**POLYA036: (2003 Putnam B-Problems, 12/6/03). **

**[Putnam 2003:B1] ****Do there exist polynomials a(x), b(x), c(y),
d(y) such that 1+xy+x^2y^2 = a(x)c(y)+b(x)d(y) holds identically?**

**[Putnam 2003:B2] ****Let
n be a positive integer. Start with the sequence 1, 1/2, 1/3, ..., 1/n,
form a new sequence of n-1 entries 3/4, 5/12, ..., (2n-1)/(2n(n-1)), by taking
the average of two consecutive entries in the first sequence. Repeat the
averaging of neighbors on the second sequence to obtain a third sequence
of n-2 entries and continue unitl the final sequence produced consists of
a single number x_n. Show that x_n < 2/n.
**

**[Putnam 2003:B3] Show that for each positive integer n, n! = Product_{i=1}^n
lcm(1,2,..., [n/i]).
Here, lcm denotes the least common multiple, and [x] denotes the greatest
integer <= x).**

[Putnam 2003:B4] ** Let f(z) = az^4+bz^3+cz^2+dz+e =a(z-r_1)(z-r_2)(z-r_3)(z-r_4)
where a, b, c, d, e are integers, a nonzero. Show that if r_1+r_2 is a rational
number, and if r_1 + r_2 is not equal to r_3+r_4, then r_1r_2 is a rational
number.
**

**[Putnam 2003:B5] Let A, B and C be equidistant
points on the circumference of a circle of unit radius, centered at
O, and let P be any point in the circle's interior. Let a, b, c be the distances
from P to A, B, C respectively. Show that there is a triangle with side lengths
a, b, c, and that the area of this triangle depends only on the distance
from P to O.
**

**int_0^1(int_0^1 |f(x)+f(y)|dxdy >= int_0^1 |f(x)|dx.
**

*************************************************************************************************
**

**[Putnam 2003:B1] ****Do there exist polynomials a(x), b(x), c(y),
d(y) such that 1+xy+x^2y^2 = a(x)c(y)+b(x)d(y) holds identically?
**

**[PY](12/7): ** No. Assume such a relation. Differentiation with
respect to x gives (1+2xy)y = a'(x)c(y) + b'(x)d(y). By putting x = 0 and
x=1, we see that c(y) and d(y)$ are both multiples of y. Similarly, a(x)
and b(x) are both multiples of x. Now, it is clear that the relation does
not hold, (by putting x=y=0).

**[JPE] (12/7):** Let a(x)=sum(a[i]x^i),... Consider the vectors C=[c[0],c[1],c[2]]
and D=[d[0],d[1],d[2]] We have a[0]C+b[0]D=[1,0,0]; a[1]C+b[1]D=[0,1,0] and
a[2]C+b[2]D=[0,0,1] which is clearly impossible because the three vectors
on the left sides of = are coplanar and the three vectors on the right side
of the equation are independent.

**[Putnam 2003:B2] ****Let
n be a positive integer. Start with the sequence 1, 1/2, 1/3, ..., 1/n,
form a new sequence of n-1 entries 3/4, 5/12, ..., (2n-1)/(2n(n-1)), by taking
the average of two consecutive entries in the first sequence. Repeat the
averaging of neighbors on the second sequence to obtain a third sequence
of n-2 entries and continue unitl the final sequence produced consists of
a single number x_n. Show that x_n < 2/n.
**

**[PY] (12/7): **We show that x_n = (2^n-1)/(n. 2^{n-1}). From this
it is clear that x_n < 2/n.

If we begin with an arbitrary sequence a_1, a_2, ..., a_n, the above
construction actually yields x_n = (1/2^{n-1}) Sum_{j=1}^n Binomial(n-1,j-1)
a_j.

With a_j = 1/j for j = 1, ..., n, we have x_n = (1/2^{n-1})
Sum_{j=1}^n (1/j) Binomial(n-1,j-1).

Consider more generally the sum

Sum_{j=1}^n Binomial(n-1,j-1)(x^j/j) = Sum_{j=1}^n Binomia(n-1,j-1) Integral
x^{j-1}dx

= Integral Sum_{j=1}^n Binomial(n-1,j-1) x^{j-1}dx

= Integral Sum_{j=0}^{n-1} Binomial(n-1,j) x^j dx

= Integral (1+x)^{n-1}dx

= (1/)n((1+x)^n-1).

It follows, by putting x=1, that x_n = (2^n-1)/(n.2^{n-1}).

**[TS](12/8): [pdf]
**

**[Putnam 2003:B3] Show that for each positive integer n, n! = Product_{i=1}^n
lcm(1,2,..., [n/i]).
Here, lcm denotes the least common multiple, and [x] denotes the greatest
integer <= x).**

**[JPE](12/8): **Let p be a prime number and v_p(m) the multiplicity
of p in the prime decomposition of m. There exists [n/p^k]-[n/p^(k+1)] values
of i for which v_p(lcm(1,2,..[n/i])) = k, because this means

n/p^(k+1) <i <= n/p^k. Thus v_p(product(lcm(1,2,..,[n/i]),i = 1..n)
= sum(k([n/p^k]-[n/p^(k+1)]),k>0) = sum([n/p^k],k>0) = v_p(n!) and
the result.

[Putnam 2003:B4] ** Let f(z) = az^4+bz^3+cz^2+dz+e =a(z-r_1)(z-r_2)(z-r_3)(z-r_4)
where a, b, c, d, e are integers, a nonzero. Show that if r_1+r_2 is a rational
number, and if r_1 + r_2 is not equal to r_3+r_4, then r_1r_2 is a rational
number.
**

**[JPE](12/8): **Let s = r_1+r_2; s' = r_3+r_4; p = r_1r_2; p' = r_3r_4.
By identification in az^4+bz^3+cz^2+dz+e = a(z^2-sz+p)(z^2-s'z+p'), we get
a(p+p'+ss') = c and a(ps'+p's) = -d. Hence p = (cs+d-as's^2)/(a(s-s')).
As s' = -s-b/a is rational, the result follows

**[Putnam 2003:B5] Let A, B and C be equidistant
points on the circumference of a circle of unit radius, centered at
O, and let P be any point in the circle's interior. Let a, b, c be the distances
from P to A, B, C respectively. Show that there is a triangle with side lengths
a, b, c, and that the area of this triangle depends only on the distance
from P to O.**

**[JPE](12/8): **Put
AP = a; BP = b; CP = c. Suppose that ABC is equilateral direct; consider the
point Q such as PAQ is equilateral direct; then the lengths of the sidelines
of the triangle BQP are a,b,c.

Now suppose that A,B,C,P have respective affix 1,exp(2iPi/3), exp(-2iPi/3),
r exp(it) with 0<=r<1. We have

a^2 = r^2-2r cos(t)+1

b^2 = r^2+(cos(t)-root(3)sin(t))r+1

c^2 = r^2+(cos(t)+root(3)sin(t))r+1

Hence U = a^2+b^2+c^2 = 3(r^2+1) and V = b^2c^2+c^2a^2+a^2b^2 = 3(r^4+r^2+1).
It follows from Heron's formula that the area is S = root(4V-U^2)/4 = root(3)(1-r^2)/4.

**[Putnam 2003:B6] ****Let f(x) be a continuous real-valued function
defined on the interval [0,1]. Show that
**

[TS](12/8): [pdf]

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