WordPlay- 550 Words You Need to Know - PDF eBook Free DOWNLOAD WordPlay: 550 Words You Need to Know Publisher: Barron's Educational Series; 2 edition ISBN: 2004 Mp3 64kbps 157MB This popular vocabulary-building. Write a Review of Wordplay: 550+ Words You Need to.

Table of contents 1. Some of these acronyms could be considered 'politically incorrect', including a few which could be interpreted to be rude or offensive. As such be careful how and where you use them. And if you are easily offended please don't read the page. This free acronyms and abbreviations finder is a dictionary of useful acronyms and abbreviations for training, learning, teaching, etc. This collection is also a study in language and communications.

This acronyms list contains acronyms and abbreviations, and (acronyms constructed restrospectively to fit a word), with origins in the armed forces, healthcare, IT and various other business and training fields, including funny lifestyle and social acronyms and abbreviations. The acronyms and abbreviations in this listing can therefore be used for various purposes: for simple amusement; for finding unknown meanings; for illustrating and emphasising points that you wish to make in training or speaking or presentations; and for examples of how language and expressions develop and evolve. Whatever, acronyms and abbreviations add colour and texture to the written and spoken word, and to life in general.

They are a fascinating reflection of the development of communications, language and social attitudes. Strictly speaking, acronyms are words formed from the abbreviations of others, but as you'll see, many of these acronyms aren't words at all, and even some of the best known acronyms like LASER and RADAR have bent the acronym rules. 'backronyms' (or 'bacronyms' - reverse acronyms) Some acronyms, usually amusing and ironic, are formed in reverse, i.e., by starting with a word, especially a brand name, or an existing acronym, and finding new words to fit each of the letters, for example the 'bacronyms' made from and.

The amusing term for these types of acronyms is 'backronyms' (or 'bacronyms'). Backronyms feature strongly in the acronyms created from and make names, and in corporate name backronyms such as, and more recently.

The bacronym or reverse acronym device is not new, as seen in the old interpretation. Two notable and quite different examples of bacronyms being used to positive effect are and. Bacronyms are often created as hoax explanations for the origins of certain swear words. The word 'bacronym' is a - which is a combination of two others, in this case combining the words 'back' and 'acronym'. Acronyms and abbreviations Acronyms, whether true acronyms or not, and abbreviations, add colour, fun and interest to our language, and thereby they act as mnemonics, or memory devices.

Many technical and process-related acronyms and abbreviations greatly assist in memory retention and learning. Many acronyms and abbreviations when used properly can certainly enhance communications, because they act as 'short-hand' and therefore increase the efficiency of communications; in other words, more meaning is conveyed in less time and fewer words. Many acronyms and abbreviations are also motivational and inspirational for training, because they contain a special theme, and because the acronym or abbreviation itself is a mnemonic device (a memory aid). Some of these acronyms and abbreviations originated as far back as the 1940s (notably the 2nd World War), and a few probably the early 1900s (notably the 1st World War). Many other abbreviations listed here are far more recent. Many older acronyms provide fascinating examples of the development of language and changing cultural attitudes. Latterly similar dark and cynical humour is evidenced in the development of acronyms and abbreviations relating to the field of customer service, especially in the contexts of IT and healthcare, for which an additional appears below separately, due to its richness and diversity.

Also, increasingly, lifestyle groupings and demographics profiles are providing fertile subject matter for acronyms - a separate listing of appears below. When using acronyms and abbreviations for serious and intentionally open communications ensure that definitions and meanings are understood or explained, or the acronym defeats its own purpose. It's advisable that if using acronyms in reports and other important communications, such as instructions, manuals, procedures, and training materials, you should include a glossary of acronyms and abbreviations, which hopefully enables the audience to understand the meanings involved. (actually these are 'backronyms', since they are retrospective constructions), and are listed in separate sections below. Some of these acronyms and 'backronyms' also appear with more details and explanations in the main acronyms listing. Acronyms and abbreviations main listing acronyms, abbreviations, menmonics, bacronyms for learning and amusement A3 Any time, Any place, Anywhere. Popular texting abbreviation (ack J Lewis).

The expression actually originated from a 1960s/70s Martini TV advert in which the song went: Any time, any place, anywhere, There's a magical world we can share (??), It's the right one, it's the bright one, It's (Thats?) Martini. Other variations of lyrics following the opening line were used in more recent years (It's a new world, Me and you girl.).

Suggestions on a postcard please as to the original words (and ad agency, composer, etc), and later versions. The word Martini in the context of media has now assumed an amusing additional modern meaning, referring to mobile and on-demand communications and media, and is also used as an ironic reference to someone exhibiting particularly flexible or pragmatic tendencies, especially politicians who pander to views and support for personal advantage above ethical considerations. A2O Apples To Oranges. Acronym to highlight any inappropriate comparison; a modern shorthand for 'chalk and cheese'. A2HS Age, Build, Clothes, Distinguishing marks, Elevation, Face, Gait, Hair, Sex. Acronym used by UK armed forces and services staff for identifying people involved in incidents or crime. AAA Alive, Alert, Aggressive. One of very many triple-A acronyms. Somewhat macho but catchy nevertheless, and not a bad rallying call for self or team in a variety of situations (visit to the dentist, d.

If you enjoy this page you can support it by shopping at amazon through this link. The year is 1994. The man on the left is my first husband, Alexander Goncharov. Although we were out of touch for a decade, when I married my third husband, Joseph Bernstein (on the right), Goncharov started visiting us. It wasn't me he was interested in: he wanted to talk mathematics with my husband. I found this situation hilarious, so I took this photo.

But that's not all. My second husband, Andrey Radul, is not in the picture. But all four of us were students of Israel Gelfand. In short, my three ex-husbands and I are mathematical siblings — that is, we are all one big happy mathematical family. The Best Writing on Mathematics 2016. I am happy that my paper The Pioneering Role of the Sierpinski Gasket is included.

The paper is written jointly with my high-school students Eric Nie and Alok Puranik as our -2014 project. At the end of the book there is a short list of notable writings that were considered but didn't make it. The 'short' list is actually a dozen pages long. And it includes two more papers of mine: • Cookie Monster Plays Games—another PRIMES project written jointly with my high-school student Joshua Xiong.

•, written jointly with PRIMES Chief Research Advisor Pavel Etingof and PRIMES Program Director Slava Gerovitch. To continue bragging, I want to mention that my paper A Line of Sages was on the short list for 2015 volume. And my paper Conway's Wizards was included in the. Which One Doesn't Belong?

I like Odd-One-Out puzzles that are ambiguous. That is why I bought the book Look at the cover: which is the odd one out?

