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Monty Hall, we have a PROBLEM
- Thread starter owslachief
- Start date
Handyman
Hug someone you care about...
Handyman
Hug someone you care about...
Re: Monty Hall, we have a PROBLEM
Ok you take your time worrying about all of that...I will take both boxes and do no worse than you
Ok you take your time worrying about all of that...I will take both boxes and do no worse than you
Re: Monty Hall, we have a PROBLEM
This problem probably could also be solved empirically....
We were discussing whether the rate of change of length of day was the same throughout the year.
We know that the shape of the earth's orbit is an ellipse. At the "pointier" ends of the ellipse, we have the solstices, and at the "flatter" sides of the ellipse, we have the equinoxes.
I was thinking that the rate of change of the length of day would be greater at the equinoxes and smaller at the solstices: the earth in effect has to slow down more to round the pointier end and speed up more slowly after it rounds the pointier ends, does it not? and it goes through the "flatter" sides more quickly, no?
PS for some reason, I also find it interesting that, even though the summer solstice is the longest day of the year, in this part of the country at least, the earliest sunrise comes before the solstice and the latest sunset comes after the solstice.
This problem probably could also be solved empirically....
We were discussing whether the rate of change of length of day was the same throughout the year.
We know that the shape of the earth's orbit is an ellipse. At the "pointier" ends of the ellipse, we have the solstices, and at the "flatter" sides of the ellipse, we have the equinoxes.
I was thinking that the rate of change of the length of day would be greater at the equinoxes and smaller at the solstices: the earth in effect has to slow down more to round the pointier end and speed up more slowly after it rounds the pointier ends, does it not? and it goes through the "flatter" sides more quickly, no?
PS for some reason, I also find it interesting that, even though the summer solstice is the longest day of the year, in this part of the country at least, the earliest sunrise comes before the solstice and the latest sunset comes after the solstice.
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Re: Monty Hall, we have a PROBLEM
The rate of change of day length is purely to do with the geometry of the earth's tilt and position relative to the sun. The angles are a bit complicated to work out, but nothing that was beyond the capabilities of the Greeks 2500 years ago.
You're correct that the rate of change of day length is greatest at the equinoxes and slowest at the solstices, but wrong about why. The difference in rates would be true even if the earth's orbit were a perfect circle (which is possible - it's just a special case of an ellipse). Additionally, the earth's speed is greatest at its perigee (when it is closest to the sun) and slowest at its apogee (farthest from the sun), so the earth is speeding as it moves toward the perigee and slowing down as it moves toward the apogee.
The rate of change of day length is purely to do with the geometry of the earth's tilt and position relative to the sun. The angles are a bit complicated to work out, but nothing that was beyond the capabilities of the Greeks 2500 years ago.
Hammer
We'll be back.
The biggest circle you could make would have a 4' diameter, so a 2' radius. Cut the circle in half, leaving you with 2 equal semicircles, each with a 2' radius.
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unofan
Well-known member
I'm 90 percent to the answer... I have the diagram of the largest area and the basic measurements. My trig is just failing me.
(15 minutes later) I think I've got the answer, but I'll probably need to wait til next week when they publish it to know for sure. And I ended up brute forcing it a bit, but given the answer I got, there's probably a more elegant way of doing it.
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unofan
Well-known member
Nope, there's a way to get bigger pieces.
ScoobyDoo
NPC
Re: Monty Hall, we have a PROBLEM
I'm not going to do the math but if you cut the rectangle into two equal Right Triangles and then make the largest semi circle you can inside the right triangles I believe you will end up with the biggest pieces.
I'm not going to do the math but if you cut the rectangle into two equal Right Triangles and then make the largest semi circle you can inside the right triangles I believe you will end up with the biggest pieces.
unofan
Well-known member
That's how I did it. I brute forced the math, but as I said, the answer leads me to think there was a more elegant way to get there.
St. Clown
Ideas Posted are Likely Not My Own
Re: Monty Hall, we have a PROBLEM
You're cutting circles, not semi-circles. And you're likely going to cut into the 8' edge of the board, accounting for the curve of the semi-circles, you'll have slightly more than a 4' diameter, but I don't know the math - I hated trig.
The Sicatoka
Kicizapi Cetan
Re: Monty Hall, we have a PROBLEM
I pulled out the 1/4" grid paper and the circle template and the scale (too lazy to do the math on a Friday afternoon).
