One need not dig into the realms of General Relativity to establish space-time into a single continuum —

100 spare hours is almost equally unaffordable as 100 sq. ft.

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For a heart-broken mathematician, calculating the smoothness of Lipschitz boundary in Sobolev space is directly proportional to the roughness of Lip-stitched boundaries in the “Sob-of-love” space.

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My gay pal is mostly busy solving “Ram-man-Hillbutt problem for homo-morphic functions” while I beat my head against the ” Riemann-Hilbert problem for holomorphic functions”

]]>There are two different kinds of layers here, as it alternates between centered hexagonal numbers (e.g. the top layer is the first centered hexagonal number, i.e. 1, the third layer from the top is the second centered hexagonal number, i.e. 7, etc.), and layers where the sides alternate between lengths n and n+1 (e.g. the second layer from the top of the pyramid is a ‘hexagon’ with sides 1 and 2, the fourth layer is a hexagon with sides 2 and 3, etc.

So, the numbers of tennis balls in all the odd layers of this pyramid is 9^3=729

For the other layers: The formula for the figure with sides n and n+1 is 3n^2.

Since you have 8 of those layers, you have 3 times the sum of the first 8 square numbers, and the general sum of the first n square numbers is n(n+1)(2n+1)/6, so that gives you 3⋅8(8+1)(2⋅8+1)/6=612 more tennis balls, for a total of 729+612=1341

]]>Alice and Bob take off on the scooter together. Charlie starts walking.

1.65hrs later:

Alice and Bob arrive at Bob’s house at 1.65hrs.

Charlie has traveled 8.25m, leaving 24.75m between them.

Alice takes off alone back for Charlie.

0.825hrs later:

Charlie walks 1/6 the remaining distance, 4.125m. (5/(5+25)) = 1/6

Alice meets up with Charlie and heads back to Bob’s

Alice and Charlie are 20.625m away from Bob’s house.

1.03125hrs later:

Alice and Charlie arrive back at Bob’s.

Total time = 3.50625hrs or 3hrs 30 minutes and 22.5 seconds.

…I think?

]]>w = walking speed

s1 = speed of scooter with one passenger

s2 = speed of scooter with two passengers

express the fraction of the total distance at which the scooter should turn back. ]]>

The interesting thing is, one can partition the route arbitrarily in segments and as long as this strategy is followed in every segment the solution is optimal, so there are infinite valid and different solutions.

(That includes the case “partition in a single segment” of Jonathan above)

]]>1. A+B take the scooter for 72 minutes (6/5 hours). C walks. At this point C has travelled 6 miles and A and B have travelled 24.

2. B takes the scooted back to meet C while A and C continue walking. B and C meet after 36 minutes. At this point B and C are at 9 miles while A is at 27 miles.

3. Spend the final 72 minutes with B and C on the scooter and A walking all in the direction of B’s house. They all arrive after another 72 minutes.

In coming up with this I thought about “go that way for 2 hours and that way for 1” and then rescaled at the end.

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