The Girl Who Loved Math - Melanie Wood is only female to represent US in International Mathematical Olympiad

Polly Shulman

EVEN IN THE REMOTE MOUNTAINS OF CARPATHIA, IN A SMALL TOWN CALLED SINAIA--WHICH very few Americans have ever heard of or are likely to visit--one is rarely far from a peculiar piece of hardware that seems to have captivated teenagers around the world: a basketball hoop. [paragraph] In Carpathia, the teens scrambling under the hoops tend to be Romanian, of course, but on this particular day in Sinaia there also happen to be some American teens who have been invited to play. But they have something they like to play a whole lot better than basketball. [paragraph] The Americans, six teens aged 15 to 17, are here to practice for a different sport. In 10 days, on July 16, they will gather with 427 other teens from around the world in Bucharest to face off against six mathematics problems so difficult that many college professors would find them taxing. This competition is called the International Mathematical Olympiad, and the American contestants have been selected from high schools across the United States.

Before those grueling days in Bucharest, though, the Americans have decided to practice--in exhibition games, if you will--against the Romanian math team. Now, after hours of work, the Romanian team is taking a much needed break playing basketball, but the Americans can't let go.

"The Americans are upstairs," says Titu Andreescu, their coach. "Our students are not very athletic. Well, they are athletes of the mind."

In a stuffy dorm room festooned with dirty laundry, Andreescu's mental athletes are sprawled on beds and linoleum. One is lost in thought, a place he finds comfortable for hours at a stretch. Another stares intently at handouts of problems from the last Olympiad. The other four have gathered around a notebook, brandishing the mathematician's weapon--number 2 pencils--to attack a particularly thorny brain boggler. Melanie Wood, tall and green-eyed with blond hair, is the only girl in the room. In fact, Melanie is the only female ever chosen to represent the United States. She turns to the five boys in the room and offers a way to solve the geometry problem at hand. She suggests inversion, a strategy for turning circles into lines to see if the simpler relationship of lines to lines will open up a solution to the problem posed by the circles. The other pencil-wielders nod their heads and join in with gusto.

What might seem ridiculous to many people--that mathematics can be more engaging to teens than basketball or video games or even dating--could stand for absolute truth in this room at this moment.

But mathematical fun does not come without stress. The U.S. team has an impressive history--three wins in 26 years of competition. And in 1994 all six Americans were awarded perfect scores. The competition is never easy: The Chinese and the Russian teams are always a threat, as are teams from countries with strong mathematical traditions such as Romania, Iran, and Hungary. Individual futures are also at stake. Although not one of these teens will have any difficulty getting into college, winning a gold medal at the Olympiad could earn a full scholarship.

Melanie may be under greater stress than anyone on the U.S. team. This is her second Olympiad. The previous year, in Taiwan, she won a silver medal. "Once you win," she says, "you have attention on you. Particularly because I'm the only American girl."

AFTER A PRACTICE EXAM THAT THE ROMANIANS finish with troubling speed, the Americans huddle. What does the Romanian victory mean? Perhaps only math can tell them. Melanie and her friends start with definitions used for solving problems involving inequalities. Do the Romanians "dominate us?" one asks. In math terms, dominate would mean the Romanians' worst player has beaten the best U.S. player. "Do they majorize us?" another asks. This term is more complicated: It would be true if the Romanians' top player got a higher score than the top U.S. player, and if the sum of their top two players' scores beats the sum of our top two, and the sum of their top three beats the sum of our top three, and so on. "What will it mean if there are teams that we beat or tie but don't majorize?" asks Melanie, who then answers her own question: "It'll mean we're clumpier than they are."

"What do you mean by clumpy?" asks Lawrence Detlor, from New York City.

"Our scores are closer together."

"That's the opposite of what I thought you meant," he says. "I thought clumpy meant `containing separate clumps.'"

