johnc@reed.UUCP (Cavanaugh) (04/16/84)
Has anyone out there gotten a hold of the answers to this year's AIME (American Invitational Mathematics Examination) yet? I have spent a few hours figuring out a few questions and am curious to see if any of my solutions are correct. If anyone out there has the answers or knows where I could get them please mail me. Thanks in advance, -John Cavanaugh ..tektronix!reed!johnc
stan@hare.DEC (Stanley Rabinowitz) (04/22/84)
1984 AIME
The answers to the 1984 AIME (American Invitational Mathematics Examination)
recently posted by sdcsvax!hu are correct. He omitted the answer to problem
15 which is 36. I give a solution at the end of this note.
I am one of the people that helped make up this examination. (I am on the
MAA's committee on high school contests.) I was glad to see so much interest
in these problems on the USENET. We would be interested in hearing
comments about this exam or any of our other examinations, such as the AHSME
(American High School Mathematics Examination), and the USAMO (U.S.A.
Mathematical Olympiad). So if you took either of these this year, or
if you took one of them many years ago, please send me your comments
and suggestions. Was the exam a rewarding mathematical experience for
you? Was it too easy? too hard? too long? too short? etc. General comments
that you think others might wish to further comment on can be posted to
NET.MATH; specific comments can be mailed to me at the electronic address
given below.
Each school that participated in the AIME should have received a copy of
the solutions pamphlet by now. (Check with your teacher.)
Some of your solutions were "better" than our "official" solutions.
I will share some of them with my colleagues. We are always interested
in seeing better or more novel solutions; so if you think you have
a real neat solution to one of these, feel free to send it to me.
If you did not see a copy of the solution pamphlet, or if you want to order
copies of previous year's exams, they can be ordered from
Dr. Walter Mientka, Executive Director
American High School Mathematics Examination
Department of Mathematics and Statistics
University of Nebraska
Lincoln, Nebraska 68588-0322
AHSME's are 40 cents each, AIME's are 50 cents each and OLYMPIADS are
50 cents each. Make checks payable to MAA Committee on High School Contests.
Free copies can also be found in your library. These exams as well as other
math contests are regularly printed in "Mathematics Magazine" and in
"Crux Mathematicorum".
Let me congratulate those who qualified for this year's USAMO.
Here is a copy of last year's USAMO problems for you to practice on.
I think that people on this network will enjoy the USAMO problems too.
1983 U.S.A. Mathematical Olympiad
1. On a given circle, six points A, B, C, D, E, and F are chosen at random,
independently and uniformly with respect to arc length. Determine the
probability that the two triangles ABC and DEF are disjoint, i.e.
have no common points.
5 4 3 2 2
2. Prove that the roots of x +ax +bx +cx +dx+e=0 cannot all be real if 2a <5b.
3. Each set of a finite family of subsets of a line is a union of two closed
intervals. Moreover, any three of the sets of the family have a point in
common. Prove that there is a point which is common to at least half of the
sets of the given family.
4. Six segments, S1, S2, S3, S4, S5, and S6 are given in a plane. These are
congruent to the edges AB, AC, AD, BC, BD, and CD, respectively, of a
tetrahedron ABCD. Show how to construct a segment congruent to the
altitude of the tetrahedron from vertex A with straight-edge and compass.
5. Consider an open interval of length 1/n on the real number line where
n is a positive integer. Prove that the number of irreducible fractions
p/q, with 1 <= q <= n, contained in the given interval is at most
(n+1)/2.
Don't get discouraged if you have trouble doing these. Remember that the
USAMO is intended for the top 100 or so students in the country!
Appendix:
Solution to problem 15 from the 1984 AIME. (Answer is 36)
Consider the equation:
2 2 2 2
x y z w
---------- + ---------- + ---------- + ---------- = 1 (1)
t - 1 t - 9 t - 25 t - 49
The four given equations tell us that this equation is satisfied
by t=4, 16, 36, and 64. Multiplying out, we find that this equation
is equivalent to
(t-1)(t-9)(t-25)(t-49)-x^2(t-9)(t-25)(t-49)-y^2(t-1)(t-25)(t-49)- (2)
z^2(t-1)(t-9)(t-49)-w^2(t-1)(t-9)(t-25)=0.
This is a 4th degree polynomial in t. Since t=4, 16, 36, 64 are known to
be roots, and a 4th degree polynomial has at most 4 roots, these must
be all of them. It follows that equation (2) is equivalent to
(t-4)(t-16)(t-36)(t-64)=0. (3)
Since the coefficient of t^4 is 1 in both (2) and (3), it follows that
the coefficients of the other powers of t must be the same. Equating
the coefficients of t^3 gives
1+9+25+49+x^2+y^2+z^2+w^2=4+16+36+64
so x^2+y^2+z^2+w^2=36.
Yeah, we pulled a rabbit out of the hat in producing this solution.
There are more straight-forward ways of getting at the answer, but
they take longer. Anyhow, we like to have at least one real difficult
problem on each exam. - STAN -
Stanley Rabinowitz
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