From rusin@math.niu.edu Mon Dec 15 18:20:59 2003 Received: from vesuvius.math.niu.edu (vesuvius [131.156.3.93]) by sanford.math.niu.edu (8.12.10/8.12.10) with ESMTP id hBG0Kx25029051; Mon, 15 Dec 2003 18:20:59 -0600 (CST) Received: (from rusin@localhost) by vesuvius.math.niu.edu (8.9.3/8.9.3) id SAA04502; Mon, 15 Dec 2003 18:21:01 -0600 (CST) Date: Mon, 15 Dec 2003 18:21:01 -0600 (CST) Message-Id: <200312160021.SAA04502@vesuvius.math.niu.edu> From: rusin@math.niu.edu To: jscholes@kalva.demon.co.uk Subject: Putnam 1941 A3 Cc: rusin@math.niu.edu X-Spam-Status: No, hits=1.1 required=5.0 tests=NO_REAL_NAME version=2.55 X-Spam-Level: * X-Spam-Checker-Version: SpamAssassin 2.55 (1.174.2.19-2003-05-19-exp) Status: O Content-Length: 1729 1941 A3. A circle radius a rolls in the plane along the x-axis. The envelope of a diameter is the curve C. Show that we can find a point on the circumference of a circle radius a/2, also rolling along the x-axis, which traces out the curve C. Answer: When the circle has rolled through an angle t, the position of the center of the circle will be ( a t, a ) (since the x-coordinate advances linearly with t and by a distance 2 pi a when t reaches 2 pi). A point originally at a position u units to the right of the center ( -a <= u <= a ) will rotate to a position ( u cos(t), -u sin(t) ) relative to the center at that time, that is, it will be at the point f(t,u) = (a t + u cos(t), a - u sin(t) ) The sweep of the initially-horizontal diameter of the circle is then the image of f as u ranges over [-a, a] and t ranges over the real line. We may then compute the envelope as the image of the singular points of f. Since f' is, in matrix form, ( a - u sin(t) cos(t) ) ( - u cos(t) -sin(t) ) we find det(f')=0 when - a sin(t) + u = 0, that is, along the curve where u = a sin(t). The image of such a point (t,u) is the point (x,y) = f(t,u) = (a t + (a sin(t)) cos(t), a - (a sin(t)) sin(t) ) = a (t + sin(2t)/2, 1 - ( 1 - cos(2t))/2 ) = (a/2) (2t + sin(2t), 1 - cos(2t) ) = a' ( t' + sin(t'), 1 - cos(t') ) with the obvious substitutions for a', t'. But in exactly the same way as in the first paragraph, this is the location of a point on a disk of radius a' after it has rolled through an angle of t' along the x-axis, if it starts on the perimeter of the disk and directly above the center at time t'=0. From rusin@math.niu.edu Fri Dec 12 16:16:24 2003 Received: from vesuvius.math.niu.edu (vesuvius [131.156.3.93]) by sanford.math.niu.edu (8.12.10/8.12.10) with ESMTP id hBCMGN25007299; Fri, 12 Dec 2003 16:16:23 -0600 (CST) Received: (from rusin@localhost) by vesuvius.math.niu.edu (8.9.3/8.9.3) id QAA26338; Fri, 12 Dec 2003 16:16:25 -0600 (CST) Date: Fri, 12 Dec 2003 16:16:25 -0600 (CST) Message-Id: <200312122216.QAA26338@vesuvius.math.niu.edu> From: rusin@math.niu.edu To: jscholes@kalva.demon.co.uk Subject: Putnam 1941 A6 Cc: rusin@math.niu.edu X-Spam-Status: No, hits=0.2 required=5.0 tests=BAYES_30,NO_REAL_NAME version=2.55 X-Spam-Level: X-Spam-Checker-Version: SpamAssassin 2.55 (1.174.2.19-2003-05-19-exp) Status: O Content-Length: 3366 1941 A6. f is defined for the non-negative reals and takes positive real values. The centroid of the area lying under the curve y = f(x) between x = 0 and x = a has x-coordinate g(a). Prove that for some positive constant k, f(x) = k g'(x)/(x - g(x))^2 e^{\int 1/(t - g(t)) dt} . Answer: I can prove this assuming f is continuous, so that we may apply the Fundamental Theorem of Calculus. (Certainly