Easy math puzzles

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July 28th, 2021 at 12:14:48 PM permalink

Going by the repeater rules, a 12 is like a 2 (must be repeated twice), an 11 three times etc

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ThatDonGuy

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July 28th, 2021 at 1:20:04 PM permalink

Quote: Ace2
Going by the repeater rules, a 12 is like a 2 (must be repeated twice), an 11 three times etc

I get about 1 in 185,100

Question for the calculus boffins: I know the probability of not rolling a 2 in time t is 1 / e^(t/36), but what is the probability of not rolling a 2 at least twice in time t?

Ace2

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July 28th, 2021 at 2:18:52 PM permalink

1 / e^(t/36) + (t/36) / e^(t/36)

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Ace2

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July 28th, 2021 at 4:28:19 PM permalink

Quote: ThatDonGuy
I get about 1 in 185,100

I agree with that estimate. The exact answer is:

228,296,650,211,142,223,842,235,175,926,562,819,565,310,943,733,017,295,518,075,498,704,612,736,041,415,500,474,174,790,373,335,885,790,742,403,815,919,759,376,329,015,547,908,450,301,690,458,558,920,417,797,769,549,693,735,380,351,101,698,849,553,401,942,470,326,707,347,462,797,442,543,279,964,206,576,929,140,810,851,250,499,537,038,847,980,599,155,971,380,821,671,855,091,570,073,172,919,198,633,489/

42,257,698,361,772,482,521,904,922,230,217,663,263,013,990,621,490,761,617,970,468,605,585,232,439,497,131,274,911,233,268,994,399,835,192,106,236,921,993,503,595,172,067,323,864,679,436,391,989,136,613,954,326,164,471,151,500,091,945,731,257,099,763,259,926,492,493,587,351,085,501,547,293,599,982,107,240,564,537,220,360,657,324,137,622,979,804,472,478,269,440,000,000,000,000,000,000,000,000,000,000,000,000

Does anyone want more time before I post the method?

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teliot

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July 28th, 2021 at 4:41:53 PM permalink

These are such small numbers, but I do appreciate multi-line integers. I'm personally a big fan of the fast growing hierarchy. :)

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unJon

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July 28th, 2021 at 8:21:36 PM permalink

Quote: Ace2
I agree with that estimate. The exact answer is:

228,296,650,211,142,223,842,235,175,926,562,819,565,310,943,733,017,295,518,075,498,704,612,736,041,415,500,474,174,790,373,335,885,790,742,403,815,919,759,376,329,015,547,908,450,301,690,458,558,920,417,797,769,549,693,735,380,351,101,698,849,553,401,942,470,326,707,347,462,797,442,543,279,964,206,576,929,140,810,851,250,499,537,038,847,980,599,155,971,380,821,671,855,091,570,073,172,919,198,633,489/

42,257,698,361,772,482,521,904,922,230,217,663,263,013,990,621,490,761,617,970,468,605,585,232,439,497,131,274,911,233,268,994,399,835,192,106,236,921,993,503,595,172,067,323,864,679,436,391,989,136,613,954,326,164,471,151,500,091,945,731,257,099,763,259,926,492,493,587,351,085,501,547,293,599,982,107,240,564,537,220,360,657,324,137,622,979,804,472,478,269,440,000,000,000,000,000,000,000,000,000,000,000,000

Does anyone want more time before I post the method?

That is a surprising number of zeros at the end of the denominator. Are you sure it didn’t round that?

The race is not always to the swift, nor the battle to the strong; but that is the way to bet.

Ace2

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July 28th, 2021 at 9:26:40 PM permalink

Quote: unJon
That is a surprising number of zeros at the end of the denominator. Are you sure it didn’t round that?

Yes I’m pretty sure. Even if it was rounded, there’s still about 300 significant digits

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July 29th, 2021 at 9:49:13 AM permalink

The answer is 1 minus the integral from zero to infinity of:

(1 - ((1 - ((5x/36)^5/120 + (5x/36)^4/24 + (5x/36)^3/6 + (5x/36)^2/2 + (5x/36) + 1) * e^(-5x/36)) * (1 - ((x/9)^4/24 + (x/9)^3/6 + (x/9)^2/2 + (x/9) + 1) * e^(-x/9)) * (1 - ((x/12)^3/6 + (x/12)^2/2 + (x/12) + 1) * e^(-x/12)) * (1 - ((x/18)^2/2 + (x/18) + 1) * e^(-x/18)) * (1 - ((x/36) + 1) * e^(-x/36)))^2) * e^(-x/6) * 1/6 dx

The integral evaluates the probabilities at all time (x) that at least one repeater bet has not been won and a seven has not been rolled. Multiply by 1/6 for the chance that the next roll is a seven, causing the bet to lose.

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ThatDonGuy

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July 30th, 2021 at 9:14:41 AM permalink

Quote: Ace2
The answer is 1 minus the integral from zero to infinity of:

(1 - ((1 - ((5x/36)^5/120 + (5x/36)^4/24 + (5x/36)^3/6 + (5x/36)^2/2 + (5x/36) + 1) * e^(-5x/36)) * (1 - ((x/9)^4/24 + (x/9)^3/6 + (x/9)^2/2 + (x/9) + 1) * e^(-x/9)) * (1 - ((x/12)^3/6 + (x/12)^2/2 + (x/12) + 1) * e^(-x/12)) * (1 - ((x/18)^2/2 + (x/18) + 1) * e^(-x/18)) * (1 - ((x/36) + 1) * e^(-x/36)))^2) * e^(-x/6) * 1/6 dx

The integral evaluates the probabilities at all time (x) that at least one repeater bet has not been won and a seven has not been rolled. Multiply by 1/6 for the chance that the next roll is a seven, causing the bet to lose.

