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Wizard
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Wizard
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October 5th, 2021 at 5:07:39 PM permalink
I apologize if discussion of the semi-sphere problem isn't finished. I also apologize if I asked this one before. I do know I asked a similar one.

Consider the following figure. All pieces are squares. The area of the white square is 25. What is the area of the entire figure?

It's not whether you win or lose; it's whether or not you had a good bet.
aceside
aceside
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October 5th, 2021 at 6:11:58 PM permalink
Here is my answer:
Set L-yellow=x, then
L-green=2x+5
L-pink=2x+5+5
L-red=2x
L-blue=3x-5
L-top=L-green+2*L-pink=2x+5+4x+20=6x+25
L-bottom=L-red+3*L-blue=2x+9x-15=11x-15
Let L-top=L-bottom, solve the equation to find x=8
Width=L-green+L-yellow+L-red=45
Length=L-top=73
Area=Width * Length=45 * 73=3285
Last edited by: aceside on Oct 5, 2021
Wizard
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Wizard
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October 5th, 2021 at 7:31:20 PM permalink
Quote: aceside

Here is my answer:
Set L-yellow=x, then
L-green=2x+5
L-pink=2x+5+5
L-red=2x
L-blue=3x-5
L-top=L-green+2*L-pink=2x+5+4x+20=6x+25
L-bottom=L-red+3*L-blue=2x+9x-15=11x-15
Let L-top=L-bottom, solve the equation to find x=8
Width=L-green+L-yellow+L-red=45
Length=L-top=73
Area=Width * Length=45 * 73=3285
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    I agree!
    It's not whether you win or lose; it's whether or not you had a good bet.
    Gialmere
    Gialmere
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    October 6th, 2021 at 8:25:46 AM permalink
    From Riddler 538 comes this pair of football puzzles...



    Problem #1
    You and two friends are in a three-person fantasy football league, drafting just three positions each for your teams: quarterback, running back and wide receiver. Yes, this is a simplified version of fantasy football.

    The following table shows the top three athletes in each position, as well as the number of fantasy points they are expected to earn over the course of the season. You and your friends must each select exactly one player from each position.




    The draft is a “snake draft.” If person A drafts first, B drafts second and C drafts both third and fourth, thus the order of the picks is as follows: A-B-C-C-B-A-A-B-C.

    Your friends — being the kind people that they are — agree that you can choose your pick number. Which draft position should you choose to maximize your expected fantasy score?


    ---------------

    Problem #2
    Hames Jarrison has just intercepted a pass at one end zone of a football field, and begins running — at a constant speed of 15 miles per hour — to the other end zone, 100 yards away.

    At the moment he catches the ball, you are on the very same goal line, but on the other end of the field, 50 yards away from Jarrison. Caught up in the moment, you decide you will always run directly toward Jarrison’s current position, rather than plan ahead to meet him downfield along a more strategic course.

    Assuming you run at a constant speed (i.e., don’t worry about any transient acceleration), how fast must you be in order to catch Jarrison before he scores a touchdown?
    Have you tried 22 tonight? I said 22.
    charliepatrick
    charliepatrick
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    Thanks for this post from:
    Gialmere
    October 6th, 2021 at 9:27:15 AM permalink
    Quote: Gialmere

    ......Which draft position should you choose to maximize your expected fantasy score?...

    Pick B.

    A's best strategy is to pick QB400. B has a choice of QB350 or RB300.
    (i) In the former C is now guaranteed QB300 so picks RB300 and WR250, B will choose WR225 over RB225. A=800, B=775, C=850.
    (ii) In the latter C chooses QB350 and RB250, B then chooses WR225 and will eventually get QB300. A=800, B=825, C=800.

    So by choosing B and RB300 you get 825.
    aceside
    aceside
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    October 6th, 2021 at 1:12:06 PM permalink
    Quote: aceside

    Quote: charliepatrick

    The sphere has two options, it either lands with the curved surface landing first (chance = 1/2) and so falls onto the curved surface, or one of the tips of the cut surface touches first.

    By symmetry just look at a semi-circle being the vertical part of the sphere, rather like a D dropped randomly in 2-D space. If the centre of gravity (assumed to be 3/8 R) lies to the right of the tip, then the sphere falls towards the curved surface, otherwise it doesn't. So just consider the angle of the straight line to the vertical and a line from the centre of gravity to the tip.