The book doesn't include answers, but it has nine more examples in each of which there are several possible odd-one-outs. A Waterfall of my Feelings I married an American citizen and moved to the US in 1990. At the time I was a very patriotic Russian. It took me a year of pain to realize that some of my ideas had been influenced by Soviet propaganda. After I washed away the brainwashing, I fell in love with the US.

For 25 years I thought that America was great. For the last several months I've been worried as never before in my life. I feel paralyzed and sick. To help myself I decided to put my feelings in words. My mom was 15 when World War II started. The war affected her entire life, as well as the lives of everyone in the USSR.

Every now and then my mom would tell me, 'You are lucky that you are already 20 and you haven't witnessed a world war.' I moved to the US while my mom stayed back in Russia. From time to time I tell myself something like, 'I am lucky that halfway through my expected lifetime, I haven't had to live through a world war.' It's been more than 70 years since WWII ended. To maintain the peace is a difficult job. Everything needs to be in balance. Trump is disrupting this balance.

I am worried sick that my children or grandchildren will have to witness a major war. The Red Button. I've noticed that, as a true showman, Trump likes misdirecting attention from things that worry him to fantastic plot twists that he invents. What's the best way to make people forget about his tax returns? It's the nuclear button. Dropping a nuclear bomb some place will divert people from thinking about his tax returns. As his plot twists are escalating, is he crazy enough to push the button?

The year 2014 was the warmest on record. The year 2015 was even warmer. And last year, 2016, was even warmer than that. I remember Vladimir Arnold's class on differential equations. He talked about a painting that had been hanging on a wall for 20 years. Then it unexpectedly fell off.

Mathematics can explain how such catastrophic events can happen. I keep thinking about our Earth: melting ice, dead reefs, fish eating plastic, and so much more. My grandchildren might not be able to enjoy beaches and forests the way I did.

What if, like the fallen painting, the Earth can spiral out of control and completely deteriorate? But Trump is ignoring the climate issues. Does he care about our grandchildren? I am horrified that Trump's policies will push climate catastrophe beyond the point of no return. Wiretapping is not wiretapping. Phony jobs numbers stopped being phony as soon as Trump decided that he deserved the credit.

The news is fake when Trump doesn't like it. Trump is a pathological liar; he assaults the truth.

Being a scientist I am in search of truth, and Trump diminishes it. I do not understand why people ignore his lies. Two plus two is four whether you are a democrat, or a republican, or whomever. Facts are facts, alternative facts are lies. I am scared that lies have become acceptable and no one cares about the truth any more.

I lived in Russia for the first unhappy half of my life, and in the US for the second happy half. I do not want to go back. There is something fishy between Putin and Trump. Whether it is blackmail or money, or both, I do not know the details yet, But Trump is under Putin's influence. Trump didn't win the elections: Putin won. This horrifies me. I do not want to go back to being under Russian rule.

Gender Issues. I grew up in a country where the idea of a good husband was a man who wasn't a drunkard.

That wasn't enough for me. I dreamed of a relationship in which there would be an equal division of work, both outside and inside the home. I could not achieve that because in Russian culture both people work full-time and the wife is solely responsible for all the house chores.

Moreover, Russia was much poorer than the US: most homes didn't have washing machines; we never heard of disposable diapers; and there were very long lines for milk and other necessities. The life in USSR was really unfair to women.

Most women had a full-time job and several hours of home chores every day. When I moved to the US, I thought I was in paradise. Not only did I have diapers and a washing machine, I was spending a fraction of the time shopping, not to mention that my husband was open to helping me, and didn't mind us paying for the occasional babysitter or cleaner.

For some time I was blind to gender issues in the US because it was so much better. Then I slowly opened my eyes and became aware of the bias.

For some years it has felt like gender equity was improving. Now, with a misogynistic president, I feel that the situation might revert to the dark ages.

When women are not happy, their children are not happy, and they grow up to be not happy. If the pursuit of happiness is the goal, the life has to be fair to all groups. But Trump insults not only women but also immigrants, Muslims, members of the LGBTQ community, as well as the poor and the sick. The list is so long, that almost everyone is marginalized. This is not a path towards a happy society. Trump attacks the press and attempts to exclude them.

Trump has insulted the intelligence community and the courts. He seems to be trying to take more power to the presidency at the expense of the other branches of government. He ignores his conflicts of interest. Trump disregards every rule of democracy and gets away with it. I am horrified that our democracy is dying. Trump's tax returns could either exonerate him or prove that he is Putin's puppet.

The fact that he is hiding the returns makes me believe that the latter is more probable. Why the Republicans refuse to demand to see his returns is beyond my understanding. Trump does so many unethical things. Most of his decisions as president seem to be governed by Trump trying to get richer. Let us consider his hotel in Azerbaijan—a highly corrupt country. Having lived in a highly corrupt country myself, I know how it works. For example, an Azerbaijani government official who has access to their country's money can make a deal that involves a personal kickback.

This means that their government is paying more than necessary for a service or product in order to cover that kickback. This is how national money makes its way into individual pockets. Since all the deals in Azerbaijan are reputed to be like that, I imagine that when Trump built his hotel there, the Trump organization was overpaid in order to cover the bribe to local officials. Will our country become as corrupt as Azerbaijan? The biggest shock of the election was that so many people were so gullible and actually voted for Trump. They didn't see that his agenda is focused on his own profit, and that he lies and makes promises he doesn't plan to deliver. It really terrifies me that there are some people who are not gullible but still voted for Trump.

Is there hope? Winning Nim Against a Player who Plays Randomly I recently wrote about my way of playing Nim against a player who doesn't know how to play. If my move starts in an N-position, then I obviously win. If my move starts in a P-position, I would remove one token hoping that more tokens for my opponent means more opportunity for them to make a mistake. But which token to remove? Does it make a difference from which pile I choose?

Consider the position (2,4,6). If I take one token, my opponent has 11 different moves. If I choose one token from the first or the last pile, my opponent needs to get to (1,4,5) not to lose. If I choose one token from the middle pile, my opponent needs to get to (1,3,2) not to lose.

But the first possibility is better, because there are more tokens left, which gives me a better chance to have a longer game in case my opponent guesses correctly. That is the strategy I actually use: I take one token so that the only way for the opponent to win is to take one token too. This is a good heuristic idea, but to make such a strategy precise we need to know the probability distribution of the moves of my opponent.

So let us assume that s/he picks a move uniformly at random. If there are n tokens in a N-position, then there are n − 1 possible moves. At least one of them goes to a P-position.

That means my best chance to get on the winning track after the first move is not more than n/( n−1). If there are 2 or 3 heaps, then the best strategy is to go for the longest game. With this strategy my opponent always has exactly one move to get to a P-position, I win after the first turn with probability n/( n−1). I lose the game with probability 1/( n−1)!! Something interesting happens if there are more than three heaps.

