- Membership
- MAA Press
- Meetings
- Competitions
- Community
- Programs
- Students
- High School Teachers
- Faculty and Departments
- Underrepresented Groups
- MAA Awards
- MAA Grants

- News
- About MAA

August 3, 2000

Today we celebrate the 100th Math Chat column. The first column, "Father's Day Puzzler," appeared in *The Christian Science Monitor* on June 14, 1996.

US TEAM won six medals at the International Mathematical Olympiad (IMO) held in Taejon, South Korea, July 19 and 20, 2000. The top twelve teams and their scores out of a possible 252 points were:

China (218), Russia (215), USA (184), South Korea (172), Bulgaria (169), Vietnam (169), Belarus (165), Taiwan (164), Hungary (156), Iran (155), Israel (139), Romania (139).

The 2000 USA IMO Team Members:

Reid Barton, Homeschooled, Arlington, MA, GOLD Medalist

George Lee, Jr., Aragon High School, San Mateo, CA, GOLD Medalist

Ricky Liu, Newton South High School, Newton, MA, SILVER Medalist

Po-Ru Loh, James Madison Memorial HS, Madison, WI, SILVER Medalist

Oaz Nir, Monta Vista HS, Saratoga, CA, GOLD Medalist

Paul Valiant, Milton Academy, Belmont, MA, SILVER Medalist

Next year the competition will be held in Washington, DC.

As this week's **New Challenge,** we present a question from this year's IMO competition, submitted by Sandor Dobos, Hungary:

A magician has one hundred cards numbered 1 to 100. He puts them into three boxes, a red one, a white one and a blue one, so that each box contains at least one card. A member of the audience selects two of the three boxes, chooses one card from each and announces the sum of the numbers on the chosen cards. Given this sum, the magician identifies the box from which no card has been chosen. How many ways are there to put all the cards into the boxes so that this trick always works? (Two ways are considered different if at least one card is put into a different box.)

**Old Challenge** (John Shonder). A small creature needs to find its way from a random point on a mile-wide strip of land to the sea on either side, by going distance r in a random direction and then (if it has not yet found the water), turning once t degrees to the right and continuing thereafter in a straight line. What choices of r and t minimize the expected journey length to the sea?

**Answer.** David Freeman and Andrew Cotton note that you can be sure of reaching the sea by going 1 mile and then turning 90 degrees to the right. Of course if you started near one shore and headed almost directly toward the other shore, not quite reaching it before turning right, you might go a long way. Joe Shipman notes that you get the shortest worst-case distance to the sea by walking or about 1.15 miles and then turning 120 degrees to the right. The minimum expected journey length to the sea, according to a complicated numerical calculus computation by the proposer, is given by going about 1.043267 miles and then turning about 101.3046 degrees to the right.

**Two Different Videos** on Wiles's proof of Fermat's last theorem were confused in the previous Math Chat. The brief excerpt was from the marvelous BBC/NOVA TV program "The Proof." The MSRI video on "Fermat's Last Theorem" starts with an even more moving interview with Wiles and gives a more serious presentation of the mathematics.

MATH CHAT is now available directly at MathChat.org.

Send answers, comments, and new questions by email to Frank.Morgan@williams.edu, to be eligible for* Flatland *and other book awards. Winning answers will appear in the next Math Chat. Math Chat appears on the first and third Thursdays of each month. Prof. Morgan's homepage is at www.williams.edu/Mathematics/fmorgan.

THE MATH CHAT BOOK, including a $1000 Math Chat Book QUEST, questions and answers, and a list of past challenge winners, is now available from the MAA (800-331-1622).

Copyright 2000, Frank Morgan.