## Poll

18 votes (50%) | |||

13 votes (36.11%) | |||

5 votes (13.88%) | |||

2 votes (5.55%) | |||

11 votes (30.55%) | |||

3 votes (8.33%) | |||

6 votes (16.66%) | |||

5 votes (13.88%) | |||

11 votes (30.55%) | |||

8 votes (22.22%) |

**36 members have voted**

Let A and B be the endpoints of the "base" of the larger semicircle, O the midpoint of AB, C and D the endpoints of the "base" of the smaller semicircle, and M the midpoint of CD.

Since the orientation of the smaller semicircle is not specified, assume its base is parallel to the base of the larger one. (This may be a requirement in order for the smaller semicircle's endpoints to touch the larger one as well as the smaller one to touch the midpoint of the larger base; I'll look into that...)

Edit:Actually, its orientation is irrelevant - since CD is a chord of AB, the line from the larger semicircle's center to the midpoint of a chord is always perpendicular to the chord, since the two triangles formed are SSS-congruent)

This means CMO is a right angle.

CM, DM, and OM are all radii of the smaller semicircle, so CM = OM, which makes OMC an isosceles right triangle with hypotenuse sqrt(2) x CM.

This is also the radius of the larger semicircle.

Since the ratio of the radii of the two semicircles is sqrt(2), the ratio of their areas is (sqrt(2))^2 = 2, which means the area of the larger semicircle outside of the smaller semicircle must be the same as the area of the smaller semicircle.

Therefore the area of the darker region = 7.

When you've googled the answer, then if you can understand the answer, then you are still high in my estimation.

If you can prove the answer is 1/6, then you are Alan M

The question....

There are 10 pieces of paper face down on a table in front of you. Each piece of paper has a number on it. Literally ANY REAL NUMBER. Can be a decimal fraction, can be tiny, can be massive, can be rational or irrational, though irrational would make it messy, can be ANY real number.

You didn't write the numbers and you have no idea what they are!

You will turn the pieces of paper over, 1 by 1. You must stop when you think you have the highest number. Simple enough?

The final piece of paper you choose to flip over is your final number; you can’t flip 6 pieces of paper over and then go back and choose the 3rd piece of paper as your number. What you opick, you stick with as your final answer.

What strategy would you use in this game to maximise the chances of having the highest number?

If you use the best strategy, what is your probability of being right.

Clue: Use a random number generator to play this game.!!!!!

There is a fun corollary to this question...

There are ten hot women at a party and you decide, because you're an old mathematician, to marry the most sexually aggressive one. Till death do you part.

You can have a day of unbridled sex with as many of them as you want, individually, of course. But only once per woman (or not at all), You will rate them yourself out of 100.

The last one you have sex with, you will marry and live out the rest of your days with.

What is the best strategy to give you the best probability of marrying the 'best' one.

What is the probability you do actually marry the 'best' one.

Who says maths has no practical value?

Quote:OnceDearHere's a cripplingly hard maths puzzle. If you can work it out, you are a better man than I.

When you've googled the answer, then if you can understand the answer, then you are still high in my estimation.

If you can prove the answer is 1/6, then you are Alan M

The question....

There are 10 pieces of paper face down on a table in front of you. Each piece of paper has a number on it. Literally ANY REAL NUMBER. Can be a decimal fraction, can be tiny, can be massive, can be rational or irrational, though irrational would make it messy, can be ANY real number.

You didn't write the numbers and you have no idea what they are!

You will turn the pieces of paper over, 1 by 1. You must stop when you think you have the highest number. Simple enough?

The final piece of paper you choose to flip over is your final number; you can’t flip 6 pieces of paper over and then go back and choose the 3rd piece of paper as your number. What you opick, you stick with as your final answer.

What strategy would you use in this game to maximise the chances of having the highest number?

If you use the best strategy, what is your probability of being right.

link to original post

This has been asked already.

Multiple times.

I never could decipher how the solution was determined

Reject the first N / e values, where N is the total number (in this case, 100), then accept the first value that is higher than all of the values revealed up to that point. For N = 10, reject the first 3, then, starting with the fourth, accept the first number that is higher than all of the numbers before it.