Now this is all by eyeball (pencil and paper), but I could see something like a 2.6875' radius working. (I did the two equal right triangles thing as a starting point.)
EDIT: Answer is here in white; highlight to read.
OK, I sucked it up and did the math. It's really just algebra. Assume the board is on a coordinate plane with x-axis the 8 foot side (y axis, 4 foot). To make two right triangles I made a "working cut" from (0,4) to (8,0) for a line defined as y(x) = (-1/2)x + 4
Now find a where y(a) = a. That means you are out 'a' on the x axis, up 'a' on the y axis, and on the cut line; thus, 'a' will be your radius. I came up mathematically at 2.6666 feet (8/3 feet) for the radius. --> That's a table with a 16/3 foot diameter, a 5.3333 foot table.
How'd I do?
PS - My eyeball on grid paper was pretty close. I'll blame the width of the 0.5 mm lead, er, I mean graphite.
I pulled out the 1/4" grid paper and the circle template and the scale (too lazy to do the math on a Friday afternoon).
Now this is all by eyeball (pencil and paper), but I could see something like a 2.6875' radius working. (I did the two equal right triangles thing as a starting point.)
EDIT: Answer is here in white; highlight to read.
OK, I sucked it up and did the math. It's really just algebra. Assume the board is on a coordinate plane with x-axis the 8 foot side (y axis, 4 foot). To make two right triangles I made a "working cut" from (0,4) to (8,0) for a line defined as y(x) = (-1/2)x + 4
Now find a where y(a) = a. That means you are out 'a' on the x axis, up 'a' on the y axis, and on the cut line; thus, 'a' will be your radius. I came up mathematically at 2.6666 feet (8/3 feet) for the radius. --> That's a table with a 16/3 foot diameter, a 5.3333 foot table.
How'd I do?
PS - My eyeball on grid paper was pretty close. I'll blame the width of the 0.5 mm lead, er, I mean graphite.
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Re: Monty Hall, we have a PROBLEM
I was working along those lines too but I'm not completely sure that the tangent line to the two semi-circles where they touch each other is necessarily also the diagonal of the rectangle. If I knew that for sure it would simplify the math greatly, and there is probably some elegant Euclidian proof.
I was working along those lines too but I'm not completely sure that the tangent line to the two semi-circles where they touch each other is necessarily also the diagonal of the rectangle. If I knew that for sure it would simplify the math greatly, and there is probably some elegant Euclidian proof.
Re: Monty Hall, we have a PROBLEM
Ah, did you use the hypotenuse as your diameter with offset semi-circles, or did you use the 8' side as your diameter with offset semi-circles, or does it even make a difference either way? it is late in the day and late in the week. maybe tomorrow morning when I am fresh.
Ah, did you use the hypotenuse as your diameter with offset semi-circles, or did you use the 8' side as your diameter with offset semi-circles, or does it even make a difference either way? it is late in the day and late in the week. maybe tomorrow morning when I am fresh.
The Sicatoka
Kicizapi Cetan
Re: Monty Hall, we have a PROBLEM
Yes, ultimately that. The hypotenuse edges of the wood will end up being the table diameter (seam).
Yes, ultimately that. The hypotenuse edges of the wood will end up being the table diameter (seam).
FlagDUDE08
Banned
Re: Monty Hall, we have a PROBLEM
I had actually thought about doing the latter. You could easily get a maximum radius above 2, just figure out where the two quarter-circles (because you could ignore the quarter-circles closest to the edges) become tangent. Sadly, although quick arithmetic is still there, logic has taken over the maths.
I may take some time to calculate it out in a while...
I had actually thought about doing the latter. You could easily get a maximum radius above 2, just figure out where the two quarter-circles (because you could ignore the quarter-circles closest to the edges) become tangent. Sadly, although quick arithmetic is still there, logic has taken over the maths.
I may take some time to calculate it out in a while...
unofan
Well-known member
I pulled out the 1/4" grid paper and the circle template and the scale (too lazy to do the math on a Friday afternoon).
Now this is all by eyeball (pencil and paper), but I could see something like a 2.6875' radius working. (I did the two equal right triangles thing as a starting point.)
EDIT: Answer is here in white; highlight to read.