In this case it is possible to have more than one winning move from a N-position. It is not obvious that I should play the longest game.

Consider position (1,3,5,7). If I remove one token, then my opponent has three winning moves to a position with 14 tokens. On the other hand, if I remove 2 tokens from the second or the fourth pile, then my opponent has one good move, though to a position with only 12 tokens. What should I do?

I leave it to my readers to calculate the optimal strategy against a random player starting from position (1,3,5,7). The Hidden Beauty It is rare when a word equation coincides with a number equation. A store sells letter magnets. The same letters cost the same and different letters might not cost the same. The word ONE costs 1 dollar, the word TWO costs 2 dollars, and the word ELEVEN costs 11 dollars. What is the cost of TWELVE?

Can You Solve My Problems? Wrote a puzzle book The book contains a mixture of famous puzzles and their solutions. Some of the puzzles are not mathematical in the strictest sense, but still have an appeal for mathematicians. For example, which integer comes up first when you alphabetize all the integers up to a quadrillion? Recognize the puzzle on that book cover? You're right! That's my puzzle.

Doesn't it look great in lights on that billboard in London? Mine isn't the only terrific puzzle in the book. In fact, one of the puzzles got my special attention as it is related to our current. Here it is: A Sticky Problem. Dick has a stick. He saws it in two. If the cut is made [uniformly] at random anywhere along the stick, what is the length, on average, of the smaller part?

Of course you can. Mathematicians always find new way to look at things. In 2012 RSI student, Kevin Garbe, did some new and cool research related to the triangle. Consider Pascal's triangle modulo 2, see picture which was. A consecutive block of m digits in one row of the triangle modulo 2 is called an m-block. If you search the triangle you will find that all possible binary strings of length 2 are m-blocks.

Will this trend continue? Yes, you can find any possible string of length 3, but it stops there. The blocks you can find are called accessible blocks. So, which blocks of length 4 are not accessible? There are only two strings that are not accessible: 1101 and 1011. It is not surprising that they are reflections of each other.

Pascal's triangle respects mirror symmetry and the answer should be symmetric with respect to reflection. You can't find these blocks on the picture, but how do we prove that they are not accessible, that is, that you can't ever find them? The following amazing property of the triangle can help.

We call a row odd/even, if it corresponds to binomial coefficients of n choose something, where n is an odd/even number. Every odd row has every digit doubled.

Moreover, if we take odd rows and replace every double digit with its single self we get back Pascal's triangle. Obviously the two strings 1101 and 1011 can't be parts of odd rows. What about even rows? The even rows have a similar property: every even-indexed digit is a zero. If you remove these zeros you get back Pascal's triangle.

The two strings 1101 and 1011 can't be part of even rows. Therefore, they are not accessible. The next question is to count the number of inaccessible blocks of a given length: a(n). This and much more was done by Kevin Garbe for his RSI 2012 project. (I was the head mentor of the math projects.) His paper is published on the. The answer to the question can be found by constructing recurrence relations for odd/even rows. It can be shown that a(2r) = 3a(r) + a(r+1) − 6 and a(2r+1) = 3a(r) + 2a(r+1) − 6.

As a result the number of inaccessible blocks of length n is n 2 − n + 2. I wonder if there exists a direct proof of this formula without considering odd and even rows separately. This RSI result was so pretty that it became a question at our entrance PRIMES test for the year 2013. In the test we changed the word accessible to admissible, so that it would be more difficult for applicants to find the research. Besides, Garbe's paper wasn't arxived yet. The pretty picture above is from the stackexchange, where one of our PRIMES applicants tried to solicit help in solving the test question. What a shame.

My Favorite Problems from the Moscow Math Olympiad 2016 I picked four problems that I liked from the Moscow Math Olympiad 2016: Problem 1. Ten people are sitting around a round table.

Some of them are knights who always tell the truth, and some of them are knaves who always lie. Two people said, 'Both neighbors of mine are knaves.' The other eight people said, 'Both neighbors of mine are knights.' How many knights might be sitting around the round table?

Today at least three members of the English club came to the club. Following the tradition, each member brought their favorite juice in the amount they plan to drink tonight. By the rules of the club, at any moment any three members of the club can sit at a table and drink from their juice bottles on the condition that they drink the same amount of juice. Gabest Filters Download Youtube here. Prove that all the members can finish their juice bottles tonight if and only if no one brings more than the third of the total juice brought to the club. Three piles of nuts together contain an even number of nuts. One move consists of moving half of the nuts from a pile with an even number of nuts to one of the other two piles. Prove that no matter what the initial position of nuts, it is possible to collect exactly half of all the nuts in one pile.

N people crossed the river starting from the left bank and using one boat. Each time two people rowed a boat to the right bank and one person returned the boat back to the left bank. Before the crossing each person knew one joke that was different from all the other persons' jokes.

While there were two people in the boat, each told the other person all the jokes they knew at the time. For any integer k find the smallest N such that it is possible that after the crossing each person knows at least k more jokes in addition to the one they knew at the start. Spoiler for Problem 2. I want to mention a beautiful solution to problem 2. Let's divide a circle into n arcs proportionate to the amount of juice members have. Let us inscribe an equilateral triangle into the circle.

In a general position the vertices of the triangle point to three distinct people. These are the people who should start drinking juices with the same speed. We rotate the triangle to match the drinking speed, and as soon as the triangle switches the arcs, we switch drinking people correspondingly.

After 120 degree rotation all the juices will be finished. My Last Visit to Smullyan. I had a distant relative Alla, who was brought up by a single mother, who died in a car crash when the girl was in her early teens. Alla was becoming a sweet and pleasant teenager; she was taken in by her aunt after the accident. Very soon the aunt started complaining that Alla was turning into a cheater and a thief.

The aunt found a therapist for Alla, who explained that Alla was stealing for a reason. Because the world had unfairly stolen her mother, Alla felt entitled to compensation in the form of jewelry, money, and other luxuries. I was reminded of Alla's story when I was reading by Dan Ariely. Ariely discusses a wide range of reasons why honest people cheat. But to me he neglects to look at the most prominent reason.

Often honest people cheat when they feel justified and entitled to do so. One of Ariely’s experiments went like this. One group was asked to write a text avoiding letters x and z. The other group was asked to write a text avoiding letters a and n. The second task is way more difficult and requires more energy. After the tasks were completed the participants were given a test in which they had a chance to cheat.

