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35

They are:


34

EDIT: The description is not very intuitive, so I took the liberty of creating an image to make clear what this solution intents. Sorry I'm new to this forum... I'm going to try to post an answer. Unfortunately I'm bad with computers so I can't draw a picture... Please get out a pencil and paper! The summary is: The details: And: And: Now before I ...


20

The trick to this puzzle is to: (And here are those tilings: the center was already given, and the rest are obtainable from these by rotation.)


17

This question is not about packing tetrahedra, it is about how many tetrahedra can share a common vertex. The constraints of this question are that the side length of the tetrahedra is equal to the radius of the sphere and all tetrahedra share a common vertex at the center of the sphere. Given these constraints, all vertices are either at the center of ...


11

Here is a CAD drawing I created to show the layout:


11

A small addition to phenomist's excellent answer: Finally, here's a photo of all the pieces in the box. The tricky packing isn't visible.


10

Sure. Method: With some fiddling, it's also possible to get all the columns to add up to 25: And here's a magic square (with duplicates, unavoidably) followed by a row of fives: And finally:


9

Here's one solution. The second solution can be achieved by: .


9

Note from /r/puzzles: The T in the top-right piece should be an I. Solved: Nice puzzle, thanks for posting!


8

Here are some statistics as to the diameter of the smallest circle that $n$ diameter-$1$ circles can be packed into: $$\begin{matrix} 2 & 2 \\ 3 & 2.1547 \\ 12 & 4.0296 \\ 13 & 4.2361 \\ 31 & 6.2915 \\ 32 & 6.4295 \\ 50 & 7.9475\end{matrix}$$ The best-known packing of $31$ puts $1$ in the inner $2.1$-circle, $6$ others in the $4....


8

I believe the answer is as shown in the flower kind figure below (This is the previous answer): The idea behind this is Here is the best answer: as you will see below:


8

Here's the MSPaint to the rescue answer. The math logic works in my head as well. The top packing is tighter. each triangle after the 1st one saves (edit: $1-\sqrt{{{1-(1-\sqrt{\frac{3}{4}}}}})^2$ ) ~0.00901523343248242768377221117738 diameters. Multiply by 132 and get 1.190010813087680454257931875414. The first triangle is on the bottom left. The 133rd is ...


8

The tricky packing interests me. The height of each part appears to be 1, 2, or 3 units. This gives a total volume of $(4 \times 3) + (12 \times 2) + (16 \times 1) = 52$ units. But the volume of the box seems to be only $4 \times 4 \times 3 = 48$ units. So how could it be done? My guess: Edit: it turned out that the solution of @JaapScherphuis is better


7

Here is one solution Thanks for the fun! Bonus shapes:


7

According to https://matthewkahle.wordpress.com/2010/11/05/packing-tetrahedra/ this is a open problem so there is no solution yet


7

EDIT: A new submission in which I have high confidence. Method Use an online program to densely pack the element names Dump the contents into Excel Use my previously-written VBA script to highlight the element names Remove all non-highlighted characters Manually search for any element symbols already highlighted. Make sure they're not fully contained ...


7

Your question falls into the area of "circle packing" problems. Even some highly restricted special cases of circle packing are very hard and messy, and there exists no general theory for attacking them. A good survey on the current knowledge of the area is given at http://www.packomania.com/


7

Is it okay if my solution has a If so, As per Edward's comment, let me try to prove that it can't be done with a smaller train car. (It seems like there might be a better way to answer this though)


7

I mostly used a computer to solve. A random brute force process. I added as much logic as I could determine from the hint and reduced which tiles could be placed in the surrounding cells (3x3 form the corners). I started the process from cell A7 and followed the blue arrow up where I determined cell A6 only had two that could fit, thus reducing the number ...


7

For Question 2: Explanation:


6

Another way to look at it is to project the base of each tetrahedron onto the sphere. The length of each segment is $\pi/3$. By using the law of cosines, each angle is $arcsec(3)$ and the area of one of the spherical triangles is $3\ arcsec(3)-\pi$. The area of the sphere is $4\pi$. Therefore, the max number of spherical triangles we can theoretically fit ...


5

With regards to the uniqueness of the solution: For other solutions:


5

A bit late to the party and cannot beat the excellent answer from @Bass. I worked out the number of distinct solutions bearing in mind they can be further permuted by ordering each row and the row sequence. I found I did this by first finding all the sets of four digits which sum to 20. I then permuted them for each row so that each digit is used the ...


5

Okay, here's a badly-drawn way to do it:


4

Playing around in paint (400 height, 100 radii circles), I made a slight stagger. The width of the three circles on top is less than if they were directly side by side, so this trapezoid can be flipped and repeated.


3

No. The rectangle has an area of $75$ square units. The only other rectangle with integer sides that would have the same area would be $3*25$ (or $1*75$ technically, I guess). The rectangles you have are $3*6, 2*3, 3*2, 2*4, 4*2, 10*2$ and $9*1$. The $9*1$ needs a $9*2$ paired with it to fill those rows and prevent leaving a $1*x$ space that cannot be ...


3

This one works Method :


3

What can I say I am a sucker for scratch. though I had great difficulty with registering drags on most objects before I went into inspect code mode then it functioned great, the ball and plane were the only ones I could drag


2

or


2

Without attempting to answer the general question let me show a small example where the hexagonal packing is not optimal. Choosing $b=1.6$ and $l=2.6$ won't allow any of those setups to fit in more than 2 circles. However, 3 circles do have space in that with the following setup: Again, this does not really answer the general case, but shows that even ...


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