Rectangle inscribed in a sphere

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!!!

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....

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.

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.

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.

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!

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.

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.

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.

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.

Quote: gordonm888

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.

Yes. The most emotionally heated math thread I can recall since teliot quit posting. For some good drama, look up the thread on 0^0.

Quote: USpapergames

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!

I'm going to lift your suspension, which I think was given by a secret admin. Welcome back. I look forward to your thoughts on my solution to the average volume of a random box in a sphere.

Here is a more more formal solution (PDF) to the question on the area of a random rectangle in a circle. I changed the radius of the circle to 1.

The only correction I see is in the beginning. Angle A is any random degree from 0-45°. If the angle could be 90° there would only be 4 internal triangles but your counting twice the data since when angle A is > 45 the rectangles long side flips in the opposite direction.

Nice work, the best solution to the misinterpreted word problem yet. I'll see if I can come up with a clever solution using nothing but equations, I haven't even given this question any thought. I much prefer 4r²/π to calculus integration.

Quote: Wizard

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.

Answer

16,000 / (9*pi^2)

Solution

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/(9*pi^2)

Multiply by 10^3 if the sphere has radius 10: 16000/(9*pi^2).

Can I get an amen?

I'm not sure where you went wrong but this surely can't be the right answer. With a sphere of radius 10, the volume is 4,188 ft³ with the maximum volume of a cube inscribed being 1,539 ft³. Your answer is proposing the average volume of the rectangle box to be 180 ft³. I think you must have been trying to solve for the average area and not average volume but even the average area doesn't look right since the maximum cube area is 800 ft³ and the average needs to be greater than half the maximum since the length increases as width & depth decrease. So without really analyzing your math since your not using equations & your solution is difficult to follow (especially comparing a rectangle prism to a cylinder) I can safely assume if you're solving for the average area the answer needs to be between 333 & 500 ft³. If you're solving for volume the answer needs to be between 385 to 550 ft³.

Quote: USpapergames

The only correction I see is in the beginning. Angle A is any random degree from 0-45°. If the angle could be 90° there would only be 4 internal triangles but your counting twice the data since when angle A is > 45 the rectangles long side flips in the opposite direction.

Nice work, the best solution to the misinterpreted word problem yet. I'll see if I can come up with a clever solution using nothing but equations, I haven't even given this question any thought. I much prefer 4r²/π to calculus integration.

Welcome back. Can I ask you to restate the word problem you meant us to try to answer?

Quote: USpapergames

I'm not sure where you went wrong but this surely can't be the right answer. With a sphere of radius 10, the volume is 4,188 ft³ with the maximum volume of a cube inscribed being 1,539 ft³.

Wouldn't the maximum volume be a cube inside the sphere, which would have volume 1000*2^(3/2) = 2828.4.

Quote:

Your answer is proposing the average volume of the rectangle box to be 180 ft³.

Yes, that does feel a little low. Let me think about it.

Quote: Wizard

Yes, that does feel a little low. Let me think about it.

I incorrectly squared the 3 in the denominator. My new answer is 540.3796.

A sphere of radius 10 has volume 4188.7092.

A random inscribed box thus has 12.9% of the volume of the sphere, which seems reasonable to me.

Here is my revised solution.

Quote: Wizard

I incorrectly squared the 3 in the denominator. My new answer is 540.3796.

A sphere of radius 10 has volume 4188.7092.

A random inscribed box thus has 12.9% of the volume of the sphere, which seems reasonable to me.

Here is my revised solution.

Solution

[spoiler=Solution]
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

I like this solution a lot better & I was definitely leaning my opinion of the answer to the greater side of the estimated value since there presumably is a lot of space gained once the length edge increases in value. But I am going to need some time to give the solution a through evaluation, like it deserves, & it looks like my weekend just got filled up :/

Quote: unJon

Welcome back. Can I ask you to restate the word problem you meant us to try to answer?

What is the area of the average rectangle inscribed in a circle?

Think of the question what is the area of the average rectangle inscribed in a square. If all 4 corners need to touch the perimeter of the square than the average rectangle is going to be half the area of the square since halfway is the average size a rectangle can be drawn if you were to start drawing ever possible me rectangle from the smallest to the largest and average their perimeters you would find the median equal to the mean. So if the square has a length of 10 the average rectangle has a length of 5 & 10.

