Math Puzzle Wiki - User contributions [en]

Pizza split

2010-11-25T06:51:29Z

Vampire Library:

You order a pizza from A1-Discount Pizza Delivers. They are incredibly cheap, but the pizza quality leaves something to be desired. The pizza you get is completely deformed (not anything close to a circle, although it does not have any sharp corners at least). What's more, there is only a single piece of pepperoni, placed somewhere on the pizza.

Is it possible to divide the pizza between you and your roommate with a single cut through the pizza in such a way that both pieces have the same amount of pizza, and exactly half of the pepperoni?

[[Category: Calculus]]

User talk:Ocafaromy

2010-11-24T07:56:17Z

Vampire Library: Created page with "Why? ~~~~"

Why? [[User:Vampire Library|Vampire Library]] 02:56, 24 November 2010 (EST)

Talk:Houses and utilities

2010-11-24T00:20:37Z

Vampire Library: Undo revision 655 by Ocafaromy (talk)

This should be put into a nicer story, and should have a graphic.

Frog staircase

2010-11-24T00:19:22Z

Vampire Library: Undo revision 654 by Ocafaromy (talk)

First saw this gem in PProblem SSSolving.

A frog must climb a 10 step staircase. He can hop up either one step or two steps at a time. How many ways are there for him to make it to the top?

==Help==

{{Hint|Start with smaller staircases and look for a pattern.}}

{{Answer|89 ways.}}

[[Category:Pattern recognition]]
[[Category:Induction]]
[[Category:Sequences]]

Double Russian roulette

2010-11-24T00:19:14Z

Vampire Library: Undo revision 653 by Ocafaromy (talk)

Here is a puzzle similar to one that was on Car Talk.

==Puzzle==

As often happens to time traveling spies, you have been captured by the Cold War era Russians. They have decided to release you, but only if you survive their game of Russian roulette: they load bullets into the first and second chambers of an old six-shooter revolver. The other four chambers are empty. They spin the wheel and take a shot. You hear a click. But you are not out of the woods yet, as you must pull the trigger one more time. However, your captors offer you a choice: you can either pull the trigger right now, or spin the wheel first. What should you do?

==References==

{{Car talk}}

[[Category: Probability]]

Splitting ten

2010-11-24T00:19:05Z

Vampire Library: Undo revision 652 by Ocafaromy (talk)

==Puzzle==

Ten people are sitting around a round table. The sum of ten dollars is to be distributed among them according to the rule that each person receives one half of the sum that his two neighbors receive jointly. Is there a way to distribute the money?

==Help==

{{Hint| This one sounds harder than it is. Read it carefully}}

{{Answer| Yes}}

{{Solution| Give everyone one dollar. Then everyone's neighbor on the right will have a dollar, and neighbor on the left also a dollar. That makes two dollars, and half of that is one dollar, which is what everyone has.}}

[[Category: Algebra]]
[[Category: Short puzzles]]
[[Category: Positional]]

Five towns

2010-11-24T00:16:53Z

Vampire Library: Undo revision 651 by Ocafaromy (talk)

A puzzle similar to the [[Houses and utilities]] puzzle.

==Puzzle==

Back in the days of yore, five small towns decided they wanted to build roads connecting each pair of towns. While the towns had plenty of money to build roads as long as they wished, it was very important that the roads not intersect with each other (as stop signs had not yet been invented). Also, tunnels and bridges were not allowed. Is it possible for each of these town to build a road to each of the four other towns without creating any intersections?

{{solution | Here's one way to prove that it's not possible.

Label the towns 1 to 5. If there's a solution, then for every possible ordering of the five towns, there is a circuit route that visits all five of them in that order and returns to the first, with no backtracking.

In particular, there must be a circuit route that visits the towns in the order 1, 2, 3, 4, 5, 1 (call this Circuit A) and a circuit route that visits the towns in the order 1, 3, 5, 2, 4, 1 (call this Circuit B).

You can quickly show that Circuit B cannot be entirely inside Circuit A, nor can it be entirely outside Circuit A. Therefore, there must be at least one town (call it N) where Circuit B crosses from being outside Circuit A to being inside Circuit A.

But, as can be seen by sketching a diagram, this necessarily seperates town N-1 from town N+1 so that no road is possible between the two. }}

{{Needs math}}

[[Category:Graph theory]]

Ten bags of gold

2010-11-24T00:16:47Z

Vampire Library: Undo revision 650 by Ocafaromy (talk)

Here is a weighing puzzle I remember from Mind-quest.