For this experiment, the participants were compensated financially according to the number of questions they solved. Not surprisingly, the second group cheated more. The book concludes that when people are tired, their guard goes down and they cheat more. I do not argue with this conclusion, but I think another reason also contributes to cheating. Have you ever tried to write a text without using the letters a and n? I did: I should try it here.

But this is so difficult. My son, Alexey, was way better than me: First, God brought forth the sky with the world. The world existed without form. Gloom covered the deep. The Spirit of God hovered over the fluids. Quoth God: let there be light.

Thus light existed. Fun as it is, this is cruel and unusual punishment. The request is more difficult than most people expect at an experiment. It could be that participants cheated not only because their capacity for honesty was depleted, but because they felt entitled to more money because the challenge was so difficult. In another experiment, the participants received a high-fashion brand of sunglasses before the test. Some of them were told that the sunglasses were a cheap imitation of the luxury brand (when they really were not). This group cheated more than the group who thought that they got a real thing.

The book concludes that wearing fake sunglasses makes people feel that they themselves are fake and so they care less about their honor. Unfortunately, the book doesn't explain in detail what was actually promised. It looks like the participants were promised high-fashion sunglasses. In this case, the fake group would have felt deceived and might have felt more justified to cheat. Dear Dan Ariely: May I suggest the following experiment. Invite people and promise them some money for a 15-minute task. Pay them the promised minimum and give them a test through which they can earn more.

Construct it so that they can earn a lot more if they cheat. Then make the non-control group wait for half an hour.

If I were in this group, I would have felt that I am owed for the total of 45 minutes—three times more than what I was promised. I do not know if I would cheat or just leave, but I wouldn't be surprised that in this group people would cheat more than in the control group. Alternators: People and Coins If like me, you fancy and his books, then you've heard about knights and knaves. Knights always tell the truth and knaves always lie. In addition to knights and knaves, there are normal people who sometimes tell the truth and sometimes lie. Here is a puzzle. How, in one sentence, can a normal person prove that they are normal?

We can draw a parallel between people and coins. We can say that knights correspond to real coins, and knaves to fake coins that are lighter than real ones. Inspired by normal people, my coauthor Konstantin Knop invented chameleon coins. Chameleon coins can change their weight and behave like real or fake coins. I just wrote a post about.

Normal people are too unpredictable: they can consistently pretend to be knights or knaves. So logicians invented a simpler type of person, one who switches from telling the truth in one sentence to a lie in the next and then back to the truth. Such people are called alternators. Here is another puzzle: Puzzle. You meet a person who is one of the three types: a knight, a knave, or an alternator.

In two questions, find out which type they are. Continuing a parallel between people and coins we can define alternator coins: the coins that switch their weight each time they are on the scale from weighing as much as real ones to weighing as much as fake ones. For the purposes of this essay, we assume that the fake coins are lighter than real ones. Unlike the chameleon coin, which might never reveal itself by always pretending to be real, the alternators can always be found. How do you find a single alternator among many real coins?

There is a simple strategy: repeat every weighing twice. This strategy allows us to find an alternator among 9 coins in four weighings. Can we do better? I used the alternator coins as a research project for my program where we do math research with students in seventh and eighth grade.

The students started the alternator project and immediately discovered the strategy above. The next step is to describe a better strategy. For example, what is the maximum number of coins containing one alternator such that the alternator can always be found in four weighings? But first we count possible outcomes.

Suppose there is a strategy that finds an alternator. In this strategy we can't have two unbalanced weighings in a row. To prove that, let us suppose there was an unbalanced weighing.

Then the alternator switches its weight to a real coin and whether or not the alternator is on the scale, the next weighing must balance. The beauty of it is that given a strategy each outcome has to point to a particular coin as an alternator. That means the number of outcomes bounds the total number of coins that can be processed. Counting the number of possible outcomes that do not have two unbalances in row is a matter of solving a recurrence, which I leave to the readers to find. The result is Jacobshtal numbers: the most beautiful sequence you might never have heard of. For example, the total number of possible outcomes of four weighings is 11.

Since each outcome of a strategy needs to point to a coin, the total number of coins that can be processed in four weighings is not more than 11. But 11 is better than 9 in our previous strategy. Can we process 11 coins in four weighings?

I will describe the first part of the strategy. So we have 11 coins, one of which is an alternator. In the first weighing we compare 5 coins against 5 coins. If the weighing unbalances, the alternator is on a lighter pan. Our problem is reduced to finding the alternator among five coins when we know that it is in the real state.

If the weighing balances, then we know that if the alternator is among the coins on the scale it must now be in the light state. For the second weighting, we pick two sets of three coins out of this ten coins and compare them against each other. Notice that 3 is a Jacobsthal number, and 5, the number of coins outside the scale, is also a Jacobsthal number. If the second weighing balances, the alternator must be among 5 coins outside the scale. All but one of these coins are in the light state, and I leave it to the readers to finish the strategy. If the weighing unbalances, we need to find the alternator among 3 coins that are in the real state now.

This can be done in two weighings, and again the readers are to the rescue. It appears that Jacobsthal numbers provide the exact lower bound of the number of coins that can be processed.

This is what my middle-schoolers discovered and proved. The strategy in the paper is adaptive. That means it changes depending on the results of the previous weighings. Can we find an oblivious strategy? I will tell you in later posts. A Random Scale Suppose we have 3 n identical-looking coins.

One of the coins is fake and lighter than the other coins which all are real. We also have a random scale.

That is a scale that at each weighing behaves randomly. Find the fake coin in the smallest number of weighings. That won't work! It is impossible to find the fake coin. The scale can consistently misbehave in such a way as to blame a specific real coin for being fake. Let's try something else.

Suppose we have two scales: one normal and one random. Find the fake coin. What am I thinking?

The normal scale can point to one coin and the random scale can point to another coin and we are in a 'she said, he said' situation which we can't resolve. Now, in my final try, I'll make it right. We actually have three scales, one of which is random. So here we go, with thanks to my son Sergei for giving me this puzzle: Puzzle. We have 3 n identical-looking coins.

One of the coins is fake and lighter than the other coins, which all are real. We also have three scales: two normal and one random. Find the fake coin in the smallest total number of weighings.

Let's start with this strategy: repeat every weighing on all three scales and have a majority vote. At least two of the scales will agree, thus pointing to the true result. This way we can use a divide-into-three-equal-groups strategy for one scale to find the fake coin.

It will require 3 n weighings. Can we do better?

Of course, we can. We can repeat every weighing on two scales. If they agree we do not need the third scale. If they do not agree, one of the scales is random and lying, and we can repeat the weighing on the third scale to 'out' the random scale.

After we identify one normal scale, the process goes faster. In the worst case we will need 2 n + 1 weighings. Can we do even better? I will leave it to the readers to find a beautiful solution that is asymptotically better than the previous one. Update on Dec 24, 2016. The total number of coins should be 3 2 n, not 3 n.