I assume that ((5x/36)^5/120 + (5x/36)^4/24 + (5x/36)^3/6 + (5x/36)^2/2 + (5x/36) + 1) * e^(-5x/36) is the probability of not rolling a 6 at least 6 times in time x. How did you work that out?

Quote: unJon
Quote: Ace2
I agree with that estimate. The exact answer is:

228,296,650,211,142,223,842,235,175,926,562,819,565,310,943,733,017,295,518,075,498,704,612,736,041,415,500,474,174,790,373,335,885,790,742,403,815,919,759,376,329,015,547,908,450,301,690,458,558,920,417,797,769,549,693,735,380,351,101,698,849,553,401,942,470,326,707,347,462,797,442,543,279,964,206,576,929,140,810,851,250,499,537,038,847,980,599,155,971,380,821,671,855,091,570,073,172,919,198,633,489/

42,257,698,361,772,482,521,904,922,230,217,663,263,013,990,621,490,761,617,970,468,605,585,232,439,497,131,274,911,233,268,994,399,835,192,106,236,921,993,503,595,172,067,323,864,679,436,391,989,136,613,954,326,164,471,151,500,091,945,731,257,099,763,259,926,492,493,587,351,085,501,547,293,599,982,107,240,564,537,220,360,657,324,137,622,979,804,472,478,269,440,000,000,000,000,000,000,000,000,000,000,000,000

That is a surprising number of zeros at the end of the denominator. Are you sure it didn’t round that?

My Markov chain result is the same, right down to the same number of zeroes in the denominator.

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July 30th, 2021 at 9:29:19 AM permalink

Quote: ThatDonGuy
I assume that ((5x/36)^5/120 + (5x/36)^4/24 + (5x/36)^3/6 + (5x/36)^2/2 + (5x/36) + 1) * e^(-5x/36) is the probability of not rolling a 6 at least 6 times in time x. How did you work that out?
.

It’s just Poisson.

e^(-5x/36) * (5x/36)^5/120 is the chance of rolling five 6s in time x, e^(-5x/36) * (5x/36)^4/24 is the chance of rolling four 6s in time x, etc. Sum these for values 0 - 5 and that’s the probability of rolling less than six 6’s.

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Ace2

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August 3rd, 2021 at 5:39:52 PM permalink

When someone wins the Fire Bet, how many total pass bets will they have won, on average?

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Ace2

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August 4th, 2021 at 9:46:20 AM permalink

Quote: Ace2
When someone wins the Fire Bet, how many total pass bets will they have won, on average?

Rephrase: All fire bet winners will have won at least six points before sevening out. When someone wins the fire bet, what is the expected number of total points won?

Last edited by: Ace2 on Aug 4, 2021

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ThatDonGuy

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August 4th, 2021 at 2:06:41 PM permalink

Quote: Ace2
Rephrase: All fire bet winners will have won at least six points before sevening out. When someone wins the fire bet, what is the expected number of total points won?

Tough one!

I tried a Markov chain, where the probability of making a particular point was in proportion to establishing that point, but I got something like 17 - while a simulation returns around 7.4.

Presumably, it's low because the higher the number of points that are established, the more likely it is that you will seven out.

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August 7th, 2021 at 2:24:26 PM permalink

Two ordinary six-sided dice are rolled until one of the following events happen:

A total of seven is rolled
A total of three is rolled 3x
A total of five is rolled 5x
A total of nine is rolled 5x
A total of 11 is rolled 3x

What is the probability the ending event is any of the last four listed?

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Ace2

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August 7th, 2021 at 3:43:42 PM permalink

1 - 30026163533 / 31381059609 =~ 4.32%

Last edited by: Ace2 on Aug 7, 2021

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ThatDonGuy

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August 8th, 2021 at 3:04:06 PM permalink

Quote: Ace2
1 - 30026163533 / 31381059609 =~ 4.32%

How about that - I actually managed to figure out the Poisson-based method for this one:

Only rolls of 3, 5, 7, 9, and 11 are concerned here, so they have probabilities of 1/9, 2/9, 1/3, 2/9, and 1/9 respectively

Note that all of these are derivatives from 0 to positive infinity, dx
In time x:
P(< 3 3s) = P(< 3 11s) = e^(-x/9) * (1 + x/9 + (x/9)^2 / 2)
P(< 5 5s) = P(< 5 9s) = e^(-2x/9) * (1 + 2x/9 + (2x/9)^2 / 2 + (2x/9)^3 / 6) + (2x/9)^4 / 24)
P(no 7s) = e^(-x/3)
P(all five) = e^(-x/3) * (e^(-x/9) * (1 + x/9 + (x/9)^2 / 2) * e^(-2x/9) * (1 + 2x/9 + (2x/9)^2 / 2 + (2x/9)^3 / 6) + (2x/9)^4 / 24))^2
= e^(-x) * ((1 + x/9 + (x/9)^2 / 2) * (1 + 2x/9 + (2x/9)^2 / 2 + (2x/9)^3 / 6 + (2x/9)^4 / 24))^2
P(rolling a 7 first) = P(all five, and then a 7) = e^(-x) * ((1 + x/9 + (x/9)^2 / 2) * (1 + 2x/9 + (2x/9)^2 / 2 + (2x/9)^3 / 6 + (2x/9)^4 / 24))^2

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August 9th, 2021 at 5:28:07 AM permalink

Quote: Ace2
1 - 30026163533 / 31381059609 =~ 4.32%

I agree! I show a brief solution of all the difficult Repeater Bet Plus bets here.

