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Problem 1.  Find all functions f   : (0,∞) → (0,∞) such thatf x +

√ x) ≤ x ≤ f (x) +  f (x) for all   x ∈ (0,∞).

Problem 2.  Find all distinct pairs (x, y) of integers that aresolutions of the equation

x2 − xy + y2 = x + y.

Problem 3.   Find the largest subset  A ⊂ {1, 2, . . . , 2003}such that for all  a,  b ∈ A,  a + b  is not divisible by  a − b.Problem 4.   Let  x1,  x2, . . . ,  x2004  be positive real numberssuch that

1

2003 + x1+

  1

2003 + x2+ · · · +   1

2003 + x2004> 1.

Prove that  x1x2· · ·

x2004  <  1.

Problem 5.   Four points are given in the plane. If the dis-tance between any two of them is not less then

√ 2 and not

greater than 2, prove that these points are the vertices of asquare.

Problem 6.   Find the maximum value of the area of a triangleABC  that has vertices on three circles centered at the samepoint with radii 1,

√ 7, and 4, respectively.

Send your solutions to  π   in the Sky :  Math Challenges.

Solutions to the Problems Published in the Septem-ber, 2002 Issue of  π   in the Sky :

Problem 1.   Let  n   be a fixed positive integer and consider the moregeneral problem of solving

xy

x + y= n,

where  x  and  y  are positive integers. Then y  =   nxx−n

. In particular, y   is

a positive integer if and only if   x− n   is a positive integer that dividesnx. But  nx  =  n2 +n(x−n), so we see that  x−n divides  nx  if and onlyif it divides  n2. Consequently, the number of  x   values yielding positiveintegers   y   is precisely equal to the number of positive divisors of   n2.Indeed, for each positive divisor  d  of  n2, we let  x  =  n + d. For example,when  n  = 100, we get

n2 = 1002 = 24 · 54.Thus, the positive divisors of 1002 are precisely the numbers of the form2a · 5b with  a  = 0, 1, 2, 3, or 4, and   b = 0, 1, 2, 3, or 4. It follows thatthere are 5 · 5 = 25 divisors and hence there are 25 positive integers  xthat yield positive integer y .

Problem 2.   If 2003+n =  m(n+ 1), then 2002 = m(n+ 1)− (n+ 1) =(m − 1)(n + 1) and  n + 1 ≥  1 is a divisor of 2002 = 2 · 7 · 11 · 13. Inparticular,

n + 1 = 2a · 7b · 11c · 13dwith each a,  b,  c,  d  being 0 or 1. Since there are 24 = 16 possible choicesfor the exponent, there are 16 possible choices for  n.

The above solutions of the problems 1 and 2 were presented to

π   in the Sky    by Jeganathan Sriskandarajah  from Madison, WI.

These problems were also correctly solved by  Robert Bilinski

from Montréal and  Edward T.H. Wang  from Waterloo.

Problem 3.   (Solution by Wieslaw Krawcewicz) The picture belowillustrates the solid   P  that is the intersection of the unit cube with acopy that is rotated 30 degrees. This solid can be obtained by cutting offfrom the initial cube six identical tetrahedrons, one of which, denotedby OABC , is indicated in the picture.

       x

     x

       x

       x

x       x

A

C 

B

O

Th triangle ABC  is right-angled and it has sides x,  x, and√ 

2x. Sincthe length of the edge of the cube is one, we get

1 =  x +√ 

2x + x   ⇐⇒   x =   12 +

√ 2.

Therefore the volume of the tetrahedron   OABC   is   16x2, and conse

quently we find that the volume  V    of the solid  P  is given by

V    = 1 − 6

1

6 · 1

(2 +√ 

2)2

 =

  5 + 4√ 

2

6 + 4√ 

2.

Madison Area Technical College Math

Club Celebrated π  Day on March 14, 2003

The main event of the   πDay celebration was the mathcompetition featuring team

from six different two-year colleges in Wisconsin. Among thother activities, there was alsan informative presentation on“The Calculation of Pi,” aposter competition and a pieeating contest. At an awardceremony the winning teamand individuals were presentedwith their prizes. The picture above shows several con

testants at the beginning of the math competition.

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