To try everything Brilliant has to offer�free�for a full 30 days, visit https://brilliant.org/ZachStar/ . The first 200 of you will get 20% off Brilliant's annual premium subscription.
STEMerch Store: https://stemerch.com/
?Follow me
Odysee: https://odysee.com/@ZachStar:0
Instagram: https://www.instagram.com/zachstar/
Twitter: https://twitter.com/ImZachStar
Support the Channel: https://www.patreon.com/zachstar
PayPal(one time donation): https://www.paypal.me/ZachStarYT
Join this channel to get access to perks:
https://www.youtube.com/channel/UCpCSAcbqs-sjEVfk_hMfY9w/join
2D Graphing Software: https://www.desmos.com/calculator
Animations: Arkam Khan (For contact info go to https://www.arqum333.com/)
Check out my Spanish channel here: https://www.youtube.com/channel/UCnkNu2xQBLASpj6cKC8vtpA
?My Setup:
Camera: https://amzn.to/2RivYu5
Mic: https://amzn.to/35bKiri
Tripod: https://amzn.to/2RgMTNL
?Check out my Amazon Store: https://www.amazon.com/shop/zachstar
Now do it for 3D. And after that for n dimentions.
@charlieb8735 Says:
The points would have to be arranged either an equilateral or infinitesimally close to. Thus it has to be 6 or 5 if ties aren’t allowed.
@eide_ball Says:
Now i know how snow work :)
@floriskleinestaarman3200 Says:
Is it not that the maximum connection a point can have is infinite, if multiple points are the closest? Such that the point is simply the middle point of a circle and the other points are on the line of the circle.
@MylesGmail Says:
I enjoyed that
@bhavyajain638 Says:
Could we add another 5 points? Closer to the centre point then the outer points?
Like if the radius of the initial circle is 5, at 2.4999..?
So we will have the degree 10.
Edit- the distances are different, say by 0.01
@epictoast Says:
1:21 Trying to make a triangle is the same as AB > BC > CA > AB, which is a contradiction.
@Archimedes115 Says:
Coincidentally I was just doing something like this a few months ago with NFL stadiums. The resulting graph was not even close to what I expected to see when I started thinking about it.
@alexharvey9721 Says:
Somehow seems related to the six degrees of separation concept.. though I guess it's just a lot of coincidentally similar variables...
@ZealanTanner Says:
Somehow my brain immediately knew the answer was 5. I thought, if it was 7, then others would be closest to others, but for 6 every adjacent point could be equal from each other, but equal is illegal, so 5
@cicik57 Says:
right i also think about the hexagon so the max amount of neighbours is 6, where distance of points are equal. If you move a vertex of hexagon, either it becomes closer to center or to next vertex
@Lucifer_Nihilum Says:
5 is my guess.
@susulpone Says:
I solved it this way: draw a circle around P. the distance between the points on the circle must be more than the radius of the circle, otherwise they would connect. Then look at the conversion to radians and you can fit 6 points. With slight permutation of the distances you fit the "no same length" rule. I had no way of proving this solution is unique tho
@FireyDeath4 Says:
Looked at the thumbnail and just assumed you could have infinity before the explanation. Once he drew AB above P I shortly realised it was 5. Well, very well, I have two other ideas:
1. Go to hyperbolic space and start drawing points an arbitrarily large distance away from P
2. Go to Hilbert space and just repeatedly add progressively closer nodes to P on every axis
@carpcarpbread Says:
"There cannot be triangles"
Laughs in equilateral triangles
@noahnaugler7611 Says:
I'd expect 5 or 6
@Rot8erConeX Says:
I assumed the solution would depend on the total number of points that were in the graph, like "n-2" or something. But as soon as he transitioned from "AB must be the longest line in the triangle" to "the corresponding angle must be the largest angle", I *immediately* realized the solution was a hard 5, he hadn't even mentioned that the angle had to be greater than 60 yet.
@teamcyeborg Says:
My first guess was 3 because _mumble mumble_ π, or 6 because _mumble mumble_ τ.
@aikorrana9455 Says:
Just by feeling it out a bit I think it’s 8? Just my gut feeling
@lizzycoax Says:
5
@Shad0wWarr10r Says:
A problem i find here, what if 2 points are equal distance from a point? Like the hexagon.
Wouldnt the answear be an line in form of a circle around a point, infinite points all the same distance away from origo
@matthewmilewski1452 Says:
Equilateral triangles. 'Nuff said.
@balijosu Says:
Sneaky of you to assume a Euclidean plane.
@heathbrinkman3126 Says:
Trying without watching the video, the limit will be when a chord of a circle between two points on the boundary is less than the radius, which napkin math doesn't give an easy answer with my skill
@RoderickEtheria Says:
There is an utmost limit based on ((n-2)180)/n=120.
This makes n 6. Any more than that many connections would block further connections by not having equilateral triangles. Given unique shortest distances, 5 becomes the most.
