| 4 |
riddles / easy / Re: Geometric puzzle, two triangles |
Apr 15th, 2024, 8:24pm |
| Started by rmsgrey | Last post by rloginunix |
That, clearly, means that that trajectory of the crossing point, shown in red, belongs to a (smaller) circular arc, as a I tried to capture that fact in the animation below.
The center of the said circle is located trivially: just construct the perpendiculars to the sides of the given equilateral triangles, etc. |
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| 6 |
riddles / easy / Re: Symbol |
Mar 29th, 2024, 6:03am |
| Started by alien2 | Last post by alien2 |
| Clue (possible spoiler): Too bad that there is no toilet inside the room... Or, perhaps, that fact just might spark a lifesaving idea! |
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| 8 |
riddles / easy / Re: The Spider-Man of Paris |
Jan 17th, 2024, 7:54am |
| Started by alien2 | Last post by rmsgrey |
For the third variation, your expected return from taking 6 paintings is always the same, so, if you're just concerned with the monetary return, the only question is how risk-averse or risk-seeking you are. If you're risk-averse, taking 3 of each type guarantees a 3-painting payout; if you're risk-seeking, taking 6 and 0 gives you either all or nothing.
You can get more interesting by delving into utility curves - if, rather than 2 paintings being twice as valuable to you as 1, 3 paintings being triple, and so on, if you instead have some other relative values for different numbers of paintings, then that changes your expected value for various strategies.
You have four options, each of which gives you a 50% chance of each of two possible outcomes - taking all the paintings from one set gives you 0 or 6; taking five from one and one from the other gives you 1 or 5; taking four and two gives you 2 or 4; and taking three of each gives you 3 or 3, which is just a guaranteed 3. Whichever pair gives you the highest total utility across the two possibilities would be your personal best pick. |
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| 10 |
riddles / hard / Re: power of 3 close to power of 2 |
Nov 13th, 2023, 10:10am |
| Started by jollytall | Last post by jollytall |
If I remember well, I proved few month ago that indeed this aL is the best sequence, but that was never validated by anyone.
Nonetheless, I am more interested in the smallest absolute difference (i.e. not ratio) between 3^n and 2^m, where 2^m > 3^n.
I came up with a related question to prove:
2^m - 3^n > 2^(m-n) if 2^m > 3^n > 3.
With other words, if we write 3^n in base 2 then the first (i.e. highest value) n bits have at least one 0 among them.
Yet with other words, we proved earlier that 3^n/2^m > 1-e can be solved for any e, but even that solution is not that close, i.e.
3^n/2^m < 1-2^-n |
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