something akin to continuity is necessary, since derivatives satisfy the Intermediate Value Property, and therefore the given expression for f implies f is not TOO strangely behaved. Note that we were not even told f is _measurable_, which is necessary for the centroid to be well defined!) The x-coordinate of the centroid is (defined to be) g(a) = N(a)/D(a), where N(a) = \int_0^a x f(x)dx D(a) = \int_0^a f(x)dx As functions of a these are differentiable, with derivatives N'(a) = a f(a) D'(a) = f(a) So we may eliminate N and D in favor of f, g, and their derivatives: from N(a) = g(a) D(a) we obtain N'(a) = g'(a) D(a) + g(a) D'(a), i.e. a f(a) = g'(a) D(a) + g(a) f(a), so that D(a) = (a - g(a)) f(a) / g'(a). (The division can be justified by noting that g'(a) is never zero: if it were, then we'd have a f(a) = g(a) f(a), which means g(a)=a since f(a) is never zero by hypothesis. But it is easy to prove that the centroid must be in the interior of a region defined by a continous function f. Note that since g'(a) is now both a continuous function of a and never zero, it is always of one sign; since g(a) >= 0 for all a, and g(a) --> 0 as a --> 0, only g'(a) > 0 is possible by the Mean Value Theorem.) Now, if f were differentiable, N, D, and thus g would be twice differentiable and we can differentiate again to obtain a polynomial relation in a,f(a),f'(a),g(a),g'(a), and g''(a), which may be integrated without much difficulty. In the absence of the differentiability of f, however, we must proceed as follows. Note that our most recent equation shows that q(a):=(f/g')(a) = D(a)/(a-g(a)) is a differentiable function of a; its derivative is q'(a) = f(a)/(a-g(a)) - D(a)(1-g'(a))/(a-g(a))^2 = q(a) g'(a)/(a-g(a)) - (1-g'(a))/(a-g(a)) q(a) = q(a) ( (2 g'(a) - 1)/(a-g(a)) ), using both the definition of q and the last expression for D(a). We have noted that both f and g' are positive, so q is as well; thus our equation may be written ( log(q(a)) )' = (2 g'(a) - 1)/(a-g(a)) = -2(1-g'(a))/(a-g(a)) + 1/(a-g(a)) i.e. log(q(a)) + 2 log(a-g(a)) = \int 1/(a - g(a) ) da (meaning, as usual, that the function on the left is AN antiderivative of the integrand on the right). If we select any one antiderivative on the right, we then have log( f(a)/g'(a) (a-g(a))^2 ) = \int 1/(t - g(t)) dt + C for some constant C, i.e. f(a) = k g'(a)/(a-g(a))^2 exp( \int 1/(t-g(t)) dt ) where k = exp(C). (Note that \int_0^a 1/(t-g(t)) dt is NOT well-defined whenever f is smooth at the origin! Thus our choice of antiderivative will have to vanish somewhere else besides the origin.) I imagine the test-takers were not expected to dwell overly much on the technical conditions needed to guarantee differentiability, positivity, uniqueness, etc. at each step. I hope not, because I don't think I have addressed these fully myself! dave From rusin@math.niu.edu Mon Dec 15 16:22:00 2003 Received: from vesuvius.math.niu.edu (vesuvius [131.156.3.93]) by sanford.math.niu.edu (8.12.10/8.12.10) with ESMTP id hBFMM025028308; Mon, 15 Dec 2003 16:22:00 -0600 (CST) Received: (from rusin@localhost) by vesuvius.math.niu.edu (8.9.3/8.9.3) id QAA04216; Mon, 15 Dec 2003 16:22:02 -0600 (CST) Date: Mon, 15 Dec 2003 16:22:02 -0600 (CST) Message-Id: <200312152222.QAA04216@vesuvius.math.niu.edu> From: rusin@math.niu.edu To: jscholes@kalva.demon.co.uk Subject: 1942 B2 Putnam Cc: rusin@math.niu.edu X-Spam-Status: No, hits=0.2 required=5.0 tests=BAYES_30,NO_REAL_NAME version=2.55 X-Spam-Level: X-Spam-Checker-Version: