## Poll

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8 members have voted

USpapergames Joined: Jun 23, 2020
• Posts: 66
June 24th, 2020 at 7:00:26 PM permalink
To think the actual answer is (40R)/π is ridiculous, but very plausible since I know the most difficult problems have very simplified solutions. The only realson why I'm going to double-check your work is that I realize my answer is adding the Sagitta value when the length increase but it's not the true value since the 90° angle of the Sagitta is off from the angle of the long edge of the rectangle. Still, your guy's answers are off from mine by < 5 ft², neither of you two have provided any proofs & I know my answer is correct besides the Sagitta value being off. I'm going to trying and correct the Sagitta portion and see if we get the same answers. Without proofs for trigonometry, I have to agree with your logic and I don't agree with all of the reasoning. It's very easy to think you have the answer to this problem lol
Last edited by: USpapergames on Jun 24, 2020
USpapergames Joined: Jun 23, 2020
• Posts: 66
June 24th, 2020 at 7:02:07 PM permalink
q=ChesterDog]Thanks for the new word to me, "sagitta." I had to look it up because our high school trig course didn't teach that. Was sagitta used in your high school trig course?

Nope, I graduated high school sophomore year and went and got my AA at a community college for my junior& senior years.
Last edited by: USpapergames on Jun 24, 2020
USpapergames Joined: Jun 23, 2020
• Posts: 66
June 24th, 2020 at 9:05:20 PM permalink
Quote: ThatDonGuy

For the sphere problem, assuming the volume is desired, and the radius of the sphere is 1,

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
The volume is 8 (1 - d2) sin t cos t = 4 (1 - d2) sin 2t

The integral from 0 to PI/2 of 4 (1 - d2) sin 2t dt
= 4 (1 - d2) * the integral from 0 to PI of sin t dt
= 4 (1 - d2) * ((-cos PI) - (-cos 0))
= 8 (1 - d2)

The integral from 0 to 1 of 8 (1 - d2) dd
= 8 - 8 * the integral from 0 to 1 of d2 dd
= 8 - 8 * (13 / 3 - 0)
= 16/3

The mean volume = (16/3) / (1 * PI/2) = 32 / (3 PI)

Maybe USpapergames doesn't know how to post spoiler tags?

[spoiler=This is a spoiler tag]Spoiler text goes here[/spoiler]
creates this:
Spoiler text goes here

Your solution makes no sense and is way off from my answer! If the radius of the sphere is 1 then the sphere volume is = 4.19 and the maximum cube inscribed in a sphere is 1.54 your answer is larger than the largest volume of a rectangular prism inscribed in a sphere with the radius = 1. Your answer 32÷3π = 3.4 can't possibly be the correct answer.
USpapergames Joined: Jun 23, 2020
• Posts: 66
June 24th, 2020 at 9:15:50 PM permalink
Quote: Wizard

Quote: ThatDonGuy

Wait a minute...I submitted this after 5:00, but I didn't want to double post, so I edited an earlier post, and it didn't change the posting time.

Let ABCD be the rectangle
Note that AC and BD are both diameters of the circle
Choose A arbitrarily, and B uniformly along the circle other than A or with AB being a diameter
Let t be the measure of angle AOB in radians; BOC has measure (PI - t)
Also note that t is a uniform random number in (0, PI)

AB has length 2 (10 sin (t/2))
BC has length 2 (10 sin ((PI - t)/2)) = 2 (10 cos (t/2))
The area of the rectangle = 4 (10 sin (t/2)) (10 cos (t/2))
= 200 (2 sin (t/2) cos (t/2)) = 200 sin t

The integral from 0 to PI of 200 sin t dt = 200 (-cos PI) - (-cos 0))
= 200 (1 - (-1)) = 400

Since t is in (0, PI), the mean area of the rectangle = 400 / PI.

I deem this correct and a worthy solution! Please add one more beer to the count I owe you. Here is my solution, which I think is simpler.

[spoiler=Wiz solution]

Let's make the radius 1 and then multiply by 10^2 at the last step to get the answer for a circle of radius 10.