OK, I sucked it up and did the math. It's really just algebra. Assume the board is on a coordinate plane with x-axis the 8 foot side (y axis, 4 foot). To make two right triangles I made a "working cut" from (0,4) to (8,0) for a line defined as y(x) = (-1/2)x + 4
Now find a where y(a) = a. That means you are out 'a' on the x axis, up 'a' on the y axis, and on the cut line; thus, 'a' will be your radius. I came up mathematically at 2.6666 feet (8/3 feet) for the radius. --> That's a table with a 16/3 foot diameter, a 5.3333 foot table.
How'd I do?
PS - My eyeball on grid paper was pretty close. I'll blame the width of the 0.5 mm lead, er, I mean graphite.
That's the answer I get too, and you got it far more elegantly than I did. I'm just struggling with conceptually how your solution gives you the radius.
And now that I drew it out, makes sense.
The Sicatoka
Kicizapi Cetan
Re: Monty Hall, we have a PROBLEM
Algebraic constraint equations are your friend.
(This and that have to be equal but meet on that constraint line, without giving up too much.)
Better? I did my first pencil and scale work on a 1":1' ratio. Now look at that, my eyeball answer, the answer, and my comment about 0.5 mm lead.
What I just said in white:
I got 2.6875 using a 1" = 1' scale and eyeballs. The answer is 2.6666'. So, the difference between my scale 2.6875" and the real answer (2.6666", at that scale) is 0.0209". Now, knowing there are 25.4 mm in an inch --> 0.0209 x 25.4 = 0.5 mm ... and that's the diameter of graphite, 0.5 mm, in the mechanical pencil I used!
Algebraic constraint equations are your friend.
(This and that have to be equal but meet on that constraint line, without giving up too much.)
Better? I did my first pencil and scale work on a 1":1' ratio. Now look at that, my eyeball answer, the answer, and my comment about 0.5 mm lead.
What I just said in white:
I got 2.6875 using a 1" = 1' scale and eyeballs. The answer is 2.6666'. So, the difference between my scale 2.6875" and the real answer (2.6666", at that scale) is 0.0209". Now, knowing there are 25.4 mm in an inch --> 0.0209 x 25.4 = 0.5 mm ... and that's the diameter of graphite, 0.5 mm, in the mechanical pencil I used!
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FlagDUDE08
Banned
Re: Monty Hall, we have a PROBLEM
Trying Fishy's latter suggestion:
Using the formula for a circle: (x - h)^2 + (y - k)^2 = r^2
Center of Circle 1 is (r, 0); (x - r)^2 + y^2 = r^2
Center of Circle 2 is (8 - r, 4); (x - 8 + r)^2 + (y - 4)^2 = r^2
Find a place where:
(x - r)^2 + y^2 = (x - 8 + r)^2 + (y - 4)^2
We can even out the y values with a +2.
x is a little trickier:
r^2 - 2rx + x^2 = r^2 + 2r(x - 8) + (x - 8)^2
These can even out with a 4.
Therefore, the place where we can get these circles to be tangent to each other is at (4, 2), ironically.
Knowing this, let's solve for y; good ol' Pythagorean theorem:
We know from the center to any point on the circle, the distance is r. So let's do a triangle with the x axis to the point we found.
(4 - r)^2 + 2^2 = r^2
16 - 8r + r^2 + 4 = r^2
-8r + 20 = 0
We can take an absolute value for radius; r = 2.5
Trying Fishy's latter suggestion:
Using the formula for a circle: (x - h)^2 + (y - k)^2 = r^2
Center of Circle 1 is (r, 0); (x - r)^2 + y^2 = r^2
Center of Circle 2 is (8 - r, 4); (x - 8 + r)^2 + (y - 4)^2 = r^2
Find a place where:
(x - r)^2 + y^2 = (x - 8 + r)^2 + (y - 4)^2
We can even out the y values with a +2.
x is a little trickier:
r^2 - 2rx + x^2 = r^2 + 2r(x - 8) + (x - 8)^2
These can even out with a 4.
Therefore, the place where we can get these circles to be tangent to each other is at (4, 2), ironically.
Knowing this, let's solve for y; good ol' Pythagorean theorem:
We know from the center to any point on the circle, the distance is r. So let's do a triangle with the x axis to the point we found.
(4 - r)^2 + 2^2 = r^2
16 - 8r + r^2 + 4 = r^2
-8r + 20 = 0
We can take an absolute value for radius; r = 2.5