Contest Registration and Books of Past Contests Please mail next year's registration (enclosed) soon. You may ask us to bill you in the Fall. We also sponsor an Algebra Course 1 Contest and contests for grades 4, 5, 6, 7, & 8. Use the enclosed form to register for contests or to order books of past contests.
Next Year's Contest Dates fall on the following Tuesdays: Oct 26, Nov 30, Jan 11, Feb 8, Mar 7, and Apr 11.
End-of-Year Awards and Certificates Symbols identify winners (we shipped plaques to the advisor). Errors? Write to: Math Plaques, P.O. Box 720, Tenafly, NJ 07670-0720. Identify the award, the contest level, your name, and the school's name and address. The Contest #5 envelope contained Certificates of Merit for the highest scoring student overall, and on each grade, for the year. Do you need extra certificates for ties? If so, send a self-addressed stamped envelope large enough to hold certificates (you need to use * DOUBLE * postage) to: Certificates, P.O. Box 2117, Miller Place, NY 11764-8896. (Allow 2-4 weeks.)
Comments About This Year's Contests Linda Chambers asked if we could have the contests end in March. She loses too many kids to Spring sports. Next year's calendar was already announced-but we will try a new schedule the following year. Do you concur or not? Ted Heavenrich called contest #6 "a lot of fun." Adrienne Irwin said "Thanks for another stimulating season of problem-solving." Ray Whipple said "Thanks for supplying a very exciting and challenging math activity over the years." Gail Ellenbaum wrote "Loved this contest. All levels had fun." Sheryl Cabral Said "Thanks for a great competition. We'll definitely participate again next year." Dennis Gwirtz said "We did not perform well this year, but the students who participated enjoyed the contests." Jerry Detweiler thinks that "all schools should be required to participate." Sue Forster said "We had fun this year and look forward to doing this again next year." Dave Stouffer said "Some of our kids take these contests very seriously. I do hope that all school advisors are protecting the integrity of this wonderful endeavor." Elmer Delventhal remarked "Once again, Dan and Steve put together a great group of problems. I was hoping for a 6, but everyone missed at least one." Dave Ollar said "This contest marks the first time we have had more than one 6 in one month. Our kids show up 40 minutes before school to take the contest and we average about 40 each time. I love these contests. One of our 6's is a 9th grade student." Mike Buonviri said "We teachers like taking these contests too. This was the first time I got all 6 right in less than 15 minutes. Feels good!" Matt Zeitz "loved" contest 6. This year, Robert Morewood had one student from each grade level get a 6, all on different contests.
Problem 6-1: Comment and an Appeal (Denied) Sheryl Cabral said "Most students just guessed 1999. Two students appealed, asking us allow the answer 28, which is the sum of the 4 digits 1+9+9+9. We asked "What final four-digit sum will result," and that means the sum is a four-digit number; it does not ask for the sum of its digits.
Problem 6-3: Comments and Alternate Solutions Robert Morewood said "almost everyone correctly identified the angle x in the diagram, but few calculated it correctly. Some thought the triangle inside the pentagon was equilateral. Others thought they saw vertical angles at the vertices of the inner pentagon. We need to review basic angle geometry!" Ted Heavenrich "liked the combination of trig and geometry." Student Cameron Freer used the law of cosines to compute that a side of the large pentagon is 2cos24°. Student Mark Losego used the law of sines. Two advisors thought that students that got this one correct "were lucky as x was not in the diagram." We wanted students to discover which angle had to be x. Students who answered this correctly were able to show that the length of a side of the large pentagon is 2cos24°. When they equate this to 2cosx°, they get x = 24. It's possible to label several angles with an x, but that would have to be done by the student, since the information was not provided in the problem statement.
Problem 6-4: Comments Ted Heavnerich said this problem was "quite accessible despite the fact that most students did it the long way." Most of Dave Farber's students use the format x2-(sum)x+product = 0 to help them. It's a great approach, Dave!
Problem 6-5: Nice Alt Sol, Comments Student Mathieu Loisell sent the following beautiful solution: From the given, f -1(f(1)) = 1 / f(f(1)). From the definition of an inverse, f -1(f(1)) = 1. Hence 1 = 1 / f(f(1)), or f(f(1)) = 1. Apply the inverse to both sides to get f(1) = f -1(1). The given implies that f -1(1) = 1 / f(1), so f(1) = 1 / f(1) becomes (f(1))2 = 1. Since f(1) > 0, it follows that f(1) = 1. Gail Ellenbaum and Jerry Detweiler both prefer that we avoid questions easy to guess but difficult to handle without guessing. One person thought we had said that f(x) = 1. We did not. We said that f(1) = 1.
Problem 6-6: Comments Ted Heavnerich thought it was quite straightforward, "but clearly my students did not agree; in any event it was a very good problem." Elmer Delventhal "especially liked #6-6; it required knowledge of sequences, exponents, factors, and estimation."
Problem #, % Correct (top 5 each school)
|6-1 96%||6-4 79%|
|6-2 88%||6-5 90%|
|6-3 54%||6-6 31%|