Now if you try this same method with a rectangle inscribed in a circle you will see that the length of the rectangle increases due to the perimeter of the circle, aka circumference, is not a constant & gradually escalating in height. So the average rectangle inscribed in a circle has half the width of the largest possible rectangle (a square) & meet halfway between the distance on the circumstance arch from a coordinate that is the corner of the maximum square to the midpoint on the circumference that meets between both corner ends of the maximum square. The entire problem is about finding there length of the rectangle.

For the average surface area of a rectangular prism inscribed within a sphere of radius 10, I get 400 / PI

Here's how I did it - hopefully, my text-based terminology can be deciphered:

Quote: ThatDonGuy

For the average surface area of a rectangular prism inscribed within a sphere of radius 10, I get 400 / PI

Here's how I did it - hopefully, my text-based terminology can be deciphered:

Yay Math

First, choose a "uniformly random" Rectangular Prism
Assume the center of the sphere is at the origin of 3-dimensional coordinate space
Let the vertices be (+/- x, +/- y, +/- z), where x, y, and z are nonnegative

Let t be in (0, PI/2) and w be in (1/2, 1); p = arccos (2w - 1)
(t represents the angle counterclockwise (from the +x ray to the +y ray) in the z = 0 plane; p represents the angle measured from the -z ray; only values where all three coordinates are positive are considered, which is why -1/2 <= w <= 1)
x = R cos t cos p
y = R sin t cos p
z = R sin p

Note that cos p = cos (arccos (2w - 1)) = 2w - 1
and sin p = sin (arccos (2w - 1)) = sqrt(1 - (2w - 1)^2) = 2 sqrt(w - w^2)

The dimensions of the rectangular prism are 2x by 2y by 2z
The surface area = 4 (xy + xz + yz)
= 4 R^2 (cos^2 p sin t cos t + sin p cos p sin t + sin p cos p cos t)
= 4 R^2 ((2w - 1)^2 sin t cos t + 2 (2w - 1) sqrt(w - w^2) sin t + 2 (2w - 1) sqrt(w - w^2) cos t)
The sum of all surface areas is
4 R^2 INTEGRAL(1/2,1) {INTEGRAL(0,PI/2) {((2w - 1)^2 sin t cos t + 2 (2w - 1) sqrt(w - w^2) sin t + 2 (2w - 1) sqrt(w - w^2) cos t) dt} dw
= 4 R^2 INTEGRAL(1/2,1) {INTEGRAL(0,PI/2) {((2w - 1)^2 sin t d(sin t) + 2 (2w - 1) sqrt(w - w^2) sin t dt + 2 (2w - 1) sqrt(w - w^2) cos t dt)} dw
= 4 R^2 INTEGRAL(1/2,1) {INTEGRAL(0,PI/2) {1/2 (2w - 1)^2 sin^2 t - 2 (2w - 1) sqrt(w - w^2) cos t + 2 (2w - 1) sqrt(w - w^2) sin t} dw
= 4 R^2 INTEGRAL(1/2,1) {INTEGRAL(0,PI/2) {(1/2 (2w - 1)^2 + 4 (2w - 1) sqrt(w - w^2)) dw}
= 4 R^2 INTEGRAL(1/2,1) {(1/2 (2w - 1)^2 + 4 (2w - 1) sqrt(w - w^2)) dw}
= 2 R^2 INTEGRAL(1/2,1) {(4w^2 - 4w + 1) dw} - 8 R^2 INTEGRAL(1/2,1) {sqrt(w - w^2) ((1 - 2w) dw)}
= 2 R^2 INTEGRAL(1/2,1) {(4w^2 - 4w + 1) dw} - 8 R^2 INTEGRAL(1/2,1) {sqrt(w - w^2) d(w - w^2)}
= 2 R^2 ((4/3 * 1^3 - 2 * 1^2 + 1) - (4/3 * (1/2)^3 - 2 * (1/2)^2 + 1/2)) - 16/3 R^2 ((1 - 1^2)^(3/2) - (1/2 - (1/2)^2)^(3/2))
= 2 R^2 (4/3 - 2 + 1 - 1/6 + 1/2 - 1/2) - 16/3 R^2 (0 - 1/8)
= R^2 * 1/3 + R^2 * 2/3
= R^2

The mean surface area = R^2 / (PI/2 * 1/2) = 4 R^2 / PI
If R = 10, then the mean surface area = 400/PI

Without checking the math your answer can't be right. So how did I come up with the minimum estimate I gave Shackleford for the average surface area? Simple I found the average surface area of a rectangle box inscribed in the maximum size cube that's inscribed in a sphere and then you would add the area that extends to the arch of the sphere. So if the maximum cube inscribed in a sphere with radius 10 then the length of the cube is 11.547. So to find the average surface area of a rectangle box inscribed in a cube take the length 11.547 and multiply it by half the length which is 5.7735 & then multiply it by 4 for all 4 congruent sides and then add the product of 5.7735² and 2 for both congruent ends of the rectangular prism. You get 333.33 ft² as the answer so you know that the average surface area of the rectangular prism must be greater than the average surface area of a rectangular prism inscribed in a cube that's inscribed in the same radius sphere with a 10 ft radius.