"For with much wisdom comes much sorrow." -- Ecclesiastes 1:18 (NIV)

Ace2

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August 16th, 2021 at 5:39:22 PM permalink

Quote: Ace2
Rephrase: All fire bet winners will have won at least six points before sevening out. When someone wins the fire bet, what is the expected number of total points won?

The exact answer is:

51,194,997,861,765,388,097,526,966,955,556,874,130,535,332,020,426,690,698,014,074,625,318,137,025,571,849,255 /

6,923,708,358,532,324,948,598,712,995,072,892,018,553,301,246,118,708,475,186,839,062,876,563,686,189,522,216

=~7.39415862291138014289 average total points won

Would any of our calculus ninjas like more time before I post the solution?

Last edited by: Ace2 on Aug 16, 2021

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ThatDonGuy

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August 16th, 2021 at 7:12:29 PM permalink

Quote: Ace2
The exact answer is:

51,194,997,861,765,388,097,526,966,955,556,874,130,535,332,020,426,690,698,014,074,625,318,137,025,571,849,255 /
6,923,708,358,532,324,948,598,712,995,072,892,018,553,301,246,118,708,475,186,839,062,876,563,686,189,522,216

=~7.39415862291138014289 average total points won

Would any of our calculus ninjas like more time before I post the solution?

Don't look at me; I'm surprised there is a rational number solution - I thought it would end up being an infinite series (calculate the probabilities of winning with 6, 7, 8, 9, ... points, then add them together).

Ace2

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August 16th, 2021 at 8:19:07 PM permalink

Okay let’s see if the Wizard wants to try.

If you know how to calculate the house edge of the fire bet using calculus, then you already have almost the entire solution. It’s just a slightly different approach

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Ace2

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August 17th, 2021 at 11:47:04 AM permalink

Quote: Ace2
Rephrase: All fire bet winners will have won at least six points before sevening out. When someone wins the fire bet, what is the expected number of total points won?

First you calculate the probability of winning the fire bet. Though this has already been posted, I'll recap as follows. Out of 3,960 possible outcomes/decisions, there are 165 ways to win a 4 (same to win a 10), 264 ways to win a 5 (same to win a 9), 375 ways to win a 6 (same to win an 8) and 2,352 ways to seven out.

The "indirect" way (counting all winning strings regardless of when they win) to calculate the edge is by integrating the following from zero to infinity:

((1 - e^(-165x/3960))^2 * (1 - e^(-264x/3960))^2 * (1 - e^(-375x/3960))^2 * e^(-2352x/3960) * 2352/3960 dx

The "direct" way (counting all strings at the exact time they win) to calculate the edge is by integrating the following from zero to infinity:

((1 - e^(-165x/3960))^2 * (1 - e^(-264x/3960))^2 * (1 - e^(-375x/3960)) * e^(-375x/3960) * 375/3960 + (1 - e^(-165x/3960))^2 * ( 1 - e^(-375x/3960))^2 * ( 1 -e^(-264x/3960)) * e^(-264x/3960) * 264/3960 + (1 - e^(-375x/3960))^2 * (1 - e^(-264x/3960))^2 * (1 - e^(-165x/3960)) * e^(-165x/3960) * 165/3960) * e^(-2352x/3960) * 2 dx

Both methods give the same probability of winning of ~0.0001624.

Next, take the same direct integral from zero to infinity but also multiply by x. Using the indirect integral will not give the correct answer. So :

((1 - e^(-165x/3960))^2 * (1 - e^(-264x/3960))^2 * (1 - e^(-375x/3960)) * e^(-375x/3960) * 375/3960 + (1 - e^(-165x/3960))^2 * ( 1 - e^(-375x/3960))^2 * ( 1 -e^(-264x/3960)) * e^(-264x/3960) * 264/3960 + (1 - e^(-375x/3960))^2 * (1 - e^(-264x/3960))^2 * (1 - e^(-165x/3960)) * e^(-165x/3960) * 165/3960) * e^(-2352x/3960) * 2 * x dx

Since this integral sums the individual probabilities of winning the fire bet at all times(x) and then multiplies by time (x), it gives us the weighed average of ~0.001201 decisions to win the fire bet. That is the average all of all outcomes, including the losing ones. We are only interested in outcomes where the fire bet was won, so take 0.001201 divided by 0.0001624 (the probability of winning) to get ~7.39 average total decisions (points won) when the winning the fire bet. The exact answer is:

51,194,997,861,765,388,097,526,966,955,556,874,130,535,332,020,426,690,698,014,074,625,318,137,025,571,849,255/

6,923,708,358,532,324,948,598,712,995,072,892,018,553,301,246,118,708,475,186,839,062,876,563,686,189,522,216

Last edited by: Ace2 on Aug 17, 2021

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ThatDonGuy

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August 18th, 2021 at 9:27:05 AM permalink

Who wants a rather lengthy explanation of how to calculate the exact (rational number) solution to these integrals?

