Tanya Khovanova's Math Blog

I stumbled on a Russian linguistics competition called The Russian Little Bear. Most of the puzzles are Russian-specific; but some of them can be translated. I concentrated on puzzles for grades six through nine and used Unicode for uncoding strange characters.

Problem 1. Here are some Latin words with their English translations:

• amo — I love
• amat — He loves
• invitor — I am invited
• invitaris — You are invited
• rogas — You ask
• rogatur — He is asked

Pick the line of words from A to E that best translates these phrases into Latin: You are loved, I ask, He invites.

• (A) amas, rogo, invitat;
• (B) amaris, rogo, invitat;
• (C) amaris, rogor, invitas;
• (D) amaris, rogat, invitatur;
• (E) amaris, rogo, invito.

Problem 2. The first astronauts from India (I), Hungary (H), France (F) and Germany (G) were Bertalan Farkas (1), Sigmund Jähn (2), Rakesh Sharma (3) and Jean-Loup Chrétien (4). Match the astronauts to the countries:

• (A) I2, H1, F4, G3;
• (B) I3, H1, F4, G2;
• (C) I3, H1, F2, G4;
• (D) I1, H4, F3, G2;
• (E) I3, H2, F4, G1.

Problem 3. You do not need to know Russian to solve this puzzle. It is enough to know the modern Russian alphabet: А, Б, В, Г, Д, Е, Ё, Ж, З, И, Й, К, Л, М, Н, О, П, Р, С, Т, У, Ф, Х, Ц, Ч, Ш, Щ, Ъ, Ы, Ь, Э, Ю. Before XVIII century, numbers in Russian were denoted by letters, for example: ТЛЕ — 335, РМД — 144, ФЛВ — 532.

How was 225 written in old Russian?

(A) ВВФ; (B) ВВЕ; (C) СКЕ; (D) СКФ; (E) ВНФ.

Problem 4. Here are several Turkish words and phrases with their English translations:

• ada — an isle
• adalar — isles
• iki tas — two cups
• adam — a man
• otuz adam — thirty men
• taslar — cups

Pick the line of words from A to E that best translates these phrases into Turkish: thirty isles, men?

• (A) otuz adalar, adamlar;
• (B) otuz ada, adam;
• (C) otuz adalar, adam;
• (D) otuz ada, adamlar;
• (E) ikilar ada, adamlar.

Due to the popularity of my previous posting of linguistics puzzles, I’ve translated some more puzzles from the online book Problems from Linguistics Olympiads 1965-1975. All of them are from the phonetics section and I’ve kept the same problem number as in the book. I’ve used the Unicode encoding for special characters.

Problem 20. In the table below there are numerals from some Polynesian languages. Note that I couldn’t find the proper English translation for one of the languages, so I used transliteration from Russian. The language sounds like “Nukuhiva” in Russian.

Your task is to find the words that should be in the empty cells. Note that wh, ‘, and ŋ denote special consonants.

Problem 21. Below you will find words in several relative languages. You can group these words into pairs or triples of words with the same origin and the same or a similar meaning.

āk, dagr, bōk, leib, fōtr, waʐʐar, buoh, dæʒ, plōgr, hām, wæter, hleifr, pfluog, eih, heimr, fuoʐ, plōʒ.

Task 1. Divide the words into groups so that the first group has words from the same language, the second group has words from another language and so on.

Task 2. (optional) List your suggestions about the meanings of the words and about the identity of the languages.

Problem 22. These words from the Aliutor language are followed by their translations. The stresses are marked by an apostrophe in front of the stressed vowel.

• t’atul — fox
• nətɣ’əlqin — hot
• nur’aqin — far away
• ɣ’əlɣən — skin
• n’eqəqin — fast
• nəs’əqqin — cold
• tapl’aŋətkən — He sews shoes
• k’əmɣətək — to roll up
• ʔ’itək — to be
• paq’ətkuk — gallop
• n’ilɣəqinat — white (they both)
• p’unta — liver
• qet’umɣən — relative
• p’iwtak — to pour
• nəm’itqin — skillful
• t’umɣətum — friend
• t’ətka — walrus
• k’əttil — forehead
• qalp’uqal — rainbow
• kəp’irik — hold (a baby in the hands)
• təv’itatətkən — I work
• p’intəvəlŋək — attack (each other)