Next time I post a solution I hope someone reviews it.

Quote: USpapergames

What is the area of the average rectangle inscribed in a circle?

I'm puzzled by the concensus definition of "The average rectangle"
USPapergames seems to apply average perimeter, but what's to say we can't use 'average length' or 'average area' which will yield different results?
If asking for some function of the area of the average rectangle, wouldn't average area be more appropriate than average perimeter?

Quote: USpapergames

Quote: ThatDonGuy

For the average surface area of a rectangular prism inscribed within a sphere of radius 10, I get 400 / PI

Without checking the math your answer can't be right. So how did I come up with the minimum estimate I gave Shackleford for the average surface area? Simple I found the average surface area of a rectangle box inscribed in the maximum size cube that's inscribed in a sphere and then you would add the area that extends to the arch of the sphere. So if the maximum cube inscribed in a sphere with radius 10 then the length of the cube is 11.547. So to find the average surface area of a rectangle box inscribed in a cube take the length 11.547 and multiply it by half the length which is 5.7735 & then multiply it by 4 for all 4 congruent sides and then add the product of 5.7735² and 2 for both congruent ends of the rectangular prism. You get 333.33 ft² as the answer so you know that the average surface area of the rectangular prism must be greater than the average surface area of a rectangular prism inscribed in a cube that's inscribed in the same radius sphere with a 10 ft radius.

Next time I post a solution I hope someone reviews it.

I did find a couple of errors. My new answer is 800 / PI, which, according to you, still can't possibly be right as it is less than 333.33 - but I am not entirely convinced that your reasoning as to why it has to be greater than this is correct.

Then again, my solution is based on the assumptions that (a) my method of finding "all rectangular prisms inscribed in a sphere" is correct, and (b) taking the integrals results in a valid sum, which may not necessarily be true, especially when polar coordinates are being used. If I try a different method, the solution will probably be different...
...like when I try to simulate a solution using a Monte Carlo approach, and somehow get an average of 849 using "my" method and 1041 using a strictly rectangular coordinate one (select x in (0,10); select y in (0, sqrt(100-x^2)); z = sqrt(100-x^2-y^2)).

Here is my revised solution:

I updated my solution to the question on the rectangle to include the perimeter, which I find to be 16/pi = apx. 5.093.

The perimeter of the whole circle is 2*pi = apx. 6.28, so I'm a little worried I am too high.

Quote: OnceDear

I'm puzzled by the concensus definition of "The average rectangle"
USPapergames seems to apply average perimeter, but what's to say we can't use 'average length' or 'average area' which will yield different results?
If asking for some function of the area of the average rectangle, wouldn't average area be more appropriate than average perimeter?

Yes, the wording is a little tricky but here is how you would analyze the question. The area of an object does not equal a shape (especially there average area of something) so there is no way to infer area for the average rectangle. If you're going to find the average length you also need to solve for the average width since the width has a direct correlation with the length of the rectangle. Therefore the only logical conclusion you could make from a mathematical perspective is average size or perimeter. It's very possible the average area of a rectangle could potentially not even form a rectangle shape since the answer is a none repeating number because your basically averaging areas of the circumstance of a circle which has π in it & it's already been proven that no square has the same area as a circle!

Quote: Wizard

I updated my solution to the question on the rectangle to include the perimeter, which I find to be 16/pi = apx. 5.093.

The perimeter of the whole circle is 2*pi = apx. 6.28, so I'm a little worried I am too high.

I concur with your assessment & find the answer very unrealistic. Please post your solution & I will try and review whenever I get free time. Apparently, I have 2 other solutions to asses, including your solution for volume which I think has a strong potential for being correct.

Quote: ThatDonGuy

I did find a couple of errors. My new answer is 800 / PI, which, according to you, still can't possibly be right as it is less than 333.33 - but I am not entirely convinced that your reasoning as to why it has to be greater than this is correct.

