Editor, Experienced Forum User, Published Author, Expert player
(2159)
Joined: 6/15/2005
Posts: 3303
p4wn3r wrote:
Therefore, solving p^4+(1-p)^4 = 1/3, which does give a number between 0 and 1, you can make your method work.
One of the solutions is p = 1/2 - sqrt(12*sqrt(6)-27)/6 ≈ 0.24213; you can solve it by substituting p = 1/2 + x and expanding to get 2x4 + 3x2 - 5/24 = 0, which can be solved easily as a quadratic in x2.
p4wn3r wrote:
About Pascal's triangle, it seems that you're right. I made a program to compute it mod 4 and did not find a row for the first 100 iterations. Maybe that property only works for prime n, I don't remember.
It's not possible for all elements of a row of Pascal's triangle (other than the ends) to be a multiple of 4. The ith row of Pascal's triangle sums to 2i, a multiple of 4. Since the two 1's at the ends only add up to 2, there has to be at least one other number in the row that isn't a multiple of 4.
p4wn3r wrote:
So, even if the rows in Pascal's triangle are not all divisible, you can pick a huge m, and set person A at the extremes, and for every row you put A on m mod (N-1) combinations. If m is large enough, these new combinations should be very small, and the equation still solvable.
Good! This is what I eventually came up with when trying to show it is possible for all N. (I got tripped up on "m mod (N-1) combinations" for a while; I eventually figured out that you meant "m combinations of 'mod (N-1)'", though I think it should be m+1 rather than m.)
I use C(m,i) to mean the binomial coefficient m choose i. Assign, for each i between 0 and m inclusive, C(m,i) mod (N-1) of the C(m,i) outcomes of probability pi(1-p)m-i on A; everything else can be divided evenly among the N-1 others. Then the probability of A winning is:
f(p)=sum[0≤i≤m] (C(m,i) mod (N-1))*pi(1-p)m-i.
As mentioned before, f(0)=1 and f(1/2) = sum[0≤i≤m] (C(m,i) mod (N-1))*(1/2)m < (m+1)(N-2)(1/2)m. So to get 1/N as a possible output of f(p), we need f(1/2)<1/N so we can use the intermediate value theorem. So we just need f(1/2)<(m+1)(N-2)(1/2)m<1/N, or (m+1)(1/2)m<1/(N(N-2)), and this can clearly be satisfied for large enough m, since (m+1)(1/2)m goes to 0 as m goes to infinity.
p4wn3r wrote:
So, the equation now is this, which also works to get a fair probability for everyone.
This equation p8+2p4(1-p)4+(1-p)8 = 1/5 works out pretty nicely. The left-hand side is just (p4+(1-p)4)2, so after taking square roots, you can solve it similarly to p4+(1-p)4 = 1/3 above.
Note that for N=5, 8 flips is not the minimum number of flips required; it is possible to do it in 6 flips since the polynomial f(p) above with m=6 and N=5 can be made equal to 1/5. Though it would be a lot harder to solve even with the p = 1/2 + x substitution (you'd have a cubic in x2 now), especially if you don't have WolframAlpha.
Editor, Experienced Forum User, Published Author, Expert player
(2159)
Joined: 6/15/2005
Posts: 3303
Riddler Classic this week has an interesting question. I'll paraphrase it below:
----
You need to fairly choose one of N people to be the winner, using a coin. If N=2, then you just flip a fair coin to decide the winner; heads for one person winning and tails for the other. If N=4, then flip the coin twice; the outcomes are HH, HT, TH, TT with 1/4 probability each, and just assign each outcome to each person.
However, if N=3, it is not possible to assign the outcomes of a finite number of flips to the three people to fairly choose one of them with probability 1/3; for k flips, every outcome has probability 1/2^k, so no matter which outcomes you take, you can never get their probabilities to add to 1/3.
However, you can choose to use an unfair coin (probability p of landing heads, 1-p of landing tails) and select any p between 0 and 1 that you wish.
The question is: Is it possible to choose such a p so that you can flip the coin a finite number of times and assign the outcomes to fairly choose one of three people, and if so, how?
How would the question change if there were five people (N=5)? Or for the case of N people in general?
(Feel free to use WolframAlpha; I used it to help with this problem.)
Editor, Experienced Forum User, Published Author, Expert player
(2159)
Joined: 6/15/2005
Posts: 3303
MrWint wrote:
FractalFusion wrote:
How many rerecords do you estimate it took to bot this?