Your task is to put the stresses in the following words: sawat ‘lasso’, pantawwi ‘fur boots’, nəktəqin ‘solid’, ɣətɣan ‘late autumn’, nəminəm ‘bouillon’, nirvəqin ‘sharp’, pujɣən ‘spear’, tilmətil ‘eagle’, wiruwir ‘red fish’, wintatək ‘to help’, nəmalqin ‘good’, jaqjaq ‘seagull’, jatək ‘to come’, tavitətkən ‘I will work’, pintətkən ‘he attacks (someone)’, tajəsqəŋki ‘in the evening’.

Note that the vowel ə is similar to many unstressed syllables in English words, such as the second syllable in the words “taken” and “pencil”. This vowel is shorter than other vowels in the Aliutor language.

Last time Sue refinanced her mortgage was six years ago. She received a 15-year fixed loan with 5.5% interest. Her monthly payment is $880, and Sue currently owes $38,000.

Sue is considering refinancing. She has been offered a 5-year fixed loan with 4.25% interest. You can check an online mortgage calculator and see that on a loan of $38,000, her monthly payments will be $700. The closing costs are $1,400. Should Sue refinance?

Seems like a no-brainer. The closing costs will be recovered in less than a year, and then the new mortgage payments will be pleasantly smaller than the old ones. In addition, the new mortgage will last five years instead of the nine years left on the old mortgage.

What is wrong with this solution? What fact about Sue’s old mortgage did I wickedly neglect to mention? You need to figure that out before you decide whether Sue should really refinance.

So many people liked the puzzles I posted in Subtraction Problems, Russian Style, that I decided to present a similar collection of multiplication and division puzzles. These two sets of puzzles have one thing in common: kids who go for speed over thinking make mistakes.

Humans have 10 fingers on their hands. How many fingers are there on 10 hands?

This one is from my friend Yulia Elkhimova:

Three horses were galloping at 27 miles per hour. What was the speed of one horse?

Here is a similar invention of mine:

Ten kids from Belmont High School went on a tour of Italy. During the tour they visited 20 museums. How many museums did each kid go to?

Another classic:

How many people are there in two pairs of twins, twice?

Can you add more puzzles to this collection?

A stick has two ends. If you cut off one end, how many ends will the stick have left?

This pre-kindergarten math problem was given to me by Maxim Kazarian who lives in Moscow, Russia. That got me thinking about math education in the US. Actually, just about anything can get me thinking about education in our country. One of our math education patterns is to provide simplified templates and to train kids to plug numbers into them without thinking.

Math education should be about thinking. We need to give kids a lot of math problems that do not fit into standard templates, in order to encourage creative thinking. Here is another puzzle from Maxim:

A square has four corners. If we cut one corner off, how many corners will the remaining figure have?

I invite my readers to invent additional problems that sound as if a subtraction by one is needed, when, in fact, it is not. Here is my contribution:

Anna had two sons. One son grew up and moved away. How many sons does Anna have now?

To get to the national swimming championship, you need to win the state running championships.

What? Is that a joke? Perhaps you’re having the same reaction. Because this is exactly what is happening with math competitions. The official USA math competition has three rounds: AMC, AIME and USAMO.

AMC is a multiple-choice competition with 25 problems in 75 minutes. To be good at it, you need speed, accuracy and the ability to guess well.

AIME is 3 hours long and has 15 problems. The problems are a different level of difficulty and guessing will not help you. Though AIME is also multiple-choice, unlike AMC where you choose out of 5, in AIME you choose out of 1,000. But you still need speed and accuracy. A small arithmetic mistake will cost you the whole problem.

USAMO is a competition that runs for 9 hours and has 6 problems. The problems are much harder and you have to write proofs. Proofs? What proofs? Where are the proofs coming from? It is like you got to the national swimming championship because you are a great runner, but you do not know how to swim.

As the result of this system of selection, the USA team at the International Math Olympiad has diverse skills: these kids are good at all three types of the math competitions. It is like taking an Olympic triathlon team to the Olympic swimming event.

