
The Problems:

The Solvers: 
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Question: 20031
submitted by Thomas Mildorf 
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1) Pinto Sandtelle, Stuyvesant
2) Brian Rice, SWVA Governor's
3) Jon Chu, AAST
4) Ricky Biggs, TJHSST
5) Brian Jacokes, TJHSST
6) Jaemin Bae, AAST
7) Jesse Geneson, TJHSST 
...... 

Question: 20032
submitted by Thomas Mildorf 
......
1) Jaemin Bae, AAST
2) Jesse Geneson, TJHSST
3) Jon Chu, AAST
4) Pinto Sandtelle, Stuyvesant
5) Yifei Chen, WWP North
6) Shrenik Shah, Monta Vista
7) John Schulman, Great Neck S 
...... 

Question: 20033
submitted by Jaemin Bae 
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1) Thomas Mildorf, TJHSST
2) Jesse Geneson, TJHSST
3) Jon Chu, AAST
4) Ricky Biggs, TJHSST
5) Thomas Pollom, Cathedral HS
6) Alexei Harvard, Great Neck South
7) John Schulman, Great Neck South 
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Question: 20034
submitted by John Schulman 
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1) Thomas Mildorf, TJHSST
2) Ricky Biggs, TJHSST
3) Brian Rice, SWVA Governor's
4) Pinto Sandtelle, Stuyvesant
5) Thomas Pollom, Cathedral HS
6) Jon Chu, AAST
7) Shrenik Shah, Monta Vista 
...... 

Question: 20035
submitted by Thomas Pollom 
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1) Thomas Mildorf, TJHSST
2) Pinto Sandtelle, Stuyvesant
3) Yifei Chen, WWP North
4) Luke Joyner, Roxbury Latin
5) Jaemin Bae, AAST
6) Ricky Biggs, TJHSST
7) Jeff Chen, TJHSST 
...... 

Question: 20036
submitted by Shrenik Shah 
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1) Jesse Geneson, TJHSST
2) Brian Rice, SW VA Gov
3) Thomas Mildorf, TJHSST
4) Luke Joyner, Roxbury Latin
5) Ricky Biggs, TJHSST
6) Pinto Sandtelle, Stuyvesant
7) Erdem Varol, Stuyvesant 
...... 

Question: 20037
submitted by Jesse Geneson 
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1) Thomas Mildorf, TJHSST
2) Shrenik Shah, Monta Vista HS
3) Albert Ni, IMSA
4) Pinto Sandtelle, Stuyvesant
5) Luke Joyner, Roxbury Latin
6) Ricky Biggs, TJHSST
7) Anthony Kim, TJHSST 
...... 

Question: 20038
submitted by Brian Rice 
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1) Thomas Mildorf, TJHSST
2) Shrenik Shah, Monta Vista HS
3) Jaemin Bae, AAST
4) Pinto Sandtelle, Stuyvesant
5) HyunSoo Kim, AAST
6) Jesse Geneson, TJHSST
7) Michael Druker, TJHSST 
...... 

ANSWERS TO 2003
PROBLEMS
Question 20031
 the minimum of 5/4 is attained if and only if
a=2, b=8, and c=2. To prove this split the third term
up into 1/2a + c/4a + c/4a, then use AMGM on the resulting
five terms.
Question 20032  there
are many possible ways to find AD; one elegant method
involves reflecting point A over line BD to a point A'. One
then argues that A' lies on line CD. After working out
various other details, one finds that CE = EA' = ED + DA' =
ED + AD, which gives AD = 12  2*sqrt(14).
Question 20033  the
answer is L = 1/6, as shown in the
solution sent by the
proposer.
Question 20034  there
are many different solutions; one simple method is to
assign to the point (x,y) the number x. (So that the
middle row consists entirely of 0's, the next higher row is
all 1's, and so on.) To prove that no such numbering can be
periodic, suppose otherwise, and then consider a maximal
number in the array.
Question 20035  one such
circle is the one with center (0, 6.25) and radius
6.25, passing through the points (0,0), (5,10), (5,10),
(6,8), and (6,8). The proposer also mentions Schinzel's
Theorem, which states that a circle passing through n
lattice points is (x(1/2))^2 + y^2 = (1/4)*5^(k1) for n =
2k, and (x(1/3))^2 + y^2 = (1/9)*5^(2k) for n = 2k+1.
Question 20036  write
n = t_0 + 3*t_1+9*t_2+...+3^k*t_k in base three, so that
each number t_i is either 0, 1, or 2. One then argues that
f(n) = t_0 + 2*t_1+4*t_2+...+2^k*t_k. The question then
amounts to finding all numbers less than 2003 = 2102022_3
(base three) with only 0's and 1's in their base three
representation. All numbers with 7 or less digits and with
only 0's and 1's work, so the answer is 128.
Question 20037  the
remarkable fact is that one ultimately obtains 1 upon
rationalizing the denominator. This may be proven by
induction.
Question 20038  First
argue that a_(n+1) = 2a_n + a_(n1). We now proceed by
induction. The base cases n=1 and n=2 are trivial. The
inductive step is as follows:
a_(n+1) = 2a_n + a_(n1) <= 2*9*(2.5)^(n2) +
9*(2.5)^(n3)
= 9*(2.5)^(n3)*(6) < 9*(2.5)^(n3)*(6.25) =
9*(2.5)^(n1), as desired.
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