I stopped trying to track this after it has been pointed out that these numbers are wholly incomparable to anything else and just skew the statistics.
Are you referring to TASVideos statistics? Because if you are, then I could care less about whether its statistics are skewed. What I liked most about your submissions were the rerecord counts running in the hundreds of millions, if only to clearly demonstrate how much more optimized your TASes are over just about everything else.
MrWint wrote:
No. Pokérus doesn't do anything useful in Gen II.
OK. I wasn't sure whether the 2x boosted EV/Stat Exp. from Pokerus would have made a difference at any point since Cyndaquil is mostly at low levels (boosted EV is unlikely to change the stat at low levels) and when you get Raikou at LV40, you OHKO the vast majority of Pokemon from then on without using criticals. Though now that we're talking about Pokerus, I'm wondering whether the 2x boosted EV make a difference in other games. The only thing I know for sure is that Pokerus doesn't matter for the current route in Pokemon Diamond/Pearl glitched.
----
By the way, I noticed that even though Pokemon Crystal reduces the number of mandatory battles compared to Gold/Silver (by 9, if I counted correctly), Crystal is still a pretty long game. There are 81 mandatory battles in the pre-league portion of the game alone (23 in post-league), which is a pretty large number (though Diamond/Pearl glitchless has 97, counting double battles as 2). Especially long is the Goldenrod Tower section; that section alone has 24 mandatory battles, more than the entire post-league!
So even though I do appreciate all the new strategies that were implemented since my really outdated TAS of Pokemon Gold, I find Gen 2 TASes rather boring, especially now that catching Raikou is clearly the way to go. I'm saying this in case anyone was wondering why I never bothered with improving my outdated TAS from 13 years ago.
And yes, I agree that my outdated TAS should be obsoleted in favor of this TAS. This TAS is far better optimized, on a far better emulator (the one I made mine on was so broken that the system time affected sync, seriously), and on a version of the game which in my opinion is better.
Editor, Experienced Forum User, Published Author, Expert player
(2159)
Joined: 6/15/2005
Posts: 3303
p4wn3r wrote:
Interestingly, even though it makes no sense for the sequence to be periodic, by reading Conway's book where he treats these kind of games, he proves that if you compute the sequence long enough, the computation itself is proof that it'll keep periodic forever!
So you do know about Winning Ways for your Mathematical Plays. It's one of my favorite mathematical books (or rather, set of volumes); it covers not just impartial games, but the angel problem (among other various combinatorial problems), as well as Conway's Game of Life.
By the way, computation itself may be used to prove periodicity of nim sequences, but this is assuming of course that the nim sequence is periodic. It is conjectured that all octal games have periodic nim sequence, but that is not proven, and there are more than a few octal games whose periodicity has not been determined even when searching up to 233.
p4wn3r wrote:
Prove that sin(10o)cos(20o)cos(40o) = 1/8
p4wn3r wrote:
Another solution is to multiply and divide by sin(20o), use sin(10o)=cos(80o) and use the double angle formula: sin(2x)=2sin(x)cos(x)
You can also multiply and divide by cos(10o), use the double angle formula, and use cos(10o)=sin(80o). Same idea. Though I never knew there was a way to do this without using cubics (casus irreducibilis was my first thought for this question).
p4wn3r wrote:
Find all pairs of positive integers n and m, such that n/m + m/n is also an integer.
Suppose n,m is such a pair. Without loss of generality, assume n≥m. If n>1, then there exists a prime p dividing n. Since n/m + m/n = (n2+m2)/(mn), it follows that if p divides n, then p divides n2+m2, and therefore m, as well. Then we can write n=pc, m=pd, and so n/m + m/n = (pc)/(pd) + (pd)/(pc) = c/d + d/c, and so c,d is a smaller pair satisfying the above statement. So we can cancel common prime factors in n and m until it is no longer possible to do so, which is the case n=1. It is easily checked that n=m=1 satisfies the statement, and so solutions occur exactly when n=m.
Editor, Experienced Forum User, Published Author, Expert player
(2159)
Joined: 6/15/2005
Posts: 3303
Looks technically good. A few questions though:
- How many rerecords do you estimate it took to bot this?
- Did Pokerus ever factor into your route? I don't see a mention of it anywhere.
- Did you consider poison deathwarping? There is one spot I can think of, and that is in the Goldenrod Radio Tower after beating the impostor manager. That would also allow you to fight a grunt with one Pokemon (around 1:18:58 in the video) instead of the scientist with three.
