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8 members have voted

USpapergames Joined: Jun 23, 2020
• Posts: 66
June 25th, 2020 at 10:02:06 AM permalink

Choose a random distance d in (0,1) from the origin to side ABCD along the X-axis
The circle containing ABCD has radius sqrt (1 - d2)

Choose a random angle t in (0, PI/2) from the XZ-plane to A
A is at (d, sqrt(1 - d2) cos t, sqrt(1 - d2) sin t)
B is at (d, sqrt(1 - d2) cos t, -sqrt(1 - d2) sin t)
C is at (d, -sqrt(1 - d2) cos t, -sqrt(1 - d2) sin t)
D is at (d, -sqrt(1 - d2) cos t, sqrt(1 - d2) sin t)
E through H are the same as A through D except that the x-value is -d

AB has length 2 sqrt(1 - d2) sin t
AD has length 2 sqrt(1 - d2) cos t
AE has length 2d (not 2 - otherwise the prism doesn't fit in the sphere, now does it?)
The volume is 8d (1 - d2) sin t cos t

The integral from 0 to PI/2 of 8d (1 - d2) sin t cos t dt
= the integral from 0 to PI/2 of 8d (1 - d2) sin t d (sin t)
= the integral from 0 to 1 of 8d (1 - d2) x dx
= 8d (1 - d2)

The integral from 0 to 1 of 8x (1 - x2) dx
= 8 * the integral from 0 to 1 of (x - x3) dx
= 8 * ((12/2 - 14/4) - 0)
= 8 * 1/4 = 2

The mean volume = 2 / (PI/2 * 1) = 4 / PI

This is a lot better work & makes more sense how you derived the answer!!! Very good, the problem is my equation for the volume of the rectangular prism is off since my equation for the rectangle inside of a circle is wrong :( I double-checked your work on 2D area & your equation is wrong! 400/π might be the right answer but the actual equation for the average area of the rectangle inscribed in a circle is actually 4R²/π! I like how you guys can give me the answer but not the equation lol. So 4R/π as the solution doesn't exactly make a whole lot of sense as the solution for the volume in 3D. Keep working on it, I think you're close!

Wizard
I had to look up the Sagitta. I don't see how that would help you. This is a very textbook integral calculus problem.

I understand how to solve this using calculus but I purposely wanted to avoid that by using a more elegant solution from geometric principles. The fact that you don't see how the sagitta would help solve the problem just tells me you guys are just good and making calculations & not just sticking to math tricks that can be done without the need of a calculator. I am going to make another video of this problem and explain how I would solve this problem using nothing but pen/paper & my mind. It can be done sir, I know it can!
Last edited by: USpapergames on Jun 25, 2020
ThatDonGuy Joined: Jun 22, 2011
• Posts: 4112
Thanks for this post from: June 25th, 2020 at 7:12:53 PM permalink
I am not as confident of my answer as I was...
I did a simulation doing the following:
Each vertex of the prism can be expressed as (+/- x, +/- y, +/- z), where x, y, and z are positive, and x^2 + y^2 + z^2 = 1
Select a random number x in (0, 1)
Select a random number y in (0, 1 - sqrt(1 - x^2))
z = sqrt(1 - x^2 - y^2)
Since four parallel edges go from -x to +x, four from -y to +y, and four from -z to +z, the volume = (2x) (2y) (2z) = 8xyz.
However, when I do this, I get a mean volume of 2/3.
USpapergames Joined: Jun 23, 2020
• Posts: 66
June 25th, 2020 at 7:50:29 PM permalink
Quote: ThatDonGuy

I am not as confident of my answer as I was...
I did a simulation doing the following:
Each vertex of the prism can be expressed as (+/- x, +/- y, +/- z), where x, y, and z are positive, and x^2 + y^2 + z^2 = 1
Select a random number x in (0, 1)
Select a random number y in (0, 1 - sqrt(1 - x^2))
z = sqrt(1 - x^2 - y^2)
Since four parallel edges go from -x to +x, four from -y to +y, and four from -z to +z, the volume = (2x) (2y) (2z) = 8xyz.
However, when I do this, I get a mean volume of 2/3.