Then again, my solution is based on the assumptions that (a) my method of finding "all rectangular prisms inscribed in a sphere" is correct, and (b) taking the integrals results in a valid sum, which may not necessarily be true, especially when polar coordinates are being used. If I try a different method, the solution will probably be different...
...like when I try to simulate a solution using a Monte Carlo approach, and somehow get an average of 849 using "my" method and 1041 using a strictly rectangular coordinate one (select x in (0,10); select y in (0, sqrt(100-x^2)); z = sqrt(100-x^2-y^2)).

Here is my revised solution:

Yay More Math

First, choose a "uniformly random" Rectangular Prism
Assume the center of the sphere is at the origin of 3-dimensional coordinate space
Let the vertices be (+/- x, +/- y, +/- z), where x, y, and z are nonnegative

Let t be in (0, PI/2) and w be in (1/2, 1); p = arccos (2w - 1)
(t represents the angle counterclockwise (from the +x ray to the +y ray) in the z = 0 plane; p represents the angle measured from the -z ray; only values where all three coordinates are positive are considered, which is why -1/2 <= w <= 1)
x = R cos t cos p
y = R sin t cos p
z = R sin p

Note that cos p = cos (arccos (2w - 1)) = 2w - 1
and sin p = sin (arccos (2w - 1)) = sqrt(1 - (2w - 1)^2) = 2 sqrt(w - w^2)

The dimensions of the rectangular prism are 2x by 2y by 2z
The surface area = 4 (xy + xz + yz)
The dimensions of the rectangular prism are 2x by 2y by 2z
The surface area = 8 (xy + xz + yz)
= 8 R^2 (cos^2 p sin t cos t + sin p cos p sin t + sin p cos p cos t)
= 8 R^2 ((2w - 1)^2 sin t cos t + 2 (2w - 1) sqrt(w - w^2) sin t + 2 (2w - 1) sqrt(w - w^2) cos t)
The sum of all surface areas is
8 R^2 INTEGRAL(1/2,1) {INTEGRAL(0,PI/2) {((2w - 1)^2 sin t cos t + 2 (2w - 1) sqrt(w - w^2) sin t + 2 (2w - 1) sqrt(w - w^2) cos t) dt} dw
= 8 R^2 INTEGRAL(1/2,1) {INTEGRAL(0,PI/2) {((2w - 1)^2 sin t d(sin t) + 2 (2w - 1) sqrt(w - w^2) sin t dt + 2 (2w - 1) sqrt(w - w^2) cos t dt)} dw
= 8 R^2 INTEGRAL(1/2,1) {INTEGRAL(0,PI/2) {1/2 (2w - 1)^2 sin^2 t - 2 (2w - 1) sqrt(w - w^2) cos t + 2 (2w - 1) sqrt(w - w^2) sin t} dw
= 8 R^2 INTEGRAL(1/2,1) {INTEGRAL(0,PI/2) {(1/2 (2w - 1)^2 + 4 (2w - 1) sqrt(w - w^2)) dw}
= 8 R^2 INTEGRAL(1/2,1) {(1/2 (2w - 1)^2 + 4 (2w - 1) sqrt(w - w^2)) dw}
= 4 R^2 INTEGRAL(1/2,1) {(4w^2 - 4w + 1) dw} - 32 R^2 INTEGRAL(1/2,1) {sqrt(w - w^2) ((1 - 2w) dw)}
= 4 R^2 INTEGRAL(1/2,1) {(4w^2 - 4w + 1) dw} - 32 R^2 INTEGRAL(1/2,1) {sqrt(w - w^2) d(w - w^2)}
= 4 R^2 ((4/3 * 1^3 - 2 * 1^2 + 1) - (4/3 * (1/2)^3 - 2 * (1/2)^2 + 1/2)) - 32/3 R^2 ((1 - 1^2)^(3/2) - (1/2 - (1/2)^2)^(3/2))
= 4 R^2 (4/3 - 2 + 1 - 1/6 + 1/2 - 1/2) - 32/3 R^2 (0 - 1/8)
= R^2 * 2/3 + R^2 * 4/3
= 2 R^2
The mean surface area = 2 R^2 / (PI/2 * 1/2) = 8 R^2 / PI
For R = 10, this is 800 / PI

I just scanned over and saw you have 2 different solutions for surface area. I wish you would summarize your solutions better because it would take me some time to review all of this.

Quote: USpapergames

I concur with your assessment & find the answer very unrealistic. Please post your solution & I will try and review whenever I get free time. Apparently, I have 2 other solutions to asses, including your solution for volume which I think has a strong potential for being correct.

Here is my solution. I am feeling more comfortable that I am right about the average perimeter.

