hu@sdcsvax.UUCP (04/19/84)
[Don't even think about truncating this message!]
Hey folks!
I made the USAMO! That's right, I've been invited to participate
in the 1984 USA Mathematical Olympiad! I am, like, totally stoked!
I want to thank all you people out there for keeping a stimulating
discussion in this group. I think solving those problems gave me
a better idea of how to attack problems in general. Keep up the good
work!
I received several responses to my posting of the 1984 AIME. Here is
a copy of the original problems. Answers and solutions follow.
This year, approximately 50 students qualified. Students who scored
a 12 or better on the AIME, or those who scored an 11 on the AIME and
a 113 or better on the AHSME were invited to participate. The median
score was 7; the mean, 6.9. Over 600 students participated in the AIME.
(I got a 12 with the help of two lucky guesses.) (Note: I made a horrible
TYPO! On number 10, the formula should be score=30 + 4c - w, not -2.)
For those of you who are wondering what this is, you might
recall taking tests in high school put out by the Mathematical Association
of America (MAA), the top-scorers of which would go to the United States
Math Olympics (United States of America Mathematical Olympiad (USAMO)).
It was a 90 minute test with 30 multiple-choice questions score on a scale
from 0-150. Last year, that system was replace by a new one. Now, you
take the AHSME (American High School Mathematics Examination). The test
is in the same format as the old MAA test except that it's easier. People
who score above 95 on the AHSME proceed to take the AIME (American Invitational
Mathematics Examination). The AIME is a 2.5 hour test with 15 questions.
The answer to each question is an integer from 0 to 999. Participants
record their answers on computer cards. Top scorers from this examination
are invited to participate in the USAMO. To help you judge the difficulty
of these tests, here are some stats: Last year, the mean score on the
AIME was 3.8 (out of 15) and the median was 3. 53 students scored a 10
or above and were invited to participate in the USAMO. Remember that
the students who took this exam were already screened as being among the
top math students in the country.
Without further ado, and without permission from the MAA (They
print and distribute for a nominal fee old tests, so I don't think they'll
mind. I'll post a list of tests and an address at the end of this article.)
Here are the problems. You have 2.5 hours. All answers are integers from
0 to 999. Begin
1. Find the value of a2 + a4 + a6 + ... + a98 if a1,a2,a3,... is an arithmetic
progression with common difference 1, and a1 + a2 + a3+ ... + a98 = 137.
[Sorry, I can't put subscripts on the terminal.]
2. The integer n is the smallest positive multiple of 15 such that every
digit of n is either 0 or 8. Compute n/15.
3. A point P is chosen in the interior of triangle ABC so that when lines
are drawn through P parallel to the sides of triangle ABC, the resulting smaller
triangles, t1, t2, t3 in the fugure, have areas 4, 9 and 49, respectively.
Find the area of triangle ABC.
C.
/ \
/ \
/ \
/ \
/\ /\ [Figure not drawn to scale.
/ \ / \ Not bad for ascii graphics!]
/ t2 \ / t1 \
/ \ / \
/--------P--------\
/ / \ \
/ / \ \
/ / \ \
/ / t3 \ \
/ / \ \
-------------------------------
A B
4. Let S be a list of positive integers--not necessarily distinct--in
which the number 68 appears. The average (arithmetic mean) of the numbers
in S is 56. However, if 68 is removed, the average of the remaining numbers
drops to 55. What is the largest number that can appear in S?
2 2
5. Determine the value of ab if log a + log b = 5 and log b + log a = 7.
8 4 8 4
6. Three circles, each of radius 3, are drawn with centers at (14,92)
(17,76) and (19,84). A line passing through (17,76) is such that the
total area of the parts of the three circles to one side of the line is
equal to the total area of the parts of the three circles to the other
side of it. What is the absolute value of the slope of this line?
7. The function f is defined on the set of integers and satisfies
f(n) = / n-3 if n >= 1000
\ f(f(n+5)) if n < 1000
Find f(84).
8. The equation z^6 + z^3 + 1 = 0 has one complex root with argument
(angle) theta between 90 degrees and 180 degrees in the complex plane.
Determine the degree measure of theta.
9. In tetrahedron ABCD, edge AB has length 3 cm. The area of face ABC
is 15 square cm and the area of face ABD is 12 square cm. These two faces meet
each other at a 30 degree angle. Find the volume of the tetrahedron in cubic
centimeters.
