Which Calculators Are NOT Allowed? Our contests do not permit the TI-89 or any calculator with a QWERTY keyboard, such as the TI-92.
Future Contests and Rescheduling Contests Contest dates are March 9 and April 13. Our annual Algebra Course I Contest is in mid-April. If circumstances (such as 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.
What Do We Publish? Wonder why a solution you sent wasn't mentioned? We discuss everything we have at the time we write the newsletter. But the newsletter is the first thing we prepare, so we may use your score report yet not use your solution. We try to be efficient! Sorry to those whose solutions were too "late" to use in our newsletter.
Contest Books Make A Great Resource Have you bought our books for your math team? Collections of past contests are a great way to work with your team.
General Comments About Contest #4 Nancy Ni wrote "I'm a student who takes your contests. The program is well organized and operates effectively. The Math League is a great way to encourage students. Job well done!" Thank you! Ted Heavenrich and Elmer Delventhal both said this was their team's best score ever. Ted said "Many thanks for making that possible." Student Joseph Young "really enjoyed the math contests. They give some of my friends and me real challenges compared to the normal everyday classes. But, my friends and I feel that the recent contests have been fairly easy . . . . Please bring back some hard questions because the contest isn't the same without them." Our aim is to make #5 difficult and #6 very difficult. Joe Holbrook said "The students loved this one!" Tim Butler, Elmer Delventhal, Dave Ollar, and Jerry Detweiler wrote "Great contest!" Tim said "The contest illuminated a few misconceptions!" Dave Farber called it "another great contest" and said his two perfect scores were from sophomores. Dave Ollar's students found 4-3 and 4-4 easier than 4-2. Mike Brown thought that "contest 4 was hard enough to separate the truly gifted, but doable enough that most students felt successful. Kudos to your test writers." That's us. Thanks!
Problem 4-1: Comments & Alt. Sol'n. Jerry Detweiler, with students who wrote "40," or "root (40)," was amazed "that no one just put root(45) - root(5) on a calculator and squared the result."
Problem 4-2: Appeal (Accepted) and a Comment Two of my students wrote "$50," and I bet many sponsors would give credit without even thinking of checking." Credit was awarded by our appeals judge. Teacher Jerry Detweiler said #4-2 was "a wonderful question for algebra teachers who try to get students to define their variables and write down steps."
Problem 4-3: Alternate Solution David Sutliffe and Dave Farber wrote the numerator as 1998(1+18) and reduced. Mike Clancy noted that this can be computed directly on a TI-85.
Problem 4-4: Alternate Solution Dave Farber put the small square of our solution into the lower right quarter of the large square, and put in a small circle centered at the lower right vertex of the large square. Then, 2r + 2 = 2 root(2), so r = root(2) - 1.
Problem 4-5: Alternate Solutions David Sutliffe began as we did in Method II. When he got x4 - x2 + a = 0, he said that, for real roots, the discriminant = 1 - 4a ³ 0; so a £ 1/4. Student Jen Lindsay and teachers Nola Forbes, Del Delventhal, David Doster, Richard Muller, Ken Welsh, Phil Shafer, Helen Manning, and Dave Farber found the four points of tangency of the hyperbolas xy = ± root(a) and the circle x2 + y2 = 1.
Problem 4-6: Comments and Alt Solutions "Problem #4-6 was one of my all-time favorites," said Jerry Detweiler. Tim Butler said the problem reminded him of this problem: "If a hen and a half lays an egg and a half in a day and a half, how many eggs will a dozen hens lay in a dozen days?" My answer is 96, Tim. Dave Farber said he read the problem 3 times "before I finally decided how I was going to do it. Without loss of generality, assume that the hound leap is 1 unit. Let h be the number of hound leaps. Since each hound leap covered a distance of 1, the total distance covered by the hound was h. Similarly, the number of leaps for the fox was (5/4)h, and the length of a fox leap was only 3/4 that of a hound leap. The total fox distance covered is (3/4)(5/4)h = (15/16)h. Since h+90 = (15/16)h, h = 1440." Student Kevin Storm identified a unit time and a unit distance to solve the entire problem arithmetically. Excellent indeed! On the lighter side, student Steph Durkin said "The hound doesn't catch the fox because he decided it never did anything to him, and let him alone. He had better things to do than catch foxes." Classmate Christina Conklin said "I don't think it is morally right to have a hound hunt a fox." and suggested an example such as kids chasing each other. She added "Plus I couldn't get it."
Problem #, % Correct (top 5 each school)
|4-1 94%||4-4 72%|
|4-2 77%||4-5 76%|
|4-3 83%||4-6 33%|