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Warp wrote:
So why is it not a platonic solid? My hypothesis is that the resulting triangles (the ones that use the previously-existing vertices as one of their vertices) are not equilateral anymore.
This is correct.
Warp wrote:
So, the problem:
A side of an icosahedron is divided into four equal triangles, as depicted above. The three new vertices are moved away from the center of the circumscribed sphere, to its surface. What are the lengths of sides of these four new triangles?
If the radius of the circle is 1 then the edge length (A) of an inscribed icosohedron is csc 2pi/5 or 1.05146...
The new vertexes are equidistance from the old ones, so we can use a few triangles to find the length of b, which is the length of one of the outer edges. The angle of the triangle formed by the radius, and the perpendicular bisector of A can be found to be 0.55357. B is the chord across this angle, which is also the length of one of the outer edges. B = 2*r sin (theta/2) = 0.54653...
The internal triangle is a bit more tricky. The internal triangle bordered by the internal edges (C) is an equilateral triangle, because the A to B transformation is symmetric on all three corners. If we visualize looking directly into the middle of a face head-on the path of our eye to the center of the icosahedron and call that leg D We can find the length of D without the triforce extruded using the length from the middle of the face to the outer edge (0.6071...). From this we find D = 0.794654...
Now we can calculate D' the length from the center to the middle of the extruded triforce face. When looking down at the extruded face we can see that the middle triangle has grown slightly, and a projection of it face on very slightly overhangs the original triangle. By how much?
Well, we can get the distance from the edge A to the corner of the interior triangle with another Pythagorean argument (A/2)^2 + E^2 = B^2 => E = 0.149349..., but this isn't the shortest distance between the faces. We need to know the angle between the center of a face and the center of an edge. We can find this through the dihedral angle, which is arccos(sqrt(5)/3) = 2.4119... radians, which we can chop in half and subtract from pi/2 to get the angle we're looking for: 0.36486...
Finally, we can use this angle to get the distance between the old and new faces (E'). It's 0.139518...
Now we're armed with enough information to find C. D' = D + E' = 0.934172... So the distance from the center of the extruded face to a corner is 0.356822... which makes C = 0.618034...
B = 0.54653... which isn't C, so it's not equilateral.
EDIT: Performing a sanity check C*2 should be > A, but it's not, I'll review my work and see if I can find the error.
EDIT2: Located a calculator error, fixed. Now passes sanity checks.
EDIT3: Apparently, I cannot Pythagoras, fixed more calculation errors.
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I was given an unfair percentage of the credit for work that was done by a large team of talented members.
Honestly, that comes with the territory of being the "face" of an operation. If you're not going out of your way to take credit then I don't see a foul.
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Warp wrote:
Warp wrote:
Speaking of the Riemann zeta function, I have had for some time now an interest in understanding the Riemann hypothesis. Just out of curiosity, as a little "hobby project". I was wondering if someone could help me with it.
In order to understand the hypothesis, I first need to understand the Riemann zeta function. In order to understand said function, I need to understand analytic continuations. And the topic already becomes really complicated (not to talk about how complicated the zeta function itself is.)
While that is indeed a good starting point for understanding qualitatively what the concept of an analytic continuation is, it's not sufficient. What you really need is to take a course in complex analysis.
I don't know your mathematical background, but you only need up until Multivariable Calculus and the basics of Complex Numbers (polar form, roots of unity, etc) before you can tackle Complex Analysis. I've been eyeing taking the Complex Analysis course at my university, so I've watched a lecture series on Complex Analysis on Youtube. Specifically, the one starting from this video: https://www.youtube.com/watch?v=exTsBIQoxcI
You should probably start with this one though, it's more indepth and has what you're interested in (analytic continuations): https://www.youtube.com/watch?v=BruPj2mUGMo&list=PLun8-Z_lTkC5wjZ-8TH99y3htILDwlji5
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The exact value of the probability for the cards is:
4610507544750288132457667562311567997623087869 /
284025438982318025793544200005777916187500000000
to 30 decimal places:
0.0162327274671946367481295565133
I solved it by using the OEIS, apparently this problem is related to the Dinner-Diner matching problem.
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Because 2100 *isn't* a leap year, we can have a 40 year gap in the otherwise regular 28 year cycle of the calendar. Which means the only possible years are 2072, 2076, 2080, 2084, 2088
Of those 5, we can check each and eliminate the ones that cycled in 1900, 1800, etc in the same way.
2080 is the same as 1872 which cycled 40 years through 1900.
2084 is the same as 1876 which cycled 40 years through 1900.
2088 is the same as 1880 which cycled 40 years through 1900.
2076 is the same as 1772 which cycled 40 years through 1800.
2072 is unique in that it couldn't cycle through 1900, 1800. It normally would have done the 40 year cycle in 1700, but because England adopted the Gregorian calendar in 1752 I guess it doesn't count.
This is what 2072's calendar looks like:
https://www.timeanddate.com/calendar/?year=2072&country=1
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Xander wrote:
I originally planned to do a 120 star TAS except I just didnt want to have to deal with estimating starbit collection. There are 2 points in an rta run where you have to stop and grind to 999 starbits except it is possible that in a TAS you might not have to. That's only a maybe and if I ran the risk of not grinding and i get to the end and it turns out I did need to grind, I would have to redo most of the TAS.
There's an easy way to determine if it's needed. Go through an RTA run and count the number of star bits available for pickup (or easily obtainable) outside of the normal grinding segments.
This will take only a few hours, and as mentioned it will save you a hell of a lot of time.