I see your problem. So here is how you do the solution using calculus. You want an inverse relationship between -X³, +Y³, -Z³ so that when you take the edge value of the maximum volume of a cube inscribed in a sphere and then divide the width & depth values by 2 the height value will increase in value. Without this relationship defined your calculations will never be correct. This is why I like working with equations so that why I can always test your solution ;) I like how you're also unsure of your work. I'm sorry I don't have a lot of time to spend this. I wanted to do another video but instead, I ended up writing a draft of a business contract :/

So I think the solution would need the starting coordinates to be the corners of the cube with the maximum volume inscribed in a sphere and use the equation X̄A❙³◯³ = (X²Z²)/4 (2S+Y²). Just remember that the sagitta is at an angle and needs to be rotated to the Y³ angle. The difference in length when rotated is 58% of the Sagitta.
Last edited by: USpapergames on Jun 26, 2020
USpapergames Joined: Jun 23, 2020
• Posts: 66
June 25th, 2020 at 7:59:46 PM permalink
Do you know how to create your own form on here? I have more to difficult questions I could post if you're really up for a challenge and it doesn't take much time to post a question, especially if I already know the answer. Btw I still think I can solve this problem using just pen & paper ;)
ThatDonGuy Joined: Jun 22, 2011
• Posts: 4112
June 26th, 2020 at 6:54:47 AM permalink
Quote: ThatDonGuy

I am not as confident of my answer as I was...
I did a simulation doing the following:
Each vertex of the prism can be expressed as (+/- x, +/- y, +/- z), where x, y, and z are positive, and x^2 + y^2 + z^2 = 1
Select a random number x in (0, 1)
Select a random number y in (0, 1 - sqrt(1 - x^2))
z = sqrt(1 - x^2 - y^2)
Since four parallel edges go from -x to +x, four from -y to +y, and four from -z to +z, the volume = (2x) (2y) (2z) = 8xyz.
However, when I do this, I get a mean volume of 2/3.

I am not entirely sure these points are uniformly random.
Using this method to select a random point on the sphere:
(a) Choose a random angle t in (0, 2 PI)
(b) Choose a random number x in (0,1), and set angle p = arccos(2x-1)
(c) The coordinates of one of the vertices is (sin t cos p, cos t cos p, sin p)
(d) The volume = 8 sin t cos t sin p cos^2 p
This gives me a mean volume of 1/2...
USpapergames Joined: Jun 23, 2020
• Posts: 66
June 26th, 2020 at 10:11:22 AM permalink
Quote: ThatDonGuy

I am not entirely sure these points are uniformly random.
Using this method to select a random point on the sphere:
(a) Choose a random angle t in (0, 2 PI)
(b) Choose a random number x in (0,1), and set angle p = arccos(2x-1)
(c) The coordinates of one of the vertices is (sin t cos p, cos t cos p, sin p)
(d) The volume = 8 sin t cos t sin p cos^2 p
This gives me a mean volume of 1/2...

There is no need for random points. Your brute force method is just not the right approach to solving this. I think this is a mathematical flaw within your reasoning. There actually is a rectangular prism that has the exact same volume as the average of all rectangular prisms inscribed within a sphere. I gave you the exact points & how to solve for this rectangular prism. Please review my last reply over again. The process in which you would calculate the average by actually recreating all the possible coordinates is nuts ;)
unJon Joined: Jul 1, 2018
• Posts: 1698
Thanks for this post from: June 26th, 2020 at 11:15:03 AM permalink
Quote: USpapergames

There is no need for random points. Your brute force method is just not the right approach to solving this. I think this is a mathematical flaw within your reasoning. There actually is a rectangular prism that has the exact same volume as the average of all rectangular prisms inscribed within a sphere. I gave you the exact points & how to solve for this rectangular prism. Please review my last reply over again. The process in which you would calculate the average by actually recreating all the possible coordinates is nuts ;)

I find it amusing to hear integration referred to as brute force.
The race is not always to the swift, nor the battle to the strong; but that is the way to bet.
USpapergames Joined: Jun 23, 2020
• Posts: 66
June 26th, 2020 at 11:33:54 AM permalink
So this is the equation so for

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 :/
USpapergames Joined: Jun 23, 2020
• Posts: 66
June 26th, 2020 at 12:29:17 PM permalink
Sounds to me like your using some software simulator. So we know that the X&Z coordinates = .5775 so place coordinates +.5775, Y, +.5775 | +.5775, Y, -.5775 | -.5775, Y, -.5775 | -.5775, Y, -.5775 & connect the Y-axis to the circumference & use your program to give you the length of the rectangular prism. You can figure out the rest from there.
USpapergames Joined: Jun 23, 2020