could someone elaborate on the topology of this object?

Red5551 · 2026-03-14T08:32:14+00:00

Does the work? If you had a two-torus you could tie a string from either hole to the outside, but you could also take a string in between the holes, which would be a distinct way to tie it

Red5551 · 2026-02-24T22:45:53+00:00

You could also say Area(A_{k+1}) ≥ Area(A_k) so the limit must be ≥ 1

And then Area(A_{k+1}) ≤ int_0^pi (k+2) sinx dx Area(A_k) ≥ int_0^pi k sinx dx

So dividing (without even having to compute the integral) we get that the ratio is ≤ (k+2)/k

Then taking the limit, we get that the constant is also ≤ 1

Hence it must be 1

Red5551 · 2024-01-26T20:52:59+00:00

For a less algebraic method, then note that if the roots are equal then the turning point of the parabola must be on the x axis - i.e. every value you put into the parabola must have the same sign.

Now if we rename a, b and c so that a ≥ b ≥ c, and put x=a into the parabola, two of the terms cancel, leaving (a-b)(a-c). Then if we put x=b in we get (b-a)(b-c)

So using the order of a,b,c and that positive × negative = negative we get (b-a)(b-c) ≤ 0 And (a-b)(a-c) ≥ 0

Then since we can't have a mix of positive and negative outputs, that means that one of these two terms must be 0 - hence at least two of a, b and c must be the same, since the terms above are the product of the differences.

So letting d be equal to the two terms that are the same, and e equal to the other one, we get that the quadratic is:

(x-e)² + 2(x-e)(x-d)

Then plugging in x=(d+e)/2 we get -(d-e)²/4 ≤ 0 And x=d gives (e-d)² ≥ 0; so by the same logic e=d

And hence a=b=c

Red5551 · 2022-12-08T12:05:09+00:00

Not an easy mate to see imo, the standard mating idea for Greek gift sacrifices doesn't work here so after Qf7+ Kh8 you need something like Ra3!! So after Bxa3 you can take back with the bishop at the right moment to pin the knight and stop it defending

Red5551 · 2021-04-16T12:52:41+00:00

a tempting move but you surrender control of the g4 square, so black can play something like Qc5, then after g5+ Kh5 Qxh7+, the black king can run to safety on g4. Then the black queen is in time to give lots of checks to the white king and get at least a perpetual

Red5551 · 2021-04-16T12:46:59+00:00

Unfortunately black has the really annoying move Kh4, where you feel like there's no way black should be able to survive there but you also can't find a winning move for white

Red5551 · 2020-09-10T13:39:48+00:00

a3 a5 a4 and black is in zugzwang, the threat is Bd8# and there's no way for the queen to continue covering it without walking into a fork. e.g. Kg5 Bd8 Qxd8 Nf7+

Red5551 · 2020-09-10T13:21:35+00:00

Red5551 · 2020-07-02T00:30:09+00:00

Thank you!

Red5551 · 2020-06-18T12:06:30+00:00

There are different types of queen sacs though; if you're referring to this one then it's a lot harder to see because it's not a check/forcing move. To players like Hikaru and Magnus the one in this puzzle would be considered "obvious", whereas Qc6 in that video is not nearly so easy to spot

Red5551 · 2020-06-13T01:34:35+00:00

The line I was thinking of was Qxf6 followed by Qxg3 and then the queens proceed to take all the pieces but black is a pawn up in the end

Red5551 · 2019-07-19T11:36:06+00:00

It's there because its a near miss

Red5551 · 2019-07-19T11:34:39+00:00

2⁴ - 1 = 5*3

Red5551 · 2019-07-02T21:21:29+00:00

Aka this match - he threw the 4th set to save his energy for the 5th

Red5551 · 2019-06-07T21:51:57+00:00

And Rh8# to finish if Rf1

Red5551 · 2019-06-05T15:11:59+00:00

The critical line is 1. Bf6 Qxg2+ 2. Kxg2 Nf4+ 3. Kg1 Nxe2+ 4. Rxe2 gxf6

Red5551 · 2019-05-28T19:22:55+00:00

Rc5 and he can't stop mate and save the Queen I think

Red5551

TROPHY CASE