We are looking at the worst case scenario, when the random scale is adversarial. Chameleon Coins We all have played with problems in which we had real coins and fake (counterfeit) coins. For this post I assume that the fake coins are always lighter than the real coins. My coauthor invented a new type of a coin: a chameleon coin. This coin can mimic a fake or a real coin. It can also choose independently which coin to mimic for each weighing on a balance scale.

You cannot find the chameleon coin in a mix with real coins if it does not want to be found, because it can consistently behave as a real coin. Let's add classic fake coins to the mix, the ones that are lighter.

Still the task of identifying the chameleons using a balance scale cannot be achieved: the chameleons can pretend to be fake coins. We can't identify the fake coins either, as the chameleons can mess things us up by consistently pretending to be fake. What we can do is to find a small number of coins some of which are guaranteed to be fake. Consider the simplest setup, when we have one fake coin and one chameleon in our mix of N coins.

That is we have N − 2 real coins. Our task now is to find TWO coins, one of which has to be fake. As usual we want to do it in the smallest number of weighings that guarantees that we'll find the two coins. Let me give you a fun problem to solve: Puzzle. The total number of coins is four. And as above we have one chameleon and one classic fake. In two weighings find two coins so that one of them is guaranteed to be fake.

If you want to learn more, we just wrote a paper titled. Even More Jokes * * * —Honey, we are like two parallel lines. —Why do you say that? —The intersection of our life paths was a mistake. * * * —Q: Why did the obtuse angle go to the beach?

—A: Because it was over 90 degrees. * * * Ancient Roman in a clothing store: How come XL is larger than L? * * * —Which is the odd one out: one, three, six, seven? —Well, three of them are odd ones. * * * When I am with you, I solve integrals in my head, so that blood can come back to my brain.

* * * There are two kinds of people in this world: Those that can extrapolate from incomplete data. * * * Seven has the word even in it, which is odd. Russian Honor Code I have always wanted to be an honest person and have followed my honor code. Soviet Russia had two honor codes. One code was for dealing with people and the other for dealing with businesses and the government. I remind you that in Soviet Russia, all businesses were owned by the government.

The money paid for work didn't have any relation to the work done. The government paid standard salaries and the businesses did whatever. Generally, that meant that they were doing nothing. Meanwhile, the government got its income from selling oil.

All people were being screwed by the government, so they had no motivation to play fair. Just as American workers do, Soviet Russians might use the work copier to make personal copies. The difference was that we didn't feel guilty at ripping off the government. We wouldn't just make a few necessary copies; we would make copies for our friends, our family, for strangers—as many copies as possible. I moved to the United States in 1990.

Several of my friends took time to explain to me the difference between Soviet Russia and the US. One of the friends, let's call her Sarah, was working as staff at Harvard University. She told me the story of a recent visit by a famous Russian professor. After he left, the department received a bill. It appears that the professor, in a short visit, used several months-worth of the department's budget allocation for copying and phone calls. I was impressed, in a good way, by the professor who I assume spent a good deal of his time making a lot of copies of papers unavailable in Russia, presumably not only for himself but also for students and colleagues.

On the other hand, it was clearly wrong. My new friend Sarah told me that in the US money does not come from nowhere, and I should include Harvard University in my honor code for people. Actually, not only Harvard, but also any place of work and the government too. Sarah also told me that since that budget problem, she was asked to talk to every incoming Russian visitor and explain to them how capitalism works. Most Russian visitors were ready to accept the rules. I too was delighted with that. It is much easier to follow one honor code than two.

I was also very happy for my son, Alexey. He was eight when we moved to Boston.

Before our move I had a dilemma. Should I tell him that Lenin and Stalin were bad guys and killed millions of people? If I gave him a truthful explanation, I would also have to teach him to lie.

Otherwise, if he mentioned this at school or on the street, we would be at risk of going to prison. If I didn't teach him the truth, he might become brainwashed and grow up believing in communism, which would be very, very bad.

I was so lucky that I moved to the US: I didn't have to teach my son to lie. Centaurs, manticores, and minotaurs roam their planet. Their society is very democratic: any two animals can become sex partners. When two different species mate, their orgasm is so potent that they merge into one creature of the third species. For example, once one centaur and one manticore make love, they are reborn as one minotaur.

At the beginning of the year 2016 there were 2016 centaurs, 2017 manticores, and 2018 minotaurs. They mated non-stop and at the end of the year only one creature was left on the planet. This is one of those puzzles that I love-hate. I hate it because it is easy to answer this puzzle by inventing a specific mating pattern that ends with one animal. It is possible to get the correct answer without seeing the beauty. I love it because there is beauty in the explanation of why, if the mating ends with one animal, it has to be a specific animal.

The solution is charming, but being a mathematician this problem makes me wonder if ii is always possible to end with one animal. So I add another puzzle.

Describe the sets of parameters for which it is impossible to end up with one creature. A Faulty Scale Today I have two new coin puzzles that were inspired by my son,: You have N >2 identical-looking coins. All but one of them are real and weigh the same. One coin is fake and is lighter than the real ones. You also have a balance scale which might or might not be faulty.

A faulty scale differs from a normal one in reversing the sense of unbalanced weighings—it shows a lighter pan as heavier and vice versa (but still shows equal-weight pans as weighing the same). What is the smallest number of weighings you need in order to figure out whether the scale is faulty? If you think about it, this problem is isomorphic to a: You have N ≥ 2 identical-looking coins.

All but one of them are real and weigh the same. One coin is fake and is either lighter than the real ones or heavier than the real ones. You also have a normal balance scale. What is the smallest number of weighings you need in order to figure out whether the fake coin is lighter or heavier? To make things more interesting let's mix the problems up. You have N >2 identical-looking coins.

All but one of them are real and weigh the same. One coin is fake and is lighter than the real ones. You also have M >1 identical-looking balance scales. All but one of them are normal and one is faulty. The faulty scale differs from a normal one in reversing the sense of unbalanced weighings—it shows a lighter pan as heavier and vice versa (but still shows equal-weight pans as weighing the same). What is the smallest number of total weighings needed to figure out which scale is faulty?

How Many Triangles? The following problem was at a 2016 entrance test for the.

I drew several triangles on a piece of paper. First I showed the paper to Lev and asked him how many triangles there were. Lev said 5 and he was right.

Then I showed the paper to Sasha and asked him how many triangles there were. Sasha said 3 and he was right. How many triangles are there on the paper? The intended answer was 8: there were 5 triangles on one side of the paper and 3 on the other. Most of the students didn't think that the paper might be two-sided, but they came up with other inventive ideas. Below are some of their pictures, and I leave it to you to explain why they work. All the students who submitted these pictures got a full credit for this problem on the test.