Melanie's mind has already raced ahead to a consideration of how the size of a team's home country might affect clumpiness. "It would seem like big countries would be more clumpy than little countries, because if you take the top six people in a big country, they'll probably be good. But small countries will tend to be clumpy, too, because they might not have anybody good. So they're clumpy, but not in an interesting way."

Later, when Melanie asks if anyone has a pamphlet from a previous Olympiad, Lawrence answers: "I do. Wait! What do you mean by pamphlet? It's probably the opposite of what I mean by pamphlet."

WHEN MELANIE ARRIVED AT HER FIRST AMERICAN TRAINING program for the Math Olympiad, in the summer of 1996, she was thrilled to find a group of kids who were not only as smart as she but who also saw the world through the same prism: "I had finally found peers. For the first time in my life, I was an average student. That meant fast-paced classes, being bombarded with exciting math I'd never seen before, never being bored. Before, I'd always thought maybe I was just weird. Here nobody is ever, ever embarrassed to be doing math on Friday night, at six in the morning--whenever you get the urge."

The urge has been with her since childhood, a fact that brings up interesting questions about whether genes are more important than environment in a child's life. After Melanie's father, Archie Wood, a middle-school math teacher, died of cancer when his daughter was just 6 weeks old, Melanie's mother decided to keep him alive through the subject he loved most. She began teaching Melanie math at age 3. "By the time I was 4, when I got bored walking around the mall, my mother would give me linear equations to solve in my head," Melanie recalls. (For example, if 3x + 2y =12 and x = 2, then what is y?) Melanie's mother, Sherry Eggers, who was then a language teacher, also tried to teach Melanie French and Spanish, but the languages didn't stick. Eggers's genius was to let her daughter explore at her own pace. She never pushed, never made choices for her; instead, she asked Melanie to decide for herself what she wanted to do.

So Melanie did. The result was both amazing and, at times, sad. In seventh grade, for example, she entered a national middle-school competition called MathCounts and quickly realized, "I wasn't just the best student in my school, or as good as any other best student in her own school. I was probably in the top handful in the nation. That separated me from my school friends, who didn't understand the problems and what they meant to me."

Melanie's mother didn't understand the math her seventh-grader knocked off with ease either. Suddenly Melanie was alone in her world, and her sense of isolation was heightened by an awareness of what she had lost: the father with whom she might have shared these very triumphs. "That's when my dad became an important figure in my life, instead of just someone I'd never met. I have a picture of him standing in front of a blackboard; (mod 5) was written on it, a concept not a lot of people would understand. I thought about how amazing it could have been to come home and have someone I could talk to about this huge thing in my life."

What makes Melanie different from a lot of people is that she has accepted her loss, and accepted it in youth rather than in maturity. She achieved what her mother had hoped--Melanie's father is a part of her. "He is with me in the competitions, or even when I am just thinking about math," she says. "His spirit and memory are there in my mind."

Because she was able to forge a profound connection with her father, and because her mother gave her so much room to grow on her own, Melanie seems to be that rare teenage girl who is not stymied by self-doubt. When she encounters a problem that appears insoluble, she moves ahead deliberately and methodically, searching for the solution with unusual certainty that she will be able to find it, and a conviction that if she fails, her failure is not a reflection of her self-worth. "I try to understand all the mathematical structures involved in a problem, even when they're not necessary to solve it, because that helps me to understand the problem better."

Melanie also thinks of herself as cooperative, not competitive, which makes her an ideal study partner. "Mathematicians work together," she says, an attitude that has helped meld the team into a cohesive unit. The students enjoy one another's company keenly and, when not working on math, they often play made-up mind games. For example, they might plunge into a variation of chess in which each piece moves as if it were the piece to its left, so that kings slide around the board like bishops and bishops hop around like knights. They delight in a game called "No Fifth Symbol," in which players must speak without using words that contain the letter e; or "Silent Football," a game with unstated rules that newcomers must deduce by watching.