I will use the "direct method fire bet" integral in the example:
(
(1 - e^(-165x/3960))^2 * (1 - e^(-264x/3960))^2 * (1 - e^(-375x/3960)) * e^(-375x/3960) * 375/3960
+ (1 - e^(-165x/3960))^2 * ( 1 - e^(-375x/3960))^2 * ( 1 -e^(-264x/3960)) * e^(-264x/3960) * 264/3960
+ (1 - e^(-375x/3960))^2 * (1 - e^(-264x/3960))^2 * (1 - e^(-165x/3960)) * e^(-165x/3960) * 165/3960
) * e^(-2352x/3960) * 2 dx

Let y = e^(-x/3960): the integral becomes
2 *
(
(1 - y^165)^2 * (1 - y^264)^2 * (1 - y^375) * y^375 * 375/3960
+ (1 - y^165)^2 * (1 - y^375)^2 * (1 - y^264) * y^264 * 264/3960
+ (1 - y^375)^2 * (1 - y^264)^2 * (1 - y^165) * y^165 * 165/3960
)
* y^2352 dx

Expand all of the squares, and take 1/3960 out of the sum:
2 / 3960 *
(
(1 - 2 y^165 + y^330) * (1 - 2 y^264 + y^528) * (1 - y^375) * y^375 * 375
+ (1 - 2 y^165 + y^330) * (1 - 2 y^375 + y^750) * (1 - y^264) * y^264 * 264
+ (1 - 2 y^375 + y^750) * (1 - 2 y^264 + y^528) * (1 - y^165) * y^165 * 165
)
* y^2352 dx

I will not bore you with calculating the expansion (hint: use Excel, and a lot of cutting and pasting), but this is:
1 / 1980 *
(
165 y^2517
+ 264 y^2616
- 165 y^2682
+ 375 y^2727
- 528 y^2781 - 330 y^2781
- 264 y^2880
- 750 y^2892 - 330 y^2892
+ 264 y^2946 + 330 y^2946
- 750 y^2991 - 528 y^2991
+ 528 y^3045 + 165 y^3045
+ 375 y^3057 + 330 y^3057
- 375 y^3102
+ 1500 y^3156 + 1056 y^3156 + 660 y^3156
- 264 y^3210 - 165 y^3210
+ 375 y^3255 + 528 y^3255
+ 750 y^3267 + 165 y^3267
- 750 y^3321 - 528 y^3321 - 660 y^3321
+ 750 y^3366 + 264 y^3366
- 750 y^3420 - 1056 y^3420 - 330 y^3420
- 375 y^3432 - 165 y^3432
- 1500 y^3531 - 528 y^3531 - 330 y^3531
+ 375 y^3585 + 528 y^3585 + 330 y^3585
- 375 y^3630 - 264 y^3630
+ 750 y^3696 + 264 y^3696 + 330 y^3696
+ 750 y^3795 + 528 y^3795 + 165 y^3795
- 375 y^3960 - 264 y^3960 - 165 y^3960
)

"Now what?"
d/dx (e^(nx)) = e^(nx) / n
y^n = e^(x * (-n/3960)), so d/dx (y^n) = e^(x * (-n/3960)) * (-3960/n)
For positive n, at x = +INF, this is 0; at x = 0, this is -3960/n, so the integral over x = 0 to +INF = 3960/n

Replace each "a y^b" with "3960 * b / a"; you can take 3960 out of each numerator and multiply the 1/1980 term by 3960 to get 2.
The solution becomes
2 *
(
165 / 2517
+ 264 / 2616
- 165 / 2682
+ 375 / 2727
- 858 / 2781
- 264 / 2880
- 1080 / 2892
+ 594 / 2946
- 1278 / 2991
+ 693 / 3045
+ 705 / 3057
- 375 / 3102
+ 3216 / 3156
- 429 / 3210
+ 903 / 3255
+ 915 / 3267
- 1938 / 3321
+ 1014 / 3366
- 2136 / 3420
- 540 / 3432
- 2358 / 3531
+ 1233 / 3585
- 639 / 3630
+ 1344 / 3696
+ 1443 / 3795
- 804 / 3960
)

Easy, right?
The sum is done one term at a time (i.e. start with the first two, then add the third, then add the fourth, and so on), using:
a / b + c / d = (ad + bc) / bd
No, you can't do this in Excel, as you will soon run out of digits of precision.

Wizard
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August 24th, 2021 at 3:19:59 PM permalink

My apologies if business is not concluded on Don's problem. Here is something easier.

Questions:
1. What is the area of the red rectangle*?
2. What is the least possible area of the entire figure?

In case it's not clear, the height of the yellow triangle is 4 and the base of the blue triangle is 5.

* Corrected from square

Last edited by: Wizard on Aug 24, 2021

"For with much wisdom comes much sorrow." -- Ecclesiastes 1:18 (NIV)

ThatDonGuy

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August 24th, 2021 at 3:26:44 PM permalink

Quote: Wizard
My apologies if business is not concluded on Don's problem. Here is something easier.

Questions:
1. What is the area of the red square?
2. What is the least possible area of the entire figure?

In case it's not clear, the height of the yellow triangle is 4 and the base of the blue triangle is 5.

Clarify: the red portion is supposed to be a square, and not just a rectangle. The way it appears, it is not a square.
Or perhaps question 1 requires that it be a square, but question 2 requires only that it be a rectangle?

Also, my "problem" was not a problem, but an explanation as to how to calculate the exact rational number solution of a Poisson-based probability problem, rather than depending on, e.g., Integral Calculator to do it for you.

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August 24th, 2021 at 3:39:41 PM permalink

Quote: ThatDonGuy
Clarify: the red portion is supposed to be a square, and not just a rectangle.

