The Uninteresting Chronicles of a High School Student

March 17, 2010

28th Annual Invitational Mathematics Examination (AIME I)

UPDATE: All answers have been obtained.

2010 March 16 AIME I problems are transcribed as below. Confirmed solutions to the AIME will be collected and incorporated into the post.

Due to…er, legal reasons I was unable to post this until 12:00 AM March 17 EDT. My apologies.

© 2010 Mathematical Association of America


1. Maya lists all the positive divisors of 20102. She then randomly selects two distinct divisors from this list. Let p be the probability that exactly one of the selected divisors is a perfect square. The probability p can be expressed in the form m/n, where m and n are relatively prime positive integers. Find m + n.


2. Find the remainder when

is divided by 1000.


3. Suppose that

and

.

The quantity x+y can be expressed as a rational number r/s, where r and s are relatively prime positive integers. Find r+s.


4. Jackie and Phil have two fair coins and a third coin that comes up heads with probability 4/7. Jackie flips the three coins, and then Phil flips the three coins. Let m/n be the probability that Jackie gets the same number of heads as Phil, where m and n are relatively prime positive integers. Find m + n.


5. Positive integers a, b, c, and d satisfy a > b > c > d, a + b + c + d = 2010, and a2 – b2 + c2 – d2 = 2010. Find the number of possible values of a.


6. Let P(x) be a quadratic polynomial with real coefficients satisfying

for all real numbers x, and suppose P(11) = 181. Find P(16).


7. Define an ordered triple (A, B, C) of all sets to be minimally intersecting if |A ∩ B| = |B ∩ C| = |A ∩ C| = 1 and A ∩ B ∩ C = Ø. For example, ({1,2}, {2,3}, {1,3,4}) is a minimally intersecting triple. Let N be the number of minimally intersecting ordered triples of sets for which each set is a subset of {1,2,3,4,5,6,7}. Find the remainder when N is divided by 1000.
Note: |S| represents the number of elements in the set S.


8. For a real number a, let ⌊a⌋ denote the greatest integer less than or equal to a. Let R denote the region in the coordinate plane consisting of points (x,y) such that ⌊x⌋2 + ⌊y⌋2 = 25.

The region R is completely contained in a disk of radius r (a disk is the union of a circle and its interior). The minimum value of r can be written as , where m and n are integers and m is not divisible by the square of any prime. Find m + n.


9. Let (a,b,c) be a real solution of the system of equations

The greatest possible value of a3 + b3 + c3 can be written in the form m/n, where m and n are relatively prime positive integers. Find m + n.


10. Let N be the number of ways to write 2010 in the form

where the ai‘s are integers, and 0 ≤ ai ≤ 99. An example of such a representation is 1·103 + 3·102 + 67·101 + 40·100. Find N.


11. Let R be the region consisting of the set of points in the coordinate plane that satisfy both |8-x| + y ≤ 10 and 3y – x ≥ 15. When R is revolved around the line whose equation is 3y – x ≥ 15, the volume of the resulting solid is

,

where m, n, and p are positive integers, m and n are relatively prime, and p is not divisible by the square of any prime. Find m + n + p.


12. Let m ≥ 3 be an integer and let S = {3,4,5,…,m}. Find the smallest value of m such that for every partition of S into two subsets, at least one of the subsets contains integers a, b, and c (not necessarily distinct) such that ab = c.
Note: a partition of S is a pair of sets A, B such that A ∩ B = Ø, A ∪ B = S.


13. Rectangle ABCD and a semicircle with diameter AB are coplanar and have nonoverlapping interiors. Let R denote the region enclosed by the semicircle and the rectangle. Line l meets the semicircle, segment AB, and segment CD at distinct points N, U, and T, respectively. Line l divides region R into two regions with areas in the ratio 1:2. Suppose that AU = 84, AN = 126, and UB = 168. Then DA can be represented as m√n, where m and n are positive integers and n is not divisible by the square of any prime. Find m + n.


14. For each positive integer n, let

.

Find the largest value of n for which f(n) ≤ 300.
Note: ⌊x⌋ is the greatest integer less than or equal to x.


15. In ΔABC with AB = 12, BC = 13, and AC = 15, let M be a point on AC such that the incircles of ΔABM and ΔBCM have equal radii. Let p and q be positive relatively prime integers such that

.

Find p + q.


AIME answers: (Highlight to see)
1. 107
2. 109
3. 529
4. 515
5. 501
6. 406
7. 760
8. 132
9. 158
10. 202
11. 365
12. 243
13. 069
14. 109
15. 045

*Unavailable answers are answers which I cannot confirm 100% as correct due to large discrepancies in the answers I have received. They will become available throughout Wednesday and Thursday.

(Sorry about the extremely lazy formatting; I would have LaTeX’d the whole post if I could have done so in a reasonable amount of time.)


I was surprised to see that the instructions for this year’s AIME have changed: evidently there is now a USAJMO contest open to the AMC 10 test takers (meanwhile effectively barring them from the actual USAMO).

There is a provision for the contest to accept AMC 10 students who score more than 11 points on the AIME, but that does not amend the main issue with MOSP qualification to any appreciable extent. The way I see it, the introduction of the USAJMO (and the subsequent shrinkage of the USAMO qualifier pool) causes this progression:

-Underclassmen are discouraged from taking the AMC 10.
-Talented 9th and 10th grade students swarm to take the AMC 12 in order to qualify for the USAMO.
-A net drop in the average constitution of the USAJMO qualifiers (relative to previous years’ USAMO underclassmen participants) occurs.
-A net increase in the difficulty of USAMO qualification also results, as a greater population of students compete for what is ultimately fewer slots.

Result: More and more people who would have qualified for the USAMO under the previous system are deprived of the opportunity. See illustration:

Assume 500 combined USAMO qualifiers; left rectangle = top 500 indexed AMC 12 takers, right rectangle = top 500 indexed AMC 10 takers.

This is something that no one would want, ostensibly, since under a strictly meritocratic ranking system the 200 “new” USAJMO qualifiers would not have ranked in the top 500. Oh well, guess that just makes it twice as hard for us ;).

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5 Comments »

  1. Well, the answer for Q12 is 243. 242 can be partitioned such that neither of the subsets contain numbers a,b,c such that ab = c as follows.

    Set 1: 3,4,5,6,7,8 and 81,82,83,…,242
    Set2: 9,10,11,…80

    So m > 242. Now, note that 3 and 9 must be in different sets. (otherwise 3^2 = 9). Also, 81 must be with 3 (otherwise 9^2 = 81). Furthermore, 27 now must be with 9 (otherwise 27*3=81). So we have

    Set 1: 3, 81
    Set 2: 27,9

    If we put 243 in set 1, then we have 3*81=243. If ewe put 243 in set 2 then 9*27 = 243. So we are done.

    Comment by Anonymous — March 17, 2010 @ 6:18 am

    • anyone know the answers to 7, 9, 13, and 15?

      Comment by MF — March 17, 2010 @ 10:51 am

      • 7: 760
        9: 158
        13: 365
        15: 45

        also, i believe
        -Underclassmen are discouraged from taking the AMC 10.
        -Talented 9th and 10th grade students swarm to take the AMC 12 in order to qualify for the USAMO.
        is incorrect. Though USAJMO is flawed, along with MOSP, why would talented 9/10th graders swarm to take the 12, when they know the USAJMO will be easy MOP?

        You are aware that you can make Red MOP via USAJMO right?

        Comment by someone — March 17, 2010 @ 10:06 pm

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