Should You Apply to PRIMES? If you are a high-school student who wants to conduct research in mathematics, you should check out the MIT PRIMES program.

If you enjoy solving the problems in our entrance test, that's the first indication that you might want to apply. But to determine if the program is right for you, and you are right for the program, please read the following questions and answers which have been prepared for you by Tanya Khovanova, the PRIMES Head Mentor. (This only addresses applications to PRIMES Math, and only to the research track) Question: I do not like math competitions. Should I apply? Answer: Math competitions are completely separate from research in mathematics. If you enjoy thinking about mathematics for long periods of time and are fascinated by our test questions, you should apply.

Question: I am good at math, but I really want to be a doctor. Should I apply? PRIMES requires a huge time commitment, so math should really be your most significant interest. Question: I want to get into Harvard, and PRIMES looks good on a resume.

Should I apply? Answer: PRIMES does look good on a resume. But if you are more passionate about, say, climate change than math, what would Harvard's admission committee see?

Our experience in the program is that if math isn't your top interest, your math student may not be sufficiently impressive to be accepted at Harvard as a math researcher. At the same time, you will not be accepted as the top climate change student as you didn't invest your time in that. Math research is a hard way to earn points for college. See also, the essay, by Simon Rubinstein-Salzedo. Question: My parents want me to apply.

Should I apply? Answer: Your parents will not be accepted to the program. Do not apply if you do not really, really want to. Question: Your website suggests that I should spend ten hours a week on the PRIMES project. I can only spend five.

But I am a genius and faster than other people. Answer: We already assume that you are a genius and faster than anyone else you know. Five hours a week are not enough for a successful project. Question: I looked at the past PRIMES projects and nothing excites me as much as my current interest in Pascal's triangle. I doubt I should apply. Answer: When you start working on a project, you will learn a lot about it.

You will understand why, for example, Cherednik algebras are cool. The excitement comes with knowledge and invested time. Not yet being excited about Cherednik algebras is not a good reason not to apply. Besides a lot of exciting mathematics is done between several different fields.

Question: I really want to do nothing else than study Pascal's triangle. Answer: We try to match our projects to students' interests as much as we can. But we almost never can fulfill a specific request as above.

You might get a project related to Young diagrams, which are connected to quantum Pascal's triangle. If this connection doesn't excite you, you shouldn't apply.

Question: I think I will be better positioned for research if I spend five more years studying. Answer: There is nothing wrong with this approach. For many years the standard was to start research in graduate school. Our program is innovative.

At PRIMES we are trying a different model. It may sound scary, but you will learn everything you need to know in order to do your project.

If the project is in representation theory, for example, you will only learn what you need—not the whole theory. Our hope is that eventually you will take a course in representation theory and expand your grasp of it and see the bigger picture behind your project. We have a reading track for people like you who reside in Boston area. Question: I love math, but I am not sure that I want to be a mathematician. Should I apply?

Answer: Many people start loving math early in life and then discover that there are many other things that require a similar kind of brain: computer science, cryptography, finance, and so on. We do not require from our students a commitment to become mathematicians. If you want to try research in math, you should apply. If students decide that they do not want to do research in math after finishing our program, we do not consider that a negative result. One way or another, the experience of PRIMES will help you understand better what you want to do with your life. Question: I want to get to the International Math Olympiad.

I am afraid that the time the research project takes prevents me from preparing for competitions. Should I apply? Answer: People who are good at Olympiads often have fantastic brain power that helps in research. On the other hand, research requires a different mind set and the transition might be painful. It is possible, but not trivial to succeed in both. It is up to you to decide how you want to spend your time. Question: I like number theory, but I do not see past PRIMES projects in number theory.

Answer: Doable number theory projects are hard to come by and we have fewer number theory projects than students who want to do number theory. There are many high-school programs that teach number theory including PROMYS and Ross programs. Our applicants like number theory because they were exposed to it. During PRIMES you will be exposed to something else and might like it as much.

Question: I found a local professor to work with on a research project. Should I apply to PRIMES? Answer: PRIMES requires that you devote 10 hours a week to research for a year.

It is unrealistic to do two research projects in parallel. Working with someone in person may be better than by Skype at PRIMES. Also, usually our mentors are not professors, but rather graduate students. On the other hand, they are MIT grad students and projects are often suggested by professors. Our program is well structured. We guarantee weekly meetings in the Spring, we give extra help with your paper, and we have a conference.

It is up to you to decide. Alexander Shapovalov Crosses a River Alexander Shapovalov is a prolific puzzle writer. He has a (in Russian). Here is one of these puzzles. Three swindlers have two suitcases each. They approach a river they wish to cross.

There is one boat that can carry three objects, where a person or a suitcase counts as one object. No swindler can trust his suitcase to his swindler friends when he is away, but each swindler doesn't mind his suitcases left alone at the river shore. Can they cross the river? 3-Pile Nim as an Automaton In my paper with Joshua Xiong,, we produced a bijection between P-positions in the three-pile Nim and a three-branch Ulam-Warburton automaton. We also defined a parent-child relationship on games that is induced by this bijection. Namely, two consecutive P-positions in a longest optimal game of Nim are the ones that correspond to a parent-child pair in the automaton. A cell in the Ulam-Warburton automaton has exactly one parent.

That means, if ( a, b, c) is a Nim P-position, then exactly one of ( a − 1, b − 1, c), ( a − 1, b, c − 1), and ( a, b − 1, c − 1) must be a P-position and a parent of ( a, b, c). (See our paper for more details.) Now I want to explicitly write out the rules of an automaton which will generate the Nim P-positions in 3D. Let me restrict the evolution of the automaton to the non-negative octant. That is, we consider points ( a, b, c) in 3D, where each coordinate is a non-negative integer. We define the neighbors of the point ( a, b, c) to be the points that differ from ( a, b, c) in two coordinates exactly by 1. So each point strictly inside the octant has 12 neighbors. (There are three ways to choose two coordinates, and after that four ways to choose plus or minus 1 in each of them.

There is a geometric interpretation to this notion of neighborhood. Let us correspond a unit cube to a point with integer coordinates. The center of the cube is located at the given point and the sides are parallel to the axes. Then two points are neighbors if and only if the corresponding cubes share one edge. Now it becomes more visual that a cube has 12 neighbors, as it has 12 edges.

