Archive for April 2009

## Oleg Kryzhanovsky’s Problems

A long ago my son Sergei went to the Streamline School Olympiad. Some of the problems were really nice and I asked the organizer, Oleg Kryzhanovsky, where he took the problems from. It seems that he himself supplied all the problems, many of which are his original creations. He told me that he can invent a math Olympiad problem on demand for any level of difficulty on any math topic. No wonder that he is the author of almost all math problems at the Ukraine Olympiad.

The following is a sample of his problems from the Streamline Olympiad. For my own convenience I have chosen problems without figures and equations. Note: I edited some of them.

1998 (8th - 9th grade). Find three numbers such that each of them is a square of the difference of the two others.

1999 (9th - 10th grade). The positive integers 30, 72, and N have a property that the product of any two of them is divisible by the third. What is the smallest possible value of N?

1999 (9th - 10th grade). You have 6 coins weighing 1, 2, 3, 4, 5 and 6 grams that look the same. The number (1, 2, 3, 4, 5, 6) on the top of each coin should correspond to its weight. How can you determine whether all the numbers are correct, using the balance scale only twice?

1999 (11th - 12th grade). In how many ways can the numbers 1, 2, 3, 4, 5, 6 be ordered such that no two consecutive terms have a sum that is divisible by 2 or 3.

2000 (6th - 7th grade). Let A be the least integer such that the sum of all its digits is equal to 2000. Find the left-most digit of A.

2000 (8th grade). You have six bags with coins that look the same. Each bag has an infinite number of coins and all coins in the same bag weigh the same amount. Coins in different bags weigh 1, 2, 3, 4, 5 and 6 grams exactly. There is a label (1, 2, 3, 4, 5, 6) attached to each bag that is supposed to correspond to the weight of the coins in that bag. You have only a balance scale. What is the least number of times do you need to weigh coins in order to confirm that the labels are correct?

## A Killer Puzzle

I’ve been translating a lot of linguistics puzzles lately. Now it is my turn to create a new linguistics puzzle. Here are some English phrases with their Russian translations:

• John killed Mary — Джон убил Мэри
• Mary killed Sam — Мэри убила Сэма
• Sam killed John — Сэм убил Джона

• John killed Sam
• Mary killed John
• Sam killed Mary

Bonus question. Have you noticed any signs that I am getting tired of linguistics?

## Linguistics Puzzles for Middle School

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
• 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:

• iki tas — two cups
• 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?

## Brown Sharpie of Courtney Gibbons

Courtney Gibbons gave me her permission to add her webcomics to my collection of Funny Math Pictures.

## Abstruse Goose’s Million Dollar Idea

Abstruse Goose gave me his permission to add his webcomics to my collection of Funny Math Pictures.

## Is There Hope for a Female Fields Medalist?

Until the introduction of the Abel prize, the Fields medal was the most prestigious prize in mathematics. The medal has been awarded 48 times and all of the recipients have been men. Can we conclude that women are inferior to men when it comes to very advanced mathematics? I do not think so.

The Fields medal was designed for men; it is very female-unfriendly. It is the prize for outstanding achievement made by people under age 40. Most people start their research after graduate school, meaning that people have 10-15 years to reach this outstanding achievement. If a woman wants to have children and devote some time to them, she needs to do it before she is 40. That puts her at a big disadvantage for winning the medal.

Recently the Abel prize for mathematics was introduced. This is the math equivalent of a Nobel prize and nine people have received the prize, all of them male. The Wolf prize is another famous award: 48 people have received it so far and they too have all been male.

On the grand scale of things, women have only recently had the option of having a career in mathematics. Not so long ago it was considered quite exceptional for a woman to work in mathematics. The number of female mathematicians is increasing, but as this is a new trend, they are younger people. At the same time, Abel prizes and Wolf prizes are given to highly accomplished and not-so-young people. That means the increase in the percentage of women PhDs in mathematics might affect the percentage of females getting the prize, but with a delay of several dozen years.

There are other data covering extreme math ability. I refer to the International Math Olympiad. The ability that is needed to succeed in the IMO is very different from the ability required to succeed in math research. But still they are quite similar. The IMO data is more interesting in the sense that the girls who participate are usually not yet distracted by motherhood. So in some sense, the IMO data better represents potential in women’s math ability than medals and prizes.

