Old Putnam problems I have not finished working on. The wording is taken from John Scholes's web site (he hasn't finished with these problems either; send responses to him at jscholes@kalva.demon.co.uk ) except for the 1942 problems which I took from the May, 1942 American Mathematical Monthly (vol 99? pp 348-351) courtesy of JSTOR; AMM printed no solutions in those days. Problems remaining here are: 41B7, 42A5, 42A6, 49A2, 49A4, 49B6 ================ 41B7 B7. Do either (1) or (2): (1) f is a real-valued function defined on the reals with a continuous second derivative and satisfies f(x + y) f(x - y) = f(x)^2 + f(y)^2 - 1 for all x, y. Show that for some constant k we have f ''(x) = +- k^2 f(x). Deduce that f(x) is one of +-cos kx, +-cosh kx. (2) a_i and b_i are constants. Let A be the (n+1) x (n+1) matrix A_{ij}, defined as follows: A_{i1} = 1; A_{1j} = x^{j-1} for j <= n; A_{1 (n+1)} = p(x); A_{ij} = a_{i-1}^{j-1} for i > 1, j <= n; A_{i (n+1)} = b_{i-1} for i > 1. We use the identity det A = 0 to define the polynomial p(x). Now given any polynomial f(x), replace b_i by f(b_i) and p(x) by q(x), so that det A = 0 now defines a polynomial q(x). Prove that f( p(x) ) is a multiple of \prod (x - a_i) , plus q(x). [I'm not sure I even understand what part (2) is saying. I hope I rewrote it properly.] [Presumably this just means f(p(x)) = q(x) mod x - a_i, assuming the a_i are distinct (Chinese Remainder Theorem), which we can indeed assume by treating the a_i as variables, i.e. working in the polynomial ring Z[x, a_1, a_2, ...]. So we need only check that f(p(a_i)) = q(a_i). ...] ================ 42 A5 A5. A circle of radius a is revolvd through 180\degrees about a line in its plane, distant b from the center of the circle, where b > a . For what value of the ratio b/a does the center of gravity of the solid thus generated lie on the surface of the solid? [C is a circle radius a whose center lies a distance b from the coplanar line L. C is rotated through pi about L to form a solid whose center of gravity lies on its surface. Find b/a.] ================ 42 A6 A6. Any circle in the XY (horizontal) plane is "represented" by a point on the vertical line through the center of the circle, and at a distance "above" the plane of the circle equal to the radius of the circle. Show that the locus of the representations of all circles which cut a fixed circle at a constant angle is a (portion of a) one-sheeted hyperboloid. By consideration of suitable families of circles in the plane, demonstrate the existence of two families of rulings on the hyperboloid. [P is a plane and H is the half-space on one side of P. K is a fixed circle in P. C is a circle in P which cuts K at an angle alpha. Let C have center O and radius r. f(C) is the point in H on the normal to P through O and a distance r from O. Show that the locus of f(C) is a one-sheet hyperboloid and that it has two families of rulings in it.] ================ 49 A2 A2. Take points O, P, Q, R in space. Let the volume of the parallelepiped with edges OP, OQ, OR be V. Let V' be the volume of the parallelepiped which has O as one vertex and which has OP, OQ, OR as altitudes to three faces. Show that V V' = OP^2 OQ^2 OR^2. Generalize to n dimensions. ================ 49 A4 A4. Take P inside the tetrahedron ABCD to minimize PA + PB + PC + PD. Show that , tero@in.gr (Tero) wrote: >I am a high-school student in Greece and I have read in a Geometry >book about Fermat's problem - that is, given a triangle ABC, to find a >point P, so that >PA + PB + PC = minimum . The book mentions that P is so, so that the >angles > ^ ^ ^ >APB = BPC = CPA = 120 degrees. However, this doesn't seem right to me. >What happens if the triangle has an obtuse angle greater than 120 >degrees? As was mentioned by Sylvain Croussette, the Fermat point of an obtuse triangle is the vertex with the obtuse angle. For a generalization of this theorem and a proof of an algorithm to find the Fermat point of a finite set of points, see http://www.whim.org/nebula/math/fermatpt.html There it is proven that if the points are not in a line, their Fermat point is unique. It is also shown that if, for some k, | --- p_k - p_j | | > ----------- | <= 1 [1] | --- |p_k - p_j| | j<>k then p_k is the Fermat point. This corresponds to the vertex with the obtuse angle in an obtuse triangle. Otherwise, iterating the function --- p_k > --------- --- |p - p_k| k M(p) = ------------- [2] --- 1 > --------- --- |p - p_k| k converges to the Fermat point except possibly on a countable set. Note that [2] is a positively weighted mean of the {p_k}. If [1] does not hold, then if p is the Fermat point of {p_k}, we have --- p - p_k 0 = > --------- [3] --- |p - p_k| k Equation [3] is the generalization of the 120 degree condition mentioned above. Rob Johnson take out the trash before replying