^2$ $= T_n T_{n-2} - T_{n-1}^2,$
and then check base cases to get the result.

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

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