Why did I spend all this time calculating stuff 
Well thanks anyway I’ll think of something 
(why i’m the only one posting)
Okay let me try this
Centres of the chords lie on the side of the triangle
So perpendeculars to the chords in those centers will intersect inside the triangle
Because distance from center of the circle to the end of the chords is the same, so projecting centre on the chord it will end up in the centre of the chord. If those projections lie on the side of the chord and not on their continuations then the center is inside the triangle.
Omg it’s so hard to explain in english…
I’ll let that count, it’s a good enough explanation. Another way you could think about it is the following:
The distance from chords with the same length to the centre of a circle is the same. So let’s imagine a second circle with the same centre whose radius is exactly that distance. That circle would have the three chords as its tangents, and thus the centre is inside the triangle, since we know that the three chords form a triangle. So you see, the lengths of the sides of the triangle were actually unimportant, the important part was that the chords have equal length. Coincidentally, that whole “distance to centre is equal if chords have equal length” is how I knew on a glance that both your constructions were wrong.
That’s what I’m asking myself as well.
Puzzle 22 will come soon.
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Yeah I realized earlier that constructing it will be very tedious and stupid, and I imagined no one would make a puzzle with tedious construction methods and such, so it has to be something more simple. Atleast I can sleep well, because I spent all night thinking 
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After seeing akafa’s first post on the matter and you not denying that he was on the right path, I threw the towel.
Maybe I’ll chime back in now that you did deny it.
How do the gingerbreads need to be layed out so that that distance between any two is minimal?
Hm, seems interesting. One question: can he lay down one gingerbread on another?
Because it feels more like math/geometry exam questions than puzzles to me? 
Your old ones required a lot less math sometimes.
Yes, he can.
Also by the distance you mean the shortest path from one gingerbread to another, like if they stand right next to each other distance would be 0 or if they touch each other then the distance is 0?
And I guess the distance between any gngerbreads should be the same?
He chewed all gingerbreads and spit them out so the distance would be minimal
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The latter, if two gingerbreads are touching each other in one way or another, then the distance between them is 0.
Creative solution, but sadly wrong.
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Without hunger it cannot be seen.
Hm, I’m pretty sure it’s impossible to do it without putting one on another. The maximum amount you can put without stacking is 3 which forms a triangle with all equal sides and there is nothing to go from there. So stacking is required.
I’ll try this. I don’t know if having one contigous angle is considered touching tho (god I redrew it 3 times lol)
For Puzzle 22, they should form a cylinder. That way they all (nearly) touch in the middle.
Edit: Scratch that, placing them in a “flat cube shape” (if we imagined an 8th gingerbread added to it) would literally have them all touch in the middle. Basically putting two gingerbread on top of one another on the table, and then arranging those in a square shape. They should all touch the very center of the construction.
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Kari’s edit is the “intended” solution, but akafa’s follows the same idea of “Every gingerbread touches each other”, so I could see that as well (although that one needs a hell of a lot of balancing).
So I’ll post Puzzle 23 soon.
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Highfives Kari
Damn that was even easier using Kari’s way, why I didn’t think about it 
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Wow that seems hard. I wonder if the number being palindromic is done on purpose or not
Also if the number is too long, what if we write it down in Hexadecimal so it’s shorter or something, just a wild guess cause Idk how to find the number in the first place.
I reckon the number is probably all 1s and 0s, else it’d be too hard to make with 7 matches moved.
As it has an odd number of factors, it must be a square number. So I’m going to check the amount of divisors for consecutive square numbers, see if I can see a pattern.
1x1 = 1, 1 divisor
2x2 = 4, 3 divisors
3x3 = 9, 3 divisors
4x4 = 16, 5 divisors
5x5 = 25, 3 divisors
6x6 = 36, 7 divisors
7x7 = 49, 3 divisors
8x8 = 64, 7 divisors
9x9 = 81, 3 divisors
10x10 = 100, 9 divisors
11x11 = 121, 3 divisors
12x12 = 144, 11 divisors
13x13 = 169, 3 divisors
14x14 = 196, 9 divisors
15x15 = 225, 9 divisors
16x16 = 256, 9 divisors
17x17 = 289, 3 divisors
18x18 = 324, 15 divisors
19x19 = 361, 3 divisors
20x20 = 400, 15 divisors
Anyone get any hints from that list? All I can tell is that a squared prime has 3 divisors, though I knew that already. Anything else you guys can see?
Well the easiest way to find a number with 10201 divisors is to take a prime number to the power of 10200, obviously that result is way too insane though. The number of divisors a number has is the number of distinct subsets of the set of its prime factors + 2 I believe. What’s the number of subsets in a set again? 2 to the power of the set size was it?