@datguiser Says:
Let’s gooooo I actually predicted the answer before actually watching to see the answer. Similar logic, but I certainly got it.
@Althemor Says:
I would say 5 or 6 (with 6 being the edge case where you would have to decide how to define the question), because at 6 you have a hexagon surrounding a middle point, with two adjacent corners and the middle point forming equilateral triangles. Any more corners and the distance between them will shrink below the radius of the circle on which the corners are placed around the middle point.
Ah, a unique distance.
5 then.
@galacticgaming3186 Says:
Before the video goes im gonna put my guess and reasoning
So if its the shortest edge that connects that means nothing smaller than an isometric triangles could be formed from the points since all points are the same distance away in an isometrictriangle, however any triangle with one angle wider than 60° would have at least one edge longer than the other 2 which means any number of of trangles could be formed as long as no single triangle has a center angle of 60° 360/60 is 6 which is all isometric triangles so 5 points would be the maximum before it would all be the same distance away
@ugielka Says:
with a hexagon with a pint in the middle the middle point would have 6 connections. any other points would rather connect to the outside 6 or would break established connections. in a hexagon the outside points are the same distance to their neighbors as to the center point. in a polygon with more sides the outside points would rather connect to their neighbors than to the center point because they’re closer. because equal distanced points are not allowed the most points possible are 5
@Blackfromstickworld Says:
make equilateral triangle
@NotYourAverageNothing Says:
4:39 Why not? A is the closest to P, P is the closest to B, and B is the closest to this other point.
@infectetgost7384 Says:
The sketches and the math just doesn’t add up for some reason
@ristolahdelma2591 Says:
5 is incorrect. Right answer is 6. Regular hexagon with point in center, each corner has shortest distance to center. Of course same distance to adjacent corner, but choosing the center point gives the maximum.
@Hannah-cb7wr Says:
Great exam question, lol
@danimyte3021 Says:
Btw, for those who wonder about the rule of all lengths being different. It's there to rule out the P=6 case which is ambiguous without the rule.
@randomnobody660 Says:
My guess would be 5, because hexagons are bestagons.
@TheCodaCrew Says:
Can you cover Animation vs Math?
@feierlord Says:
is it 5?
@the4spaceconstantstetraqua886 Says:
Fun fact:
The answer would be infinity on a non-euclidean plane.
@maaikevreugdemaker9210 Says:
Very elegant to identify the triangle inequality and apply it!
@rotervogel1523 Says:
You sound like the guy that always plays as god on that one comedy channel
@TheJmax04 Says:
So confused to discover the guy who does wacky zany skits is the same guy talking about interesting maths puzzles.
@ysgramornorris2452 Says:
Now imagine solving the same puzzle but on a convex surface :p
@PrometheusMMIV Says:
Based on intuition I figured 6 connections would be an upper limit. A single point surrounded by 6 evenly spaced points would create 6 equilateral triangles, and any more points would cause the outer points to be closer to each other than the center.
However, since the problem specifies unique distances, increasing the distance of any of the 6 points from the center would also cause the outer points to become closer to each other, given a 60° angle between them. Therefore, I believe 5 connections would be the most, with each outer point at a slightly different distance from the center point.
@ghasternoxost7540 Says:
What happens when two points are eqaully distant away from a point
@Robinsonero Says:
I think I missed the step that showd why side AB being the longest makes the angle 60°. Where does 60° come from here? Intuitively it makes sense give. The geometry of hexagonal or tiranvular tilings of the plane but I don't see how it follows from the argument layed out here
@FallenSpike Says:
I thought about this similar to a sphere/circle packing problem so it's pretty easy to see that with 6 you would have everything be the same distance, whereas with 5 it's possible to not have the outside spheres/circles touching. Also works in higher dimensions
@TheFoxMaster101 Says:
Let’s say you had a system made only of points connected to the single point no matter how much you increase the distance if you don’t change the angle at which the two lines connected to the center form you will not change the connections therefore the maximum number of points would be equal to the maximum number of points on a perfect polygon with points inscribed around a circle centered at the point which we want to max the amount of connections to in which the radius is less than the length of each side which is 6 however due to the fact each set of points has to be equal to each other it would need to be 5 because this rule is broken if it is 6
@beirirangu Says:
I got the same answer from a completely different thought process: each point has a circle with it as it's center and the closest point as the point on its edge, such that no circle can have points inside of them. Therefore because the maximum number of points on a circle can be is 6 (hexagons) , and the parameters of the "world" dictates no two distances are the same, the maximum number must be less than 6: 5
@scoutgaming737 Says:
I thpught about it like this
In a regular hexagon the side length and longest radius are equal. If p is the center of that hexagon, there are equal distances, so you have to move the center point, but then the side lengths will be greater that the distance from the other points to p, which will break the connections
In a pentagon that is not a problem, so the maximum number of connections is 5
LATEST COMMENTS