Yes, I should have said red rectangle.

"For with much wisdom comes much sorrow." -- Ecclesiastes 1:18 (NIV)

ThatDonGuy

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August 24th, 2021 at 4:29:09 PM permalink

Well, the area of the red rectangle could be anything, so #1 doesn't have a single answer. Assuming for just problem #1 that it is a square:

Let b be the "missing" length of a leg of the blue triangle, and y the "missing" length of a leg of the yellow one.

Both problems depend on the fact that the blue triangle, the yellow triangle, and the large triangle are all similar; 4 / y = b / 5 = (4 + b) / (5 + y).

For #1, y = b, so 4 / b = b / 5, which means b = 2 sqrt(5), and the area of the large triangle = 1/2 (4 + 2 sqrt(5)) (5 + 2 sqrt(5)) = 20 + 9 sqrt(5)

For #2, 4 / y = b / 5, so y = 20 / b. The area of the triangle is 1/2 (4 + b) (5 + y) = 1/2 (4 + b) (5 + 20/b) = 20 + 5/2 b + 40/b.
Let A denote the area; the first derivative with respect to b, dA / db = 5/2 - 40/b^2. This equals zero when b^2 = 16; since b > 0, b = 4.
Note that the second derivative, d^2A / db^2 = 80/b^3, which is positive when b = 4, so A is a minimum at b = 4.
b = 4, so y = 5, and the minimum area = 1/2 (4 + 4) (5 + 5) = 40.

Note that there is no maximum area, as 20 + 5/2 b + 40/b approaches infinity as b approaches infinity.

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August 24th, 2021 at 4:40:03 PM permalink

Quote: ThatDonGuy
Well, the area of the red rectangle could be anything

I disagree. It has a specific answer.

If the area of the rectangle were a billion, you couldn't inscribe it a right triangle with the 4 and 5 distances given. If you feel otherwise, please give me the dimensions of a the rectangle with area one billion.

"For with much wisdom comes much sorrow." -- Ecclesiastes 1:18 (NIV)

ssho88

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August 24th, 2021 at 5:01:02 PM permalink

Quote: Wizard
I disagree. It has a specific answer.

If the area of the rectangle were a billion, you couldn't inscribe it a right triangle with the 4 and 5 distances given. If you feel otherwise, please give me the dimensions of a the rectangle with area one billion.

1) x= width of rectangle, y = height of rectangle, 4/x=y/5, xy = 20

2) Entire figure area, A = (4+y) * (5 +x)/2 = (20 + 4x + 5y + xy)/2, and xy = 20

So, A = 20 + 2x + 50/x

dA/dx = 2 - 50/x^2 = 0

x= 5

Minimum A = 20 + 2(5) + 50/5 = 40

What if x=0, y= 0 ? still applicable to your question ?

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August 24th, 2021 at 5:01:22 PM permalink

Quote: Wizard
I disagree. It has a specific answer.

If the area of the rectangle were a billion, you couldn't inscribe it a right triangle with the 4 and 5 distances given. If you feel otherwise, please give me the dimensions of a the rectangle with area one billion.

Oh, you're looking for the area of the rectangle. I misread it - I thought you were looking for the area of the large triangle.
In that case...

As stated in my earlier solution, if b is the length of the undetermined leg of the blue triangle and y the length of the undetermined length of the yellow triangle, then 4 / y = b / 5, so by = 20. Since the rectangle has height b and width y, its area is yb, or 20.

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August 24th, 2021 at 5:49:31 PM permalink

Quote: ssho88

1) x= width of rectangle, y = height of rectangle, 4/x=y/5, xy = 20

2) Entire figure area, A = (4+y) * (5 +x)/2 = (20 + 4x + 5y + xy)/2, and xy = 20

So, A = 20 + 2x + 50/x

dA/dx = 2 - 50/x^2 = 0

x= 5

Minimum A = 20 + 2(5) + 50/5 = 40

What if x=0, y= 0 ? still applicable to your question ?

I agree to both!

"For with much wisdom comes much sorrow." -- Ecclesiastes 1:18 (NIV)

ssho88

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August 24th, 2021 at 7:16:37 PM permalink

Quote: Wizard
I agree to both!

What if x=0, y= 0 ? still applicable to your question ?

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August 24th, 2021 at 7:28:07 PM permalink

Quote: ssho88
What if x=0, y= 0 ? still applicable to your question ?

If you want to be pedantic, if x and y = 0, it's no longer a rectangle, but a point. Even if just one of them = 0, it's a line segment.

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August 24th, 2021 at 8:51:06 PM permalink

Quote: ThatDonGuy
Who wants a rather lengthy explanation of how to calculate the exact (rational number) solution to these integrals?

