Rectangle inscribed in a sphere

Quote: USpapergames

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.

All hail the mighty USPapergames:
King of Math.
Prince of English comprehension.
Knight Errant in his spelling of 'Forum' and 'From'

$:o)

I'm reminded of Ted Kaczynski, a mathematics prodigy whose total immersion into that world I believe contributed to his winding up half crazy. That he wasn't totally crazy was something he insisted on in his defense at his trial.

Quote: odiousgambit

I'm reminded of Ted Kaczynski, a mathematics prodigy whose total immersion into that world I believe contributed to his winding up half crazy. That he wasn't totally crazy was something he insisted on in his defense at his trial.

There's a good mini-series about him on Netflix right now. In my opinion a textbook example of "genius is usually touched by madness." Some say I'm starting to look like him with my refusal to cut my hair or shave.

Quote: odiousgambit

I'm reminded of Ted Kaczynski, a mathematics prodigy whose total immersion into that world I believe contributed to his winding up half crazy. That he wasn't totally crazy was something he insisted on in his defense at his trial.

Honestly, you might be on to something there. Maybe I should drop the whole company image and create a signature name for myself. Become a symbol of power & fear like Batman did. Maybe I should go by the "Mad Mathematician"?

Just to let everyone know I don't have any mental illness, I've had to provide a full mental evaluation for court by a Ph.D. psychologist & she cleared me of any misconceived notions of psychosis. So everything chemically is balanced and no abnormal brain functions where found but even she admitted that a much longer mental evaluation would be required to identify personality disorders. Personality disorders can take sometimes years to diagnose and the overwhelming majority of homeless people suffer from these disorders and they can even become life-threatening when not in check.

Most CEOs have many personality disorders like narcissism or antisocialism. But illness can only really be treated with medication we're as personality disorders can only really be fixed without medication & I've never had a therapist who has prescribed me medication, even if I asked for it they didn't want to. I hope this clears things up, I don't want people thinking I'm crazy if I act weird. I wish you guys would have known me in 2014....

In a way, I wish my problems we're from illness because then I could just start taking some pill & everything would be fixed :/ I suffer from depression & anxiety really badly. It doesn't help that I'm an introvert & can't stand large groups of people.

If we could measure the amount of stress in each others lives mine would dominate the sum of everyone on the form. Not even a week ago I had to pull my sword out because a group of homeless people tried attacking me all because I was asking them if they knew who stole my wallet. I was on the phone with my friend & software developer Mike for over an hour just venting my frustrations after that happened to me. It was only my driver's license, prepaid card, and $27 but I was determined to find whoever lifted it out of my bag. You guys don't have a clue the struggles I deal with, I can't even use my local gym to shower because of Covid 19. If you look at my hair closely I have so many grey hairs I think I will lose all my pigment when I'm in my 40's!

Assuming the current problem is to find the median perimeter of a rectangle inscribed in a circle of radius 10,
here is my answer, although to be honest, it doesn't pass the "smell test":

Let ABCD be the rectangle, and O the center of the circle.
Choose A arbitrarily; AC is a diameter of the circle.
Choose B uniformly at random on the circle other than at A or C; BD is also a diameter of the circle.
Let t be the measure of angle AOB; 0 < t < PI.
If t > PI/2, then it is congruent to the rectangle where t is replaced by (PI - t), but with the side lengts switched (e.g. AB in one = AC in the other).
We can thus limit this to finding the median perimeter for a rectangle over t where 0 < t < PI/2.
AB and CD each have length 20 sin (t/2), and BC and AD each have length 20 cos (t/2).
The perimeter = 40 (sin (t/2) + cos (t/2)).
If P(t) is the perimeter with angle t, then dP/dt = 40 (sin (t/2) - cos (t/2)).
Since cos x > sin x for all x in (0, PI/2), dP/dt is always negative for 0 < t < PI/2, so P(t) is a strictly decreasing function for 0 < t < PI/2.
Thus, each value of P(t) is unique and decreasing for all t such that 0 < t < PI/2.
Therefore, the median of the range will be at the midpoint of the domain, which is t = PI/4.
The median perimeter = 40 (sin (PI/8) + cos (PI/8))
Note that cos (PI/8) = sqrt(2 + sqrt(2)) / 2 and sin (PI/8) = sqrt(2 - sqrt(2)) / 2
Thus, the median perimeter = 40 (sqrt(2 + sqrt(2)) / 2 + sqrt(2 - sqrt(2)) / 2)
= 20 (sqrt(2 + sqrt(2)) + sqrt(2 - sqrt(2)))
= about 52.2625

