What course you planning on doing (if you want to say)?@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
That was an interesting read, never thought about it that way.For the chord in a circle problem, see also https://en.wikipedia.org/wiki/Bertrand_paradox_(probability).
I sure hope that geometry doesn't have more of a weight in this exam, convoluted geo questions are the absolute worst.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.
YesIs the answer to that problem though?
Wait the one just posted or the one before that which I said was ? 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.TheOnePheeph can u work out the previous one if possible
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.The latter. Take your time man
A solution based on elementary geometry will just be more convoluted.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?
We construct such that is an equilateral triangle with those points going around the triangle anti-clockwise. Then, we note that , and (we're using directed angles here). Therefore, by the SAS criterion, so . Similarly, , so is a parallelogram, and the midpoint of is the midpoint of , namely .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.
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No, DM me - I'll pay $100 for it!Can anyone help me on this one?
Consider the Riemann zeta function , defined as
where 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 .
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 GUYSCan anyone help me on this one?
Consider the Riemann zeta function , defined as
where 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 .
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).
First q: Note thatWas this one solved?
If you like to play with induction, you may also try
Having said that, it is in fact easier to do it without induction.