relick
Welcome to Eientei!
Posts: 935
Favorite Game: Double Dealing CharacterFavorite Character: Rin KaenbyouCustom Title: Welcome to Eientei!Mini-Profile Background: {"image":"http://i.imgur.com/DE6ayQP.png","color":"e32222"}Mini Profile Text Color: 000000Mini-Profile Name Color: 000000Mini-Profile Text Border: WhiteOverride Avatar (Auto-Extended Mini-Profile): Enable
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Post by relick on May 6, 2015 21:24:14 GMT
EDIT: First two solutions updated: Chain Linking:Cut all three rings in one of the three-ring chains. Hook each of them between two other chains, and then weld them back again. A colourful picture: 60 cents. Bridge Crossing:17 minutes: Rows of Trees:I showed this one to my parents - my Mum googled the answer straight away. My Dad spent an hour or so trying before eventually getting it (and then teasing me because I hadn't got the solution yet). Interesting to see the different ways my parents approached the problem (and how much I'm like my Dad :V). Decided to go with the Arya style of cute answer =3= Here's some (fairly simple) problems for everyone to try:
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Post by Deleted on May 7, 2015 8:32:05 GMT
...alright, geometry problems. GeometryQuestion 4 Question 5 Question 6
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Proto
It looks kind of like a big fluffy muffin!
Posts: 343
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Post by Proto on May 7, 2015 9:43:51 GMT
EDIT: First two solutions updated: Chain Linking:Cut all three rings in one of the three-ring chains. Hook each of them between two other chains, and then weld them back again. A colourful picture: 60 cents. Bridge Crossing:17 minutes: Rows of Trees:I showed this one to my parents - my Mum googled the answer straight away. My Dad spent an hour or so trying before eventually getting it (and then teasing me because I hadn't got the solution yet). Interesting to see the different ways my parents approached the problem (and how much I'm like my Dad :V). Decided to go with the Arya style of cute answer =3= Yup, all three correct. I'm really liking these solution presentations. ...alright, geometry problems. GeometryQuestion 4 Question 5 Question 6 All three of these also correct. Lemme try the ones Relick the Relick posted: Points of IntersectionEach set intersects with all other sets. Number of intersections between two sets of lines is the product of the number of lines in each set.
Therefore, Total Number of Points of Intersection = 2(3 + 4 + 5) + 3(4 + 5) + 4(5) = 2(12) + 3(9) + 4(5) = 24 + 27 + 20 = 71 points StarpeziumsFirst, let's get the "other" angle of the trapezium (the two equal internal angles at the top corners). I'm gonna call it ß.
ß = 180° - α = 180° - 83 1/3° = 96 2/3°
Let's say the number of tiles required is n. By observing the arrangement in Figure 2, we can see that the resulting star polygon has n sides as well. This also means that the polygon has n internal angles. Among those n angles, half of them are acute angles (for the points that are "outside") and the other half are reflex angles (for the points that are "inside").
From Fig 2, it can be seen that the acute angle is made up of lines parallel to the lightly shaded trapezium. Therefore, the acute angle = α = 83 1/3° Also from Fig 2, it can be seen that the reflex angle the external angle of the top-right corner of the trapezium (based on Fig 1 orientation). Therefore, reflex angle = 360° - ß = 263 1/3°
We also know that the sum of all internal angles in a polygon with n sides = 180°(n - 2) => (n/2)(83 1/3°) + (n/2)(263 1/3°) = 180°n - 360° => n(83 1/3° + 263 1/3°) = 2(180°n - 360°) => n(346 2/3°) = 360°n - 720° => n(346 2/3° - 360°) = -720° => n = -720°/(346 2/3° - 360°) = 54 tiles
Since 54 is an even number, we know that the construction of this 27-point star polygon is feasible in this set-up No changeSince the number of people in the queue is fixed (m+n) and the number of people with $1 (m) and the number of people with $2 (n) is also fixed, I'm gonna treat the randomness as being like picking up a ball from a bag with a fixed number of balls in two colors. Also, rather than get a general solution (which probably involves crazy converging recursions), I'll take the easy way out and take n = 3 from the start. The total number of people is thus, m+3 I drew a lame tree diagram for this: Letters are just labels. Numbers inside the circle indicate whether the person has $1 or $2. The brown sets in parantheses are (number of people remaining with $1, number of people remaining with $2, current change in dollars). If the current change is negative (red), that means you can't actually provide change and you'll quit. When the number of $2 people remaining is less than the current change (blue), that means you have enough change to satisfy all the $2 customers regardless of where the rest of them are. The blue highlights the probabilities we want. Let P(state) indicate the probability of the argument state occurring, and V(state) the probability of the argument state being able to satisfy all customers. P(state) is calculated by getting the number of people carrying the specified about of money divided by the total number of people left. V(L) = V(H) = V(B) = 0 V(K) = V(I) = V(E) = 1 Probability = V(start) = (m - 2)/(m + 1)I bet it was a Canadian that came up with this question.