==Puzzle==

You have ten bags filled with ten gold coins each -- or so you thought. A mysterious note arrives informing you that one of the bags contains fake coins, which weigh slightly less than the real coins. In fact, each real coin weighs one ounce, while the fake coins each weigh 0.9 ounces. You happen to have a scale which can display the weight accurate to a tenth of an ounce. However, the batteries in the scale are almost out, so you can only use it once. How can you find the bag of fake gold?

[[Category: Weighing puzzles]]

Prisoners in red hats

2010-11-24T00:16:02Z

Vampire Library: Undo revision 649 by Ocafaromy (talk)

Here is different version of the [[Green eyed dragons]] puzzle.

==Puzzle==

Twenty-eight tongueless prisoners awake to see that they are all wearing hats (although they cannot see their own). Although they do not know it, it turns out that all the hats are red. The guard comes in and announces that from now on, once a day, he every prisoner will write on a piece of paper either that he does not know what color hat he has, or the color of his hat. If the prisoner correctly writes down his own hat color, her goes free. If he guesses wrong, he is executed.

This continues for many months, with no prisoner ever guessing his hat color (or even attempting a guess, for fear of his life). Eventually the guard decides to give the prisoners a hint. He announces, “at least one of you is wearing a red hat.” Twenty-eight days later, the prisoners all go free. How did this happen?

==See also==

*[[Green eyed dragons]]
*[[Three prisoners]]
*[[Three logicians]]
*[[Nine prisoners]]

[[Category:Logic puzzles]]
[[Category:Induction]]

Bags of gems

2010-11-24T00:15:51Z

Vampire Library: Undo revision 648 by Ocafaromy (talk)

Here is another one I found in Smullian's book.

==Puzzle==

You have ten bags of gems. Each bag contains three gems, which are either diamonds, rubies, or emeralds. As it happens, no two of the ten bags have exactly the same combination of these gems. For instance, one bag contains three diamonds, another contains two rubies and an emerald, another contains one of each.

You reach into one of the bags, and happen to pull out a diamond. You are now allowed to pull one stone out of any of the ten bags. If that second stone is also a diamond, you get to keep it. Otherwise, you get nothing. Are you better off taking the second stone from the bag you first pulled the diamond out of, or out of one of the other nine bags?

==References==

{{Smullyan riddle}}

[[Category: Probability]]

Angels and demons

2010-11-24T00:14:55Z

Vampire Library: Undo revision 647 by Ocafaromy (talk)

Here is a modification of a question asked on the [http://forums.xkcd.com/viewforum.php?f=3 xkcd logic puzzle forum]. I've modified it slightly so there is (I believe) an answer.

==Puzzle==

Faced with a particularly vexing moral dilemma, two semitransparent miniature versions of you appear on either shoulder, each offering advise on the situation. From your experience with cartoons, you know that one of these characters is an angel, and the other a demon. Angels always tell the truth if they can, but demons sometimes lie and sometimes tell the truth. Of course, you wish to determine which is which, but you only have time to ask one question of one of the two. What should you ask?

==Help==

{{Hint| You can ask a single yes/no question. If assume that whoever you ask will answer with either a "yes" or "no," unless they are unable to.}}
{{Answer| "If I were to ask your friend there if he was an angel, would he say yes?"}}
{{Solution| The question can not be answered by the angel, since there is no way to know whether the demon would answer truthfully or not. The demon would either say yes or no, depending on whether he wished to lie or tell the truth.}}

==See also==

[[Two guards]]

[[Category:Liar puzzles]]
[[Category:Logic]]

Chests of logic

2010-11-24T00:14:46Z

Vampire Library: Undo revision 646 by Ocafaromy (talk)

You are an archaeologist that has just unearthed a long sought pair of ancient treasure chests. One chest is plated with silver, and the other is plated with gold. According to legend, one of the two chests is ﬁlled with great treasure, whereas the other chest houses a man-eating python that can rip your head oﬀ. Faced with a dilemma, you then notice that there are inscriptions on the chests:

Silver Chest: This chest contains the python.

Gold Chest: Exactly one of these two inscriptions is true.

Which should you open?
==Variation==

Same story, but this time all you know is that in each chest there is either a treasure or a man eating snake. You know that either both boxes are true, or both are false. The chests read:

Silver chest: At least one of these boxes contains a treasure

Gold chest: The silver chest contains a man eating snake.

Which should you open?

[[Category: Liar puzzles]]
[[Category: Logic puzzles]]
[[Category: Logic]]

Nine weights

2010-11-24T00:14:38Z

Vampire Library: Undo revision 645 by Ocafaromy (talk)

A classic logic puzzle.