The range is 4 to 4*sqrt(2) = 4 to 5.6569, so an average of 5.093 doesn't seem so unreasonable.

Quote: Wizard

I updated my solution to the question on the rectangle to include the perimeter, which I find to be 16/pi = apx. 5.093.

The perimeter of the whole circle is 2*pi = apx. 6.28, so I'm a little worried I am too high.

The easiest way to think about it is look at the range of rectangles whose points are on the unit circle such that their sides are parallel to the X and Y axes.

You can start at 12 o'clock (H) where the rectangle is a vertical line, its area is zero and its perimeter is 4.
At 1 o'clock the rectangle is A A' A'' A''' and the top is 2 Sin (15 o), the side is 2 Cos (15 o), the area 4 Sin Cos.
At 2 o'clock the rectangle is B B' etc.
You can continue until you get to a horizontal line.

Putting these into a spreadsheet you get average area = 1.273 (4/Pi) and perimeter = 5.093 (16/Pi).

Degrees	Sine	Cosine	Area	Perimeter
0.00	0.000 000	1.000 000	0.000 000	4.000 000
9.00	0.156 434	0.987 688	0.618 034	4.576 491
18.00	0.309 017	0.951 057	1.175 571	5.040 294
27.00	0.453 990	0.891 007	1.618 034	5.379 988
36.00	0.587 785	0.809 017	1.902 113	5.587 209
45.00	0.707 107	0.707 107	2.000 000	5.656 854
54.00	0.809 017	0.587 785	1.902 113	5.587 209
63.00	0.891 007	0.453 990	1.618 034	5.379 988
72.00	0.951 057	0.309 017	1.175 571	5.040 294
81.00	0.987 688	0.156 434	0.618 034	4.576 491
90.00	1.000 000	0.000 000	0.000 000	4.000 000

Thank you, Charlie, for the confirmation. Nice graphic.

So I don't see you guys trying to answer my original question so here are the answers for the surface area & volume of the average rectangular prism! The surface area is 487.901 ft² & volume is 607.999 ft³! I'll try and post solutions but I have no idea when I'm going to get the time for that. The length = 18.24 ft & the width & depth = 5.77 ft. You can confirm my answers by finding the length that corresponds to the median width & depth of an inscribed rectangle prism. Notice how the volume & area of the average rectangle or rectangular prism is always greater than the average volume or area. Also, notice how the question for the surface area of the average rectangular prism vs the average surface area of a random rectangular prism are identical question.

charliepatrick - Best solution yet, it's almost self-evident! I still want an algebraic equation, theorems are so much more superior since they can reverse the flow of data to analyze another variable. Imagine being given the average area of a random rectangle & being asked to find the radius of the circle that is inscribed in the rectangle lol.

Quote: USpapergames

So I don't see you guys trying to answer my original question so here are the answers for the surface area & volume of the average rectangular prism! The surface area is 487.901 ft² & volume is 607.999 ft³!

I'm still trying to figure out how to construct an "average" - well, a uniformly random - inscribed rectangular prism. I don't think selecting a random point in one quadrant and reflecting them through the x = 0, y = 0, and z = 0 planes is correct.

Quote: ThatDonGuy

I'm still trying to figure out how to construct an "average" - well, a uniformly random - inscribed rectangular prism. I don't think selecting a random point in one quadrant and reflecting them through the x = 0, y = 0, and z = 0 planes is correct.

No that's not the correct way to determine the average of a shape. Start from the center axes & create coordinates that are half of the edge lengths. Remember that the x & y coordinates are equal to half the distance of the coordinates for a cube inscribed & the center axes; the z coordinate is where the math gets tricky. So coordinates should be:

x = 2.89,...y = 2.89,...z = 9.12
x = 2.89,...y = -2.89,..z = 9.12
x = -2.89,..y = 2.89,...z = 9.12
x = -2.89,..y = -2.89,..z = 9.12
x = 2.89,...y = 2.89,...z = -9.12
x = 2.89,...y = -2.89,..z = -9.12
x = -2.89,..y = 2.89,...z = -9.12
x = -2.89,..y = -2.89,..z = -9.12

I've got 2 new problems that are harder than all the others! The area of the average rectangle in a sphere & the average area of a random rectangle in a sphere! Notice how the scope of all possible answers increases by a factor of 4 and its much more difficult to plot since the rectangle can cross the space diagonal of a cube inscribed in a sphere.

AX̄❙²◯³ = ?

X̄AΣ❙²◯³ = ?

Quote: USpapergames

No that's not the correct way to determine the average of a shape. Start from the center axes & create coordinates that are half of the edge lengths.

