Future Contests and Rescheduling Contests  Contest dates are Feb. 8, Mar. 7, and Apr. 11. Our annual Algebra Course I Contest is in April. If circumstances (such as vacations, school closings or special testing days) require it, we permit you to give the contest on another date. If your scores are late, please attach a brief explanation, or the scores may be considered unofficial.

Regional Groupings We sometimes receive requests about regional groupings. Within guidelines, we try, when possible, to honor such requests for the next school year.

Student Cumulative Scores  Although completion of the Cumulative Column is optional, we list (and consider official) only cumulative scores reported in this column. A student whose cumulative scores are incorrect (or don't appear regularly in the Cumulative Column) may lose eligibility for recognition by the League.

T-Shirts Anyone?  We're often asked "Are T-shirts available? The logo lets us know fellow competitors." Featuring grey shirting and a small, dark blue logo in the "alligator region," MATH T-shirts are available in all sizes at a very low price. There's just one low shipping charge per order, regardless of order size. For a VISA or MasterCard order, please phone 1-201-568-6328, or place an Internet Order from our Web Site. You may fax a purchase order to 1-201-816-0125.

General Comments:  Art Wenk wrote "I don't know how you keep turning out great questions year after year, but we really enjoy the contests. We have upwards of 70 participants each time." Gary B. Hicks said contest 3 was "another great contest! Our kids really look forward to these contests. Keep on keepin' on." Eric Brohaugh said "Keep up the good work. Both my students and I enjoy your tests!" Don Stewart said that his "students did poorly on this contest. It was too soon after coming back after the holidays." John P. Wojtowicz called contest 3 a "good contest." Suzanne Moll said "We thoroughly enjoyed contest 3. Once again, you found a very creative way to use the year in a problem, and not as an answer this time."

Alternate Solution to Problem 3-1 Dave Farber and Ed Imgrund wrote 900a²- 400b² = 100(3a+2b)(3a-2b). Now, a/b = 2/3  <=>  3a-2b = 0 implies that 900a²- 400b² = 0.

Problem 3-3: Amazing Alternate Solution  Art Wenk wrote "When you asked about sums of squares of odd numbers, I figured there had to be something like this concealed in Pascal's Triangle. Sure enough, if you take alternate entries in the fourth diagonal, the numbers you're looking for turn up. So, the sum of the squares of the first 2000 odd numbers would be 4001 choose 3. This comes out to be (4001 × 4000 × 3999) ÷ (3 × 2 × 1), which clearly ends in a 0." Student Jay Crowley and advisors Dave Farber and Ed Imgrund sent us solutions using units' digits.

Problem 3-4:Comments,Appeals(Accepted&Not)  An appeal for the answer 44.444...% was accepted. In fact, there is no need to appeal such an answer, since we automatically accept as correct any answer mathematically exactly equivalent to the official answer, and 4/9 = 44.444...%. One advisor appealed that "an answer to a probability problem is generally expressed as a ratio, not a fraction." Our appeals judge, Prof. Brian Conrad, Math Dept., Harvard Univ agrees with our view that the expressions 4:9 and 4/9 represent the exact same real number. One anonymous advisor said that probability had always been a weak spot for his students. He thought that questions like this might help.

Problem 3-6: Comments, Appeal (Denied), Alt. Sols.  An appeal that ".5 or .53 should be an acceptable answer" was denied, because neither complies with our requirement, printed at the top of each contest, that "Answers must be exact or have 4 (or more) significant digits, correctly rounded." Suzanne Moll said "3-6 produced much discussion and many groans when they saw how easy it was to do (in hindsight). The one person who got it spent 5 minutes on the first 5 problems and 25 minutes on 3-6. "Dick Olson said "3-6 was brilliant. My students were beside themselves when they saw how easy it could have been solved using the Pythagorean Theorem." Gary B. Hicks said "3-6 fit right into our unit on solving radical equations." He split the region into 5 polygons by connecting the three centers and drawing perpendiculars from the center of the small circle to the nearby sides of the square and then used Hero's Formula, setting the sum equal to 4 and concluding that r² - 8r + 4 = 0, as we had also concluded. Student Keith Lo and advisors Dave Farber and Gary Dennis all connected the center of the small circle to the furthest vertex of the square and the center of one of the larger circles. The sides of that triangle are 1, 1 + r, and (2 - r)sqrt(2). One angle of the triangle is 45°. Using the law of cosines, r² - 8r + 4 = 0. Keith Culkins said "My students found 3-6 tough. One student wrote 4 ± 2sqrt(3). The 4 + 2sqrt(3)  is an interesting circle also, but it is not the one sought." Eric Brohaugh said "3-6 was a cool problem! I was playing around with the tangent lines and making no progress. I finally gave up and looked at the solution. It was very straight-forward and easy to understand - making me think `Why didn't I think of that?'"

Statistics / Contest #3