Each important math medal or prize is given to one person a year on average. So the IMO champion would be the equivalent of the Fields medal or the Wolf prize winner. While no girl was the clear best in any particular year, there were several years when girls tied for the best IMO score with several other kids. For example:

• In 1995, 14 students tied for the perfect score; two of them were girls. (Maryam Mirzakhani and Chenchang Zhu)
• In 1994, 22 students tied for the perfect score; two of them were girls. (Theresia Eisenkölbl and Catriona Maclean)
• In 1991, 9 students tied for the perfect score; one of them was a girl. (Evgenia Malinnikova)
• In 1990, 4 students tied for the perfect score; one of them was a girl. (Evgenia Malinnikova)
• In 1987, 22 students tied for the perfect score; one of them was a girl. (Jun Teng)
• In 1984, 8 students tied for the perfect score; one of them was a girl. (Karin Gröger)

In one of those years, a girl might have been the best, but because the problems were too easy, she didn’t have a chance to prove it. Evgenia Malinnikova was an outstanding contender who twice had a perfect score. In 1990, she was one out of four people, and she was younger than two of them, as evidenced by the fact they they were not present in 1991. Only one other person — Vincent Lafforgue — got a perfect score in 1990 and 1991. We can safely conclude that Evgenia was one of two best people in 1990, because she was not yet a high school senior.

This might be a good place to boast about my own ranking as IMO Number Two, but frankly, older rankings are not as good as modern ones. Fewer countries were participating 30 years ago, and China, currently the best team, was not yet competing.

Girls came so close to winning the IMO that there is no doubt in my mind that very soon we will see a girl champion. The Fields medal is likely to take more time.

## Phonetics Puzzles

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.

Languages 1 2 3 4 5 6 7 8 9 10
Hawaiian kahi lua   ha lima ono hiku walu   *****
Māori tahi rua toru wha   ono whitu waru iwa *****
Nukuhiva tahi   to’u ha   ono   va’u   *****
Rarotongan ta’i     ‘a rima ono ‘itu varu iva ŋa’uru
Samoan tasi lua     lima ono fitu   iva ŋafulu

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.

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
• 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.

## Sue’s Mortgage Puzzle

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.

## Multiplication Problems

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?

## USAMO and the Election, by John Berman

Today I have my first invited guest blogger, John Berman. John is a 2006, 2007, 2008 and 2009 USAMO qualifier. He was also selected to be on the US team at the Romanian Masters in Mathematics competition. Also, he placed 6th at the North American Computational Linguistics Olympiad. Here is his piece:

The analysis is based on the list of 2009 USAMO qualifiers.

There is a rule that if nobody naturally qualifies for the USAMO from a state, then the highest scoring individual will qualify. Unfortunately, this means that we must remove those states with only one USAMO qualifier. We have 33 states remaining. If we sort these strictly by number of USAMO qualifiers, then we find the following result.

States with at least 4 USAMO qualifiers (24 total) voted for Obama, with the following exceptions: Georgia, Texas, South Carolina, and Missouri. In addition, of the two states with 3 USAMO qualifiers, one voted for Obama and one for McCain. The remaining states with 2 qualifiers (5 total) voted Republican.

Now this is not really unexpected. States with very large populations tend to be democratic and also produce more USAMO qualifiers. The most notable exceptions are Georgia and Texas, both of which were indeed exceptions (major outliers, in fact) above. This prompts the following consideration.

States with at least 8 USAMO qualifiers per 10 million residents (25 total) voted for Obama, with the following exceptions: Florida, Wisconsin, South Carolina, Missouri, and Georgia. Of these, all but Georgia fall within 50% of the target 8 USAMO qualifiers per 10 million residents. Georgia has 18 qualifiers per 10 million residents. Note also that the entire USA has 16 qualifiers per 10 million residents.

Furthermore, if USAMO qualifiers had been used instead of population for determining electoral votes, Obama would have won with 86% of the vote rather than 68%. In general, if the Democrat can secure all those states with at least 1 qualifier per million residents (plus DC), he will win with 303 votes. He can even lose the three red states in that category (Georgia, Missouri, and South Carolina) for exactly 269.

USAMO qualifiers per 10 million residents (for states with more than one qualifier) are:

• NH — 122 (all of their qualifiers are from Phillips Exeter Academy)
• MA — 55
• ME — 30
• CA — 29
• NJ — 29
• CT — 29
• VA — 28
• MD — 25
• WA — 21
• IN — 19
• OR — 19
• NY — 18
• GA — 18
• IA — 17
• NM — 15
• MI — 15
• PA — 14
• MO — 14
• SC — 13
• IL — 13
• NC — 13
• OH — 9
• CO — 8
• MN — 8
• UT — 7
• KS — 7
• WI — 7
• KY — 7
• TX — 7
• FL — 6
• LA — 5
• AL — 4
• TN — 3

The states with only one USAMO qualifier are WY, VT, ND, AK, SD, DE, MT, RI, HI, ID, NE, WV, NV, AR, MS, OK, and AZ. The only blue one of these which falls below 8 qualifiers per 10 million is Nevada (we would expect it to have at least 2 qualifiers to fit the expected pattern). Otherwise, it is at least possible that each state fits the pattern of 8 qualifiers per 10 million residents if and only if it votes Democratic.