I am not sure what you're saying, but this sounds like taking an average of shapes. That is what everybody using calculus is doing, but averaging EVERY shape.

I would also recommend you use a little respect with the advanced mathematicians on the site like ThatDonGuy. That advice applies not just here but in life.

Quote: Wizard

I am not sure what you're saying, but this sounds like taking an average of shapes. That is what everybody using calculus is doing, but averaging EVERY shape.

I would also recommend you use a little respect with the advanced mathematicians on the site like ThatDonGuy. That advice applies not just here but in life.

Wow....1st off, there was absolutely no disrespect towards ThatDonGuy, I was agreeing with him that he is right about that not being a correct solution. 2nd, If I'm not considered an advanced mathematician because I'm not using calculus to solve problems that don't require calculus then I am just going to leave this form and let the "advanced mathematicians" try and solve the answers lol. If I can analyze calculus than I know how to use it. Like I said earlier, I'm focused on solving for area of the mean rectangle, not the mean area of a random rectangle. I don't care about finding a solution to the problems or even proofs for that matter, I want theorems that are practical which is why I'm specifically using algebraic equations. It's not my fault if you don't understand why the mean average of a shape is equal to its mean surface area.

I think you should take your own advice & show a little bit of respect to me ;)

Quote: USpapergames

Wow....1st off, there was absolutely no disrespect towards ThatDonGuy, I was agreeing with him that he is right about that not being a correct solution. 2nd, If I'm not considered an advanced mathematician because I'm not using calculus to solve problems that don't require calculus then I am just going to leave this form and let the "advanced mathematicians" try and solve the answers lol. If I can analyze calculus than I know how to use it. Like I said earlier, I'm focused on solving for area of the mean rectangle, not the mean area of a random rectangle. I don't care about finding a solution to the problems or even proofs for that matter, I want theorems that are practical which is why I'm specifically using algebraic equations. It's not my fault if you don't understand why the mean average of a shape is equal to its mean surface area.

I think the problem is that the term “mean rectangle” has different meanings for USpapergames vs Wiz and others. To be honest I don’t know what USp means by the term “mean rectangle.”

Quote: USpapergames

I've got 2 new problems that are harder than all the others! The area of the average rectangle in a sphere...

Assume the sphere has its centre at the origin and, for mathematical simplicity, has unit radius. Multiply the answer at the end if needed.

Starting with a unit radius sphere, you can rotate the rectangle so it lies in a plane that is totally flat, i.e. all the z co-ordinates are the same. Then we can look at the flat circle area in which the rectangle lies and know its points lie on the sphere.

We know the average (area/perimeter etc.) of a rectangle in a circle. If it a function of the radius of the circle R^2 or R, e.g. K*R^2 where K=4/Pi.

One could integrate, then average, the above as R = cos (angle) as the angle goes from 0 to 90 (i.e. the top half of the sphere), the problem is this isn't linear as z increases (otherwise you'd use the volume of a sphere), and I'm not sure it is as the angle increases (since there are fewer points on the spjere's surface as the circle gets smaller).

I'll leave the integration part for others!

charliepatrick - I love this outline you have provided! This is how real mathematicians think of geometry problems! You have given me a great starting point, I might actually solve this today!

unJon - "I think the problem is that the term “mean rectangle” has different meanings for USpapergames vs Wiz and others. To be honest I don’t know what USp means by the term “mean rectangle.”"

I'm talking about mean in size (or perimeter) is reflected by the mean of its surface area. Let's say we have a rectangle with a constant length of 10 & width that varies from 0 to 10. To find the mean you could take the sum of all width values and you will get the median value or 5. Similarly, you can do the same thing but calculate the area of the rectangle and you will get the median value of 50. Go ahead and try it, the maximum area would be 100 & the lowest possible value is 0 so the greatest and lowest value divided by 2 is 50. This also works with surface area & in 3D if you keep the z coordinate fixed at 10 and have x&y as variables from 0-10. The mean will be the maximum 10,10,10 divided by 4. It's ok if you don't understand it from my explanations since my linguistic skills aren't the greatest :/

I think the best way to understand the average size of an object is to find the average of the individual lengths of the object that can be drawn within the confines of the allowed space. So if I want to draw the median rectangle inscribed in a circle I need to start from the maximum size and find the halfway point from the minimum which is just the diameter line. So maximum length is 10√2 and the minimum length is 0 so the average width of this rectangle is going to be 5√2. Plot 2.5√2 and -2.5√2 & find your length that connects those points to the circumference and that's your answer for the average rectangle that can be drawn within a circle. What's great about this is you could actually do this in reverse and find the halfway point from the square coordinate on the 10.5 o'clock circumference to 12 o'clock diameter (note that this is a straight line between both points & not the halfway mark of the length of the arch) then draw a line parallel to 12 o'clock diameter & repeat for both sides. You will end up with the same width measurement.