==Puzzle==

On the table sit nine identical looking nuggets of gold. However, you know that one of the nine is a clever forgery. The only difference between the fake and the real nuggets is in weight: the fake gold weighs slightly less than the true gold. Unfortunately, this weight difference is not great enough to be noticed in human hands. Fortunately, you have a standard balance scale. Unfortunately, you may only use the balance scale twice. Fortunately, there is a way to find the fake gold even with these restrictions. How?

[[Category: Comparison puzzles]]
[[Category: Weighing puzzles]]
[[Category: Logic puzzles]]
[[Category: Pigeonhole principle]]

==See also==

[[Twelve weights]] - you can use the scale three times to find a weight that is either lighter or heavier. Quite a challenge.

[[Six weights]] - you can use the scale twice to locate the heavy weight in each of three colors.

Wrong address

2010-11-24T00:14:32Z

Vampire Library: Undo revision 644 by Ocafaromy (talk)

Found this puzzle on pzzls.com

==Puzzle==

Mr. House would like to visit his old friend Mr. Street, who is living in the main street of a small village. The main street has 50 houses divided into two blocks and numbered from 1 to 20 and 21 to 50. Since Mr. House has forgotten the number, he asks it from a passer-by, who replies "Just try to guess it." Mr. House likes playing games and asks three questions:

# In which block is it?
# Is the number even?
# Is it a square?

After Mr. House has received the answers, he says: "I still do not know, but tell me, is the digit 4 is in the number?" After hearing the answer, Mr. House runs to the building in which he thinks his friend is living. He rings, a man opens the door and it turns out that he has the wrong address. The man starts laughing and tells Mr. House: "Your adviser is the biggest liar of the whole village. He never speaks the truth!". Mr. House thinks for a moment and says "Thanks, now I know the real address of Mr. Street".

What is the address of Mr. Street?

==Help==

{{Needs answer}}

==References==

{{pzzls}}

[[Category:Logic puzzles]]
[[Category:Number theory]]

Three logicians

2010-11-24T00:14:24Z

Vampire Library: Undo revision 643 by Ocafaromy (talk)

==Puzzle==

Three logicians are chatting after teaching their classes for the day. As they talk, each notices that the other two have smeared dry-erase marker on their face, and all three begin laughing at each other. Of course, none of the logicians can see their own face. After a minute, the most senior and well trained logician stops laughing, having realized that her own face is smeared. How did she know?

==See also==

[[Green eyed dragons]] - Same idea, but with 100 dragons.

[[Category: Logic puzzles]]

Prom problem

2010-11-24T00:13:13Z

Vampire Library: Undo revision 641 by Ocafaromy (talk)

A nice, medium difficulty logic puzzle as appears in {{Problem Solving}}

==Puzzle==

After the senior prom, six friends went to their favorite restaurant, where they shared a booth. The group consisted of the senior class president, the valedictorian, the head cheerleader, a player on the school volleyball team, a player on the school basketball team, and the school principle’s only child. Their names
were Betty, Frank, Gina, Joe, Ron, and Sally, not necessarily in that order. Each of the six was in love with one of the others of the opposite sex, but no
two had crushes on the same person.

# Frank liked the cheerleader but was sitting opposite the valedictorian.
# Gina was sitting next to the cheerleader and was crazy about the class president.
# Betty was in love with the person sitting opposite her.
# Joe, who was not the valedictorian, was sitting between the volleyball player and the class president.
# Ron disliked the basketball player.
# Sally, an orphan, was sitting against the wall and had a crush on the volleyball player.
# The volleyball player sat opposite the principle’s child.

Identify each person’s claim to fame.

==Similar puzzles==

[[High school play]]

==References==

{{Problem Solving}}

[[Category: Logic puzzles]]
[[Category: Positional]]

Talk:Fifteen

2010-11-24T00:06:59Z

Vampire Library:

Very nice puzzle -- I've never seen it before. I think this would work well as a follow up to a discussion of magic squares. [[User:Oscarlevin|Oscarlevin]] 14:11, 18 November 2010 (EST)

I agree. Admittedly I don't know very much about magic squares, but I have some ideas on what we can add. First, something on how they form a Z-module (once we allow magic squares whose entries can be any integer and not just those in {1,...,n}. And actually, by considering magic squares with rational/real/complex/F_{p^n} entries a vector space. Second, how to construct odd-sized magic squares and specifically how the D4 action gives you all other 3x3 magic squares (i.e. action is transitive so that orbit=all eight 3x3 magic squares). Third, Wikipedia mentions a very interesting application of genetic algorithms to the construction of magic squares. This might be worth exploring. [[User:Vampire Library|Vampire Library]] 19:03, 23 November 2010 (EST)

