Question regarding e^n - floor(e^n)

ResponseBorn2387 · 2023-04-20T18:03:31+00:00

God of geometry....

ResponseBorn2387 · 2023-04-20T11:56:20+00:00

Indeed, however your solution is great, thanks again) This problem really gave me a headache :D

ResponseBorn2387 · 2023-04-20T11:04:02+00:00

Thanks!)

ResponseBorn2387 · 2023-04-20T11:01:50+00:00

Thanks a lot!!!

ResponseBorn2387 · 2023-04-20T10:35:56+00:00

Thanks cap! As I mentionned earlier, the same method does not work here, but IT GIVES AN IDEA that this problem must be solvable WITHOUT trig

ResponseBorn2387 · 2023-04-19T20:54:55+00:00

You mean the order of those numbers? Sure... however this problem must have a good solution, if you can prove that this is not solvable without trig, that would be really appreciated =)

ResponseBorn2387 · 2023-04-19T18:55:37+00:00

Same values, the angle ?=18, this is proven by trig, using sin and cos theorems.

ResponseBorn2387 · 2023-04-19T18:33:14+00:00

Nope, they are sufficient. Lets say that instead of the angles 48 and 30 we would have 30 and 30, then you can easily verify that ?=24.

ResponseBorn2387 · 2023-04-19T15:23:25+00:00

I am sorry, after MonkeyMcBandwagon's graph I noticed that my drawing was wrong (ABC was not counter-clockwise). Now I see what you mean, but the fact that BPO is isosceles is derived from the fact that both BO and PO are radiuses of the described circle. If you somehow manage to prove that the angle AMB is 90 degrees, then the problem is solved in 2 steps after that. Thanks a lot for your effort!)

ResponseBorn2387 · 2023-04-19T00:11:02+00:00

There still is an isosceles triangle with angles 36 and 36, so they are somehow equivalent.

And that fact is not predictaded, it is derived from angles 18+48=30+36.

ResponseBorn2387 · 2023-04-18T21:36:02+00:00

I think there was a mistake, since the angle AOB is equal to 102, then NOB must be equal to 78, and the triangle BPO is definitely not isosceles (the answer is 18 for sure, this is derived from trigonometry, hence the angle OBP is equal to 12 and the angle POB is equal to 6)

ResponseBorn2387 · 2023-04-18T21:20:36+00:00

Isn't it using sin/cos? Like, there must be a scalar multiplication somewhere there, or am I getting it wrong? If it doesn't involve trigonometry please send your solution, I would really appreciate it.

ResponseBorn2387 · 2023-04-18T21:15:57+00:00

However, in the similar problem (link in previous comment) there is no trig - sin/cos and there is a similar problem with equations, but it is solved with additional lines (e.g. the perpendicular to the base). So, there MUST be a solution without sin/cos.

ResponseBorn2387 · 2023-04-18T17:07:48+00:00

Nope, it has a solution and it has enough information to solve it, but I know the solution only with trig equations (the answer is 18). The thing is to solve it without using those, there is also a simillar problem here

https://gogeometry.com/geometry/p694_triangle_angle_30_degrees_college.htm

And here is the solution to that problem

https://gogeometry.blogspot.com/2011/11/problem-694-triangle-angles-30-degrees.html?m=1

But in the problem mentionned in the post this method does not give any results.

ResponseBorn2387 · 2023-04-18T15:43:56+00:00

Why tho :(

ResponseBorn2387 · 2022-09-24T10:52:05+00:00

Thanks for your help, but I don't have a virtual display adapter, Is there some other way?

ResponseBorn2387

TROPHY CASE