10. Mary told John her score on the AHSME, which was over 80. From
this, John was able to determine the number of problems Mary solved
correctly. If Mary's score had been any lower, but still over 80,
John could not have dtermined this. What was Mary's score? (Recall
that the AHSME consists of 30 multiple-choice problems and that one's
score, s, is computed by the formula s=30 + 4c - w, where c is the
number of correct and w is the number of wrong answers; students are
not penalized for problems left unanswered.)
11. A gardener plants three maple trees, four oak trees and five
birch trees in a row. He plants them in random order, each
arrangement being equally likely. Let m/n in lowest terms be the
probability that no two birch trees are next to one another. Find m +
n.
12. A function f is defined for all real numbers and satisfies
f(2+x)=f(2-x) and f(7+x)=f(7-x)
for all real x. If x=0 is a root of f(x)=0, what is the least number
of roots f(x)=0 must have in the interval -1000<=x<=1000?
13. Find the value of 10cot(arccot 3 + arccot 7 + arccot 13 + arccot
21).
14. What is the largest even integer which cannot be written as the
sum of two odd composite numbers? (Recall that a positive integer is
said to be composite if it is divisible by at least one positive
integer other than 1 and itself.)
15. Determine x^2 + y^2 + z^2 + w^2 if
x^2/(2^2-1^2) + y^2/(2^2-3^2) + z^2/(2^2-5^2) + w^2/(2^2-7^2) = 1
x^2/(4^2-1^2) + y^2/(4^2-3^2) + z^2/(4^2-5^2) + w^2/(4^2-7^2) = 1
x^2/(6^2-1^2) + y^2/(6^2-3^2) + z^2/(6^2-5^2) + w^2/(6^2-7^2) = 1
x^2/(8^2-1^2) + y^2/(8^2-3^2) + z^2/(8^2-5^2) + w^2/(8^2-7^2) = 1
Whew!!!! I was pretty confident of 10 of my answers at the end of the
2.5 hour period. However, I don't have my answer sheet. Please mail
me copies of your answers (together with how sure you are of them).
I would be interested in knowing your solution method, too, if you
want to mail it to me. Note that for the test, we were allowed ruler,
compass, pencil, pen, and paper. We were not allowed to use
calculators, slide rules, computers, VAXen, tutors, math professors,
textbooks, notes, USENET, etc.
Respondants were: (in the order that I received their responses)
ihuxr!lew Lew Mammel, Jr [lew]
metheus!howard [howard]
uvacs!mgp Mark Pleszkoch [mgp]
asgb!gupta Yogesh Gupta [gupta]
ihlts!rjnoe Roger Noe [rjnoe]
The answers are:
[I originally planned to include all solutions submitted together with
the person who submitted it. After spending a long time doing that, I
decided not to. Almost every respondant got an answer to almost every
problem, so kudos to all of them.]
1. 93
The odd terms are 49 numbers, each 1 less than the following even term
so 2*s - 49 = 137 and s = 93.
[Alternate]
Since an = a0 + n for some a0, summing ai from i=1 to 98 gives
98*a0 + 4851 = 137, or a0 = -2357/49. Summing a(2i) from i=1 to i=49
gives 49*a0 + 49*50 = -2357 + 2450 = 93.
2. 592
The last digit is 0, so N must be a multiple of 30.
All the digits are 8 or 0, so N/10 must be divisible by 8.
N/80 must be all 1's and 0's and divisible by 3.
Therefore, N/80 is 111.
N = 111*80 = 8880, N/15 = 592
3. 144
The area must be independent of the shape of the triangle,
so assume that the triangle is right, right-isoceles, equilateral,
or some other special case. It is then easy to determine the
lengths of the sides and the area.
[Alternate]
Working from proportions and similar triangles, you can get the
same result.
4. 649
Let S = (a[1], ... a[k]), with a[k] = 68.
Let x = a[1] + ... + a[k-1].
Then (x + 68) / k = 56, so x + 68 = 56k.
Also x / (k - 1) = 55, so x + 55 = 55k.
Subtracting, k = 13, so a[1] to a[k-1] are 12 numbers with mean 55,
and sum x = 12*55 = 660.
Taking a[2] through a[12] to be 1, we get
a[1] = x - 11 = 649 is the answer.
5. 512
Converting to base 2 logarithms, we get:
(1/3)lg a + lg b = 5 and lg a + (1/3) lg b = 7
lg a + 3 lg b = 15 and 3 lg a + lg b = 21
Solving simultaneously, lg b = 3 and lg a = 6
lg ab = 3 + 6 = 9
ab = 512
6. 24
The circle centered at (17,76) doesn't matter.