As already mentioned here, I enjoyed the normal difficulty run over the Ultimortal difficulty one. I imagine that for me it would be the same case here regarding difficulty. Though this "all items" TAS doesn't really interest me all that much, in my opinion. (I don't know the game a whole lot since I never played it.)
Editor, Experienced Forum User, Published Author, Expert player
(2159)
Joined: 6/15/2005
Posts: 3303
Thanks for giving everyone unique names this time.
The new skip actually makes it better by reducing the number of bosses where you use Slots (still a lot of Slots usage though). In particular, the fight against Materia Keeper where you manipulate its HP to 7777 and poison it so it would take 7777 damage because of the All Lucky 7s status is ingenious.
Editor, Experienced Forum User, Published Author, Expert player
(2159)
Joined: 6/15/2005
Posts: 3303
Here is an encode (don't forget to enable 720p60, Mega Man must be watched in 60fps):
Link to video
Rockman: Spirits of Hackers is a Mega Man 6 hack by Puresabe (the ROM hacker who did Rockman 4: Minus Infinity) and a number of others. The ROM hack can be downloaded here and you'll need to patch the JP version of Mega Man 6. (Note the checksum of the ROM I used doesn't match the information given at romhacking.net, but it matches the one in the movie file).
As posted here, Version 0.00 is the only version of this hack, and it is definitively final (no further updates will be made) despite its "INCOMPLETE" designation. So there should be no problem in submitting TASes for this hack with regards to which version should be used, will there be a newer version, etc.
Editor, Experienced Forum User, Published Author, Expert player
(2159)
Joined: 6/15/2005
Posts: 3303
I mentioned the endings because it is possible to save time by killing one of the players in the last screen so they don't have to get to the rope to finish the game, but you would end with the 1-player ending. This is not a strat that I would use myself (because getting the 1-player ending in a 2-player run really does look like a bad ending) but it has been used before in this 2-player speedrun.
Editor, Experienced Forum User, Published Author, Expert player
(2159)
Joined: 6/15/2005
Posts: 3303
p4wn3r wrote:
The game works like this. The dealer deals an amount N>=2 of cards face down in a straight line (the content of the cards doesn't matter). The player and the dealer take alternating turns, with the player starting. At any point, the person who's playing can either:
(a) Remove exactly two consecutive cards from any point in the interval.
(b) Remove a single card, provided that it's isolated from the rest of the pack.
The game goes on like this until it's someone's turn and there are no cards left. In that case, the person who has the turn loses.
I'll just cut straight to the answer since it would otherwise take a giant post to explain every little thing here:
This game is the octal game ".17". This OEIS sequence gives the nim values (or "Sprague-Grundy values") for this game; starting N=0:
0,1,1,0,2,1,3,0,1,1,3,2,2,3,4,1,5,3,2,2,3,1,1,0,3,1,2,0,1,1,4,4,2,6,
4,1,1,0,2,1,3,0,1,1,3,2,2,3,4,4,5,7,2,2,3,1,1,0,3,1,2,0,1,1,4,4,3,6,
4,1,1,0,2,1,3,0,1,1,3,2,2,3,4,4,5,7,2,2,3,1,1,0,3,1,2,0,1,1,4,4,3,6,
...
The starting position of N cards is a win for the player (who has the first move) if and only if its nim value is nonzero. Since it is periodic with period 34, and only 4 values in each period are zero, assuming that there is no limit to how large N can be, this gives a winning probability of 30/34 = 15/17. So a fair payoff would be 2/15 a dollar for every dollar bet.
----
PS: Every position in an impartial game under normal play convention (last to play wins) can be assigned a nim value; if the nim value is n, there exist moves to positions with nim value 0, 1, 2, ..., n-1, but no moves to positions with nim value n. Hence positions can be thought of as heaps in a slightly-modified Nim game. Positions that can be divided into independent subpositions have a nim value equal to the nim sum (binary XOR sum) of the nim values of all its subpositions.
The nim value of a position can be calculated as: a position where no move is possible has nim value 0, and for all other positions, the nim value is the smallest non-negative integer not appearing as the nim value of any position it can move to. Using this, it is possible to calculate the nim values for the octal game ".17", and it turns out to be periodic with period 34 and just 4 exceptional values; this can be verified computationally.