I am not entirely sure I am calculating "median" correctly, but since t is uniformly random in (0, PI), the perimeter is less than 52.2625 for half of the possible values of t and greater than 52.2625 for the other half.

Quote: ThatDonGuy

Assuming the current problem is to find the median perimeter of a rectangle inscribed in a circle of radius 10,
here is my answer, although to be honest, it doesn't pass the "smell test":

Let ABCD be the rectangle, and O the center of the circle.
Choose A arbitrarily; AC is a diameter of the circle.
Choose B uniformly at random on the circle other than at A or C; BD is also a diameter of the circle.
Let t be the measure of angle AOB; 0 < t < PI.
If t > PI/2, then it is congruent to the rectangle where t is replaced by (PI - t), but with the side lengts switched (e.g. AB in one = AC in the other).
We can thus limit this to finding the median perimeter for a rectangle over t where 0 < t < PI/2.
AB and CD each have length 20 sin (t/2), and BC and AD each have length 20 cos (t/2).
The perimeter = 40 (sin (t/2) + cos (t/2)).
If P(t) is the perimeter with angle t, then dP/dt = 40 (sin (t/2) - cos (t/2)).
Since cos x > sin x for all x in (0, PI/2), dP/dt is always negative for 0 < t < PI/2, so P(t) is a strictly decreasing function for 0 < t < PI/2.
Thus, each value of P(t) is unique and decreasing for all t such that 0 < t < PI/2.
Therefore, the median of the range will be at the midpoint of the domain, which is t = PI/4.
The median perimeter = 40 (sin (PI/8) + cos (PI/8))
Note that cos (PI/8) = sqrt(2 + sqrt(2)) / 2 and sin (PI/8) = sqrt(2 - sqrt(2)) / 2
Thus, the median perimeter = 40 (sqrt(2 + sqrt(2)) / 2 + sqrt(2 - sqrt(2)) / 2)
= 20 (sqrt(2 + sqrt(2)) + sqrt(2 - sqrt(2)))
= about 52.2625

I am not entirely sure I am calculating "median" correctly, but since t is uniformly random in (0, PI), the perimeter is less than 52.2625 for half of the possible values of t and greater than 52.2625 for the other half.

Suprisingling very close. I'm getting 51.5587

So when reviewing I think your t value is wrong. Your taking the angle of any 2 diameter points but the ength or width is never on the diameter. Also your angles are off for when the values of length and width switch. It's not π/2 but rather π/4 since taking point A and moving it past 12 o'clock will become point B & any lower from 9 o'clock will turn the point to point C and 10½ o'clock is the reverse point where anywhere between 10½ and 9 results in the AC turning into width instead of length.

Quote: USpapergames

In a way, I wish my problems we're from illness because then I could just start taking some pill & everything would be fixed :/ I suffer from depression & anxiety really badly. It doesn't help that I'm an introvert & can't stand large groups of people.

If we could measure the amount of stress in each others lives mine would dominate the sum of everyone on the form. Not even a week ago I had to pull my sword out because a group of homeless people tried attacking me all because I was asking them if they knew who stole my wallet. I was on the phone with my friend & software developer Mike for over an hour just venting my frustrations after that happened to me. It was only my driver's license, prepaid card, and $27 but I was determined to find whoever lifted it out of my bag. You guys don't have a clue the struggles I deal with, I can't even use my local gym to shower because of Covid 19. If you look at my hair closely I have so many grey hairs I think I will lose all my pigment when I'm in my 40's!