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relick
Welcome to Eientei!
Posts: 935
Favorite Game: Double Dealing CharacterFavorite Character: Rin KaenbyouCustom Title: Welcome to Eientei!Mini-Profile Background: {"image":"http://i.imgur.com/DE6ayQP.png","color":"e32222"}Mini Profile Text Color: 000000Mini-Profile Name Color: 000000Mini-Profile Text Border: WhiteOverride Avatar (Auto-Extended Mini-Profile): Enable
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Post by relick on May 7, 2015 16:36:21 GMT
Lemme try the ones Relick the Relick posted: Points of IntersectionEach set intersects with all other sets. Number of intersections between two sets of lines is the product of the number of lines in each set.
Therefore, Total Number of Points of Intersection = 2(3 + 4 + 5) + 3(4 + 5) + 4(5) = 2(12) + 3(9) + 4(5) = 24 + 27 + 20 = 71 points StarpeziumsFirst, let's get the "other" angle of the trapezium (the two equal internal angles at the top corners). I'm gonna call it ß.
ß = 180° - α = 180° - 83 1/3° = 96 2/3°
Let's say the number of tiles required is n. By observing the arrangement in Figure 2, we can see that the resulting star polygon has n sides as well. This also means that the polygon has n internal angles. Among those n angles, half of them are acute angles (for the points that are "outside") and the other half are reflex angles (for the points that are "inside").
From Fig 2, it can be seen that the acute angle is made up of lines parallel to the lightly shaded trapezium. Therefore, the acute angle = α = 83 1/3° Also from Fig 2, it can be seen that the reflex angle the external angle of the top-right corner of the trapezium (based on Fig 1 orientation). Therefore, reflex angle = 360° - ß = 263 1/3°
We also know that the sum of all internal angles in a polygon with n sides = 180°(n - 2) => (n/2)(83 1/3°) + (n/2)(263 1/3°) = 180°n - 360° => n(83 1/3° + 263 1/3°) = 2(180°n - 360°) => n(346 2/3°) = 360°n - 720° => n(346 2/3° - 360°) = -720° => n = -720°/(346 2/3° - 360°) = 54 tiles
Since 54 is an even number, we know that the construction of this 27-point star polygon is feasible in this set-up No changeSince the number of people in the queue is fixed (m+n) and the number of people with $1 (m) and the number of people with $2 (n) is also fixed, I'm gonna treat the randomness as being like picking up a ball from a bag with a fixed number of balls in two colors. Also, rather than get a general solution (which probably involves crazy converging recursions), I'll take the easy way out and take n = 3 from the start. The total number of people is thus, m+3 I drew a lame tree diagram for this: Letters are just labels. Numbers inside the circle indicate whether the person has $1 or $2. The brown sets in parentheses are (number of people remaining with $1, number of people remaining with $2, current change in dollars). If the current change is negative (red), that means you can't actually provide change and you'll quit. When the number of $2 people remaining is less than the current change (blue), that means you have enough change to satisfy all the $2 customers regardless of where the rest of them are. The blue highlights the probabilities we want. Let P(state) indicate the probability of the argument state occurring, and V(state) the probability of the argument state being able to satisfy all customers. P(state) is calculated by getting the number of people carrying the specified about of money divided by the total number of people left. V(L) = V(H) = V(B) = 0 V(K) = V(I) = V(E) = 1 Probability = V(start) = (m - 2)/(m + 1)I bet it was a Canadian that came up with this question. All three correct - also, I don't know why it changed it to dollars, it was originally £1 and £2 coins. Short discussion of the third problem..: It is very much a number cruncher, so there is no particular short-cut. The quickest way is to find the probabilities for the scenarios in which you would run out of change, and subtract them from 1. Good job!