Ok, let me make it real figures so no miscommunication. We start with the minimum length distance which is 10√2 or 14.14 ft add the maximum length distance which is the diameter of 20 ft, then divide by 2 to give you 17.07 which is the average length minus the sagitta. We draw a line from the minimum length to the maximum length and find the midway point of the line & draw a parallel line from 12 o'clock at a distance of 17.07 and that will equal the adjacent midway point of the opposite circumference points. Again, do this for both sides and add the sagitta to get the average way a rectangle can be drawn or the average size of a rectangle with these specific conditions.

I think there can be 2 different surface area questions and they need to be worded differently so no confusion. 1st question is the surface area of the average rectangle (perimeter) & the 2nd is the average of the sum of all surface areas of all rectangles.

I think you guys have been trying to do question #2 which I don't know what the answer is but I'm betting the answer is slightly less than my answer for question #1. Just remember that the average sum of all areas confined within a circumference as a perimeter is probably going to give you an answer that can not equal an area of any rectangle so the wording from the 1st question to the 2nd need to be changed from the average rectangle to all rectangles, or single to plural. You guys really like these integral problems.

Let me pose a question to explain why mathematicians think about the word "average" as there are several possible meanings.

• "Mean" - The mean is what most people mean when they say 'average'. It is found by adding up all of the values you have and dividing by the number of values.
• "Mode" - The mode is the number in a set of numbers which occurs the most often. This one is more useful for continuous functions as it shows what is the most common result. For instance if you threw twenty dice and counted how many sixes you threw, the "Mode" would be 3 (23.79%) whereas the "Mean" = 20/6 = 3 1/3.
• "Median" - The median of a group of values is the value in the middle, so that 50% of the numbers are greater and 50% less than the "Median". As an extension this leads to quartiles using 25% and 75%, and when considering things like normal curves, confidence levels.

Here's the problem. I have eleven blank playing cards and write the squares 0, 1, ... ,100 on them. What is the average? From a betting point of view you make a wager with me and I'll give you one random card. You win what is written on the card. What is a fair price to pay to play the game?

The answer is obtained by adding up all the squares 0 thru 100 and dividing by 11. This is called the average and in the long term is what you'll be paid out; so is is a fair price to pay to play the game.

Quote: charliepatrick

Let me pose a question to explain why mathematicians think about the word "average" as there are several possible meanings.

• "Mean" - The mean is what most people mean when they say 'average'. It is found by adding up all of the values you have and dividing by the number of values.
• "Mode" - The mode is the number in a set of numbers which occurs the most often. This average is more useful for continuous functions as it shows what is the most common result. For instance if you three twenty dice and counted how many sixes you threw, the mode would be 3 (23.79%) whereas the "Mean" = 20/6 = 3 1/3.
• "Median" - The median of a group of values is the value in the middle, so that 50% of the numbers are greater and 50% less than the "meadian". As an extension this leads to quartiles using 25% and 75%, and when considering things like normal curves, confidence levels.

Here's the problem. I have eleven blank playing cards and write the squares 0, 1, ... ,100 on them. What is the average? From a betting point of view you make a wager with me and I'll give you one random card. You win what is written on the card. What is a fair price to pay to play the game?

The answer is obtained by adding up all the squares 0 thru 100 and dividing by 11. This is called the average and in the long term is what you'll be paid out; so is is a fair price to pay to play the game.

Actually, no & I'll explain why and this is a good example to do so. So if I were to ask you what the value of the average card should be the answer needs to be a specific card and should be the median value of the 11 cards and if those values are congruent then median value should equal the mean. If not then the median value is still the average value of a card choose since then average of the sum of all values of every card will not equal a card meaning it can not possibly be the average card but it could be the average value of the sum of all cards. Do you see how this word problem is constructed? Your absolutely right there we mathematicians struggle to find the correct words to describe their intentions and there are plenty of debates on mathematical translations but I think I am right about my interpretation of the question.

I just realized your analogy lacks 1 thing. So instead of the cards having different numbers on them let's say the cards were all different sizes! Now you would definitely say that the 6th card is the average no doubt, correct? So the area of the sixth card is the answer to question #1!