Here is the rule of the automaton. Points never die. We start with the patriarch, (0,0,0), one point being alive. The non-alive point is born inside the non-negative octant if it has exactly 1 alive neighbor that is closer to the patriarch. In other words the point ( a, b, c) is born if and only if exactly one out of three points ( a − 1, b − 1, c), ( a − 1, b, c − 1), and ( a, b − 1, c − 1) is alive. It follows that the points that are born at the n-th step has a coordinate sum 2 n. Consider for example the starting growth.

At the first step the points (0,1,1), (1,0,1) and (1,1,0) are born. At the next step the points (0,2,2) and (2,0,2) and (2,2,0) are born. While the (1,1,2) will never be born as starting from the second step it has at least two live neighbors: (0,1,1) and (1,0,1) that are closer to the patriarch.

In the resulting automaton, the points that are born at step n are exactly the P-positions of Nim with the total of 2 n tokens. Only the points with an even total can be born. Now we proceed by induction on the total number of tokens. The base case is obvious.

Suppose we proved that at step n exactly P-positions with 2 n tokens are born. Consider a P-position of Nim: ( a, b, c) such that a + b + c = 2 n + 2. Remember, that bitwise XOR of a, b, and c is zero. Consider the 2-adic values of a, b, and c (aka the smallest powers of 2 dividing a, b, and c).

There should be exactly two out of these three integers that have the smallest 2-adic value. Suppose these are a and b. Then ( a − 1, b − 1, c) is a P-position, while ( a − 1, b, c − 1) and ( a, b − 1, a − 1) are not. That means by the inductive hypothesis ( a, b, c) has exactly one alive neighbor. So the position ( a, b, c) is born at time n + 1. Now we need to proof that nothing else is born.

For the sake of contradiction suppose that ( a, b, c) is the earliest N-position to be born. That means it has a live neighbor that is a P-position closer to the patriarch. WLOG we can assume that this neighbor is ( a − 1, b − 1, c). If a − 1 and b − 1 are both even, then ( a, b, c) is a P-position, which is a contradiction. Suppose a − 1 and b − 1 are both odd. Then their binary representations can't have the same number of ones at the end. Otherwise, ( a, b, c) is a P-position.

That is a and b have different 2-adic values. Suppose a has a smaller 2-adic value, Then, for ( a − 1, b − 1, c) to be a P-position a and c has to have the same 2-adic value. That means ( a, b − 1, c − 1) is a P-position too. Now suppose a − 1 and b − 1 are of different parities.

Without loss of generality suppose a − 1 is odd and b − 1 is even, then c is odd. Then ( a − 1, b, c − 1) is a P-position too. Thus we can always find a second neighbor with the same number of tokens. That is, both neighbors are alive at the same time; and the N-position ( a, b, c) is never born. □ One might wonder what happens if we relax the automaton rule by removing the constraint on the distance to the patriarch.

Suppose a new point is born if it has exactly one neighbor alive. This will be a different automaton. Let us look at the starting growth, up to a permutation of coordinates. At step one, positions (0,1,1) are born. At the next step positions (0,2,2) are born. At the next step positions (0,1,3), (1,2,3) and (0,3,3) are born.

Refx Nexus Guitar Expansion Download Music. We see that (0,1,3) is not a P-positions. What will happen later? Will this N-position mess up the future positions that are born? Actually, this automaton will still contain all the P-positions of Nim.

In the new automaton, the points that are born at step n and have total of 2 n tokens are exactly the P-positions of Nim with the total of 2 n tokens. Only the points with an even total can be born. Now we proceed by induction on the total number of tokens.

The base case is obvious. The birth of the points that have total of 2 n tokens and are born at step n depend only on the points with the total of 2 n − 2 tokens that are born at step n − 1. By the inductive hypothesis, those are the P-positions with 2 n − 2 tokens. So the points have total of 2 n tokens and are born at step n match exactly the first automaton described above.

To reiterate, N-positions with 2 n tokens are born after P-positions with 2 n tokens, so they do not influence the birth of P-positions with 2 n + 2 tokens. □ The Longest Optimal Game of Nim In the game of Nim you have several piles with tokens. Players take turns taking several tokens from one pile.

The person who takes the last token wins. The strategy of this game is well-known. You win if after your move the bitwise XOR of all the tokens in all the piles is 0.

Such positions that you want to finish your move with are called P-positions. I play this game with my students where the initial position has four piles with 1, 3, 5, and 7 tokens each. I invite my students to start the game, and I always win as this is a P-position.

Very soon my students start complaining that I go second and want to switch with me. What should I do? My idea is to make the game last long (to have many turns before ending) to increase the chances of my students making a mistake.

So what is the longest game of Nim given that it starts in a P-position? Clearly you can't play slower then taking one token at a time. The beauty of Nim is that such an optimal game starting from a P-position is always possible. I made this claim in several papers of mine, but I can't find where this is proven. Contains an indirect proof by building a bijection to the Ulam-Warburton automaton. But this claim is simple enough, so I want to present a direct proof here. Actually, I will prove a stronger statement.

In an optimal game of Nim that starts at a P-position the first player can take one token at each turn so that the second player is forced to take one token too. Consider a P-position in a game of Nim. Then find a pile with the lowest 2-adic value. That is the pile such that the power of two in its factorization is the smallest. Suppose this power is k. Notice that there should be at least two piles with this 2-adic value.

The first player should take a token from one of those piles. Then the bitwise heap-sum after the move is 2 k+1−1. Then the Nim strategy requires the second player to take tokens from a pile such that its value decreases after bitwise XORing with 2 k+1−1. All piles with the 2-adic value more than k will increase after xoring with 2 k+1−1.

That means the second player has to take tokens from another pile with 2-adic value k. Moreover, the second player is forced to take exactly one token to match the heap-sum. □ In the position (1,3,5,7) all numbers are odd, so I can take one token from any pile for my first move, then the correct move is to take one token from any other pile. My students do not know that; and I usually win even as the first player. Plus, there are four different ways I can start as the first player.

This way my students do not get to try several different options with the same move I make. After I win several times as the first player, I convince my students that I win anyway and persuade them to go back to me being the second player.

After that I relax and never lose. Replace Question Marks Replace question marks with letters in the following sequence: Y, Y,?,?, Y,?, Y,?, R, R, R, R. Geometry Puzzles from the 35-th Lomonosov Tournament Problem 1. It is true that it is possible to put positive numbers at the vertices of a triangle so that the sum of two numbers at the ends of each side is equal to the length of the side? This looks like a simple linear algebra question with three variables and three equations. But it has a pretty geometrical solution.

Prove that it is possible to assign a number to every edge of a tetrahedron so that the sum of the three numbers on the edges of every face is equal to the area of the face. Again we have six sides and four faces.