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Proto
It looks kind of like a big fluffy muffin!
Posts: 343
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Post by Proto on May 11, 2015 10:10:26 GMT
Short discussion of the third problem..: It is very much a number cruncher, so there is no particular short-cut. The quickest way is to find the probabilities for the scenarios in which you would run out of change, and subtract them from 1. Good job!
Ah, yeah, since the "running out of change" cases have less depth in total, they'd be easier to calculate than "guaranteed to have enough change". More problems, now focused on raw math as opposed to puzzle-like problems:
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Proto
It looks kind of like a big fluffy muffin!
Posts: 343
Favorite Game: Phantasmagoria of Flower ViewFavorite Character: Flandre ScarletCustom Title: It looks kind of like a big fluffy muffin!Mini-Profile Background: {"image":"https://i.imgur.com/ViZwf70.png","color":""}Mini Profile Text Color: ffffffMini-Profile Name Color: ffffffMini-Profile Text Border: BlackOverride Avatar (Auto-Extended Mini-Profile): Enable
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Post by Proto on May 19, 2015 13:52:32 GMT
A certain somebody wanted more problems related to combinatorics so here are a few. Note: these are meant to be solved without a calculator.
Also, for those who haven't solved it already, there was another interesting problem that was presented in the chat:
Have fun!
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relick
Welcome to Eientei!
Posts: 935
Favorite Game: Double Dealing CharacterFavorite Character: Rin KaenbyouCustom Title: Welcome to Eientei!Mini-Profile Background: {"image":"http://i.imgur.com/DE6ayQP.png","color":"e32222"}Mini Profile Text Color: 000000Mini-Profile Name Color: 000000Mini-Profile Text Border: WhiteOverride Avatar (Auto-Extended Mini-Profile): Enable
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Post by relick on May 24, 2015 15:55:33 GMT
Three Integers x, y, z 1, 2015, 2014 1, 2014, 2015 -1, 2015, 2016 -1, 2016, 2015 10 bucks from coins - walked through, so I can't claim any credit for solving this 2601 Colored Ball Game a) P(WIN) = 1/6, E(Earning) = -35/3p b) P(0) = 0, P(1) = 7/12, P(2) = 35/108, P(3) = 35/432, P(4) = 5/432, E(Earning) = -23/96p Worked it out by multiplying probabilities Four-digit numbers - I was also walked through this, so once again nil points for me a) 4! b) 81
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Proto
It looks kind of like a big fluffy muffin!
Posts: 343
Favorite Game: Phantasmagoria of Flower ViewFavorite Character: Flandre ScarletCustom Title: It looks kind of like a big fluffy muffin!Mini-Profile Background: {"image":"https://i.imgur.com/ViZwf70.png","color":""}Mini Profile Text Color: ffffffMini-Profile Name Color: ffffffMini-Profile Text Border: BlackOverride Avatar (Auto-Extended Mini-Profile): Enable
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Post by Proto on May 24, 2015 16:11:56 GMT
All correct~
I hope you can try out the rest when you have time.
Meanwhile, here are some puzzles that aren't really Math problems *cough* @vexman *cough*
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alzack13
Bun of Darkness
Posts: 936
Rabbits shall rule
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Post by alzack13 on May 24, 2015 16:27:07 GMT
for the second one, easier version.
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