Thanks for your candor. I think I can empathize. If you want to talk about it, let me know.

Quote: USpapergames

So when reviewing I think your t value is wrong. Your taking the angle of any 2 diameter points but the ength or width is never on the diameter. Also your angles are off for when the values of length and width switch. It's not π/2 but rather π/4 since taking point A and moving it past 12 o'clock will become point B & any lower from 9 o'clock will turn the point to point C and 10½ o'clock is the reverse point where anywhere between 10½ and 9 results in the AC turning into width instead of length.

AC is always a diameter, so AOB + BOC = PI. t is defined as AOB; BOC = PI - t.

You may be misvisualizing what I am doing. Point A is chosen arbitrarily, so it is always "at 12:00". B is chosen uniformly at random from all of the points of the circle other than A (where t would be 0) and C (where t would be PI); without loss of generality, it can be anywhere from 12:00 to 6:00 (if it is between 6:00 and 12:00, "make the clock counterclockwise").
If t = PI/2, then AOB = BOC; i.e. you have a square. It is where t > PI/2 that AB < BC, in which case the rectangle is congruent to the one for t = PI - t, but with the AB and BC lengths switched.

Yes, I understand you now. Basically point A could be the top left corner of the rectangle & point C is is the bottom right corner and vice versa for B & D. Let me re-evaluate your solution and I'll get back to you. Thank you for the improved visual explanation, I just sold my truck so m currently busy & leaving for thunder valley tonight.

Quote: Wizard

Thanks for your candor. I think I can empathize. If you want to talk about it, let me know.

Thank you, means a lot actually. I didn't share all that because I wanted pity but rather I feel severally misunderstood :)

Quote: ThatDonGuy

AC is always a diameter, so AOB + BOC = PI. t is defined as AOB; BOC = PI - t.

You may be misvisualizing what I am doing. Point A is chosen arbitrarily, so it is always "at 12:00". B is chosen uniformly at random from all of the points of the circle other than A (where t would be 0) and C (where t would be PI); without loss of generality, it can be anywhere from 12:00 to 6:00 (if it is between 6:00 and 12:00, "make the clock counterclockwise").
If t = PI/2, then AOB = BOC; i.e. you have a square. It is where t > PI/2 that AB < BC, in which case the rectangle is congruent to the one for t = PI - t, but with the AB and BC lengths switched.

So I just woke up after going to bed at 5am and sleeping outside thunder Valley. I ran into a fan of my channel surprisingly & we played poler at the same table! Had a great conversation and made $525 profit off $200 but I would have even had more if I didn't get super unlucky in a hand where my opponent can only draw to a chop!

So what in trying to say is my attention and free time and dropped farther than the norm right now but I will put your solution as #1 for analysis since you are solving the question I was going for :) My head is currently killing me

Quote: USpapergames

So I just woke up after going to bed at 5am and sleeping outside thunder Valley.

Thunder Valley, as in the California casino? I'm a little surprised it's open (as well as Jackson Rancheria); Cache Creek is still closed.

Quote: ThatDonGuy

AC is always a diameter, so AOB + BOC = PI. t is defined as AOB; BOC = PI - t.

You may be misvisualizing what I am doing. Point A is chosen arbitrarily, so it is always "at 12:00". B is chosen uniformly at random from all of the points of the circle other than A (where t would be 0) and C (where t would be PI); without loss of generality, it can be anywhere from 12:00 to 6:00 (if it is between 6:00 and 12:00, "make the clock counterclockwise").
If t = PI/2, then AOB = BOC; i.e. you have a square. It is where t > PI/2 that AB < BC, in which case the rectangle is congruent to the one for t = PI - t, but with the AB and BC lengths switched.

Sorry, poker and working out this contract have been very stressful but I think I see an error. You have the inverse from 12:00 to 6:00 but it should be from 12oclock to 1½ clock. Any B point past 1½ clock is a direct copy of a previously defined perimeter. The B point of 2 o'clock is the same perimeter of point B of 1 o'clock. You're getting the average of 4 copies of the true range of all possible rectangles inscribed in a circle.

