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USpapergames
USpapergames
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June 27th, 2020 at 5:33:14 PM permalink
Quote: ThatDonGuy

Or maybe it's because nobody except you seems to have any idea what the definition of an "average rectangular prism" is. Average as defined in what way?



Omg, guess I got to start teaching you some statistics lol. So there are 3 statistical averages to a population or sample size (also population & sample formulas differ slightly, so you could say 6 total). The mean is deemed the most accurate statistical average over the remaining two & the mode is only ever used if a mean or median can not be determined. It just so happens that the median of the average size is the statistical average of all the sizes of rectangles incribed within a circle is the mean as well. Notice how it's the average of the size (or more precisely parameter) of the rectangle, not the area!!!
Last edited by: USpapergames on Jun 27, 2020
USpapergames
USpapergames
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June 27th, 2020 at 5:45:29 PM permalink
How did nobody figure out the meaning of the question when I literally did the math to describe the answer several times????? If I was a client of your guys I would be demanding my money back lol

What's scary is if I was a client I probably wouldn't be able to tell that you gave me the wrong answer and more importantly, even if I could you guys wouldn't have listened to me....
USpapergames
USpapergames
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June 27th, 2020 at 9:01:55 PM permalink
So now that you understand the right question have you figured out the answer or is the real question also difficult? I think for 2D finding the average area of the rectangle might actually be a much easier problem than finding the area of the average rectangle.
Last edited by: USpapergames on Jun 27, 2020
unJon
unJon
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Thanks for this post from:
OnceDearcharliepatrickCrystalMath
June 28th, 2020 at 3:42:11 AM permalink
Quote: USpapergames

So now that you understand the right question have you figured out the answer or is the real question also difficult? I think for 2D finding the average area of the rectangle might actually be a much easier problem than finding the area of the average rectangle.



Your posts are deeply unpleasant, and the question you meant to ask is completely uninteresting. Bye.
The race is not always to the swift, nor the battle to the strong; but that is the way to bet.
charliepatrick
charliepatrick
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gordonm888
June 28th, 2020 at 11:22:17 AM permalink
Quote: USpapergames

Literally, everyone is wrong except me...

I don't think it helps making these kinds of assertions about fellow members of the forum.

Puzzles are nearly always posted on this forum as a challenge, posed in good faith and intended to be mathematical teasers rather that trying to trip people up. Sometimes it's not totally clear what the puzzle is and one is perfectly allowed to ask for clarifications - I did that with the zombie six bullets puzzle.
USpapergames
USpapergames
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June 28th, 2020 at 3:49:53 PM permalink
Quote: ThatDonGuy

Or maybe it's because nobody except you seems to have any idea what the definition of an "average rectangular prism" is. Average as defined in what way?



Omg, I gave you the answer earlier!!! The fact that you can't figure it out just shows how narrow-minded your math skills are. The idea of nobody but me knowing the average size of a rectangular prism is like calling me crazy which is extremely hurtful.

Quote: USpapergames

So this is the equation so for

With a radius = 10
Maximum Cube inscribed (All figures rounded to nearest thousandth:
Side = 1.155
Face Diagonal = 1.633
Space Diagonal = 2
Face Area = 94.28
Surface Area = 800
Volume = 1,539.601
Sagittal Arch (at an angle) = (5π/8)×58.21144

So here is the equation. (Face Area / 4) × (Side + 2 × Sagitta)

The area without the additional length of the Sagitta = 371.793 ft³ so we know it can't be less than that & is very close to the answer. Let me find some free time tonight & I'll solve the Sagitta problem & we will have the answer. Got to get back to work now :/



Quote: unJon

Your posts are deeply unpleasant, and the question you meant to ask is completely uninteresting. Bye.



Are you serious? All you have been saying to me has been disrespectful! You don't know what it was like to have everyone telling you your wrong when you know your right! I lost sleep over this shit, a decrease in work productivity & actually had a sick feeling in my stomach since I couldn't find the time to review your work. And for you to blame me for not writing the question better when Shackleford wrote it correctly & then edit your comment later is just ridiculous! You need to start blaming yourself for your shitty word problem skills & learn to apologize when your wrong & take some responsibility for once. But I guess that's beneath you since it's been 24 hours by now.

Btw if you don't like what I have to say its probably because you don't like hearing the truth when it hurts. Your opinion of my question doesn't mean two shits to me, especially when your not intelligent enough to solve the problem. Btw this is a very good problem on a test since people like you would probably piss themselves thinking you need a scientific calculator to solve the answer ;)

Quote: charliepatrick

I don't think it helps making these kinds of assertions about fellow members of the forum.

Puzzles are nearly always posted on this forum as a challenge, posed in good faith and intended to be mathematical teasers rather that trying to trip people up. Sometimes it's not totally clear what the puzzle is and one is perfectly allowed to ask for clarifications - I did that with the zombie six bullets puzzle.



Omg, the math problem was worded clearly & correctly. Take some responsibility for not understanding the question! I had to analyze there work & everyone ignored me when I showed them the correct equation & multiple solutions! I'm the person who made the question and multiple people what to tell me what the question means or what it should mean! Stay out of this you hear me! I'm the victim here & I'm done with the forms on this website! Everyone here is completely rude & passive-aggressive. I don't play bs drama games, I'm not going to kiss your ass when you treat me like shit so just leave me alone!
ThatDonGuy
ThatDonGuy 
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June 28th, 2020 at 5:31:49 PM permalink
Quote: USpapergames

Omg, I gave you the answer earlier!!! The fact that you can't figure it out just shows how narrow-minded your math skills are. The idea of nobody but me knowing the average size of a rectangular prism is like calling me crazy which is extremely hurtful.

