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ThatDonGuy
ThatDonGuy
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Thanks for this post from:
Gialmere
December 9th, 2020 at 12:54:32 PM permalink
Quote: Gialmere

It's toughie Tuesday. You're hexed...



A hexagon with consecutive sides of lengths 2, 2, 7, 7, 11, and 11 is inscribed in a circle.

What is the radius of the circle?


Sorry this is a little late...

Let 2a, 2b, and 2c be the measures of the angles opposite sides of 2, 7, and 11. Since there are two of each , 2 (2a + 2b + 2c) = 360 degrees, so c = 90 - (a + b).
Let R be the radius of the circle.
sin a = 2 / 2R; cos 2a = sqrt(4R^2 - 2) / 2R
sin b = 7 / 2R; cos 2b = sqrt(4R^2 - 49) / 2R
sin c = sin (90 - (a + b)) = cos (a + b) = cos a cos b - sin a sin b = 11 / 2R
(sqrt(4R^2 - 2) sqrt(4R^2 - 49) - 14) / 4R^2 = 11 / 2R
44 R^2 = 2R (sqrt(16 R^4 - 212 R^2 + 196) - 14)
22R + 14 = sqrt(16 R^4 - 212 R^2 + 196)
Square both sides, then reduce:
2R^3 - 87 R - 77 = 0 = (R - 7) (2 R^2 + 14 R + 11)
The only real root of this is R = 7, which is the solution

chevy
chevy
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December 9th, 2020 at 6:42:53 PM permalink
ThatDonGuy: Thanks for this, I could not get past my law of cosines approach to actually solve.

Comment in spoiler


I see what you did and follow it through.

Not sure if it matters as others may have moved on...but I think there may be a couple typos, not in the final equation or answer, just in typing up work.....(assuming I really am following your work)

You type cos(2a) and cos(2b) but I think you find cos(a) and cos(b) from sqrt(1-sin())...at least it seems that is what you plug in for.

And then I think cos(a) = sqrt(4R^2 - 2^2) = sqrt(4R^2-4)

That (4R^2-2) repeats two lines later

BUT it disappears because the equation after seems correct. (-212 = -4*49 - 4*4)....so I assume it was just transposing your work.


But it was clear what you intended and I thank you for ending my agony.

Gialmere
Gialmere
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December 9th, 2020 at 6:46:37 PM permalink
Quote: Joeman

Ah, I thought you meant how much starting from GO.

$800

Roll 1: start from Boardwalk and roll a hard 8 (or Park Place with a hard 10 or Short Line with a 12):
Pass Go and collect $200, then Chance:



+$200

Roll 2: 2 -- Community Chest:



+$200

Roll 3: 5 -- Chance:



+ $200 for a total of $800 all from passing Go.


Correct!

As mentioned, there are actually several ways to do this making finding an actual probability tedious. As Joeman points out, you must start on Boardwalk, Park Place or Short Line...

One Way
1) Roll doubles, pass Go ($200), land on Chance 7 and advance to boardwalk.
2) Roll a hard eight, pass Go ($200) land on Chance 7 and advance to Go ($200).
3) Roll 7, land on Chance 7 and take a ride on the Reading ($200).

Another Way
1) Roll doubles, pass Go ($200), land on Chance 7 and take a ride on the Reading ($200).
2) Roll midnight, land on Community Chest 17, advance to Go ($200).
3) Roll 7, land on Chance 7, advance to Go ($200).

I originally wanted to ask what the probability of going to jail on your first turn was (thinking it would be interesting mixing dice with cards) but quickly found it to be a can of worms.

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

If you're in a relationship and you can't quite work up the nerve to break things off, suggest a game of Monopoly.
Have you tried 22 tonight? I said 22.
chevy
chevy
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December 9th, 2020 at 8:07:22 PM permalink
Staying with Monopoly Theme

Question: What is the most money you can lose in a single turn of Monopoly?


(I have an answer, but past examples suggest I am not always right....)
Wizard
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Wizard
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December 10th, 2020 at 5:35:16 AM permalink
Quote: ThatDonGuy


22R + 14 = sqrt(16 R^4 - 212 R^2 + 196)
Square both sides, then reduce:
2R^3 - 87 R - 77 = 0 = (R - 7) (2 R^2 + 14 R + 11)
The only real root of this is R = 7, which is the solution



Since this problem is a bit dated, I'll dispense with the spoiler tags.

That said, how did you know you could make that reduction? I came to a similar equation that I didn't know how to solve without some kind of help.
It's not whether you win or lose; it's whether or not you had a good bet.
Joeman
Joeman
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December 10th, 2020 at 6:20:21 AM permalink
Quote: chevy

Thanks Joeman. As to your bonus puzzle, I can think of


Directly Specified: Illinois, ST. Charles, Boardwalk, Reading
Indirectly specified : 2 nearest Utilities, 4 nearest railroads (only adds 3 to count)

If you include where you can end up, There is also Jail and GO....but you didn't list those as choices.