I will use the "direct method fire bet" integral in the example:
(
(1 - e^(-165x/3960))^2 * (1 - e^(-264x/3960))^2 * (1 - e^(-375x/3960)) * e^(-375x/3960) * 375/3960
+ (1 - e^(-165x/3960))^2 * ( 1 - e^(-375x/3960))^2 * ( 1 -e^(-264x/3960)) * e^(-264x/3960) * 264/3960
+ (1 - e^(-375x/3960))^2 * (1 - e^(-264x/3960))^2 * (1 - e^(-165x/3960)) * e^(-165x/3960) * 165/3960
) * e^(-2352x/3960) * 2 dx

Let y = e^(-x/3960): the integral becomes
2 *
(
(1 - y^165)^2 * (1 - y^264)^2 * (1 - y^375) * y^375 * 375/3960
+ (1 - y^165)^2 * (1 - y^375)^2 * (1 - y^264) * y^264 * 264/3960
+ (1 - y^375)^2 * (1 - y^264)^2 * (1 - y^165) * y^165 * 165/3960
)
* y^2352 dx

Expand all of the squares, and take 1/3960 out of the sum:
2 / 3960 *
(
(1 - 2 y^165 + y^330) * (1 - 2 y^264 + y^528) * (1 - y^375) * y^375 * 375
+ (1 - 2 y^165 + y^330) * (1 - 2 y^375 + y^750) * (1 - y^264) * y^264 * 264
+ (1 - 2 y^375 + y^750) * (1 - 2 y^264 + y^528) * (1 - y^165) * y^165 * 165
)
* y^2352 dx

I will not bore you with calculating the expansion (hint: use Excel, and a lot of cutting and pasting), but this is:
1 / 1980 *
(
165 y^2517
+ 264 y^2616
- 165 y^2682
+ 375 y^2727
- 528 y^2781 - 330 y^2781
- 264 y^2880
- 750 y^2892 - 330 y^2892
+ 264 y^2946 + 330 y^2946
- 750 y^2991 - 528 y^2991
+ 528 y^3045 + 165 y^3045
+ 375 y^3057 + 330 y^3057
- 375 y^3102
+ 1500 y^3156 + 1056 y^3156 + 660 y^3156
- 264 y^3210 - 165 y^3210
+ 375 y^3255 + 528 y^3255
+ 750 y^3267 + 165 y^3267
- 750 y^3321 - 528 y^3321 - 660 y^3321
+ 750 y^3366 + 264 y^3366
- 750 y^3420 - 1056 y^3420 - 330 y^3420
- 375 y^3432 - 165 y^3432
- 1500 y^3531 - 528 y^3531 - 330 y^3531
+ 375 y^3585 + 528 y^3585 + 330 y^3585
- 375 y^3630 - 264 y^3630
+ 750 y^3696 + 264 y^3696 + 330 y^3696
+ 750 y^3795 + 528 y^3795 + 165 y^3795
- 375 y^3960 - 264 y^3960 - 165 y^3960
)

"Now what?"
d/dx (e^(nx)) = e^(nx) / n
y^n = e^(x * (-n/3960)), so d/dx (y^n) = e^(x * (-n/3960)) * (-3960/n)
For positive n, at x = +INF, this is 0; at x = 0, this is -3960/n, so the integral over x = 0 to +INF = 3960/n

Replace each "a y^b" with "3960 * b / a"; you can take 3960 out of each numerator and multiply the 1/1980 term by 3960 to get 2.
The solution becomes
2 *
(
165 / 2517
+ 264 / 2616
- 165 / 2682
+ 375 / 2727
- 858 / 2781
- 264 / 2880
- 1080 / 2892
+ 594 / 2946
- 1278 / 2991
+ 693 / 3045
+ 705 / 3057
- 375 / 3102
+ 3216 / 3156
- 429 / 3210
+ 903 / 3255
+ 915 / 3267
- 1938 / 3321
+ 1014 / 3366
- 2136 / 3420
- 540 / 3432
- 2358 / 3531
+ 1233 / 3585
- 639 / 3630
+ 1344 / 3696
+ 1443 / 3795
- 804 / 3960
)

Easy, right?
The sum is done one term at a time (i.e. start with the first two, then add the third, then add the fourth, and so on), using:
a / b + c / d = (ad + bc) / bd
No, you can't do this in Excel, as you will soon run out of digits of precision.

It’s very useful to see that some of these can be hashed out manually. However, if you need to take the integral of, for instance, 1-(1-(1+(x/6)+(x/6)^2/2+ (x/6)^3/6+ (x/6)^4/24+ (x/6)^5/120)*e^(-x/6))^6 or 1-(1-1/e^(x/38))^38, would it be feasible to do it manually ?

Incidentally, the result is the expected number of rolls to hit all six sides of a single die at least six times and the expected number of spins to hit all 38 numbers on a roulette wheel at least once, respectively

It’s all about making that GTA

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August 25th, 2021 at 2:59:36 AM permalink

Quote: ssho88
What if x=0, y= 0 ? still applicable to your question ?

No, because it wouldn't be inscribed in the triangle.

"For with much wisdom comes much sorrow." -- Ecclesiastes 1:18 (NIV)

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August 25th, 2021 at 10:17:01 AM permalink

Quote: Ace2
It’s very useful to see that some of these can be hashed out manually. However, if you need to take the integral of, for instance, 1-(1-(1+(x/6)+(x/6)^2/2+ (x/6)^3/6+ (x/6)^4/24+ (x/6)^5/120)*e^(-x/6))^6 or 1-(1-1/e^(x/38))^38, would it be feasible to do it manually ?

Who said anything about "feasible"? Just integrating something like x e^(-x/6) is a pain.
Integrate by parts:
u = x; dv = e^(-x/6) dx
du = dx; v = -6 e^(-x/6)
The integral = -6x e^(-x/6) - the integral of -6 e^(-x/6) dx = -6x e^(-x/6) + 6 (-6 e^(x/6)) = -(6x + 36) e^(-x/6) = -6 (x + 6) / e^(x/6)
As x approaches positive infinity, the numerator and denominator approach positive infinity, so apply L'Hopital's Rule; the limit of 1 / (6 e^(x/6)) = 0, so the definite integral over x from 0 to positive infinity = 0 - (-6 (0 + 6) / e^0)) = 36.

teliot

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August 26th, 2021 at 9:44:30 AM permalink

What is the largest solution you can find of the equation:

1 + 2 + 3 + ... + N = (N+1) + (N+2) + ... (N+K)?