You mean this?

Quote: USpapergames

Omg, guess I got to start teaching you some statistics lol. So there are 3 statistical averages to a population or sample size (also population & sample formulas differ slightly, so you could say 6 total). The mean is deemed the most accurate statistical average over the remaining two & the mode is only ever used if a mean or median can not be determined. It just so happens that the median of the average size is the statistical average of all the sizes of rectangles incribed within a circle is the mean as well. Notice how it's the average of the size (or more precisely parameter) of the rectangle, not the area!!!

First, the word you appear to be looking for is "perimeter," not "parameter." A parameter is a value you put into a function - for example, in f(x), the parameter is x.

Second, both in the text of your YouTube video, and in the video itself near the end, you ask for the "average area", but here, you say you want the perimeter.

The mean area over all possible rectangles inscribed in a circle of radius 10 is 400 PI. If you have a different answer, kindly show us (a) what it is, and (b) how you got it.

If you want the mean perimeter, that's another question.

For that matter, if your question about the "average rectangular prism" inscribed in a sphere of radius 10 is asking for something other than the mean volume, then please tell us what is being requested - mean surface area? Mean value of the sum of the lengths of the 12 edges?

Also, if there is a non-calculus solution, it might be more suitable for the IMO instead of Putnam.
USpapergames
USpapergames
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June 28th, 2020 at 7:44:49 PM permalink
Quote: ThatDonGuy

You mean this?

First, the word you appear to be looking for is "perimeter," not "parameter." A parameter is a value you put into a function - for example, in f(x), the parameter is x.

Second, both in the text of your YouTube video, and in the video itself near the end, you ask for the "average area", but here, you say you want the perimeter.

The mean area over all possible rectangles inscribed in a circle of radius 10 is 400 PI. If you have a different answer, kindly show us (a) what it is, and (b) how you got it.

If you want the mean perimeter, that's another question.

For that matter, if your question about the "average rectangular prism" inscribed in a sphere of radius 10 is asking for something other than the mean volume, then please tell us what is being requested - mean surface area? Mean value of the sum of the lengths of the 12 edges?

Also, if there is a non-calculus solution, it might be more suitable for the IMO instead of Putnam.



1st, there is nothing wrong with the wording of the question in either of my YouTube video, Mr. Shackleford's question, or anything part of this discussion. If there is any part of the misunderstanding it's your word comprehension skills. The question would need to be worded as the average area of the average rectangle within a circle, or how your word it "The mean area over all possible rectangles inscribed in a circle" and you're on glue if you think I ever implied that question. Notice how the question of the average area of a rectangle inscribed in a circle literally makes no sense because the question says nothing about the size or coordinates of the rectangle drawn within a circle.

Notice in the video I am doing nothing but speak about the size and shape of the possible rectangle. It's implied in the wording that the average perimeter is what is being asked, very rarely do word problems give you the exact terminology & it's rather common for the solver to use reasoning to deduce the meaning from word problems. Sorry my writing skills aren't as good & can't write without spell check to save my life, you gonna make fun of that about me also? Nothing but low blows from you!

Omg, the question is the volume of the average rectangle box inscribed in a sphere. Do I need to tell you this question is also about the average perimeter of the rectangle box? Also, the International Mathematical Olympiad test is a great option for this kind of question. I still think the Putnam would be a great test for this since it would throw the average test taker for a curveball lol.
Last edited by: USpapergames on Jun 28, 2020
Wizard
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Wizard
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June 30th, 2020 at 4:35:33 PM permalink
I see USpapergames got a time-out. It wasn't me who gave it, but I agree with it. As a reminder, there is a rule against personal insults.

I've been thinking about the second question about the mean area of a box inscribed in a sphere. I solved it for a sphere of radius one, but we can multiply that by 1,000 for one of radius 10.

That said, here are my answers and very brief solution.


16,000 / (3*pi^2)



Again, let me assume a radius of 1 and multiply by 10^3 at the end.

First, I solved the question of the average CYLINDER in a sphere.

Let's say the cylinder has flat sides directly at the bottom and top of the sphere.

Let t be the angle from the center of both to where the edge of the cylinder touches the sphere at the top. So, if t were 0 the cylinder would have no height and at 90 degrees it would have no area to the top/bottom.

That said, the area of the cylinder of angle t is 2*pi*cos^2(t)*sin(t)

Integrate that from 0 to pi/2 and divide by 2/pi to get the average.

Hint:
Let u = cos(t)
du = -sint(t) dt

If my calculus is right, the answer is 4/3.

Recall the area of a sphere is (4/3)*pi*r^3, so the can to sphere ratio is 1/pi.

From part 1, we established the average area of a rectangle in a circle of radius 1 is 4/pi.

So, the ratio of the rectangle to circle is 4/pi^2.

Finally, we are inscribing a box inside the cylinder. The ratio of box to cylinder should be the same at 4/pi^2.

Thus, the average area of the box is (4/3)*(4/pi^2) = 16/(3*pi^2) =~ 0.5404

Multiply by 10^3 if the sphere has radius 10: 16000/(3*pi^2) =~ 540.3796

Last edited by: Wizard on Jul 3, 2020
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gordonm888
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June 30th, 2020 at 11:35:01 PM permalink
I haven't been on this forum very much recently, but I just read this thread - and this is one of the weirdest threads I can ever remember.
So many better men, a few of them friends, were dead. And a thousand thousand slimy things lived on, and so did I.

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