Such inventiveness isn't just about fun, says Melanie--it's also helpful for solving math problems. "My sense is that we have good intuition about how to win games, so it's useful to rephrase a problem as if it were a game," she says. A problem about polygons inscribed inside a circle, for example, might be thought of as a game between two players who take turns drawing lines that connect points on the circle.

The team is also united by a fierce belief in the purity of mathematics, which can make their world seem a lot more dependable and certain. Melanie defines it this way: "You start from nothing and deduce whole worlds just from logic. You don't have to take into account arbitrary facts about the world around you. If we were in a different world where atoms could combine in different ways and sulfur were a different color, math would still be math."

AFTER SEVEN DAYS OF PRACTICING, THE TEAM FORSAKES THE SYLVAN refuge of Sinaia for sweltering Bucharest and the competition. Finally, the moment of truth and proof has arrived. Teams from 82 countries begin the first of two 4 1/2-hour sessions over two days. Contestants will be expected to find a solution to each problem and also to prove that each answer is correct.

At the end of the first session, students emerge looking grim and shaken. The next day they are even more depressed. "It made difficult look easy," says one contestant. A television crew collars Melanie on her way out--the American girl has become a media magnet. With polish and politeness, she tells the camera that the test was challenging, points out that the Romanians have a tradition of choosing challenging problems, and says the team members won't know how they did until the awards ceremony. To her teammates she whispers: "Come on, we've got to get out of here." They hurry back to the safety and respite of the dorm.

With the problem-solving done, team coaches must fight for points, a process one coach describes as "horse trading." First, students meet with their coaches and go over copies of their exams, explaining what they had in mind. Sometimes they can find a partial solution to a problem hidden among false starts. The coaches try to convince the judges to grant partial credit. While math itself can be beautiful and universal, it turns out that a contest is, nonetheless, a contest.

When the dickering finally ends, Russia and China have tied for first. Romania is fourth, and the United States places ninth. Nevertheless, Reid Barton and Paul Valiant win gold medals; Gabriel Carroll, Po-Shen Loh, and Melanie get silver; and Lawrence Detlor is awarded a bronze. Characteristically, Melanie is not rocked by the loss. "As a team, we certainly didn't do better than average for the United States," she says. "Still, all the teams that placed above us are very good."

Coach Andreescu reassures his charges: "The coordinator congratulated me for the thoroughness of your solutions. Even though they were not the simplest, they showed mathematical maturity that impressed him." Later he adds, "That test would have been a challenge for a professional mathematician."

A PROFESSIONAL MATHEMATICIAN IS WHAT MELANIE PLANS TO be, although she just might "end up directing on Broadway," because she is majoring in theater as well as math. She has finished her freshman year at Duke University, which she chose over Harvard because she thought the math department there was "cold and competitive." She has been taking graduate-level classes in real analysis, complex analysis, and algebraic number theory, along with a drama class entitled Voice and Body Gesture.

This month Melanie is working with the U.S. math team's summer program as a grader, which gives her a chance to mentor younger mathematicians bound for next month's Olympiad in Seoul. All through high school she volunteered at MathCounts, the middle-school competition in which she had competed. "One of my jobs was emceeing the Cool Down Round, which follows the official competition. I run around with a mike among students who are furiously solving problems, and I jump on tables, yelling, `Hey, hey, we have an answer over here.' It's nothing like an actual math competition. Competitions are silent. Most of the math I do isn't competition math. It's openly working with others and full of laughter." Like her father, Melanie is passing along her love of mathematics to others in an exuberant, generous way that no doubt would have made him proud.

SATISFYING PAIRS

This was one of six problems contestants were asked to solve at the International Mathematical Olympiad in Bucharest last year. This year's Olympiad will be held in Seoul, South Korea, on July 18 and 19.

