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.

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

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

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[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

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

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

Bibliography.