Talk:Fifteen

2010-11-24T00:03:55Z

Vampire Library:

Very nice puzzle -- I've never seen it before. I think this would work well as a follow up to a discussion of magic squares. [[User:Oscarlevin|Oscarlevin]] 14:11, 18 November 2010 (EST)

I agree. Admittedly I don't know very much about magic squares, but I have some ideas on what we can add. First, something on how they form a Z-module (once we allow magic squares whose entries can be any integer and not just those in {1,...,n}. And actually, by considering magic squares with rational/real/complex/F_{p^n} entries a vector space. Second, how to construct odd-sized magic squares and specifically how the D4 action gives you all other 3x3 magic squares (i.e. orbit=all eight 3x3 magic squares). Third, Wikipedia mentions a very interesting application of genetic algorithms to the construction of magic squares. This might be worth exploring. [[User:Vampire Library|Vampire Library]] 19:03, 23 November 2010 (EST)

Talk:Around the world

2010-11-23T22:28:55Z

Vampire Library:

I think Will D. told me this puzzle. At some point I should email him to get the correct version.

So I think I have a solution that uses one plane fewer than the given solution (so 3 in all, including the plane making the whole trip). Should I just replace the given solution or edit it? I would prefer to treat the trip as being on a line with endpoints identified like

O--1/8--1/4--3/8--1/2--5/8--3/4--7/8--O

where O is the island. [[User:Vampire Library|Vampire Library]] 17:28, 23 November 2010 (EST)

Talk:Wagon collision

2010-11-23T22:27:31Z

Vampire Library:

I don't really understand the problem as it is worded. I'm having trouble understanding what you mean by "wagon." [[User:Oscarlevin|Oscarlevin]] 14:16, 18 November 2010 (EST)

The wagons can really just be thought of as circles of radius R that are traveling along their centers. This is the terminology used in Arnold's book and I was just sticking to it since I couldn't think of what else to call them. Maybe we can think of something more appropriate or add pictures (the solution badly needs a picture in any case). You should be able to find the original wording of the problem in Amazon's preview of the book along with some nice pictures (I think you can preview it, Amazon appears to be down for me at the moment oddly enough). [[User:Vampire Library|Vampire Library]] 17:27, 23 November 2010 (EST)

User talk:Vampire Library

2010-11-23T22:26:38Z

Vampire Library: /* Hello and welcome! */

==Hello and welcome!==
Sorry I haven't had a chance to say hi before now, I've been busy with work this semester. I see that you have contributed already: thank you! Feel free to add to any part of the site, or to leave me suggestions on how to improve it in general. Cheers! [[User:Oscarlevin|Oscarlevin]] 14:04, 18 November 2010 (EST)

No problem, I understand! Thanks for welcoming me. My contributions won't likely be very regular because of school, but I should be able to contribute quite a bit during this coming winter break. By the way, how can I delete articles from showing up on the main page? I created two different pages for the same puzzle (minor spelling difference) and wanted to delete one of them. Thanks! [[User:Vampire Library|Vampire Library]] 17:26, 23 November 2010 (EST)

Talk:Around the world

2010-11-23T02:30:27Z

Vampire Library:

I think Will D. told me this puzzle. At some point I should email him to get the correct version.

So I think I have a solution that uses one plane fewer than the given solution (so 3 in all, including the plane making the whole trip). Should I just replace the given solution or edit the given one? I would prefer to treat the trip as being on a line with endpoints identified like

O--1/8--1/4--3/8--1/2--5/8--3/4--7/8--O

where O is the island. - Vampire Library

Talk:Wagon collision

2010-11-19T04:47:05Z

Vampire Library:

User:Vampire Library

2010-11-18T15:10:13Z

Vampire Library: /* Personal Information */

Talk:Commuter

2010-11-08T01:30:26Z

Vampire Library:

There should probably be some more information here or at least it should be more explicit. I think you need to know when the husband arrives at the station in order to solve it (i.e. does he arrive at exactly the same time as his wife, 5 minutes earlier, etc.). Good puzzle though! - Vampire Library

Talk:Commuter

2010-11-08T01:28:56Z

Vampire Library: Created page with "There should probably be some more information here or at least it should be more explicit. I think you need to know when the husband arrives at the station in order to solve it ..."

Friends or non-friends

2010-11-08T00:44:49Z

Vampire Library: /* Help */

If there are six people in a room, must there be at least three of them who are either all friends with each other or all non-friends with each other? (If A is friends with B, then B is friends with A.) Explain.