SpamAssassin 2.55 (1.174.2.19-2003-05-19-exp) Status: O Content-Length: 4240 Like many of the early questions, this one requires less cleverness and more computational patience than is typical of modern questions. Also, like many of the old ones, it is not entirely clear what the expected interpretation of the question was. dave 1942 B2. Let P_a be the parabola y= a^3 x^2/3 + a^2 x/2 - 2 a. .... (*) Find the locus of the vertices of P_a, and the envelope of P_a. Sketch the envelope and two P_a. The vertex of a parabola given in the form y = f(x) has x-coordinate at the critical point of f (memorized in high school as "-b/2a"); here this is at x = (-3/4)/a. The corresponding y-coordinate is y = (-35/16) a . When a = 0, (*) does not describe a parabola, of course. It is not entirely clear what is meant by "envelope" here. Typically the envelope of a family of curves P_a is intended to be a curve tangent to each P_a which it meets (and it is supposed to meet some!). In practice that is the boundary of the union of the curves. But in this problem, almost every point (x,y) of the plane lies in a P_a, since we need only solve for a in (*), and that is a cubic polynomial in a and so always has a real root. (The only exception occurs when x = 0; but even in that case, (*) can be solved for a, for every y!). So the union of the P_a is the entire plane. (It might help to imagine the problem posers intended a to be positive, say; by Descartes' Rule of Signs, the cubic (*) has exactly one positive real root a for every x (except zero) and every y > 0 , but for some negative y there is no positive real root a. As we will show below, the union of the P_a is precisely the region above two hyperbolic arcs.) Alternatively, if the curves are given by a parameterization (x,y) = f_a(t) (or (x,y) = f(a,t) ) we may define the envelope to be the set of singular values of the mapping f . (That is, the envelopes are the "folds in the fabric" as we lay the a-t plane atop the x-y plane.) Ours is a family of curves which is invariant under the transformation (x,y) -> (c x, y/c), so we may use, for example, the parameterization f(a,t) = (t/a, a g(t) ) where our g(t) = t^2/3 + t/2 - 2. In that case we may compute the singular points to be those where f' is singular; in matrix form we have f'(a,t) = ( 1/a -t/a^2 ) ( a g'(t) g(t) ) so that the singular points occur where g(t) + t g'(t) = 0 (and a is arbitrary). For our g(t) this condition works out to t^2 + t - 2 = 0, so the singular points form the lines t=1 and t=-2. These map respectively to the singular values f(a,1) = ( 1/a, (-7/6)a ) and f(a,-2) = ( (-2)/a, (-5/3) a ); the first set of points forms the hyperbola y = (-7/6)/x and the second forms the hyperbola y = (10/3)/x. We can check that these curves are indeed tangent to the parabolas: we may check for points of intersection by solving e.g. ( a^3 x^2/3 + a^2 x/2 - 2 a ) = (-7/6) / x and noting that there is a solution of multiplicity 2 at x = 1/a (y = (-7/6)a ). Likewise P_a meets the other hyperbola at a double point at x = (-2)/a, y = (-5/3) a . Note that when a > 0, P_a is tangent to each of the lower branches of the two hyperbolas; when a < 0 the two points of tangency occur on the upper branches. On the other hand, each P_a also meets the _other_ branches of each hyperbola, and is not tangent to it, so I would be reluctant to refer to either branch of either parabola as lying in the envelope of the entire family P_a ! So while it is not clear what the