The story goes (I'm still trying to find the source for it) that this property was discovered only about 4 years ago. Can you discover it too?
In each of these regular polygons, the sides are equal to 2 units of length. For each of these polygons, take each of the colored line segments (red, blue, and purple) and find the equation that holds for each polygon and that takes these lengths as input, as well as an additional undisclosed polygon-specific quantity.
Tell us this equation.
Hint: the equation is mathematically very simple, you don't need to use complicated operations.
Additional hints will follow as the day progresses.
Previous iteration: #713295
Hint:
If one defines the lengths of the colored segments with letters a, b, c, d, e, etc... What do they add up to when taking the sum of their squares?
Example, for the last one (with 3 segments):
Do this for each polygon and discover a pattern...
It always adds up to the number of sides!
That's fascinating. I don't think I have the time to prove it right now, though. That being said, I think my algorithm is working because it was discovering this pattern :)
I also don't know how you go about decomposing
a2+b2+c2+…=n
into the individual length components.
Hmm, I don't have time to check if my solution is accurate, but if you measure the angles at each corner (measured from the vertical line downwards), the first angle is always:
And subsequent angles follow the pattern:
So I guess the i-th vertical line always has the length:
where $\ell$ is the length of the polygon's sides.
Moreover, there are floor(n/2) such verticals so you can stop the algorithm after that many.
Will check this more carefully tomorrow, I'm on the go now. I wonder if applying your formula works out to find the pattern mentioned in my most recent comment.
That's for the red line on the triangle. It's derived from Pythagoras theorem.
yes, i think it works for all the red lines.
Yes, true
doesn't it work for all lines in all polygons? less on the blue line of the fourth polygon. 🤔 cc/ @south_korea_ln
Maybe, I haven't given it much thought tbh
Hint: sum of the squares...
As for yesterday's puzzle (probability of ending at 1), here is the solution: #714130.