MX2 Marathon (1 Viewer)

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.

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.

HeroWise · Oct 12, 2019

I have solved the Riemann zeta question. I had inspiration from the heavens. God spoke to me directly and helped me solve it. He remarked "Trivial, you should be able to solve it after you do it with me,,,"
Oh well, I can post the solution but this comment box is too small to contain it.

stupid_girl · Oct 12, 2019

HeroWise said:
I have solved the Riemann zeta question. I had inspiration from the heavens. God spoke to me directly and helped me solve it. He remarked "Trivial, you should be able to solve it after you do it with me,,,"
Oh well, I can post the solution but this comment box is too small to contain it.

Did you prove or disprove the hypothesis?

sharky564 · Oct 13, 2019

stupid_girl said:
Did you prove or disprove the hypothesis?

inb4 unprovable in ZFC

Checkmate · Oct 14, 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'm willing to take more of a risk and will pay you $10000 if you DM only me

HeroWise · Oct 14, 2019

@Checkmate k DM'd Send money rn or the Nigerian prince will die.

Checkmate · Oct 14, 2019

HeroWise said:
@Checkmate k DM'd Send money rn or the Nigerian prince will die.

No clue what you talking about

.... I was the one who solved it

HeroWise · Oct 14, 2019

Damn I got screenshots Ill send it to DMCA right this moment

Checkmate · Oct 14, 2019

HeroWise said:
Damn I got screenshots Ill send it to DMCA right this moment

Too busy claiming the fields medal and discussing my proof with Terence Tao

. Ill get back to you later.

HeroWise · Oct 15, 2019

JUst used Time machine to show KId Terry Tao how to do Question 6

stupid_girl · Dec 16, 2019

(a) Consider the graph of $y=\sqrt[3]{x^{5}}\sin\frac{1}{x}$ .
(i) Is (0,0) a stationary point?
(ii) Is (0,0) a minimum point?
(iii) Is (0,0) a maximum point?
(iv) Is (0,0) a point of inflection?

(b) Consider the graph of $y=\left|\sqrt[3]{x^{5}}\sin\frac{1}{x}\right|$ .
Is (0,0) a minimum point?

(c) Consider the graph of $y=\sqrt[3]{x^{5}}\left|\sin\frac{1}{x}\right|$ .
Is (0,0) a point of inflection?

stupid_girl · Dec 21, 2019

stupid_girl said:
(a) Consider the graph of $y=\sqrt[3]{x^{5}}\sin\frac{1}{x}$ .
(i) Is (0,0) a stationary point?
(ii) Is (0,0) a minimum point?
(iii) Is (0,0) a maximum point?
(iv) Is (0,0) a point of inflection?

(b) Consider the graph of $y=\left|\sqrt[3]{x^{5}}\sin\frac{1}{x}\right|$ .
Is (0,0) a minimum point?

(c) Consider the graph of $y=\sqrt[3]{x^{5}}\left|\sin\frac{1}{x}\right|$ .
Is (0,0) a point of inflection?

It seems no one has attempted this interesting question yet.

(a)(i) Yes. When x=0, dy/dx=0.
(ii) No. For any interval -𝛿<x<𝛿, there exists x such that y<0.
(iii) No. For any interval -𝛿<x<𝛿, there exists x such that y>0.
(iv) No. The graph is neither concave nor convex near x=0.

(b) Yes. For any interval -𝛿<x<𝛿, y≥0.
(0,0) is actually one of the global minimum points.

(c) No. The graph is neither concave nor convex near x=0.

HazzRat · Dec 14, 2023

Next term or even later in the holidays we'll bring this back. Looks fun

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