1. Chose an arbitrary point on the circle, for example, the 12:00 position. That will be one corner of the rectangle.
2. Observe that both diagonals of the rectangle will cut through the center of the circle.
3. Another corner of the rectangle will be directly across the first point, at the 6:00 position.
4. Choose a random point anywhere within 90 degrees of the point in step 1, for example anywhere from the 12:00 to 3:00 position. Please note we don't have to go all the way to the 6:00 position, because it doesn't make any difference whether this point is closer to the 12:00 or 6:00 position. This will be the another corner of the rectangle.
5. Call the points from step 1 and 4, A and B. Call the enter of the circle O.
6. Let x be the angle AOB.
7. The area of the rectangle, given x, is 8*(1/2)*sin(x)*cos(x).
8. Integrate that area, 4*sin(x)*cos(x) over the range from 0 to pi/2:

integral of 4*sin(x)*cos(x) dx
Let u = sin(x)
du = cos(x) du
Integral = 4u du
= 2u^2/2
=2*sin^2(x)

9. Plug in the limits of integration of 0 to pi/2 to get 2*(1^2 - 0^2) = 2.
10. Divide that by the range of integration pi/2, to get the average (I admit I forgot this step initially) = 2/(pi/2) = 4/pi.
11. Now, multiply by 10^2 to get the answer for a circle of radius 10: 100*4/pi = 400/pi =127.324.

[spoiler=Wiz solution]

I truly think my way of solving this problem is the best way, it's a geometry problem not a trigonometry problem so use the tools of geometric principals. I think the degree of the sagitta affects the heights which would cause my equation to be slightly off :(
Last edited by: unnamed administrator on Jun 25, 2020
Wizard Joined: Oct 14, 2009
• Posts: 21630
Thanks for this post from: June 24th, 2020 at 9:34:48 PM permalink
I ran simulations in Excel, which confirm the 400/pi answer.
It's not whether you win or lose; it's whether or not you had a good bet.
USpapergames Joined: Jun 23, 2020
• Posts: 66
June 24th, 2020 at 9:41:52 PM permalink
Wow, the difference in the distance from the angle of the Sagitta must be significant then. That's got to be what's wrong with my equation right? The rest of the math has to be correct right?
USpapergames Joined: Jun 23, 2020
• Posts: 66
June 24th, 2020 at 9:44:00 PM permalink
So basically there is no way of doing the math by hand then. This question would never be in a timed test to begin with :(
Last edited by: USpapergames on Jun 24, 2020
USpapergames Joined: Jun 23, 2020
• Posts: 66
June 24th, 2020 at 10:47:44 PM permalink
I think I just suck at geometry then. I don't know why I didn't see the angle of the sagitta being as not being a factor :(
Wizard Joined: Oct 14, 2009
• Posts: 21630
June 25th, 2020 at 5:39:18 AM permalink
Quote: USpapergames

I think I just suck at geometry then. I don't know why I didn't see the angle of the sagitta being as not being a factor :(

I had to look up the Sagitta. I don't see how that would help you. This is a very textbook integral calculus problem.
It's not whether you win or lose; it's whether or not you had a good bet.
ThatDonGuy Joined: Jun 22, 2011
• Posts: 4112
Thanks for this post from: June 25th, 2020 at 8:09:53 AM permalink
Quote: USpapergames

Quote: ThatDonGuy

For the sphere problem, assuming the volume is desired, and the radius of the sphere is 1,

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
The volume is 8 (1 - d2) sin t cos t = 4 (1 - d2) sin 2t

The integral from 0 to PI/2 of 4 (1 - d2) sin 2t dt
= 4 (1 - d2) * the integral from 0 to PI of sin t dt
= 4 (1 - d2) * ((-cos PI) - (-cos 0))
= 8 (1 - d2)

The integral from 0 to 1 of 8 (1 - d2) dd
= 8 - 8 * the integral from 0 to 1 of d2 dd
= 8 - 8 * (13 / 3 - 0)
= 16/3

The mean volume = (16/3) / (1 * PI/2) = 32 / (3 PI)

Maybe USpapergames doesn't know how to post spoiler tags?

[spoiler=This is a spoiler tag]Spoiler text goes here[/spoiler]
creates this:
Spoiler text goes here

Your solution makes no sense and is way off from my answer! If the radius of the sphere is 1 then the sphere volume is = 4.19 and the maximum cube inscribed in a sphere is 1.54 your answer is larger than the largest volume of a rectangular prism inscribed in a sphere with the radius = 1. Your answer 32÷3π = 3.4 can't possibly be the correct answer.

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