Editor, Experienced Forum User, Published Author, Expert player
(2159)
Joined: 6/15/2005
Posts: 3303
Some context:
Cowlitz Gamers 2nd Adventure is a free NES homebrew game created in 2017 for Cowlitz Gamers for Kids Expo. It is a platformer where you collect coins to open up the path upward; there are 48 screens (not counting bonus screens) in this game. You can download the ROM from the game's main site.
The only difference between difficulty levels is the number of lives you start out with and how many coins you need to get more lives.
It is faster to complete the game with 2 players since they can both collect the coins necessary to open the path to the next screen. Deathwarps are abused a lot in this TAS. There are a couple endings depending on whether you finished with both players alive or with just one player.
Note that there are collectible gems in each screen which allow you to access hidden bonus screens. Collecting all the gems (and coins) is a goal I would like to see as well.
I uploaded an encode (will take some time before it is ready):
Link to video
Note that the ending music is a few minutes long before it ends. A button press is necessary to get to the "THE END" screen with all the stats; I included it in the encode.
Editor, Experienced Forum User, Published Author, Expert player
(2159)
Joined: 6/15/2005
Posts: 3303
p4wn3r wrote:
Prove or disprove:
Let a, b, c be real numbers. Suppose that l is a real number satisfying l > max(|a-b|,|b-c|,|c-a|) and l < min(a+b,b+c,c+a), so that it's possible to form a triangle having one of the sides measuring l, and any two numbers from a, b and c. Then, in these conditions, it's always possible to construct a triangular pyramid having an equilateral triangle of side l as basis, and having a, b and c as the lengths of the remaining edges.
If false, try to come up with a condition that allows the pyramid with equilateral basis to be constructed.
Ptolemy's inequality on the tetrahedron (=triangular pyramid) with equilateral triangle base of side l and other edges a,b,c gives al+bl>cl, which gives a+b>c. Note that it is equality only if the four points lie on a circle in a plane, which is not the case here. Likewise b+c>a and c+a>b. If any one of these inequalities does not hold, then there exists no such tetrahedron. E.g. For a=1, b=1, c=2.5, l=1.75, it is not possible, even though the triangle inequality on all faces is satisfied.
Note that a+b>c, b+c>a, c+a>b is still not sufficient; for example, as explained in the posts above, l=1, a=b=c=0.55 still does not give a tetrahedron even though all necessary conditions given so far are satisfied.
Searching online reveals a paper giving a necessary and sufficient condition: Six given edge lengths determine a tetrahedron iff the triangle inequality holds for all faces and its Cayley-Menger determinant is greater than 0. Using the last formula on the Wikipedia page and substituting A=a2, B=b2, C=c2, L=l2 gives:
4ABC+(A+B-L)(B+C-L)(A+C-L)-A(B+C-L)2-B(A+C-L)2-C(A+B-L)2>0
or, after expanding and writing as a polynomial in L:
-L3+(A+B+C)L2+(AB+AC+BC-A2-B2-C2)L>0
Since L must be positive, divide by -L to get:
L2-(A+B+C)L+(A2+B2+C2-AB-AC-BC)<0
Now A2+B2+C2-AB-AC-BC=(1/2)((A-B)2+(B-C)2+(A-C)2)>0, and A+B+C is positive, so solving x2-(A+B+C)x+(AB+AC+BC-A2-B2-C2)=0 gives us positive solutions for x. We get:
x=(1/2) (A+B+C ± sqrt(3)sqrt(2AB+2AC+2BC-A2-B2-C2))
But note that 2AB+2AC+2BC-A2-B2-C2 = (a+b+c)(-a+b+c)(a-b+c)(a+b-c). And so this gives us:
x=(1/2) (a2+b2+c2 ± sqrt(3)sqrt((a+b+c)(-a+b+c)(a-b+c)(a+b-c)))
Thus:
(1/2) (a2+b2+c2 - sqrt(3)sqrt((a+b+c)(-a+b+c)(a-b+c)(a+b-c))) < L < (1/2) (a2+b2+c2 + sqrt(3)sqrt((a+b+c)(-a+b+c)(a-b+c)(a+b-c)))
and since this interval is in the positive reals, this gives for l:
One more thing: how do we know that (a+b+c)(-a+b+c)(a-b+c)(a+b-c) must be positive? Well, that just follows from a+b>c, b+c>a, c+a>b which I derived from Ptolemy's inequality above. So the necessary and sufficient conditions that allows the equilateral base to be constructed are l > max(|a-b|,|b-c|,|c-a|), l < min(a+b,b+c,c+a), a+b>c, b+c>a, c+a>b, and the bound on l above.