There should be many solutions. Can you find a geometric one? Your Deal is Better than my Deal Do you know how to cut a cake? I mean, mathematically. There is a whole area of mathematics that studies cake cutting. Mathematicians usually assume that each person has their own idea of what is the best part of the cake. Suppose three sisters are celebrating the New Year by having a cake.

Anna likes only icing, Bella likes only chocolate chips, and Carol likes only pieces of walnuts mixed into the cake. Mathematicians want to cut cakes fairly. But what is fair? Fair is fair, right? There are several different notions of fair cake division. There is a proportionate division. In such a division every sister gets at least one-third of her value of the cake.

This seems fair. Let's see an example.

Anna gets one third of the icing, Bella gets one third of the chocolate chips, and Carol gets everything else. This is a fair proportionate division. Each of the sisters believes that she got at least one-third of the cake, in their own value.

But it doesn't seem quite fair. There is a stronger notion of fairness. It is called envy-free. In this division each sister gets at least one-third of the cake and, in addition, none of the three sisters would improve their value by swapping pieces. That means, if Anna wants only icing, not only does she get at least one-third of the icing, but also no one else gets more icing than Anna. The previous example of the proportionate division is not envy-free. Carol got two-thirds of the icing, so Anna would want to switch with her.

Let's try a different division. Anna gets one third of the icing, Bella gets the chocolate chips and another third of the icing, and Carol gets all the walnuts and another third of the icing.

Formally, this is envy-free cake cutting. But poor Anna. What do you think Anna feels when she sees the smiles of contentment on the faces of her sisters? Whoever invented the name doesn't understand envy. Anna got one-third of the cake by her value, but the other sisters got the whole cake! Luckily mathematicians understand this conundrum. So they invented another name for a cake division.

They call a division equitable if everyone values all the pieces the same. So the division above is envy-free but not equitable.

Let's try again. Let's give each sister one-third of all the components of the cake. This division is very good mathematically: it is proportionate, envy free, and equitable. By the way, envy-free division is always proportionate.

This division seems fair. But is it a good division? There is another term here: Pareto-efficient division means that it is impossible to make one person feel better, without making another person feel worse. All divisions above are not Pareto-efficient.

Moving some icing from Carol to Anna, doesn't decrease the value for Carol, but increases the value for Anna. There is an even better way to divide the cake. We can give the icing to Anna, the walnuts to Bella, and the chocolate chips to Carol.

This division is proportionate, envy-free, and Pareto-efficient. It is perfect. Mathematicians even have a word for it. They call it a perfect division.

Mathematically this division is perfect. Unfortunately, sisters are not. I know an Anna who would still envy Bella. The Reuleaux Tetrahedron The manhole covers are round because the manholes are round. But the cute mathematical answer is that the round shapes are better than many other shapes because a round cover can't fall into a round hole. If we assume that the hole is the same shape as the cover but slightly smaller, then it is true that circular covers can't fall into their holes. But there are many other shapes with this property.

They are called the shapes of constant width. Given the width, the shape with the largest area is not surprisingly a circle. The shape with the smallest area and a given constant width is a Reuleaux Triangle. Here is how to draw a Reuleaux triangle. Draw three points that are equidistant from each other at distance d. Then draw three circles of radius d with the centers at given points.

The Reuleaux triangle is the intersection of these three circles. Can we generalize this to 3d? What would be an analogue of a Reuleaux Triangle in 3d? Of course, it is a Reuleaux Tetrahedron: Take four points at the vertices of a regular tetrahedron; take a sphere at each vertex with the radius equal to the edge of the tetrahedron; intersect the four spheres. Is this a shape of the constant width? Many people mistakenly think that this is the case. Indeed, if you squeeze the Reuleaux tetrahedron between two planes, one of which touches a vertex and another touches the opposite face of the curvy tetrahedron, then the distance between them is equal to d: the radius of the circle.

This might give you the impression that this distance is always d. If you squeeze the Reuleaux tetrahedron between two planes that touch the opposite curvy edges, the distance between these planes will be slightly more than d. To create a shape of constant width you need to shave off the edges a bit. Theoretically you can shave the same amount off every edge to get to a surface of constant width. But this is not the cool way to do it.

The cool way is to shave a bit more but only from one edge of the pair of opposite edges. You can get two different figures this way: one that has three shaved edges forming a triangle, and the other, where three shaved edges share a vertex. These two bodies are called Meissner bodies and they are conjectured to be shapes of the constant width with the smallest volume. On the picture I have two copies of a pair of Meissner bodies.

The two left ones have three edges that share a vertex shaved off. The very left shape gives a top view of this vertex and the solid next to it has its bottom with holes looking forward. The two shapes on the right show the second Meissner body in two different positions. I recently discovered a. It falsely claims that the Reuleaux tetrahedron has constant width. I wrote to TED-Ed, to the author, and posted a comment on the discussion page. There was no reaction.

They either should remove the video or have an errata page for it. Knowingly keeping a video with an error that is being viewed by thousands of people is irresponsible. Who is Guilty? I am running a program for middle school students, where we try to do research in mathematics.

In the fall of 2015 we decided to study the following topic in logic. Suppose there is an island where the following four types of people live: • Absolute truth-tellers always tell the truth. • Partial truth-tellers tell the truth with one exception: if they are guilty, they will say that they are innocent. • Absolute liars always lie. • Responsible liars lie with one exception: if they are guilty, they will say that they are guilty.

See if you can solve this simple logic puzzle about people on this island. It is known that exactly one person stole an expensive painting from an apartment. It is also known that only Alice or Bob could have done it.

Here are their statements: Alice: I am guilty. Bob is a truth-teller. Bob: I am guilty. Alice stole it. Alice is the same type as me. Who stole the painting and what types are Alice and Bob?

My students and I discovered a lot of interesting things about these four types of people and wrote a paper:. This paper contains 11 cute logic puzzles designed by each of my 11 students. I envied my students and decided to create two puzzles of my own.

You have already solved the one above, so here is another, more difficult, puzzle: A bank was robbed and a witness said that there was exactly one person who committed the robbery. Three suspects were apprehended. No one else could have participated in the robbery. Alice: I am innocent. Bob committed the crime.

Bob is a truth-teller. Bob: I am innocent. Alice is guilty. Carol is a different type than me. Carol: I am innocent.

Alice is guilty. Who robbed the bank and what types are the suspects?

A Travel Puzzle The following cute puzzle of unknown origins was sent to me by Martina Balagovic and Vincent van der Noort: A car travels from A to B and back again. When going uphill it goes 56 km an hour, when going downhill it goes 72 km an hour and while driving on flat surface it goes 63 km an hour. Getting from A to B takes 4 hours and getting back (over the same road) takes 4 hours and 40 minutes. What is the distance between A and B?

Really Big Numbers.