Math game (fun quizzes) ^^

[Kyriakos]

Hi, happy 2023 ^^
Maybe a math-based game can interest a few here?

You can post any math-related quiz (eg simple geometry, critical thinking etc - obviously NOT homework-type ) and whoever answers correctly, first, can post the next challenge...

Here is a relatively simple one (just involves basic geometry), I hope some will participate and upon solving it post the next:



To win, just calculate (using any method) the value of x. "O" is the center of a circle=>center of the diameter AB.

[Maximinus Thrax]

[nhytgbvfeco2]

Using the pythagorian whatever I can calculate AP, since C is a 90 degrees angle and stuff because it's on the diameter, which is the root of AC^2 + CP^2, which is 25. AOP and BOP are like twins or something, dunno the english term, so PB is also 25. This means that ya boi CB is 32, and once again we can use our bro pythagoras to find AB, which is a big thicc boi at 40, and AO is half of it so 20. Once again it's the man, the myth, the legend, Pleasetouchgrass. X=15.

[Kyriakos]

Using the pythagorian whatever I can calculate AP, since C is a 90 degrees angle and stuff because it's on the diameter, which is the root of AC^2 + CP^2, which is 25. AOP and BOP are like twins or something, dunno the english term, so PB is also 25. This means that ya boi CB is 32, and once again we can use our bro pythagoras to find AB, which is a big thicc boi at 40, and AO is half of it so 20. Once again it's the man, the myth, the legend, Pleasetouchgrass. X=15.

Correct, although I - being a Greek, as well as hipster - of course preferred to use the double Thales (the first theorem you alluded to is the circular Thales)

Here is my solution:



Feel free to post the next quiz, or declare open floor ^^

[nhytgbvfeco2]

Uh, alright. Shouldn't I be given rep or something, though?
Simplify A^95, if we know that A = PDP^-1

[Kyriakos]

Uh, alright. Shouldn't I be given rep or something, though?
Simplify A^95, if we know that A = PDP^-1

I did try to give you rep, but the system didn't allow me yet :S
The message I'd have written would be: "Well done, Anaximander will bring you a cookie"

PS: as for the new quiz, don't you think you should at least have defined what P and D are? :S Otherwise people can imagine they were just variables and then simply DD^-1=1=>PDP^-1=D=>D^95=A^95.
But looking on the web, I did find this used for matrices, eg here.

[nhytgbvfeco2]

I shall patiently await my cookie.

[Kyriakos]

I shall patiently await my cookie.

So you took the cookie but didn't feel like saying if you meant the matrices solution by induction? Typical
At least you didn't present a Taylor series quiz

edit: to clarify: I think it will be better if the quizzes mostly can be solved by people who have a general familiarity with math, not having to be majoring/having a math degree

[nhytgbvfeco2]

So you took the cookie but didn't feel like saying if you meant the matrices solution by induction? Typical
At least you didn't present a Taylor series quiz

edit: to clarify: I think it will be better if the quizzes mostly can be solved by people who have a general familiarity with math, not having to be majoring/having a math degree

To be fair, your edit wasn't there when I started writing my message, and after I posted I simply didn't notice it.

I did try to give you rep, but the system didn't allow me yet :S
The message I'd have written would be: "Well done, Anaximander will bring you a cookie"

PS: as for the new quiz, don't you think you should at least have defined what P and D are? :S Otherwise people can imagine they were just variables and then simply DD^-1=1=>PDP^-1=D=>D^95=A^95.
But looking on the web, I did find this used for matrices, eg here.

Yeah fair enough. I didn't quite have anything at the ready lol.
As for your solution, not quite. In Matrice multiplication the order matters, you cannot multiply P by P^-1 in this case. However, in this equation it becomes PAP^-1PAP^-1...PAP^-1 (95 instances), which means that all but the first P are followed by a P^-1, and in this case you can multiply them. The result therfore is P(A^95)P^-1

[Kyriakos]

Yes ^^ It's what MathWizard presented in the video - proof by induction. I suppose I could also have used proof by induction to infer you were talking about matrices, but I had no other instant of you posting about them

If you feel like it, you are welcome to post another quiz. Or it can be open floor.

[nhytgbvfeco2]

well i figured you were close enough to the answer so it's your turn.

[Kyriakos]

Ok...
But I will wait for a few hours, to see if others wish to post a quiz ^^

[Kyriakos]

Hm... Ok, for the time being I will use the other puzzle I had already presented elsewhere. But it is a little bit more difficult - and I worry that the thread will become a brief exchange between nhytgbvfeco2 and myself, before just dying :o

Regardless, here is the new puzzle:

[Kyriakos]

No one?
Ok, I guess there's not much interest. I will post the answer tomorrow...

[Septentrionalis]

An interesting puzzle. I can at least tell that if one part is longer than the two other parts put together, there is no triangle to be made. Most likely in all other cases there is. But how to calculate the chance of that happening.

Already the first cut can exclude a triangle if it hits sufficiently far from the center.

Help me out here so that Kyriakos doesn't have to reveal the answer. We can crack this.

[Kyriakos]

Here is one of the possible solutions:



I declare open floor, so if anyone feels like posting a quiz, feel free to do so ^^

(PS: hopefully I didn't butcher significantly the proof; I did correct some typos and other stuff, I am not really in my best mood currently... :S )

[Kyriakos]

An interesting puzzle. I can at least tell that if one part is longer than the two other parts put together, there is no triangle to be made. Most likely in all other cases there is. But how to calculate the chance of that happening.

Already the first cut can exclude a triangle if it hits sufficiently far from the center.

Help me out here so that Kyriakos doesn't have to reveal the answer. We can crack this.

Damn, I didn't see your post
You were certainly on the right path...
Feel like posting the next one?!!!

[Kyriakos]

By the way, a nice bit of trivia:
When Euclid presented his proof of the triangular inequality, some of the followers of Epicurus snarkily remarked that this didn't need a proof, because "even a donkey could see it"

[nhytgbvfeco2]

Already the first cut can exclude a triangle if it hits sufficiently far from the center.

I mean, not really, one side can be tiny and still it'd be a triangle.

[Kyriakos]

I mean, not really, one side can be tiny and still it'd be a triangle.

Well, yes, since if the thing nears having two right angles (not a triangle then), its two larger sides would tend to have the same length and the third would approach zero.
Still, I guess Septen meant that if the first cut is away from the center and the second cut is to the same direction the first cut went, you already have no triangle (because the first cut would create a side larger than the other two sides).
I wonder if a different proof of the 1/4 could be, then (after defining the cuts' direction) that for the first cut there is (inevitably, not counting hitting the absolute center) a 1/2 chance your cut lands to the left side (or the right), and after that there is again a 1/2 chance the second cut lands to the same side as the first= (1/2)^2=1/4. Do you think this is workable as a rigorous proof, as long as the sides are defined?