New Calculator Rule Our contests say that neither a TI-89 nor a TI-92 is permitted. That rule has changed. Since Contest 2, we have allowed any calculator without a QWERTY keyboard. New Calculator Rule Our contests say that neither a TI-89 nor a TI-92 is permitted. That rule has changed. Since Contest 2, we have allowed any calculator without a QWERTY keyboard.
Future Contest Dates and Our Algebra Contests Our final contest is Apr. 11. This is year 7 of April's Algebra Course I Contest. To participate, write for information.
Rescheduling A Contest & Mailing Results If you have school closings or testing days, our rules allow an alternate contest date. We prefer the previous week so we get results on time. Mail scores by Friday of the official contest week. If scores are late for due cause, attach a brief explanation. Late scores unaccompanied by such an explanation are not normally accepted.
Next Year's Contest Dates fall on the following Tuesdays: Oct. 17, Nov. 14, Dec. 12, Jan. 16, Feb. 13, Mar. 20. We also sponsor contests for grades 4, 5, 6, 7, 8 and Algebra Course I. Use the enclosed form for any contest or for books of past contests.
End-of-Year Awards Engraving of awards begins Apr. 25. We give awards to the 2 schools and 2 students with the highest totals in the entire League and to the school with the highest score in each region. Winning schools must postmark their results by Apr. 14. Results postmarked later cannot be used to determine winners. Completion of the cumulative column is optional, but student awards are based only on scores regularly listed in that column. (Student certificates of merit were enclosed with Contest 5.)
General Comments About Contest 5 One advisor said "it's difficult to get students together to take the contest." Dave Ollar "thought this contest was difficult. I liked all the problems except 5-5. This problem gives the students who haven't had trig no chance. I like hard ones where anyone who can think well can get them." We agree, and we don't give trig on the early contests. But, we think it wrong to totally avoid advanced topics. Suzanne Moll said "All in all it was another enjoyable contest." Michael Keyton said "very good contest, but it would be nice to omit the TI-89 and other symbolic calculators." Sue Haberger said "thank you for the effect these contests have had on our school. The problems are at a good level and the immediate marking and feedback creates an excitement and stimulates discussion." Susan Wong "was disappointed that 3 problems had such a high difficulty level. None of my students got 5-5 or 5-6 correct." Robert Morewood said "Thanks for another stimulating competition." Bryan Knight wrote "Thanks for a great contest." Joe Griesbach thanked us "for another great set of problems." Brian Balsdon thanked us "for another fine contest."
Problem 5-1: Alternate Solution Bryan Knight noted that the triangles are congruent, so the large one has the same area as the square. The small triangle, with sides half those of the larger one, has an area equal to (1/2)2 = 1/4 of the large one. Michael Keyton thought #1 was "excellent for every student."
Problem 5-2: Alternate Solutions Student Scott Kallal and advisor Bryan Knight noted that the factors are (x-3p)(x-2p), so 2p = 15 or 3p = 15. Alternatively, he noted that the root must be a factor of 6p2, so 2p2 is a multiple of 5, which is true only for the prime p = 5.
Problem 5-3: Appeal (Denied) & Comments One appeal said that 4, 6, 8, 9, 10, 12, 14, 15, 16, 18, 20, 21, 22 are 13 consecutive composite integers. Prof. Brian Conrad of Harvard University denied the appeal, claiming that such an interpretation would allow for so many such sequences as to make that interpretation incorrect. Michael Keyton said that, with a TI-89, generate the sequence, factor, and count. Suzanne Moll and Ted Heavenrich commented that very few students are familiar with the word "composite." Student Linda Wu gave a very clever, though lengthy argument which led her quickly to the solution.
Problem 5-4: Alternate Solution & Comments Suzanne Moll "was disappointed that 5-4 could be done with the graph and root options on a TI-83. Ed Imgrund solved x2+x-1 = 0 by the quadratic formula. Since x4+2x3+x2 = [x(x+1)]2, substitution of either root yields the value sought. In fact, the product is a simple "difference of two squares" factorization. Michael Keyton noted that "solving with the TI-89 and substituting produced a non-simplified expression, but, when copied to the author line, it is fully simplified."
Problem 5-5: Comments Ted Heavenrich "was very happy that kids with TI-89s who tried to solve it analytically using the solve command (and domain limits) inexplicably got only 60, 90, 240, and 300, missing two of the answers," a situation also noted by Michael Keyton.
Problem 5-6: Comments Robert Morewood said that, to check decimal approximations he "had to calculate the sum, from x = 1 to 2001 of logx to get 6.636 × 105738. Ted Heavenrich said that 5-6 "may have been too hard." His student Michael Purcaro, computed that, for the set {1,2,3,4}, f(1)+f(2)+f(3)+f(4) = 119, which he recognized as 5!-1. He then checked the pattern for the sets {1,2,3} and {1,2}. We expected that students who solved the problem would use a pattern approach. Michael Keyton called the problem "very difficult" Brian Balsdon said that 5-6 "was a beauty" since they "do this part of the theory of polynomial equations in their 12th grade honors course." R. Lee called 5-6 "pretty obscure." We should have listed the pattern approach, Method II, as our first method. That approach is not obscure at all. An advisor named "shehs" wrote "Where did you come up with 5-6 That had to be the toughest problem I've ever seen on your contests." Mike Faleski, a physics teacher, used the concept of telescoping series to develop, inductively, the formula required.
Prob #, % Correct (top 5 each school)
|
5-1 97% |
5-4 35% |
|
5-2 92% |
5-5 48% |
|
5-3 43% |
5-6 4% |