==Help==

{{Solution| Arbitrarily choose one of the six people and label him/her A. For each of the remaining five people A is either a friend or he is not. If A is a friend then draw a black line between A and that person, and if he is not then draw a white line. We get 5 distinct lines connecting A to the other people in the room, and so by the pigeonhole principle 3 of them must be the same color. With no loss of generality we may assume 3 of them are black. Denote the other ends of these lines B,C,D. This gives an additional 3 lines BC, BD,CD. If even one of these lines is black then we can form a black triangle with A, which means there are 3 people at the party who are all friends. Otherwise the lines BC,BD,CD are all white and thus the triangle BCD is white, giving a group in which no one knows each other.}}

[[Category: Graph theory]]
[[Category: Pigeonhole principle]]

Friends or non-friends

2010-11-08T00:39:15Z

Vampire Library:

If there are six people in a room, must there be at least three of them who are either all friends with each other or all non-friends with each other? (If A is friends with B, then B is friends with A.) Explain.

==Help==

{{Solution| Arbitrarily choose one of the six people and label him/her A. For each of the remaining five people A is either a friend or he is not. If A is a friend then draw a black line between A and that person, and if he is not then draw a white line. We get 5 distinct lines connecting A to the other people in the room, and so by the pigeonhole principle 3 of them must be the same color. With no loss of generality we may assume 3 of them are black. Denote the other ends of these lines B,C,D. This gives an additional 3 lines BC, BD,CD. If even one of these lines is black then we can form a black triangle with A, which means there are 3 people at the party who are all friends. Otherwise the lines BC,BD,CD are all white and thus the triangle BCD is white, giving a group for which no one knows each other.}}

[[Category: Graph theory]]
[[Category: Pigeonhole principle]]

Friends or non-friends

2010-11-08T00:38:17Z

Vampire Library:

If there are six people in a room, must there be at least three of them who are either all friends with each other or all non-friends with each other? (If A is friends with B, then B is friends with A.) Explain.

{{Solution| Arbitrarily choose one of the six people and label him/her A. For each of the remaining five people A is either a friend or he is not. If A is a friend then draw a black line between A and that person, and if he is not then draw a white line. We get 5 distinct lines connecting A to the other people in the room, and so by the pigeonhole principle 3 of them must be the same color. With no loss of generality we may assume 3 of them are black. Denote the other ends of these lines B,C,D. This gives an additional 3 lines BC, BD,CD. If even one of these lines is black then we can form a black triangle with A, which means there are 3 people at the party who are all friends. Otherwise the lines BC,BD,CD are all white and thus the triangle BCD is white, giving a group for which no one knows each other.}}

[[Category: Graph theory]]
[[Category: Pigeonhole principle]]

Fifteen

2010-11-08T00:37:16Z

Vampire Library:

Two friends, Sam and Lloyd, play the following game: each boy, on their turn, removes a number from 1,...,9 without replacement. The winner of the game is the one who first obtains 3 numbers that sum to 15. Does Sam have a winning strategy assuming he goes first?

==Help==

{{Hint| Try to reinterpret the game as taking place on a 3x3 grid.}}
{{Answer| No. At most he can guarantee a draw.}}
{{Solution| This game is actually "equivalent" to Tic Tac Toe on a 3x3 magic square (i.e. each row, column and diagonal sums to 15). Since there is no winning strategy for Tic Tac Toe neither is there for this game.}}

[[Category:Game theory]]

Fuses

2010-11-07T18:28:48Z

Vampire Library:

A very nice puzzle from {{Problem Solving}}

==Puzzle==

You have two fuses, each twelve inches long. Each fuse burns in exactly one hour, but does not necessarily burn at a uniform rate. Also, the two fuses do not necessarily burn at the same rate over corresponding segments, but a given segment on a given fuse burns in the same amount of time in either direction. How do you use these two fuses to time 15
minutes? (You are allowed as much time to prepare as you wish.)

{{Solution| Take one of the fuses, fuse 1, and bend it so that it forms a loop. Take the other fuse, fuse 2, and place it so that one end touches both ends of fuse 1. Light this area where all three ends meet. Then fuse 1 will finish burning in exactly 30 minutes, at which time fuse 2 has another 30 minutes remaining. Thus we should light the other end of fuse 2 at this time. From this point on it will take 15 minutes for what remains of fuse 2 to finish burning.}}

==Extension==

How do you time 15 minutes using only one fuse?

==References==

{{Problem Solving}}

Wagon collision

2010-11-07T18:26:06Z

Vampire Library:

This puzzle appears in V. I. Arnold's classical text on ODEs, where it's attributed to N.N. Konstantinov.