Editor, Experienced Forum User, Published Author, Expert player
(2159)
Joined: 6/15/2005
Posts: 3303
In case anyone wants an encode of my 1:46 TAS, I made one just now:
Link to video
(I didn't make one before because I never expected the sudden interest in Pitfall.)
Editor, Experienced Forum User, Published Author, Expert player
(2159)
Joined: 6/15/2005
Posts: 3303
p4wn3r wrote:
Prove that:
First, note the triple angle identity
which we can rewrite as
Using the identity on itself multiple times, we get
As k goes to infinity, we get
The left hand side is of the form theta*sin(x)/x (where x=theta/3k) as x goes to 0, for which the limit is theta. Then plugging in theta=pi gives
and dividing by 4 gives you the identity you posted.
Editor, Experienced Forum User, Published Author, Expert player
(2159)
Joined: 6/15/2005
Posts: 3303
PJ wrote:
The biggest change is the cutscene after fighting Mash in the Robot Museum. You can hold start to advance the text during that cutscene; you let it play out in the WIP.
Thanks! That explains why I wasn't gaining any time over the 100% TAS. Now I feel better about it.
Note that there are some things I did in the WIP which are actually suboptimal (I think one or two Rush Coil instances are like that), so it isn't completely well done. I used my WIP mainly to try out different strategies.
Editor, Experienced Forum User, Published Author, Expert player
(2159)
Joined: 6/15/2005
Posts: 3303
I did it incrementally. And similar to what p4wn3r did, you don't ever need to leave the world of symmetric polynomials.
● (a+b+c)2 = a2+b2+c2 + 2(ab+ac+bc) → ab+ac+bc = (12-2)/2 = -1/2
● (a2+b2+c2)(a+b+c) = a3+b3+c3 + a2b+ab2+a2c+ac2+b2c+bc2 → a2b+ab2+a2c+ac2+b2c+bc2 = 2(1) - 3 = -1
● (ab+ac+bc)(a+b+c) = a2b+ab2+a2c+ac2+b2c+bc2 + 3abc → abc = ((-1/2)(1)-(-1))/3 = 1/6
● (ab+ac+bc)2 = a2b2+a2c2+b2c2 + 2a2bc+2ab2c+2abc2 → a2b2+a2c2+b2c2 = (-1/2)2 - 2abc(a+b+c) = 1/4 - 2(1/6)(1) = -1/12
● (a2+b2+c2)2 = a4+b4+c4 + 2(a2b2+a2c2+b2c2) → a4+b4+c4 = 22 - 2(-1/12) = 25/6.
I imagine you would get higher symmetric polynomials using this method.
----
Actually, there is a method to solve this using a recursive formula that gives you all higher powers an+bn+cn for n≥4. If we define f(n)=an+bn+cn, then f(n) is the solution to a recursive formula with f(1)=1, f(2)=2, f(3)=3 and whose characteristic polynomial is:
(x-a)(x-b)(x-c) = x3 - (a+b+c)x2 + (ab+bc+ac)x - abc = x3 - x2 - (1/2)x - (1/6),
that is, f(n) = f(n-1) + (1/2)f(n-2) + (1/6)f(n-3). Plugging in gives f(4) = f(3)+(1/2)f(2)+(1/6)f(1) = 25/6, and you can use this formula to get the higher powers: f(5)=6, f(6)=103/12≈8.5833, and so on.
Editor, Experienced Forum User, Published Author, Expert player
(2159)
Joined: 6/15/2005
Posts: 3303
Riddler Express this week sounds interesting:
You purchase a new clock but are dismayed to realize that both of its hands are identical. At first, it seems it’s going to be impossible to tell the time because you don’t know which hand is for the minutes and which is for the hours.
However, you realize you don’t need to know which is which for every time — for example, when it’s 12:30, the minute hand will be exactly on the 6 and the hour hand will be halfway between the 12 and the 1. It can’t be the other way around because if the hour hand were exactly on 6, the minute hand would have to exactly on 12, which it’s not. So you know what time it is.
How many times during the day will you not be able to tell the time?
Editor, Experienced Forum User, Published Author, Expert player
(2159)
Joined: 6/15/2005
Posts: 3303
Mar. 2019 ranking in detail.