best answer is, it is probably the case that what was intended was the union of the two hyperbolas y = (-7/6)/x and y = (10/3)/x. As for the sketches: it is perhaps worth noting that all the curves mentioned are setwise invariant under the scalings (x,y) -> (c x, y/c ), which permute the family of the P_a, so that it is in some sense sufficient to sketch a single one of the P_a ! The one with a = 1 is simple to sketch and clearly shows the vertex and two points of tangency lying on three hyperbolas. (The limiting cases a --> 0 and a --> \infty are also worth a look; the cases with a < 0 are obtained by a 180-degree rotation around (0,0) from the cases with a > 0.) From rusin@math.niu.edu Thu Dec 18 14:08:04 2003 Received: from vesuvius.math.niu.edu (vesuvius [131.156.3.93]) by sanford.math.niu.edu (8.12.10/8.12.10) with ESMTP id hBIK8425023497; Thu, 18 Dec 2003 14:08:04 -0600 (CST) Received: (from rusin@localhost) by vesuvius.math.niu.edu (8.9.3/8.9.3) id OAA11920; Thu, 18 Dec 2003 14:08:05 -0600 (CST) Date: Thu, 18 Dec 2003 14:08:05 -0600 (CST) Message-Id: <200312182008.OAA11920@vesuvius.math.niu.edu> From: rusin@math.niu.edu To: jscholes@kalva.demon.co.uk Subject: Putnam 1942 B3 Cc: rusin@math.niu.edu X-Spam-Status: No, hits=0.2 required=5.0 tests=BAYES_30,NO_REAL_NAME version=2.55 X-Spam-Level: X-Spam-Checker-Version: SpamAssassin 2.55 (1.174.2.19-2003-05-19-exp) Status: O Content-Length: 5627 1942 B3. f(x, y) and g(x, y) satisfy the differential equation f_1(x, y) g_2(x, y) - f_2(x, y) g_1(x, y) = 1 ....(*). Taking r = f(x, y) and y as independent variables, and x = h(r, y), g(x, y) = k(r, y), show that k_2(r, y) = h_1(r, y). Integrate and hence obtain a solution to (*). What other solutions does (*) have? Answer: Given the informal statement of the problem ("independent variables"?) it seems probable that "cookbook" solutions are probably expected. (But see below.) On the other hand, asking for "a" solution to (*) seems rather silly since clearly e.g. f(x,y)=x, g(x,y)=y is a solution. Any solution of the original differential equation may be graphed in R^4: we obtain a set of points (x,y,r,k) where r=f(x,y), k=g(x,y); that is, we obtain a smooth surface on which F : R^4 --> R^2 vanishes, where F is given by F(x,y,r,k) = (f(x,y)-r, g(x,y)-k). The derivative of F is a 2x4 matrix, ( f_1, f_2, -1, 0 ) ( g_1, g_2, 0, -1 ) By the implicit function theorem we may (locally) solve for any pair of variables in terms of the other two, as long as the determinant of the corresponding 2x2 submatrix is nonzero, and can compute the derivative of the resulting relation: d(s,t)/d(u,v) = - (dF/d(s,t))^{-1} (dF/d(u,v)). In particular we solve for x,k in terms of r,y and get d(x,k)/d(r,y)=-(1/f_1)[[-1,0],[-g_1, f_1]] [[-1,f_2],[0,g_2]] =-(1/f_1)[[1,f_2],[g_1, -g_1 f_2 + f_1 g_2]] =-(1/f_1)[[1,f_2],[g_1,1]] so that, as desired, we have dx/dr = dk/dy (= -1/f_1). (Note that the derivative f_1 is presumed to be nonzero by the assertion that "r = f(x,y) and y are independent".) This condition makes (x,k)=(x(r,y),k(r,y)) a conservative vector field defined on the (r,y) plane; thus there is a scalar-valued function U=U(r,y) with x=dU/dy, k=dU/dr. Thus the surface in R^4 is parameterized by two variables, say s and t, as the set of points (dU(s,t)/ds, s, t, dU(s,t)/dt) and the corresponding functions f, g satisfy t = f( dU/ds, s ) dU/dt = g( dU/ds, t ) (**) (e.g., in order to define f we "solve for t" by "eliminating s".) The complete solution to (*) is