Find all pairs (n,p) of positive integers such that

* p is prime

* n [is less than or equal to] 2p

* [(p-1).sup.1] + 1 is divisible by [n.sup.p-1]

Hint: The cases P|n and p/n should be handled separately, In the latter case, consider the congruence [(p-1).sup.n] [equivalent] -1 modulo a suitable prime divisor of n.

MATH OLYMPICS SOLUTIONS

The pairs satisfying the given condition are (n,p) = (1,p) for any prime p, (2,2) and (3,3). The reader may easily check that these actually are solutions, and that there are no more solutions for p = 2, 3. So we may assume hereafter p [is greater than or equal to] 5 and n [is greater than or equal to] 2. In particular, since [(p - 1).sup.n] + 1 is odd and n divides this quantity, n must be odd.

First suppose p divides n; since n [is less than or equal to] 2p and n is odd, in fact n = p. Expanding [(p - 1).sup.p] by the binomial expansion reveals that

[(p - 1).sup.p] + 1 [equivalent] (p/1])p - (p/2)[p.sup.2] + (p/3)[p.sup.3] ... [equivalent] [p.sup.2] (mod [p.sup.3])

for p [is greater than or equal to] 3, which gives a contradiction for p [is greater than or equal to] 5. Alternatively (as noted by Lawrence Detlor), if [(p - 1).sup.p] + 1 = [kp.sup.p-1], then clearly k [is less than] p, but k must also be congruent to 1 modulo p - 1, so k = 1. On the other hand, [x.sup.y] [is greater than] [y.sup.x] for x [is greater than] y [is greater than or equal to] e (this reduces to the fact that (log x)/x is decreasing for x [is greater than or equal to] e, which may be shown by easy calculus), so [(p - 1).sup.p] [is greater than] [p.sup.p-1] for p [is greater than or equal to] 5, a contradiction.

Thus we may assume p does not divide n. We will give two proofs of the fact that [(p - 1).sup.n] [equivalent] 1 (mod n) has no solutions with p not dividing n.

* First Proof: Since n [is greater than] 1, it has a smallest prime divisor q. Since q - 1 has all its prime divisors less than q, n and q - 1 have no common prime divisor, that is, they are relatively prime. We now note that this implies [x.sup.n] [equivalent] [y.sup.n] (mod q) if and only if x [equivalent] y (mod q). This is obvious if either x or y is divisible by q; if not, we apply Fermat's little theorem in the form [x.sup.q-1] [equivalent] [y.sup.q-1] [equivalent] 1 (mod q). From Euclid's algorithm, we recall that there exist integers a, b with an + b (q - 1) = 1, and so

x [equivalent] [x.sup.an+b(q - 1) [equivalent] [x.sup.an] [equivalent] [y.sup.an] [equivalent y (mod q).

In the case of interest, this means p - 1 [equivalent] - 1 (mod q), a contradiction since p cannot equal q by our assumption that p does not divide n.

* Second Proof: Suppose the claim is false, and let n be the smallest integer greater than 1 such that [(p - 1).sup.n] [equivalent] - 1 (mod n). Let m be the smallest positive integer such that [(p - 1).sup.m] [equivalent] - 1 (mod n) and let d be the order of p - 1 modulo n. On one hand, m [is less than] d since otherwise e could have been replaced by e - d; on the other hand, d divides 2m since [(p - 1).sup.2m] [equivalent] 1 (mod n). Therefore d = 2e, and since [(p - 1).sup.2n] [equivalent] 1 (mod n), d divides 2n and so m divides n. From [(p - 1).sup.m] [equivalent] - 1 (mod n) and the fact that e divides n, we have [(p - 1).sup.m] [equivalent] - 1 (mod m). However, m [is not equal to] 1 (since n does not divide (p - 1) + 1 = p) and m [is less than] n (since m divides [Phi](n), which is less than n for n [is less than] 1), so m is a smaller counterexample to the assertion, contradicting the choice of n.

In either case, we conclude that the only solutions are those given above.

COPYRIGHT 2000 Discover
COPYRIGHT 2000 Gale Group