Runs from this site that made it on there: Magical Quest 3 starring Mickey & Donald, Tetris TGM2, Super Mario All Stars: Super Mario Bros. 3, and Indiana Jones and the Last Crusade (UBI Soft).
Editor, Experienced Forum User, Published Author, Expert player
(2159)
Joined: 6/15/2005
Posts: 3303
Oh, you actually did SoftLock%. Cool. I looked at it briefly after seeing Shenanagans do it but then I eventually forgot about it.
While testing, I made a TAS on the Japanese Red version (Pokemon Red JP V1.0) that apparently completes it 1:20. It seems as though mine is faster (just from comparing the gameplay portion of it) but I don't run my stuff off GBC firmware. I don't have time to figure out which one is legitimately faster, but I'll note that it is possible to start the Prof. Oak scene before the girl gets there, because the girl can still move to that square before your character bonks her. If you would like an encode, just ask me for it.
The other thing is that I wish you had chosen a day of the year other than this one to submit this TAS. I hate this day.
Editor, Experienced Forum User, Published Author, Expert player
(2159)
Joined: 6/15/2005
Posts: 3303
Hi PJ. Haven't seen you around for a while.
I already found out that glitch two years ago (although I never made it public). I'll tell you what I know so far.
- I came to the same conclusion as you, that the Wily Machine can be quick killed with this method (but not Wily Capsule, since it doesn't have a hitbox).
- I disassembled how Rush Search works. Rush Search has three random components. One is the item to be dug up, one is the first Rush delay (wag-tail animation) and the last one is the second Rush delay (digging animation). Note the last one does not apply when digging up important items. There is a 3/32, 3/32, 1/32, and 1/32 chance of digging up small life energy, small weapon energy, large life energy, and large weapon energy, respectively. So the chances of pulling this trick off for the Wily Machine is only 1/4.
- Using it on the Cloud Proto visit saves time, if it works. It is also possible to use on the Turbo Proto visit but is difficult to do and unnecessary because there is a full Turbo Proto visit skip which is much faster and isn't random (you have to enter the room by jumping and then exit the room while touching the ground at the same time).
- As you already found, you can use it to get the Proto Shield while Proto Man is talking. With Shade Man already done you can then leave the level. According to my calculations, it is now faster in a TAS to do Shade Man first, quick kill Turbo Man then revisit Shade Man's stage for the Proto Shield.
- For the Bass scene in Shade Man's stage, if Rush Search has very small delay and digs up a small life/weapon energy, you can save a small amount of time here. In 100% you get a shot at this for free because you need Rush Search three rooms before that for the Energy Balancer.
- I couldn't find other places where Rush Search worked.
The Rush Search before the Proto door in Shade Man is something I never found before though, so thanks for that (by the way, you need to hold left while going through the door for it to work). It definitely saves time in a TAS. However this trick doesn't seem to work for any other door.
I've looked at 100% for a while because there are a lot of possible improvements (and not just because of Rush Search glitches). Here is the test run I made two years ago (requires snes9x 1.51 v7); don't know whether it helps or not.
Editor, Experienced Forum User, Published Author, Expert player
(2159)
Joined: 6/15/2005
Posts: 3303
I watched both this and the Ultimortal 1HP version. While I don't particularly admire this game, I do appreciate that you took the time to do these modes.
That being said, I prefer this TAS for having more variety and tricks (such as damage abuse / flying across the screen) as well as having shorter boss fights.
It's too bad Iji 1.7 is not TASable though. Stuff like this is why I gave up on the notion of Windows TASing being a real thing.
Is this the first time these collections have been in 60fps? I dont think I ever saw it as such until now.
Nicovideo has allowed 60fps uploads for a long time. However, if you access the site when there is a lot of traffic, it may play the 30fps 360p "economy" encode instead of the normal one.
One more thing: how do we know that (a+b+c)(-a+b+c)(a-b+c)(a+b-c) must be positive? Well, that just follows from a+b>c, b+c>a, c+a>b which I derived from Ptolemy's inequality above. So the necessary and sufficient conditions that allows the equilateral base to be constructed are l > max(|a-b|,|b-c|,|c-a|), l < min(a+b,b+c,c+a), a+b>c, b+c>a, c+a>b, and the bound on l above.
which we can rewrite as
Using the identity on itself multiple times, we get
As k goes to infinity, we get
The left hand side is of the form theta*sin(x)/x (where x=theta/3k) as x goes to 0, for which the limit is theta. Then plugging in theta=pi gives
and dividing by 4 gives you the identity you posted.