Btw Thunder is great but a lot of sharks are here now and I underestimated them and lost 500 last night, just made it back an hour ago tho ;)

Quote: USpapergames

Sorry, poker and working out this contract have been very stressful but I think I see an error. You have the inverse from 12:00 to 6:00 but it should be from 12oclock to 1½ clock. Any B point past 1½ clock is a direct copy of a previously defined perimeter. The B point of 2 o'clock is the same perimeter of point B of 1 o'clock. You're getting the average of 4 copies of the true range of all possible rectangles inscribed in a circle.

Btw Thunder is great but a lot of sharks are here now and I underestimated them and lost 500 last night, just made it back an hour ago tho ;)

Sorry, change the 1.5 o'clock to 3 o'clock. So the inverse is anything past 3 o'clock so your counting double the sum of all perimeters except the 3 o'clock circle. I think that is why your answer is slightly off.

So my example should be that the B point of 2 o'clock should have the same perimeter as the B point of 4 o'clock.

Now that I think about it only counting 3 o'clock once would lower your average, not raise it :/

Quote: USpapergames

Sorry, change the 1.5 o'clock to 3 o'clock. So the inverse is anything past 3 o'clock so your counting double the sum of all perimeters except the 3 o'clock circle. I think that is why your answer is slightly off.

So my example should be that the B point of 2 o'clock should have the same perimeter as the B point of 4 o'clock.

I already took this duplication into account:

"Let t be the measure of angle AOB; 0 < t < PI.
If t > PI/2, then it is congruent to the rectangle where t is replaced by (PI - t), but with the side lengts switched (e.g. AB in one = AC in the other).
We can thus limit this to finding the median perimeter for a rectangle over t where 0 < t < PI/2."

t = PI/2 is at 3:00; t = PI is at 6:00.

I proposed that, since the perimeter increases as t increases from 0 to PI/2, then at t = PI/4 (i.e. the "1:30" point), half of the perimeters are greater than the perimeter at that point, and the other half are less than that, so the median perimeter is where A is at 12:00 and B is at 1:30 (or, alternatively, at 4:30).

lol, omg that is exactly what I said! Then why are our answers lightly off??? What is your length and width of the sides? Maybe it's due to rounding?

There must be something wrong with your end but I can't figure it out and its bothering me, I already checked my theorem multiple times.

Quote: USpapergames

lol, omg that is exactly what I said! Then why are our answers lightly off??? What is your length and width of the sides? Maybe it's due to rounding?

The length is 20 cos 22.5 (degrees), which is about 18.4776, and the width is 20 sin 22.5, which is about 7.6537.

Missed the edit window...

I think I know why our numbers differ.
Graph the angle of two adjacent vertices of the rectangle with the center of the circle (X-axis) against that rectangle's perimeter (Y-axis), which is the graph y = 40 (sin x + cos x) for x = 0 to 45 degrees.
If you take the median based on x, then the answer is where x = 22.5 degrees, which is 52.26252.
Since there is a one-to-one correspondence between x and y, the median based on y "should be" at the same point.
Then again, using that reasoning, since f(x) is strictly increasing, every value of f(x) from 40 to 40 sqrt(2) exists exactly once in the interval, so the median "should be" (40 + 40 sqrt(2)) / 2, which is about 48.2848.
On the other hand, if you take the median based on the length of the graph itself, then the answer appears to be around 51.39. I could come up with an exact answer if I knew how to calculate the integral of sqrt(2 - sin 2x) dx.