Btw the only reason why the variable length of the rectangle is not congruent is because the shape of the piece of the circumference that the rectangle length can be drawn from is not uniform because it's shape is drawn the degree of an arch & not a palabera :(

Quote: USpapergames

...if I were to ask you what the value of the average card should be the answer needs to be a specific card and should be the median value of the 11 cards and if those values are congruent then median value should equal the mean. If not then the median value is still the average value of a card choose since then average of the sum of all values of every card will not equal a card meaning it can not possibly be the average card but it could be the average value of the sum of all cards. Do you see how this word problem is constructed?...

I sense there is a total difference of opinion over the meaning of the word "average". You seem to think it means the "Median" whereas I think it means "Mean". I suspect others may have their opinions.

The words "Average", "Standard Deviation", "Variance" etc. have specific meanings within Statistics.

Quote: charliepatrick

I sense there is a total difference of opinion over the meaning of the word "average". You seem to think it means the "Median" whereas I think it means "Mean". I suspect others may have their opinions.

The words "Average", "Standard Deviation", "Variance" etc. have specific meanings within Statistics.

It's not that simple, 1st the mean does not reflect a single rectangle which is required of the question. 2nd this question is about the area of the average perimeter and not average area! So to reclarify your analogy the cards must be in all possible perimeter sizes and not just an area value that doesn't equal a rectangle.

My question is about the average perimeter & if you need me to change it to ha e that exact wording there then fine but I argue it's implied. Again the average card or average rectangle is asking for an actual average of all possible rectangles and not just the average of the area values. The overwhelming majority of these types of questions have the average perimeter equal to the average area but because this question is inscribed within a circ!e and does not have a congruent circumference of all possible rectangles drawn within it, these 2 figures don't match meaning they must be 2 different questions!

I think we should take a voting poll on the matter!

The average of a shape is:

A) the average perimeter of the shape (since perimeter lengths technically define shape lol)

B) the average area of the shape (which area has absolutely nothing to do with an object's shape)

I think we all know how I will vote.

P.S. The only time area can be considered the average of something is if it's shape is not defined. If the question was the average of an undefined object then yes area or volume would be a good start since the radius of 10 ft was given. And you could also argue density would also be considered a good candidate for average if there was a weight value given. But the keynote to remember is if a shape is defined and the question is asking for the average of that shape then the area, volume, density & whatever else need to fall within the confines of that shape of it's not answering the question.

So here is my theorem for the area of the average rectangle (perimeter) inscribed in a circle & this will give you the exact answer. I call it the Jacobs theorem & it doesn't require π or the sagitta!

AX̄❙²◯² = 2√(((2R-R√2)/2+¾R√2) (¼R√2+(2R-R√2)/2)) (½R√2)

2√(((2R-R√2)/2+¾R√2) (¼R√2+(2R-R√2)/2)) = Length

(½R√2) = Width

Theorems are at least 5x more difficult to solve for & have probably 10x more value to the world of mathematics than the majority of solutions that don't have a theorem. Not every problem can have a theorem define the solution so show me some respect & prove it wrong if you dare!

Now I need to do this for the surface area & volume of the average rectangular prism...

So my theorem can be simplied to just R²√7/2

Idk guys but I would think all this work should prove my math skills but maybe I'm just not smart enough for the mathletes lol.

Quote: charliepatrick

I sense there is a total difference of opinion over the meaning of the word "average". You seem to think it means the "Median" whereas I think it means "Mean". I suspect others may have their opinions.

The words "Average", "Standard Deviation", "Variance" etc. have specific meanings within Statistics.

So I think I solved where the confusion is coming from! I was hoping someone else would have argued my side and was better at putting ideas into words than myself but here goes me teaching this classroom some more.

So the only reason why everyone thinks the average rectangle refers to the average area, & not perimeter because the word average is used at the beginning of the question, but surely not everyone here is that bad at translating word problems?

So let's change the question to the perimeter lengths of the average rectangle! Now does everyone still think 🤔 somehow that the average rectangle is referring to area?? Or does the average rectangle magically turn into the average perimeter because the perimeter is used in a segment of the question???

So if we need to change the question to the perimeter lengths of the average rectangle I'm fine with that since I can take the product of that answer & find the area of the average rectangle! Let this be a lesson that if I am willing to argue about something, 99% of the time I'm right. Everyone on this form has made at least 1 error in their posts except me & nobody has been the instructor of this class more than I have. So prove me wrong if you want to dethrone me, but I might just walk away first if I don't start seeing some respect form everyone.