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MX2 Marathon (4 Viewers)

TheOnePheeph

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4/pi sounds about right. Been a while so I can't remember.

But the average definition used is:
Yeah thats the way I did it with the angle as the parameter, which gives 4/pi. I still find it weird though that it yields a different answer with distances, using:

Oh well, the angle way is more intuitive anyway lol.
 

TheOnePheeph

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I found this which offers some explanation: https://www.physicsforums.com/threads/average-chord-length-of-a-circle.929479/

They claim the answer is 4/pi. The article goes into a bit of detail, but I don't think it's necessary.

In the calculation, they use "u" as the perpendicular distance to the chord. Now, this part doesn't sit right with me: imagine the question was phrased as: calculate the average length of all horizontal chords on the circle. The calculation, and hence the answer, would be remain unchanged. I.e. the average length of all horizontal chords is the same as the average length of all chords. So, do we expect this to be the case? Intuitively I would expect it to be the same, but at the same time I'm not entirely convinced. If we were to reduce the problem to a semi-circle, the average chord length would not be the same as the average horizontal chord length since you can't have angled chords that have a length equal to the diameter. As a result, I think there's an inherent flaw in this method. Pretty interesting to think about nonetheless.

Either way, I remember verifying the 4/pi answer by writing a simple Matlab script to compute the average of a million random chords lengths.
Oh right, so it has to do with the different probability distributions of the chords. But yeah I would imagine (not 100% sure) the average of any chord would be the same as the average of all horizontal chords, because you can do the exact same infinite set of chords at any inclanation from the horizontal on a circle, which is why just doing it with the angles gives the right answer of 4pi/3. It goes quite deep lol.
 

stupid_girl

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Yeah thats the way I did it with the angle as the parameter, which gives 4/pi. I still find it weird though that it yields a different answer with distances, using:

Oh well, the angle way is more intuitive anyway lol.
h is not uniformally distributed. You need to take into account the probability density function of h.🙄
 
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TheOnePheeph

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It's essentially just saying that is uniformly distributed, resulting in in a uniform distribution of lengths.

Consider this example: Suppose you wanted to plot an equal number of points around a unit circle. If you calculated the (x,y) coordinates from equal increments of theta, you'll end up with a uniform distribution of points. Now, try doing it by generating the y coordinates from equal increments of x coordinates. You'll find that the points aren't uniformly distributed - if you were to imagine plotting a huge number of points around the circle, most of the points will be skewed away from the left and right ends, because a small change in x there will lead to a large change in y (since the circle is very steep at x = +-1). In this case, the points are not uniformly distributed because the probability of picking a point with a small y value (near zero) will be much smaller than picking a point with a large y value (near 1) since most of the points have y-values skewed towards 1.

In our case, parameterisation by theta is correct since theta results in a uniform distribution. If you try to parameterise by h, you're more likely to get a longer chord length than a shorter chord length due to the steepness of the circle when h is small (this is analogous to the example above if you try to parameterise using x). When h is small, small changes to h result in small changes in L. When h is large, small changes in h result in large changes in L. Thus, the chord lengths are more skewed towards the larger values since there's a lot more closely spaced L values when h is small. This is also evident from the calculation: parameterisation by theta results in a value of 4/pi = 1.27, whereas parameterisation by h results in a value of pi/2 = 1.57. The value is significantly larger due to this skewed distribution.
Ahhhhhhh that makes so much more sense. I was thinking it may have something to do with that uneven distribution of chord lengths with perp distance, but didn't know how to put it into words or even if that was a thing. Thanks for clearing up that misconception
 

stupid_girl

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Let's have another geometry problem. This should be simpler than the one yesterday.

O is a point inside equilateral triangle ABC such that 6∠OAB=3∠OBC=2∠OCA. Find OA:OB.
 

HeroWise

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@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

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@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

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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

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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 though?
 

stupid_girl

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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.
geometry.png
 

TheOnePheeph

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TheOnePheeph can u work out the previous one if possible
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.
 

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