Look at it this way:
Let f(x) = x for x <= 1, and f(x) = 2x - 1 for x >= 1.
What is the median of f(x) in the interval 0 <= x <= 2?
Based on x, half of the f(x) values are less than 1, and half are greater than 1, so the median = f(1) = 1.
Based on f(x), since every real number in [0,3] has an f(x) value exactly once, the median = 3/2.
The length of f(x) for 0 <= x <= 1 is sqrt(2), and the length for 1 <= x <= 2 is sqrt(5), so the median based on the length of f(x) is where the total length is (sqrt(2) + sqrt(5)) / 2.
This is where the length for x > 1 is (sqrt(5) - sqrt(2)) / 2.
The length from (1,1) to (x, 2x-1) is sqrt(5) * (x - 1); this equals (sqrt(5) - sqrt(2)) / 2 where x = (15 - 2 sqrt(5)) / 10, so the median arc length is at f(x) = (15 - 2 sqrt(5)) / 5 - 1 = 2 - 2/5 sqrt(5).

I am sticking by my answer - at least, for now - as it is based on the assumption that the angle from the center of the inscribed circle to two adjacent vertices of the rectangle is uniformly random, so the median would be where the angle = 22.5 degrees.

Very nice, I occur with your analysis but I still think there is something wrong within your solution & It's topics like this that make me want to get back into Calculus. It's true what they say about if you don't use something you lose it, I don't use calculus in game design ever but it feels nice to use it again.

So with a length of 2√(((2R-R√2)/2+¾R√2) (¼R√2+(2R-R√2)/2)) we get 18.7083 & with a width of (½R√2) we get 7.0711. So I am sticking with my answer also ;)

Quote: ThatDonGuy

Missed the edit window...

I think I know why our numbers differ.
Graph the angle of two adjacent vertices of the rectangle with the center of the circle (X-axis) against that rectangle's perimeter (Y-axis), which is the graph y = 40 (sin x + cos x) for x = 0 to 45 degrees.
If you take the median based on x, then the answer is where x = 22.5 degrees, which is 52.26252.
Since there is a one-to-one correspondence between x and y, the median based on y "should be" at the same point.
Then again, using that reasoning, since f(x) is strictly increasing, every value of f(x) from 40 to 40 sqrt(2) exists exactly once in the interval, so the median "should be" (40 + 40 sqrt(2)) / 2, which is about 48.2848.
On the other hand, if you take the median based on the length of the graph itself, then the answer appears to be around 51.39. I could come up with an exact answer if I knew how to calculate the integral of sqrt(2 - sin 2x) dx.

Look at it this way:
Let f(x) = x for x <= 1, and f(x) = 2x - 1 for x >= 1.
What is the median of f(x) in the interval 0 <= x <= 2?
Based on x, half of the f(x) values are less than 1, and half are greater than 1, so the median = f(1) = 1.
Based on f(x), since every real number in [0,3] has an f(x) value exactly once, the median = 3/2.
The length of f(x) for 0 <= x <= 1 is sqrt(2), and the length for 1 <= x <= 2 is sqrt(5), so the median based on the length of f(x) is where the total length is (sqrt(2) + sqrt(5)) / 2.
This is where the length for x > 1 is (sqrt(5) - sqrt(2)) / 2.
The length from (1,1) to (x, 2x-1) is sqrt(5) * (x - 1); this equals (sqrt(5) - sqrt(2)) / 2 where x = (15 - 2 sqrt(5)) / 10, so the median arc length is at f(x) = (15 - 2 sqrt(5)) / 5 - 1 = 2 - 2/5 sqrt(5).

I am sticking by my answer - at least, for now - as it is based on the assumption that the angle from the center of the inscribed circle to two adjacent vertices of the rectangle is uniformly random, so the median would be where the angle = 22.5 degrees.

I figured out your problem. So you're trying to find the median arc length but that won't help your solve the problem since the line segment does not pass through the circumference at a perpendicular angle! Therefore the median line segment of that angle will be off from the center point of the arch. Here is how you should look at it. What is the width length of the maximum perimeter of a rectangle inscribed in a circle? That length is a square with a side of 10√2 therefore the median width of the rectangle is 5√2. If you can figure out the length of the rectangle with that width then you got your answer.

Quote: ThatDonGuy

Missed the edit window...