However, the US loses by selecting in this way. There are many kids who are great mathematicians: they may be good at difficult problems but not that good at speed-racing problems. An arithmetic mistake costs you only one point at IMO, but a whole problem at AIME. There are kids who are deep mathematicians prone to small arithmetic mistakes. They could get a gold medal at IMO, but they can’t pass AMC or AIME.

Not only that. As many AMC problems are boring and do not require ideas, AMC might discourage kids from all math competitions at an early stage.

I will write later with my ideas about how to change AMC. Right now I would like to offer a solution to a smaller problem. I am sure that the US math team organizers know many cases in which a non-senior kid does great at USAMO and is potentially a team member for the next year’s US IMO team, but, oops, next year he can’t pass AMC.

I suggest the following: USAMO participants are allowed to go to next year’s AIME no matter what their AMC scores are. USAMO winners are allowed to go to the next year’s USAMO no matter what their AIME results are. This way kids who have proved that they are great swimmers do not need to compete in running again.

I have forgotten all the problems from the math Olympiads I participated in, except for one. That problem cost me one point, and as result I didn’t get a perfect score. Here is Problem Number 4 from the 1976 IMO:

Determine the greatest number that is the product of some positive integers, and the sum of these integers is 1976.

The solution goes like this. Consider a partition of 1976. If there is an integer x in the partition greater than 5, then replacing it with two integers x-2 and 2 gives a bigger product. Replacing 4 in the partition by 2 and 2 doesn’t change the product. On the other hand, if there is 1 in the partition, it is profitable to attach it to any other number: to replace 1 and x by x+1.

That means the maximum-product partition of a number greater than one has to consist only of twos and threes. If we have three twos in the partition, we can replace them by two threes, thus increasing the product. That means that our partition should consist of twos and threes, and the number of twos shouldn’t be greater than three.

I lost my point when I made an elementary arithmetic mistake while calculating the remainder of 1976 modulo 3.

Let’s generalize this problem for any number n. If we define the maximum product of partitions a(n) as a function of the number n, we get the sequence A000792 in the Encyclopedia of Integer Sequences. This sequence has other interesting definitions, which appear if we add some structure to partitions of n.

For example, we can suppose that our n is not just a number, but it also represents n objects that are being permuted by the symmetric group S_n. In this case, we can associate some Abelian subgroups of the symmetric group with every partition. Namely, we can divide the set of n elements into disjoint subsets of sizes corresponding to our partition and cycle the elements in every subset. The product of these cycles is an Abelian subgroup, the order of which doesn’t depend on how exactly we partition or how we cycle. This order is the product of the partition numbers. It appears that we can get all maximal Abelian subgroups of the symmetric group in this manner. Thus the sequence a(n) is the maximum size of the maximal Abelian subgroup in a symmetric group S_n.

We can do something else with our n objects. Suppose they are represented by vertices of a graph. We take any partition of n and divide our graph into groups of vertices according to this partition. Next we connect vertices of the graph that belong to different groups. If we take a vertex from every group, then the corresponding subgraph is a clique. We can see that it is a maximal clique as two vertices from the same group are never connected and we can’t add any more vertices. The number of different cliques of this size is the product of the number of elements in every group. We can prove that a(n) is the maximum possible number of maximal cliques in a graph with n vertices.

Even though I can’t return to 1976 to correct my stupid mistake, I love this problem and the corresponding sequence.

How many disjoint letters “O” can you draw on a plane? Clearly, the answer is infinitely many. Now the real question is whether this number can be uncountable. We assume that the letter “O” is just a circle. In this case, if we draw all possible concentric circles, we get an uncountable number of circles. I call letters for which you can build a disjoint uncountable set on the plane stackable letters.

If we do not allow one letter “O” to be drawn inside another, then we can prove that it is not possible to draw more than a countable number of letters “O”. Indeed, we can always find a special point inside each letter “O” with both coordinates being rational numbers. As circles do not overlap, no two circles can share the same special point. The set of ordered pairs of rational numbers is countable, thus this set of letters “O” has to be countable.

Now let us move to a more interesting question: how many disjoint letters “T” can you draw on a plane? Let us define the letter “T” as two perpendicular line segments, where the end point of one of the segments is the middle point of the other. Prove that the letter “T” is unstackable.

