## Poll

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8 members have voted

Wizard Joined: Oct 14, 2009
• Posts: 21630
June 23rd, 2020 at 7:11:24 PM permalink
Quote: USpapergames

Also, there is no way to put a corner of the rectangle at 12:00 since that would be the exact middle of the circle and all rectangles must have congruent sides of equal length so the solution Mr. Shackleford is describing would be to solve a trapezoid.

Here is what I meant by putting a corner at 12:00. The second one is at any random point on the circumference. From there, the rest of the rectangle draws itself. 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 23rd, 2020 at 7:47:03 PM permalink
Yes, that makes sense & it is a valid solution if you can get the average of that random point on the circumference. But still, your solution does not explain how you would go about getting the average area of the average random point, yet alone the value of the average circumference point. Notice how when you decrease the width of the rectangle the height of the rectangle increases. This is why the problem is so tricky. I would love to see another equation that solves this problem.

P.S. I just solved for the smallest solution for the problem, check the description of my video for latest update. It's a lot easier to solve a problem when you are already provided with a solution. And taking the problem to the 3rd dimension is much more difficult. Let's see if your form members can solve the problem. Btw, do you think I haven't provided a valid solution to the problem yet?
Last edited by: USpapergames on Jun 23, 2020
Wizard Joined: Oct 14, 2009
• Posts: 21630
Thanks for this post from: June 23rd, 2020 at 9:19:16 PM permalink
I solved this problem in about 10 minutes, hopefully correctly.

We have a process for doing math problems here at WoV. Here are the basic rules for problems deemed worthy of their own thread.

1. The one asking gives the audience a chance to answer the problem first. It would ruin the fun if I just told you the answer at the same time I posted the question.
2. The first person with a satisfactory answer AND solution is awarded a beer and are deemed a member of the "beer club."
3. Members of the Beer Club may not post answers or solutions for 24 hours after the a new question is posted, to give new people a chance to join the club. They may ask for clarification on the problem during that 24 hour waiting period.
4. Answers and solutions should be put in spoiler tags, until somebody presents a correct answer and solution.
5. Please don't post links to other sites that go over the same 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
June 24th, 2020 at 7:18:57 AM permalink
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)

Quote: Wizard

4. Answers and solutions should be put in spoiler tags, until somebody presents a correct answer and solution.

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
Last edited by: ThatDonGuy on Jun 24, 2020
Wizard Joined: Oct 14, 2009
• Posts: 21630
June 24th, 2020 at 11:27:59 AM permalink
Reminder that members of the Beer Club may jump in approximately 4.5 hours from the time of this post.
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
June 24th, 2020 at 3:40:18 PM permalink
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.

Last edited by: ThatDonGuy on Jun 24, 2020
Wizard Joined: Oct 14, 2009
• Posts: 21630
June 24th, 2020 at 6:07:02 PM permalink
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.

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.
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 6:22:28 PM permalink
Both of you are wrong. I understand why you would want to use trigonometry to solve this but you can solve this problem using only geometry.

The equation is X̄A❙◯ = ((R√2+2R)/2 + 2(R−√(R²−(2Rπ÷16)²))) (R√2)/2

Here is the equation broken down X̄A❙◯ = ((B+D)/2 + 2S) B/2

B = Base of Arch
D = Diameter
S = Sagitta

I found the most simplified solution! It's not as small as I would like it but I don't think it can get smaller with fewer characters. Here is the question you should be rewarding the answer to.

X̄A❙◯ = (R²(2√2+12 - √(8+π)(8-π)) √2) / 8
USpapergames Joined: Jun 23, 2020
• Posts: 66
June 24th, 2020 at 6:23:05 PM permalink
Actually, I messed up. You guys might be right :( I thought I found an elegant solution but no.

The only reason why my answer is wrong is because the angle on the Sagitta is wrong. Does nobody see the solution as I do? Do you need to use trigonometry or calculus to solve this problem? I don't think so, I think I should be able to solve this problem with just paper & pencil. I think I'm going to just give up now and kill myself, I hate being wrong.
Last edited by: USpapergames on Jun 24, 2020
ChesterDog Joined: Jul 26, 2010