Harder problem:

You are the second person in line to do this. The first one says that he will implement the above strategy. Before you make your choice, you are told that the first person's answer was incorrect. What is your strategy now?

Then look at the rest and pick the first one which is higher than the one you've already seen.

Firstly if the best is in the first N, then you can't win.

If the second best is in the first N then you must win (as only the best is left and no other pick will apply)

If the third best is in the first N then it's 50/50 whether you encounter the 1st or 2nd during the second part.

Similar logic for 4th best (i.e. 1 chance in 3)

Similarly for 5th best (i.e. 1 chance in 4)

Thus what is the best value for N.

Best seen | Perms | Pr(win) |

1st | 1 | none |

2nd | 1 | 1/1 |

3rd | 1 | 1/2 |

4th | 1 | 1/3 |

5th | 1 | 1/4 |

6th | 1 | 1/5 |

7th | 1 | 1/6 |

8th | 1 | 1/7 |

9th | 1 | 1/8 |

10th | 1 | 1/9 |

28.290% |

Best seen | Perms | Pr(win) |

1st | 9 | none |

2nd | 8 | 1/1 |

3rd | 7 | 1/2 |

4th | 6 | 1/3 |

5th | 5 | 1/4 |

6th | 4 | 1/5 |

7th | 3 | 1/6 |

8th | 2 | 1/7 |

9th | 1 | 1/8 |

10th | 0 | 1/9 |

36.579% |

Best seen | Perms | Pr(win) |

1st | 36 | none |

2nd | 28 | 1/1 |

3rd | 21 | 1/2 |

4th | 15 | 1/3 |

5th | 10 | 1/4 |

6th | 6 | 1/5 |

7th | 3 | 1/6 |

8th | 1 | 1/7 |

9th | 0 | 1/8 |

10th | 0 | 1/9 |

Pr of winning | 39.869% |

Best seen | Perms | Pr(win) |

1st | 84 | none |

2nd | 56 | 1/1 |

3rd | 35 | 1/2 |

4th | 20 | 1/3 |

5th | 10 | 1/4 |

6th | 4 | 1/5 |

7th | 1 | 1/6 |

8th | 0 | 1/7 |

9th | 0 | 1/8 |

10th | 0 | 1/9 |

39.825% |

I believe for N the best strategy is SQRT(N) but couldn't prove it!

Quote:charliepatrickThis is known as the wives problem (or similar name) but I get much better chances than 1 in 6.

The plan is to look at the first N items and remember the highest (so far)

Then look at the rest and pick the first one which is higher than the one you've already seen.

Firstly if the best is in the first N, then you can't win.

If the second best is in the first N then you must win (as only the best is left and no other pick will apply)

If the third best is in the first N then it's 50/50 whether you encounter the 1st or 2nd during the second part.

Similar logic for 4th best (i.e. 1 chance in 3)

Similarly for 5th best (i.e. 1 chance in 4)

Thus what is the best value for N.

Best seen Perms Pr(win) 1st 1 none 2nd 1 1/1 3rd 1 1/2 4th 1 1/3 5th 1 1/4 6th 1 1/5 7th 1 1/6 8th 1 1/7 9th 1 1/8 10th 1 1/9 28.290%

Best seen Perms Pr(win) 1st 9 none 2nd 8 1/1 3rd 7 1/2 4th 6 1/3 5th 5 1/4 6th 4 1/5 7th 3 1/6 8th 2 1/7 9th 1 1/8 10th 0 1/9 36.579%

Best seen Perms Pr(win) 1st 36 none 2nd 28 1/1 3rd 21 1/2 4th 15 1/3 5th 10 1/4 6th 6 1/5 7th 3 1/6 8th 1 1/7 9th 0 1/8 10th 0 1/9 Pr of winning 39.869%

Best seen Perms Pr(win) 1st 84 none 2nd 56 1/1 3rd 35 1/2 4th 20 1/3 5th 10 1/4 6th 4 1/5 7th 1 1/6 8th 0 1/7 9th 0 1/8 10th 0 1/9 39.825%