==Puzzle==

Suppose there are cities A and B connected to each other by two non-intersecting roads. Furthermore, suppose we know that two cars attached by a rope of length less than 2R are able to travel together on different roads from City A to City B without the rope tearing. Given this, is it possible for two circular wagons, each of radius R, each traveling along its center and each starting in a different city, to travel in opposite directions along different roads without colliding as they pass?

==Help==

{{Solution| Let x denote the distance between City A and a vehicle (so either a wagon or a car) traveling on one of the roads. Similarly, let y denote the distance between City A and the other vehicle traveling on the other road. We can imagine these values together representing points in the unit square $I=\{(x,y)\ |\ 0 \leq x,y \leq 1\}$. In the case of the cars, both vehicles start at City A and so $x=y=0$ initially. As both cars traverse their respective roads to City B a continuous curve is drawn out linking the bottom left corner of $I$ to the top right corner. In the case of the wagons, both vehicles start at different cities, say $x=1$ and $y=0$. Thus, as both wagons move in opposite directions x is tending towards 0 (to City A) and y is tending towards 1 (to City B), and so a continuous curve is traced out from the bottom right corner to the top left corner. Since both curves link different corners they must intersect at some point in $I$. At this point of intersection each wagon must be at the exact position each car was at on their respective roads. Since the length of the rope attaching both cars was less than 2R and each wagon has radius exactly R both wagons must collide at this point.
}}

==References==

[http://www.amazon.com/Ordinary-Differential-Equations-V-Arnold/dp/0262510189 Ordinary Differential Equations] - V. I. Arnold's book on ODEs.

Wagon collision

2010-11-07T18:21:18Z

Vampire Library:

This puzzle appears in V. I. Arnold's classical text on ODEs, where its attributed to N.N. Konstantinov.

==Puzzle==

Suppose there are cities A and B connected to each other by two non-intersecting roads. Furthermore, suppose we know that two cars attached by a rope of length less than 2R are able to travel together on different roads from City A to City B without the rope tearing. Given this, is it possible for two circular wagons, each of radius R, each traveling along its center and each starting in a different city, to travel in opposite directions along different roads without colliding as they pass?

==Help==

{{Solution| Let x denote the distance between City A and a vehicle (so either a wagon or a car) traveling on one of the roads. Similarly, let y denote the distance between City A and the other vehicle traveling on the other road. We can imagine these values together representing points in the unit square $I=\{(x,y)\ |\ 0 \leq x,y \leq 1\}$. In the case of the cars, both vehicles start at City A and so $x=y=0$ initially. As both cars traverse their respective roads to City B a continuous curve is drawn out linking the bottom left corner of $I$ to the top right corner. In the case of the wagons, both vehicles start at different cities, say $x=1$ and $y=0$. Thus, as both wagons move in opposite directions x is tending towards 0 (to City A) and y is tending towards 1 (to City B), and so a continuous curve is traced out from the bottom right corner to the top left corner. Since both curves link different corners they must intersect at some point in $I$. At this point of intersection each wagon must be at the exact position each car was at on their respective roads. Since the length of the rope attaching both cars was less than 2R and each wagon has radius exactly R both wagons must collide at this point.
}}

==References==

[http://www.amazon.com/Ordinary-Differential-Equations-V-Arnold/dp/0262510189 Ordinary Differential Equations] - V. I. Arnold's book on ODEs.

User:Vampire Library

2010-11-07T06:11:31Z

Vampire Library:

== Personal Information ==

Student studying math and computer science. Unfortunately, I'm completely inept when it comes to editing wikis (as you might have found out) and I apologize for any mistakes I've made. I'm basically learning as I contribute.

Wagon collision

2010-11-07T05:28:13Z

Vampire Library:

Next sequence

2010-11-07T05:12:44Z

Vampire Library:

Consider the sequence 1, 11, 21, 1211, 111221, 312211... What comes next?

==Help==

{{Answer| 13112221.}}

[[Category: Lateral thinking]]
[[Category: Sequences]]

Wagon collision

2010-11-07T04:44:22Z

Vampire Library: Created page with "This puzzle appears in V. I. Arnold's classical text on ODEs, where its attributed to N.N. Konstantinov. ==Puzzle== Suppose there are cities A and B connected to each other by..."

Eight options with six sides

2010-11-07T01:53:55Z

Vampire Library:

This one is based on a problem of the week I saw [http://hilltop.bradley.edu/~delgado/potw/potw.html here].

==Puzzle==

The problem with great restaurants is that they often have so many delicious menu items that making a decision can be near impossible. For this reason, Kirk always brings his trusty 6-sided die with him when he goes out to dinner. But one fateful night, Kirk found no fewer than eight dishes he wanted to try. How could he use his die to fairly decide between his eight options? What is the least number of die rolls he would have to make?