I have solutions so far for N = 2, 14, 84, 492, 2870, 16730 and 97512.

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teliot

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August 26th, 2021 at 10:42:38 AM permalink

Quote: teliot
What is the largest solution you can find of the equation:

1 + 2 + 3 + ... + N = (N+1) + (N+2) + ... (N+K)?

I have solutions so far for N = 2, 14, 84, 492, 2870, 16730 and 97512.

And now I just ruined it for myself by looking up the sequence on OEIS.org :) Well, it was fun anyway for a few minutes.

Climate Casino: https://climatecasino.net/climate-casino/

charliepatrick

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August 26th, 2021 at 11:47:18 AM permalink

Quote: Wizard
...I agree to both!

I didn't see the solution beforehand...

Just coming to this but happened to fall over the solution quite quickly. The way I saw it was to use similar triangles which gives 4/x = y/5 leading to xy=20.
The second idea is then to flip the 5-wide triangle to lie on top of the diagonal line. Construct a rectangle 4x5 (to the right of the 4-side). Where x is less than 5, the top triangle extends below this rectangle. Hence the total area of the original triangle exceeds 40. Similar logic applies where x exceeds 5 except the 4-side triangle extends beyond the rectangle. Hence 40 is the minimum.

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August 31st, 2021 at 7:39:40 PM permalink

Which method of stacking cannonballs is the most efficient -- a pyramid with a square base or triangular?

For full credit, I'd like to the degree of efficiency (meaning ratio of the volume of cannonballs to to the volume of the whole pyramid), both ways, as the number of cannonballs approaches infinity?

There will probably be an extra credit problem coming up, but one thing at a time.

"For with much wisdom comes much sorrow." -- Ecclesiastes 1:18 (NIV)

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August 31st, 2021 at 7:53:45 PM permalink

For efficiency calculation purposes, how is the "volume of the whole pyramid" measured - the volume of the circumscribing pyramid (I just had a 40-year-old flashback to an AMC exam problem I had in high school), or the volume of the pyramid whose vertices are the centers of the golf balls at each vertex?

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September 1st, 2021 at 6:26:21 AM permalink

Quote: ThatDonGuy
For efficiency calculation purposes, how is the "volume of the whole pyramid" measured - the volume of the circumscribing pyramid (I just had a 40-year-old flashback to an AMC exam problem I had in high school), or the volume of the pyramid whose vertices are the centers of the golf balls at each vertex?

As the number of cannonballs approaches infinity, it shouldn't matter how exactly the volume is of the pyramid is measured.

However, if you must have an answer, let's keep it simple and say the length of a side of the base is the sum of the diameters of the cannonballs in a that side. For the height and slant height, use the same logic, keeping the angles in the pyramid correct.

"For with much wisdom comes much sorrow." -- Ecclesiastes 1:18 (NIV)

ssho88

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September 1st, 2021 at 7:23:10 AM permalink

Quote: Wizard
As the number of cannonballs approaches infinity, it shouldn't matter how exactly the volume is of the pyramid is measured.

However, if you must have an answer, let's keep it simple and say the length of a side of the base is the sum of the diameters of the cannonballs in a that side. For the height and slant height, use the same logic, keeping the angles in the pyramid correct.

r = radius of golf ball.

For square base, imagine a ball put on top of 4 balls and you join the center of five balls to form a square base pyramid, the vertical height of square base pyramid = 2^0.5 * r = 1.414r, and the slant distance = 2r.

For triangular base, imagine a ball put on top of 3 balls and you join the center of four balls to form a triangular base pyramid, the vertical height of triangular base pyramid = 2/3 * 6^0.5 * r = 1.633r, and the slant distance = 2r.

From there, we can calculate the total vertical height of N layers of balls, Am I missing something ?

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September 1st, 2021 at 7:47:19 AM permalink

Assume the "volume of the shape" is the volume of the tetrahedron/pyramid whose vertices are the centers of the cannonballs at the vertices.
Also assume each cannonball has radius 1.

If there are N levels:
The number of cannonballs in the tetrahedron is N (N^2 + 3 N + 2) / 6
The combined volume of the cannonballs is N (N^2 + 3 N + 2) / 6 * 4 PI / 3
= 2 PI (N^3 + 3 N^2 + 2 N) / 9

The tetrahedron is a regular tetrahedron of side length 2 (N - 1)
The volume is (2 (N - 1))^3 / (6 sqrt(2))
= 2 sqrt(2) / 3 * (N^3 - 3 N^2 + 3 N - 1)

The number of cannonballs in the pyramid is N (2 N^2 + 3 N + 1) / 6
The combined volume of the cannonballs is N (2 N^2 + 3 N + 1) / 6 * 4 PI / 3
= 2 PI (2 N^3 + 3 N^2 + N) / 9

The pyramid has sides of length 2 (N - 1)
The area of the base is 4 (N - 1)^2, and the height is (N - 1) sqrt(2),
The volume is 4 (N - 1)^2 * (N - 1) sqrt(2) / 3
= 4 sqrt(2) * (N^3 - 3 N^2 + 3 N - 1) / 3

