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Line segments AB, CD, EF are all parallel. So triangles ACB and ADB have the same base and height, and hence the same area, as do triangles AFB and AEB. So the area of the shaded quadrilateral ACBF is the same as the square ADBE, i.e. 100.pic.twitter.com/vgNMs7b9ws
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If you were inside an orbiting space station that rotated to keep one face towards the body it orbited, some free falling objects would execute elliptical motion around the centre of the station. https://www.gregegan.net/INCANDESCENCE/Orbits/Orbits.html …pic.twitter.com/rjPYErFBfe
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[1/2] If anyone’s still bamboozled, the angle-doubling arises from pairs of non-convex quadrilaterals with pairs of equal sides. ABCD has side lengths 1/4,1/2,1/4,1/2. AFGB has side lengths 1/8,1/4,1/8,1/4. Because ∠ABG = ∠ABC, these quads are similar, and ∠BAF = ∠DAB.pic.twitter.com/eXF1SQaDfX
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[1/2] Now that the ingenious
@ChocoLinkage has posted linkages that compute both squares and cubes, I thought it would be fun to reverse-engineer the squaring one. If you want to work through this in detail yourself, note that all lengths are multiples of 1/16. The magic here…pic.twitter.com/K9xVNUuWSCPrikaži ovu nit -
English contains many words that are difficult to spell. But “corona” and “virus” are spelled exactly as they’re spoken. OK, it’s easy to make transpositions or repeat letters when typing. But most computers have a spellchecker. How hard is it to take 5 seconds to get it right?pic.twitter.com/uhXSoOQEHg
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[1/2] My new eBook, INSTANTIATION, has 11 stories: • “The Discrete Charm of the Turing Machine” • “Zero For Conduct” • “Uncanny Valley” • “Seventh Sight” • “The Nearest” • “Shadow Flock” • “Bit Players” • “Break My Fall” • “3-adica” • “The Slipway” • “Instantiation”pic.twitter.com/2uveQZgWto
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Scammers are so lazy. This happened on 4 January, and it’s taken them 3 whole weeks to get their act together and start exploiting it.pic.twitter.com/XAxFb6pQYk
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[1/4] Sometimes you do need to go backwards to go forwards. Suppose you want to match position *and* velocity with a constantly accelerating car (the blue curve), ASAP, in a car that can accelerate (or decelerate) up to twice as fast, and starts from rest either ahead or behind.pic.twitter.com/VTVQ1DRh3Z
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Although vertical and horizontal distances from the target have different costs in terms of the arrival time, the *sign* of the vertical distance makes no difference at all. Whether you start out at (x,y) or (x,–y), with the target at (0,0), you will reach it at the same time.pic.twitter.com/MVuxsgePJf
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This contour plot over the grid of initial positions shows how long each pursuer takes to reach the target. The elliptical shape of the contours means a horizontal distance from the initial target position costs you less time than the same vertical distance.pic.twitter.com/dxXhh6FEDc
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This version should make it clearer where the distinct arms of the X come from.pic.twitter.com/aIvc3cfZEr
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[2/2] The diffusion coefficients D_u, D_v control the scale, while the parameters F and k determine the dynamics. For the animation above: D_u = 2×10^{-6} D_v = 10^{-6} F = 0.04 k = 0.062 The computations were performed on a sphere, so the stripes are really all the same size.pic.twitter.com/L4bPkusmpk
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[1/2] The Gray-Scott model describes two chemicals diffusing as they interact. In 1993, John Pearson found that this system could form a variety of intricate patterns: “Complex Patterns in a Simple System” https://arxiv.org/abs/patt-sol/9304003 …pic.twitter.com/ROG05LMHLC
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[1/2] The red dot accelerates at a constant rate and direction. The black dots can accelerate faster. How can they match the red’s position *and* velocity as soon as possible? The calculus of variations shows the optimal trajectories obey a simple rule: https://www.gregegan.net/ORTHOGONAL/E3/SoftInterception.html …pic.twitter.com/8z17WcDhbm
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Take a 600-cell in 4D with 120 vertices ±v_i, i=1…60 Define F(x)=𝚷_i v_i·x Stereographically project F from a 4-sphere to 3D, then Fourier-transform from momentum space to position space. Result: this hydrogen wave function, with energy level n=61. https://www.gregegan.net/SCIENCE/SymmetricWaves/SymmetricWaves.html …pic.twitter.com/DdYMD55XsY
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Next time someone asks you to name a computable irrational number that has been proved *not* to be normal in any base (i.e. does *not* contain every finite sequence of digits with equal density) you can turn to this beautiful example from Greg Martin: https://arxiv.org/abs/math/0006089 …pic.twitter.com/1iL9fpFadm
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[5/5] There were some examples Rigby couldn’t prove by elementary methods. In 2015, an anonymous Japanese woman posted proofs for these cases online! You can read her proofs in this English translation by Hiroshi Saito: https://www.gensu.co.jp/saito/challenge/3circumcenter_en.html … Or if you want to try your hand:pic.twitter.com/QmgO6nwvwe
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[3/5] Rigby sought quadrilaterals where the angles were rational multiples of π. He realised this amounted to looking for cases where 3 distinct diagonals of a regular n-gon intersect at a single point. Solving the quad below means proving AA*, BB*, CC* intersect at 1 point, D.pic.twitter.com/WmZUcJFEYx
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[1/5] How fiendish can a quadrilateral be? There’s a series of infamous puzzles where you’re given four of the eight angles between the sides and diagonals of a quadrilateral. The puzzle is to find one of the unknown angles—using classical elementary geometry, not trigonometry.pic.twitter.com/bBuVhL8700
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