==Extra credit==

What if he had nine options? What about other numbers of options? That is, for which number of options is there a way for Kirk to fairly decide between those options using only his 6-sided die. If $n$ is a possible number of options, what is the least number of rolls needed to decide?

==Help==

{{Needs answer}}

==References==

{{Bradley PoW}}

[[Category:Probability]]
[[Category:Optimization puzzles]]

Fuses

2010-11-07T01:52:05Z

Vampire Library: /* Puzzle */

A very nice puzzle from {{Problem Solving}}

==Puzzle==

You have two fuses, each twelve inches long. Each fuse burns in exactly one hour, but does not necessarily burn at a uniform rate. Also, the two fuses do not necessarily burn at the same rate over corresponding segments, but a given segment on a given fuse burns in the same amount of time in either direction. How do you use these two fuses to time 15
minutes? (You are allowed as much time to prepare as you wish.)

{{Solution| Take one of the fuses, fuse 1, and bend it so that it forms a loop. Take the other fuse, fuse 2, and place it so that one end touches both ends of fuse 1. Light this area where all three ends meet. Then fuse 1 will finish burning in exactly 30 minutes, at which time fuse 2 has another 30 minutes remaining. Thus we should light the other end of fuse 2 at this time. From this point on it will take 15 minutes for the remains of fuse 2 to finish burning.}}

==Extension==

How do you time 15 minutes using only one fuse?

==References==

{{Problem Solving}}

Fuses

2010-11-07T01:51:04Z

Vampire Library: /* Puzzle */

A very nice puzzle from {{Problem Solving}}

==Puzzle==

You have two fuses, each twelve inches long. Each fuse burns in exactly one hour, but does not necessarily burn at a uniform rate. Also, the two fuses do not necessarily burn at the same rate over corresponding segments, but a given segment on a given fuse burns in the same amount of time in either direction. How do you use these two fuses to time 15
minutes? (You are allowed as much time to prepare as you wish.)

{{Solution| Take one of the fuses, fuse 1, and bend it so that it forms a loop. Take the other fuse, fuse 2, and place it so that one end touches both ends of fuse 1. Light this area where all three ends meet. Then fuse 1 will finish burning in exactly 30 minutes, at which time fuse 2 has another 30 minutes remaining. Thus we should light the other end of fuse 2 at this time. From the point on it will take 15 minutes for the remains of fuse 2 to burn.}}

==Extension==

How do you time 15 minutes using only one fuse?

==References==

{{Problem Solving}}

Fuses

2010-11-07T01:43:22Z

Vampire Library:

Turn off a calculator

2010-11-07T01:39:22Z

Vampire Library:

As you are finishing up your math homework, the power goes out. You realize that it might not come on for a while, so you should probably just call it quits. You reach for the calculator to turn it off, but then realize that you are not sure whether it was on or off to begin with. You know that the "on" button will turn the calculator on if it is off, but does nothing if the calculator is already on. To turn the calculator off, you must first press the "2nd" key, and then the "on" key. However, if the calculator is off, the "2nd" key has no action. Given that the calculator could be off, on, or on with the "2nd" key already pressed, is there a sequence of button presses that will guarantee the calculator is off once finished?

{{Solution| Such a sequence is given by On-->On-->2nd-->On. In the case that the calculator is already off this sequence turns it on, does nothing, and then turns it off again. If the calculator is on then this sequence does nothing, does nothing and then turns it off. Finally, if the calculator is on with the "2nd" key pressed this sequence turns it off, turns it on, and then turns it off. }}

[[Category: Computer science]]

Turn off a calculator

2010-11-07T01:37:46Z

Vampire Library:

As you are finishing up your math homework, the power goes out. You realize that it might not come on for a while, so you should probably just call it quits. You reach for the calculator to turn it off, but then realize that you are not sure whether it was on or off to begin with. You know that the "on" button will turn the calculator on if it is off, but does nothing if the calculator is already on. To turn the calculator off, you must first press the "2nd" key, and then the "on" key. However, if the calculator is off, the "2nd" key has no action. Given that the calculator could be off, on, or on with the "2nd" key already pressed, is there a sequence of button presses that will guarantee the calculator is off once finished?