Would go back three to end on New York count? (I loved owning the oranges for exactly this reason by the way)

So maybe 9 + 2 + 1 = 12??



You pretty much got it. And, yes, by "properties," I was asking specifically for those spaces that can be owned by a player.

1) Illinois Ave -- Via explicit instruction
2) St. Charles Place -- Via explicit instruction
3) Boardwalk -- Via explicit instruction
4) Reading RR -- Via explicit instruction or a "Nearest Railroad" from the chance next to Park Place
5) Pennsylvania RR -- Via a "Nearest Railroad" from the Chance next to Oriental Ave
6) B&O RR -- Via a "Nearest Railroad" from the Chance next to Illinois Ave
7) Electric Company -- Via a "Nearest Utility" from a Chance next to Oriental Ave or next to Park Place
8) Water Works -- Via a "Nearest Utility" from the Chance next to Illinois Ave

The 9th is the one that may be tricky to some folks. It's not Short Line, as SL is never the "Nearest Railroad" from any Chance space. As you stated, it is:

9) New York Ave -- Via a "Go Back 3 Spaces" from the Chance next to Illinois Ave


Quote: chevy

Staying with Monopoly Theme

Question: What is the most money you can lose in a single turn of Monopoly?



I would say $5,500, but you would get $200 for passing Go for a net loss of $5,300. Here's how:

First, you need to assume that both Boardwalk and Park Place have hotels and are owned by one of your opponents. Next, you need to be on B&O, Ventnor, Marvin Gardens, Pacific, Community Chest, or Short Line and roll doubles to land on Park Place (-$1500). Next roll would be a 2, landing on Boardwalk (-$2,000). Third roll would have to be an easy 8, landing on Chance and drawing an "Advance to Boardwalk" card (-$2,000).

This sequence would have you passing Go once, so you would collect $200 during your roll for a net loss of $5,300.


I think we need a separate "Easy Monopoly Puzzles" thread! ;)
"Dealer has 'rock'... Pay 'paper!'"
Wizard
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Wizard
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December 10th, 2020 at 6:56:25 AM permalink
Anyone know of a good site where we can play Monopoly against each other?
It's not whether you win or lose; it's whether or not you had a good bet.
Gialmere
Gialmere
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December 10th, 2020 at 8:16:58 AM permalink
Quote: Wizard

That said, how did you know you could make that reduction? I came to a similar equation that I didn't know how to solve without some kind of help.



I'll post the official solve for the hexagon puzzle later today.
Have you tried 22 tonight? I said 22.
Gialmere
Gialmere
Joined: Nov 26, 2018
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December 10th, 2020 at 8:17:44 AM permalink


"Threes Away (Low)" is a game played with five standard dice. The object is to score the fewest points.

You begin your turn by rolling all five dice. You MUST set aside at least one of them for scoring but you may set aside as many as you wish. 3s score zero points. All other dice score their face values.

You then reroll any remaining dice and repeat the process until all your dice are scored. Low score wins.

Assuming optimal play, what is the expected value of your score?


If I'm not being clear with the rules, this video explains it in less than 90 seconds...
Have you tried 22 tonight? I said 22.
ThatDonGuy
ThatDonGuy
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December 10th, 2020 at 8:32:40 AM permalink
Quote: Wizard

Since this problem is a bit dated, I'll dispense with the spoiler tags.

That said, how did you know you could make that reduction? I came to a similar equation that I didn't know how to solve without some kind of help.


When you square both sides:
484 R^2 + 616 R + 196 = 16 R^4 - 212 R^2 + 196
16 R^4 - 696 R^2 - 616 R = 0
2 R^3 - 87 R - 77 = 0

Okay, at this point I used Newton-Raphson with Excel to find the real positive root of 7.
Since 7 is a root, 2 R^3 - 87 R - 77 = (R - 7) multiplied by (a R^2 + b R + c).
"Obviously," a = 2
The R^2 coefficient = 0 = (-7 x 2) + (1 x b), which means b = 14
The R coefficient = -87 = (1 x c - 7 x b) = c - 98, which means c = 11
Check: (R - 7) (2 R^2 + 14 R + 11) = 2 R^3 + (14 - 14) R^2 + (11 - 98) R + ((-7) x 11) = 2 R^3 - 87 R - 77.

Also, there was a slight error in my statement which did not affect the solution:
2 R^2 + 14 R + 11 = 0
R = (-7 +/- 3 sqrt(3)) / 2
Both of the other roots are real, but neither is positive; 7 is the only positive solution.

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