With any luck, there will be a clear surplus, and you don't have to worry about routing star bits too much. There are clearly a massive number of star bits available, and this is the most likely option (you collected over 100 without stopping on the first star.) But this way you can confirm it.
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feos wrote:
OmnipotentEntity wrote:
When do you consider the situation in Ukraine to have started? With the annexation? Or with the election/attempted Russian assassination of Yushchenko in 2004?
Some other time?
With the events that resulted in the civil war. Which actions of the Russia before then were examples of dividing the West? Or you can put it differently. Which actions of Russia not related to Ukraine were examples of dividing the West? You can answer both questions separately if you wish.
Not related to Ukraine is different from occurring before Euromaidan. For instance, covert support for Trump and Brexit is a big one. It's hard to say either of these are directly related to Ukraine.
Especially if you consider the Foundations of Geopolitics, which is apparently used as a textbook in Russian General Staff Academy. (Link is to an ancient revision of the wikipedia article, prior to the Ukrainian "situation.")
Primarily, the Russian goal for the UK is to separate it from the EU, and for the US is to stir up racism and extremism.
Additionally, Finland should be absorbed into Russia. Just letting the Finns in the audience know the endgame here. Putin isn't going to be placated.
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feos wrote:
How many examples of that can I get, before the "situation" in Ukraine has started?
When do you consider the situation in Ukraine to have started? With the annexation? Or with the election/attempted Russian assassination of Yushchenko in 2004?
Some other time?
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I'm skeptical Putin is going to be "placated." The Russian game plan is to gain power by dividing the West. And it's succeeded so far.
So now we get to motivations. Why does Putin want the West weak? I doubt the answer to that is terribly comforting to anyone. And I really doubt that Putin is going to just sit on his hands once the EU falls apart.
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Both of those answers depend on the nature of the end of the universe.
If the universe ends in a Big Rip, the answer very strongly depends on the nature of dark energy, and whether or not the ratio of its pressure and its energy density is less than, greater than or equal to -1. The ship would not survive the Big Rip.
If the universe ends in Heat Death, this could take as much as 10^2100 years to finalize, but the universe will be exceedingly uninteresting for us after about 10 to 100 quadrillion years. The ship would survive this. It would take between 71 and 76 years to complete the journey to the uninteresting bit.
Universe ends via vacuum decay. No telling how long this might take. So no good estimate.
Big Crunch will almost certainly not happen thanks to dark energy.
But overall it just might be easier to link you this: http://nathangeffen.webfactional.com/spacetravel/spacetravel.php
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TheAngryPanda wrote:
And I don't mean for this to be a degrading or asshole-ish post, these are simply the reasons and ways I would use an input file if I had one, nothing more. That's all I'm going to say on the matter as well because I'd rather put my focus towards something more productive.
Let's be frank. Saturn has decided that he doesn't want to work with TASVideos because we didn't publish one of his Super Metroid runs once. So he's made it a goal to troll us. Please stop feeding him.
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I figured out the issue with my logic. I was conflating dy (the width of the slice) with ds (the bias of the slice when taking into account the slope in the y direction). When pulled flat, the width of the slice is ds not dy.
If we take this into account, we'll likely arrive at f^2*x*y, but I don't have time to demonstrate it fully right now.
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Bobo the King wrote:
I was about to throw down a healthy dose of skepticism, but a quick analysis below shows that your answer is plausible. I'm still troubled by the fact that you very quickly state that the area of the cross-section is f*x*dy. Can you go into more detail on how you got this? I don't see where you took into account that the villi exhibits the same ridges along the y direction.
f*x being the length is plausible. So, I believe you're taking exception with the width being dy.
So here's my logic: if you take that cross section and pull it straight you get a line with a length of f*x. But the width does not change with this action. Also, this action smooths out the undulations in both the x and the y directions for that slice. Then we just add up all of the slices.
I could, of course, be overlooking something critical. It's early now, and late when I wrote that. And I need more coffee.
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We can assume that the villi are spaced randomly and that our single cross section represents the mean cross section.
Because this is our mean, all cross sections will have on average, the same arc length.
Let's call the arc length / length = f, our constant of proportionality.
Let's consider the area of this cross section to be f*x*dy. x = length.
Therefore, the area is just integral of this value, or f*x*y. So f*A rather than f^2*A.
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Amaraticando wrote:
I made this cute geometry problem for you 😛😈 Can you find the value of this angle (DAB)?
60 + 40 + phi + x = 180 (large triangle)
30 + 15 + y = 180 (bottom triangle)
30 + phi + z = 180 (left triangle)
25 + w + x = 180 (right triangle)
y + z + w = 360 (angles about D)
5 variables, 5 functions. But the solutions are insufficiently constrained with just these equations. And phi can take any angle in (0, 80) and still be valid.
phi elem (0,80)
z=150-phi
x=80-phi
w=phi+75
y = 135
So your question, unless I'm missing a hint, is ill-formed and has no unique solution.
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We only care about extreme values because we're determining the winner. We don't care about the number of ties, just whether or not the top result it. Extreme values in 1d6 are more likely, so although ties are rare, the likelihood that the top result is uniform. As you increase the amount of players, the probability you have a tie increases as well.
On the other hand median of 3d6 you're more likely to get the middle numbers, but we care about the high ones. So when someone rolls high, that is an improbable event, and it's more likely that someone will not replicate it. But it's also less likely that you'll roll high to begin with. So with fewer people you're more likely to all roll in the middle, and result in a tie, but as you add more people you're more likely to get just one person who rolls high by chance.
It's all about which effect dominates.
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