The line must pass through the midpoint of the segment
connecting the centers of the other two circles, because
they have the same radius. Therefore, the line passes
through (17,76) and (16.5,88). Its slope is -24. Abs(-24)=24.
7. 997
f(1004) = 1001
f(1003) = 1000
f(1002) = 999
f(1001) = 998
f(1000) = 997
f( 999) = f(f(1004)) = f(1001) = 998
f( 998) = f(f(1003)) = f(1000) = 997
f( 997) = f(f(1002)) = f( 999) = 998
f( 996) = f(f(1001)) = f( 998) = 997
f( 995) = f(f(1000)) = f( 997) = 998
f( 994) = f(f( 999)) = f( 998) = 997
etc.
All even numbers < 1000 give 997, so 84 gives 997.
8. 160
Solve as a quadratic in z^3. z^3 = +- 2/3 pi.
Therefore, z = pi * (2/9, 8/9, 14/9, 4/9, 10/9, 16/9)
z = (8/9)pi is in the required range. Therefore, 160 degrees.
9. 20
The height of the 15cm^2 face with base 3cm must be 10cm.
The height of the pyramid with base 12 cm^2 is then 5cm.
1/3 * 12 * 5 = 20.
10. 119
For any score in the range given by the problem, there are
at least two ways of achieving it: 30 + 4c - w and
30 + 4(c+1) - (w+4). However, there are only 30 problems
on the test, therefore c+w<=30. Therefore, c=23, w=3.
Mary's score was 119.
11. 106
[I'm not sure. Probability isn't my strong suit. Here are all
the solutions.]
[lew]
106. there are 12!/( 3! * 4! * 5!) arrangements. Putting at least
one other between 5 birches leaves 3 to put in any of seven places.
I count 56 ways to do this. There are then 56 * 7!/(3! * 4!) arrangements
with no birches adjacent. Dividing and reducing gives 7/99.
[mgp]
I though the principle of inclusion-exclusion would work, but I did
it this way.
Consider the maple and oak trees to be non-birch (N).
Then the number of ways of planting is
12! / (5! * 7!) = 792 = 8*9*11.
To count the ways of planting without two birches (B) in a row,
add a 13-th non-birch tree at the end. This will not change the
total possible ways of planting. Now, every birch is followed
by a non-birch, so farmer is reduced to planting
5 (BN) pairs and 3 N's, in 8 (moving) slots.
The number of ways is
8! / (5! * 3!) = 56 = 8*7.
Thus the probability of no two birches together is 56/792
= 7/99, so the answer is 7+99 = 106.
[gupta]
Let the birch trees be denoted by B and the others by A.
Then, for no two birch trees to be adjacent, we have
B A B A B A B A B
as the necessary configuration with the remaining three A's being
planted anywhere in this row.
The total number of ways the trees can be planted in any random order
is:
12!
-------- = 12 * 11 * 10 * 7 * 3
5! 4! 3!
The number of different ways in which the above pattern can be done
is:
7 * 6 * 5 * 4 * 10 * 11 * 12
---------------------------- = 11 * 10 * 7 * 5 * 2
4! * 3!
Thus, the probability that no two birch trees are next to each other:
11 * 10 * 7 * 5 * 2 5
-------------------- = ---
12 * 11 * 10 * 7 * 3 18
Ans. 23
[rjnoe]
[61] The number of combinations is 12!/3!4!5! and the number of two-or-
more-birches together is 10!/3!4!3!. The probablility that two birch trees
ARE together is 5/33 so m/n = 28/33 and m+n = 61.
12. 401
The set of all required roots is the minimum set of integers
which contains 0 and is closed under reflection about both 2 and 7.
Since reflecting about 2 (or 7) is self inverse,
only alternating sequences of operations generate new roots.
Starting with 2, we get 0,4,10,-6,20,-16,30,-26,40,-36,...
Starting with 7, we get 0,14,-10,24,-20,34,-30,44,-40,54,...
These two sequences contain all required roots.
The total number of required roots is thus 401.
13. 15
Apply cot(a+b) = (cot(a)*cot(b) - 1)/(cot(a) + cot(b)) twice.
[alternate]
From each cot, we can construct a right triangle. If we
"stack" these triangles to add the angles, we can find
the cotangent by using vectors.
14. 38
9 = 0 mod 3
25 = 1 mod 3
35 = 2 mod 3
Therefore, any number greater than 38 can be expressed as
the sum of 9, 25, or 35, and a multiple of 3. 38 works,
so it must be it.
15. No one submitted a solution to this problem.
--Alan J. Hu
sdcsvax!hu