Can you analyze the whole alphabet for stackability? It doesn’t matter which alphabet — you just need to define the drawing of every letter in some reasonable manner.

My real question is, can you define the drawings of English letters in such a way that they are recognizable and the number of stackable letters is maximized? What is the biggest number of letters that you can make stackable?

I brought up this subject to my son Sergei one fine morning. As a result we came up with the following breakfast theorem: Stackable letters are topologically equivalent to a line segment or a circle.

And here is our proof. If the letter in question has a part which is topologically equivalent to “T”, then the letter is not stackable, because “T” is not stackable.

Fortunately, or unfortunately, this is not the whole story. The stackability of each letter depends on how you define its shape. There are two groups of letters in the English alphabet that you can make stackable. The first group consists of letters like “V” that can be described as a function on a segment and stacked as a parallel translation of the same letter. The second group, like letter “O”, can be described as a function of an angle in polar coordinates, and stacked as a dilation of the same letter.

It is clear that this doesn’t cover all cases. For example, a piece of a spiral can be stackable — you stack it by rotating it.

I invite my readers to define two different homeomorphic shapes for the same letter — stackable and unstackable.

Romanian Masters in Mathematics might become the most prestigious math competition for high school students. Romania invites teams from the 20 best countries from the previous year’s International Math Olympiad. This way they guarantee that all the competitors are extremely strong and they can give more difficult problems than the usual IMO problems.

This year they held their second competition and my son, Sergei Bernstein, was invited to join the USA team. Here is one of the problems:

For a finite set X of positive integers, define ∑(X) as the sum of arctan(1/x) for all elements x in X. Given a finite set S, such that ∑(S) < π/2, prove that there exists a finite set T such that S is a subset of T and ∑(T) = π/2.

The official solution involved a tedious greedy algorithm. My son, Sergei, got an extra point for his unique solution, which was very different from the official one. Here’s his solution:

To an integer x we associate a complex number x + i, which in polar coordinates is r(cosθ + i sinθ). Note that θ equals arctan(1/x). That means we can interpret ∑(X) as the angle in the polar form of ∏(x+i) — the product of (x+i) for all x in X.

We are given a set S with elements s_j such that ∏(s_j+i) has a positive real part, and we need to find other elements t_k, such that ∏(s_j+i) multiplied by ∏(t_k+i) has real part 0.

But we know that the ∏(s_j+i) = a + bi, for some integers a and b. I claim that we can always find a positive integer y, so that (a + bi)(y + i) has a smaller real part than a.

Note that ∑(S) < π/2, hence a and b are positive. By an ever so slightly modified version of the division algorithm we can find integers q and r such that b = aq + r, 0 < r ≤ a. Now simply set y equal to q + 1 (which is a positive integer). Then the real part of (a + bi)(y + i) equals ay – b = a – r, and is a non-negative number less than a.

Now we just repeat the process, which obviously has a finite number of steps.

My son’s solution wasn’t complete. The problem talked about sets of numbers and that implies that the numbers should be distinct. I leave it to my readers to finish this solution for distinct numbers.

You are visiting your girlfriend and she orders pizza. Your evil girlfriend has perfect eyesight and notices fly droppings in three places on the pizza. She is seeking revenge on you for refusing to babysit her poodle and proceeds to cut the pizza. Assuming that the droppings occurred independently and in random places, what is the probability that she will be able to cut a half-circle pizza slice with all three droppings in one half?

Now that I’ve ruined your appetite, here’s a simpler puzzle you can solve as an appetizer to the one I just gave you above. In this puzzle, you have a bread stick that is, of course, in the shape of a line segment. Your girlfriend notices that the bread stick also has three fly droppings on it and she offers to cut an exact half length out of the middle for you. What is the probability that she can cut it so that you end up with all the droppings?

Can you bear to continue this tasty discussion? If so, I’ll give you an even simpler problem. Try to solve both of the above puzzles — but with only two droppings instead of three.

If I’ve spoiled your appetite so completely that you’ll skip dinner, why not use the extra time to solve the most challenging of these problems with a meatball that has four fly droppings on it. These droppings, too, are in random places and you must calculate the probability that you are able to cut a semi-sphere with all four droppings on one side.

This final puzzle — in a less distasteful setting — was sent to me by Nick Petry.