{{Solution| Such a sequence is given by On-->On-->2nd-->On. In the case that the calculator is already off this sequence turns it on, does nothing, and then turns it off. If the calculator is on then this sequence does nothing, turns it on and then turns it off again. Finally, if the calculator is on with the "2nd" key pressed this sequence turns it off, turns it on, and then turns it off. }}

[[Category: Computer science]]

Turn off a calculator

2010-11-07T01:36:47Z

Vampire Library:

As you are finishing up your math homework, the power goes out. You realize that it might not come on for a while, so you should probably just call it quits. You reach for the calculator to turn it off, but then realize that you are not sure whether it was on or off to begin with. You know that the "on" button will turn the calculator on if it is off, but does nothing if the calculator is already on. To turn the calculator off, you must first press the "2nd" key, and then the "on" key. However, if the calculator is off, the "2nd" key has no action. Given that the calculator could be off, on, or on with the "2nd" key already pressed, is there a sequence of button presses that will guarantee the calculator is off once finished?

{{Solution| The desired sequence is On-->On-->2nd-->On. In the case that the calculator is already off this sequence turns it on, does nothing, and then turns it off. If the calculator is on then this sequence does nothing, turns it on and then turns it off again. Finally, if the calculator is on with the "2nd" key pressed this sequence turns it off, turns it on, and then turns it off. }}

[[Category: Computer science]]

Turn off a calculator

2010-11-07T01:36:24Z

Vampire Library:

As you are finishing up your math homework, the power goes out. You realize that it might not come on for a while, so you should probably just call it quits. You reach for the calculator to turn it off, but then realize that you are not sure whether it was on or off to begin with. You know that the "on" button will turn the calculator on if it is off, but does nothing if the calculator is already on. To turn the calculator off, you must first press the "2nd" key, and then the "on" key. However, if the calculator is off, the "2nd" key has no action. Given that the calculator could be off, on, or on with the "2nd" key already pressed, is there a sequence of button presses that will guarantee the calculator is off once finished?

{{Solution| The desired sequence is On->On->2nd->On. In the case that the calculator is already off this sequence turns it on, does nothing, and then turns it off. If the calculator is on then this sequence does nothing, turns it on and then turns it off again. Finally, if the calculator is on with the "2nd" key pressed this sequence turns it off, turns it on, and then turns it off. }}

[[Category: Computer science]]

Fifteen

2010-11-07T01:22:05Z

Vampire Library:

==Puzzle==

Two friends, Sam and Lloyd, play the following game: each boy, on their turn, removes a number from 1,...,9 without replacement. The winner of the game is the one who first obtains 3 numbers that sum to 15. Does Sam have a winning strategy assuming he goes first?

==Help==

{{Hint| Try to reinterpret the game as taking place on a 3x3 grid.}}
{{Answer| No. At most he can guarantee a draw.}}
{{Solution| This game is actually "equivalent" to Tic Tac Toe on a 3x3 magic square (i.e. each row, column and diagonal sums to 15). Since there is no winning strategy for Tic Tac Toe neither is there for this game.}}

[[Category:Game theory]]

Fifteen

2010-11-07T01:12:27Z

Vampire Library: /* Help */

User:Vampire Library

2010-11-07T01:10:47Z

Vampire Library: Blanked the page

Fifteen

2010-11-07T01:02:27Z

Vampire Library: Created page with "==Puzzle== Two friends, Sam and Lloyd, play the following game: each boy, on their turn, removes a number from 1,...,9 without replacement. The winner of the game is the one who..."

==Puzzle==

Two friends, Sam and Lloyd, play the following game: each boy, on their turn, removes a number from 1,...,9 without replacement. The winner of the game is the one who first obtains 3 numbers that sum to 15. Does Sam have a winning strategy assuming he goes first?

==Help==

{{Hint| Try to reinterpret the game as taking place on a 3x3 grid.}}
{{Answer| No. At most he can guarantee a draw.}}
{{Solution| This game is actually "equivalent" to Tic Tac Toe on a 3x3 magic square (i.e. each row, column and diagonal sums to 15). Since there is no winning strategy in Tic Tac Toe neither is there for this game.}}

==See also==

[[Category:Game theory]]

User:Vampire Library

2010-11-07T01:00:50Z

Vampire Library:

==Puzzle==

Two friends, Sam and Lloyd, play the following game: each boy, on their turn, removes a number from 1,...,9 without replacement. The winner of the game is the one who first obtains 3 numbers that sum to 15. Does Sam have a winning strategy assuming he goes first?

==Help==

{{Hint| Try to reinterpret the game as taking place on a 3x3 grid.}}
{{Answer| No. At most he can guarantee a draw.}}
{{Solution| This game is actually "equivalent" to Tic Tac Toe on a 3x3 magic square (i.e. each row, column and diagonal sums to 15). Since there is no winning strategy in Tic Tac Toe neither is there for this game.}}

==See also==

[[Category:Game theory]]