The tetrahedron's efficiency is
(2 PI (N^3 + 3 N^2 + 2 N) / 9) / (2 sqrt(2) * (N^3 - 3 N^2 + 3 N - 1) / 3)
= PI / (3 sqrt(2)) * (N^3 + 3 N^2 + 2 N) / (N^3 - 3 N^2 + 3 N - 1)
As N approaches infinity, this approaches PI * sqrt(2) / 6

The pyramid's efficiency is
(2 PI (2 N^3 + 3 N^2 + N) / 9) / (4 sqrt(2) * (N^3 - 3 N^2 + 3 N - 1) / 3)
= PI / (3 sqrt(2)) * (N^3 + 3/2 N^2 + 1/2 N) / (N^3 - 3 N^2 + 3 N - 1)
As N approaches infinity, this also approaches PI * sqrt(2) / 6

The two have equal efficiency - specifically, PI sqrt(2) / 6

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September 1st, 2021 at 7:56:32 AM permalink

Quote: Wizard
Which method of stacking cannonballs is the most efficient -- a pyramid with a square base or triangular?

For full credit, I'd like to the degree of efficiency (meaning ratio of the volume of cannonballs to to the volume of the whole pyramid), both ways, as the number of cannonballs approaches infinity?

There will probably be an extra credit problem coming up, but one thing at a time.

Good problem. One of my calc professors in college solved the Kepler conjecture (ie, what’s the most efficient way to pack spheres). When I was in his class was before he solved it, but he was obsessed with it. Talked about it during multiple classes. He had reduced the problem to a differential equation but was unable to integrate it. A few years later he managed to find the integral and solved it.

The race is not always to the swift, nor the battle to the strong; but that is the way to bet.

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September 1st, 2021 at 8:42:07 AM permalink

Quote: ThatDonGuy

Assume the "volume of the shape" is the volume of the tetrahedron/pyramid whose vertices are the centers of the cannonballs at the vertices.
Also assume each cannonball has radius 1.

If there are N levels:
The number of cannonballs in the tetrahedron is N (N^2 + 3 N + 2) / 6
The combined volume of the cannonballs is N (N^2 + 3 N + 2) / 6 * 4 PI / 3
= 2 PI (N^3 + 3 N^2 + 2 N) / 9

The tetrahedron is a regular tetrahedron of side length 2 (N - 1)
The volume is (2 (N - 1))^3 / (6 sqrt(2))
= 2 sqrt(2) / 3 * (N^3 - 3 N^2 + 3 N - 1)

The number of cannonballs in the pyramid is N (2 N^2 + 3 N + 1) / 6
The combined volume of the cannonballs is N (2 N^2 + 3 N + 1) / 6 * 4 PI / 3
= 2 PI (2 N^3 + 3 N^2 + N) / 9

The pyramid has sides of length 2 (N - 1)
The area of the base is 4 (N - 1)^2, and the height is (N - 1) sqrt(2),
The volume is 4 (N - 1)^2 * (N - 1) sqrt(2) / 3
= 4 sqrt(2) * (N^3 - 3 N^2 + 3 N - 1) / 3

The tetrahedron's efficiency is
(2 PI (N^3 + 3 N^2 + 2 N) / 9) / (2 sqrt(2) * (N^3 - 3 N^2 + 3 N - 1) / 3)
= PI / (3 sqrt(2)) * (N^3 + 3 N^2 + 2 N) / (N^3 - 3 N^2 + 3 N - 1)
As N approaches infinity, this approaches PI * sqrt(2) / 6

The pyramid's efficiency is
(2 PI (2 N^3 + 3 N^2 + N) / 9) / (4 sqrt(2) * (N^3 - 3 N^2 + 3 N - 1) / 3)
= PI / (3 sqrt(2)) * (N^3 + 3/2 N^2 + 1/2 N) / (N^3 - 3 N^2 + 3 N - 1)
As N approaches infinity, this also approaches PI * sqrt(2) / 6

The two have equal efficiency - specifically, PI sqrt(2) / 6

I agree!

"For with much wisdom comes much sorrow." -- Ecclesiastes 1:18 (NIV)

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September 3rd, 2021 at 3:15:52 PM permalink

The above figure contains 10 squares inscribed in a rectangle. The white square has side length of 1. What is the area of the entire rectangle?

Please, no cheating with rulers.

"For with much wisdom comes much sorrow." -- Ecclesiastes 1:18 (NIV)

charliepatrick

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September 3rd, 2021 at 3:44:03 PM permalink

Quote: Wizard
...The above figure contains 10 squares inscribed in a rectangle. The white square has side length of 1. What is the area of the entire rectangle?...

Simultaneous equations gets it...

Let Yellow = x
Green = 2x+1
Pink = 2x+2
Red = 2x
Blue = 3x-1

Then look at the widths 2 Pink + Green (6x+5) or 3 Blue + Red (11x-3) giving x=8/5
So width = 14 3/5 and Height = (Pink+Blue) 5x+1 = 9
So area = 131.4.

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September 3rd, 2021 at 8:49:24 PM permalink

Quote: charliepatrick
Simultaneous equations gets it...
Let Yellow = x
Green = 2x+1
Pink = 2x+2
Red = 2x
Blue = 3x-1

Then look at the widths 2 Pink + Green (6x+5) or 3 Blue + Red (11x-3) giving x=8/5
So width = 14 3/5 and Height = (Pink+Blue) 5x+1 = 9
So area = 131.4.

I agree!

"For with much wisdom comes much sorrow." -- Ecclesiastes 1:18 (NIV)

ThatDonGuy