I think I know why our numbers differ.
Graph the angle of two adjacent vertices of the rectangle with the center of the circle (X-axis) against that rectangle's perimeter (Y-axis), which is the graph y = 40 (sin x + cos x) for x = 0 to 45 degrees.
If you take the median based on x, then the answer is where x = 22.5 degrees, which is 52.26252.
Since there is a one-to-one correspondence between x and y, the median based on y "should be" at the same point.
Then again, using that reasoning, since f(x) is strictly increasing, every value of f(x) from 40 to 40 sqrt(2) exists exactly once in the interval, so the median "should be" (40 + 40 sqrt(2)) / 2, which is about 48.2848.
On the other hand, if you take the median based on the length of the graph itself, then the answer appears to be around 51.39. I could come up with an exact answer if I knew how to calculate the integral of sqrt(2 - sin 2x) dx.

Look at it this way:
Let f(x) = x for x <= 1, and f(x) = 2x - 1 for x >= 1.
What is the median of f(x) in the interval 0 <= x <= 2?
Based on x, half of the f(x) values are less than 1, and half are greater than 1, so the median = f(1) = 1.
Based on f(x), since every real number in [0,3] has an f(x) value exactly once, the median = 3/2.
The length of f(x) for 0 <= x <= 1 is sqrt(2), and the length for 1 <= x <= 2 is sqrt(5), so the median based on the length of f(x) is where the total length is (sqrt(2) + sqrt(5)) / 2.
This is where the length for x > 1 is (sqrt(5) - sqrt(2)) / 2.
The length from (1,1) to (x, 2x-1) is sqrt(5) * (x - 1); this equals (sqrt(5) - sqrt(2)) / 2 where x = (15 - 2 sqrt(5)) / 10, so the median arc length is at f(x) = (15 - 2 sqrt(5)) / 5 - 1 = 2 - 2/5 sqrt(5).

I am sticking by my answer - at least, for now - as it is based on the assumption that the angle from the center of the inscribed circle to two adjacent vertices of the rectangle is uniformly random, so the median would be where the angle = 22.5 degrees.

I just posted a video that I think will explain why your answer doesn't match mine! https://youtu.be/RiRdVHVrjjI

Quote: USpapergames

I just posted a video that I think will explain why your answer doesn't match mine! https://youtu.be/RiRdVHVrjjI

I understand why your answer doesn't match mine.

You are calculating the median based on the median possible length of the shorter side of the two sides.

I am calculating it based on the median possible smallest angle formed by two adjacent vertices of the rectangle and the center of the inscribed circle. This is based on the assumption that, if the rectangle is ABCD, choose A arbitrarily; AC is a diameter of the circle, and a "random" rectangle will have B equally likely at any point on the circle other than A or C. If angle AOB > 90 degrees, then the rectangle is congruent to one where B and D are switched, so AOB can always be 45 degress or less.

However, there is a third possibility; take the median value of all possible perimeters of the rectangle. Since the minimum perimeter is 40 (well, 40 plus epsilon), the maximum perimeter is 40 sqrt(2), and all values between 40 and 40 sqrt(2) are equally likely, the median would be 20 + 20 sqrt(2) = about 48.2848.

I think our main problem is, I can't find a definition of "median" that applies to a continuous function - only to a discrete number of points. As I posted earlier, for y = f(x), you can get three different values depending on whether or not the x values are equally likely, the y values are equally likely, or the points on the curve of f are equally likely.
If you uncover one online, can you post a link here?

Ya, I think you have a valid point for defining averages. See this is actually a truly unique question in mathematics that I came up with but I can see how you could get multiple interpretations but in the end, the average way a rectangle can be drawn is the width of that rectangle being half the distance of the largest possible width. If your drawing rectangle in a circle the midpoint isn't going to be the average of the circumference points but rather the average line segment of the angle that contacts the circumference. Also, it should be the average of 40√2 to 20 since the smallest rectangle is just the diameter line of 20 with a width of 0. But that's not the correct at of calculating the average since the length increases as the width decreases & vice versa, so there is no even distribution for decreasing both the length & width so the average of the min and maximum values are meaningless. Anyone who things this is a simple question is ignorant. We are truly working on a cutting edge math problem that definitely deserves analysis