I believe for N the best strategy is SQRT(N) but couldn't prove it!

link to original post

Probability of success with best strategy about 1 in 3

Strategy is to calculate N= Count of cards/e (That's e: Eulers number)

Reject the first N cards

Stop at the first, next card with a higher value than all those previously rejected.

https://www.youtube.com/watch?v=OeJobV4jJG0

Of course, according to a notorious youtube video, you don't choose a wife or even a date by how sexy they are or how great in bed they may be. The other major factor is how crazy they are. If they are too crazy, it doesn't matter how sexy they are, you will regret it forever.

I know this answer is unlikely to be completely correct but:

You choose a number by where it falls in the range defined by the previously generated numbers.

Given the stated maximum of 10 numbers, from the 5th number on, choose the number if it is higher than at least 4 of the previous numbers that were generated.

Quote:gordonm888I give OnceDear credit for writing an unusually entertaining problem statement.

Of course, according to a notorious youtube video, you don't choose a wife or even a date by how sexy they are or how great in bed they may be. The other major factor is how crazy they are. If they are too crazy, it doesn't matter how sexy they are, you will regret it forever.

I know this answer is unlikely to be completely correct but:

You choose a number by where it falls in the range defined by the previously generated numbers.

Given the stated maximum of 10 numbers, from the 5th number on, choose the number if it is higher than at least 4 of the previous numbers that were generated.

link to original post

I recognise that this has, indeed been seen before. Sorry.

Quote:charliepatrickIt's also 7.

The logic involves creating right angled triangles (one for each semi circle) that can be formed used the diameter as the hypotenuse. Consider the large one first, its hypotenuse is 2r. Now chop the triangle in half (so two of the sides of it are radii) - clearly its area is half the original one. Then rotate it by 45 degrees. This will now line up so its hypotenuse matches the diameter of the smaller semi-circle.

Thus by similar triangles within similar semi circles, the area of the smaller semi-circle is half the larger one.

link to original post

Quote:ThatDonGuy

Let A and B be the endpoints of the "base" of the larger semicircle, O the midpoint of AB, C and D the endpoints of the "base" of the smaller semicircle, and M the midpoint of CD.

Since the orientation of the smaller semicircle is not specified, assume its base is parallel to the base of the larger one. (This may be a requirement in order for the smaller semicircle's endpoints to touch the larger one as well as the smaller one to touch the midpoint of the larger base; I'll look into that...)

Edit:Actually, its orientation is irrelevant - since CD is a chord of AB, the line from the larger semicircle's center to the midpoint of a chord is always perpendicular to the chord, since the two triangles formed are SSS-congruent)

This means CMO is a right angle.

CM, DM, and OM are all radii of the smaller semicircle, so CM = OM, which makes OMC an isosceles right triangle with hypotenuse sqrt(2) x CM.

This is also the radius of the larger semicircle.

Since the ratio of the radii of the two semicircles is sqrt(2), the ratio of their areas is (sqrt(2))^2 = 2, which means the area of the larger semicircle outside of the smaller semicircle must be the same as the area of the smaller semicircle.

Therefore the area of the darker region = 7.

link to original post

Correct!!

Well done.

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At long last, Dakota Jones is close to finding the Lost Arc, a geometric antiquity buried deep in the sands of Egypt. Along the way, she discovered what she described as a “highly symmetric crystal” that’s needed to precisely locate the Arc. Dakota measured the crystal using her laser scanner and relayed the results to you. But nefarious agents have gotten wind of her plans, and Dakota and the crystal are nowhere to be found.

Locating the Arc is now up to you. To do that, you must recreate the crystal using the data from Dakota’s laser scanner. The scanner takes a 3D object, and records 2D cross-sectional slices along the third dimension. Here’s the looping animation file the scanner produced for the crystal:

What sort of three-dimensional shape is the crystal?

No pressure — Dakota Jones, nay, the entire world, is counting on you to locate the Lost Arc and ensure its place in a museum!