    The tangent of the angle is 3/8, so the angle is arctan(3/8) = about 20.556 degrees. Consider the angle of the flat surface compared to vertical. So between 0 and 20.556 it falls curved, between 20.556 and 90 it falls flat. By symmetry the same logic applies 90 thru 180.
    So chance of flat is about (90-20.556)*2/360 = 0.3858.
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    I thought about this again and come up with a slightly different answer by considering the solid angle of this cone.
    My answer is 2pi x [1-cos(90-20.556)] / 4 pi =0.324
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    I have been expecting /charliepatrick or anybody to comment on this. I brought up this solid angle consideration because the flip (rotation) axis of the hemisphere can be parallel to the flat surface or vertical, or in any direction.
    charliepatrick
    charliepatrick
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    October 6th, 2021 at 2:52:14 PM permalink
    Quote: aceside

    I have been expecting /charliepatrick or anybody to comment on this. I brought up this solid angle consideration because the flip (rotation) axis of the hemisphere can be parallel to the flat surface or vertical, or in any direction.

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    I couldn't comment on your solution as I don't really understand how you've got to your answer. (I'm guessing it's a formula looking at the surface area of the sphere and comparing it with the surface (on the sphere) formed by a cone with the appropriate angle.)

    For simplicity assume the semi-sphere is placed on a level surface and left to drop, thus there in no spin/flip/rotation. If the curved surface (the sphere part) touches the surface then it will fall face up. Otherwise consider the edge of the flat surface and consider the disc formed from the flat surface. The disc has a single point of contact with the level surface, thus one only has to consider the line from the point of contact through the centre of the disc and what that angle is.

    Thus my answer does not involve a cone but the angles between 0 and 180 where 20.556 thu 90 and 90 thru (180-20.556) will land face down.
    I think our difference is I'm looking at the angle and comparing it with 360 degrees (i.e. how much arc the angle forms on a circle) and you're looking at the area formed by the cone compared to the whole sphere's surface area.

    Perhaps someone knows why one is correct and the other is wrong. It might be that the probability distribution of the line I describe isn't constant.
    aceside
    aceside
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    October 6th, 2021 at 4:38:43 PM permalink
    Quote: charliepatrick

    Quote: aceside

    I have been expecting /charliepatrick or anybody to comment on this. I brought up this solid angle consideration because the flip (rotation) axis of the hemisphere can be parallel to the flat surface or vertical, or in any direction.

  • link to original post

    I couldn't comment on your solution as I don't really understand how you've got to your answer. (I'm guessing it's a formula looking at the surface area of the sphere and comparing it with the surface (on the sphere) formed by a cone with the appropriate angle.)

    For simplicity assume the semi-sphere is placed on a level surface and left to drop, thus there in no spin/flip/rotation. If the curved surface (the sphere part) touches the surface then it will fall face up. Otherwise consider the edge of the flat surface and consider the disc formed from the flat surface. The disc has a single point of contact with the level surface, thus one only has to consider the line from the point of contact through the centre of the disc and what that angle is.

    Thus my answer does not involve a cone but the angles between 0 and 180 where 20.556 thu 90 and 90 thru (180-20.556) will land face down.
    I think our difference is I'm looking at the angle and comparing it with 360 degrees (i.e. how much arc the angle forms on a circle) and you're looking at the area formed by the cone compared to the whole sphere's surface area.

    Perhaps someone knows why one is correct and the other is wrong. It might be that the probability distribution of the line I describe isn't constant.
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    Perfect! That is exactly what I meant. In the mean time, I am thinking about using a similar device to attach to a Craps dice to gain an edge. Thank you for your confirmation.
    Wizard
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    Wizard
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    October 7th, 2021 at 3:48:04 PM permalink
    On a windless day, Charlie takes his canoe to the river and paddles up upstream. He always paddles at the same rate (in other words if we ignore the current, he would always go the same speed). A mile* after launching, his hat falls in the river. Ten minutes after that, he realizes his hat is missing and immediately makes a u-turn to catch up to it downstream. Charlie catches up to his hat at the same place he launched.

    How fast is the current?



    Note: I originally, incorrectly, said "an hour." It should be a mile.
    Last edited by: Wizard on Oct 7, 2021
    It's not whether you win or lose; it's whether or not you had a good bet.
    teliot
    teliot
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    October 7th, 2021 at 3:57:27 PM permalink
    I saw this problem on Twitter -- it has a rather easy argument.

    Find all solutions X, Y, Z in integers to:

    Poetry website: www.totallydisconnected.com

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