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.
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% |