obtained by considering all (twice-differentiable) functions U and all possible eliminations for f and g to solve (**) in the previous paragraph. For example, if U(s,t) = s^2 t then (x,y,r,k) = (2st, s, t, s^2) and so f(x,y) = x/(2y) , g(x,y) = y^2, and indeed we check that f_1(x, y) = 1/(2y) g_2(x, y) = 2y f_2(x, y) = -x/(2y^2) g_1(x, y) = 0 so that (*) holds. When we are asked, What other solutions does (*) have? it's not clear whether we are requested to find the general solution to (*) under the condition that r and y be "independent" (which I have done above), or to find the solutions to (*) for which r and y are not independent, that is, r = f_1(x,y) depends only on y. Solutions with r=r(y) not identically zero are also covered by the discussion above, so we need only look for solutions with f_1 = 0, i.e. f is a function of y alone, say f(x,y) = a(y). In that case (*) reads - a'(y) g_1(x,y) = 1 so we may solve for g : g(x,y) = -x/a'(y) + b(y) for some other function b also of one variable only. That is, we have solutions of the form (f(x,y), g(x,y)) = ( a(y), -x/a'(y) + b(y)). It should be noted that this method is fairly formal, and there are some questions about how effectively it solves the original question. Our recipe (through (**)) need not always produce a solution to (*), and not every solution to (*) is obtained by our recipe: First, there are some potential limitations on the possible functions U which may be used. A differentiable solution to (**) exists, and is unique, by the Implicit Function Theorem again, but only as long as d^2U(s,t)/dsdt is not zero. Near any point where this second derivative vanishes, it is potentially true that (**) admits either no solution, or multiple solutions. Second, there may be solutions to the original problem which cannot be obtained in this way for any U : while we proved that dx/dr = dk/dy , as demanded in the question, our proof requires that f_1 be nonzero. The "independence" clause is probably meant to imply that f_1 does not vanish on an open set, but it may not have been intended to imply that f_1 _never_ vanishes. So there may be pairs of functions (f,g) which solve the initial ODE (*), and for which the derivative f_1 vanishes along some curve (say); then the preceding analysis applies to show that there is a suitable function U but we have no guarantee that it can be defined globally, and indeed it need not even be single-valued on the region where f_1 is nonzero. It is perhaps worth noting that, once any solution to (*) is found, we can easily find the complete set of solutions (fbar, gbar) to (*) which have the same first function (i.e., having fbar=f ). For, as a differential equation in g alone, (*) is linear, and so all other solutions gbar are obtained from an initial solution g by adding arbitrary solutions to the corresponding homoegenous equation f_2 g_1 = f_1 g_2 . Among these solutions are g(x,y) = V( f(x,y) ) where V is an arbitrary differentiable function of one variable. (These may not be the only solutions in some cases, e.g. if f = W(h(x,y)) for some smooth functions W (of one variable) and h (of two variables) then the homogeneous equation is solved by any g(x,y) = Vtilde(h(x,y)); in some cases such functions g are not of the form V(f(x,y)) for any differentiable funtion V.) From rusin@math.niu.edu Sun Dec 21 22:09:30 2003 Received: from vesuvius.math.niu.edu (vesuvius [131.156.3.93]) by sanford.math.niu.edu (8.12.10/8.12.10) with ESMTP id hBM49U25016477; Sun, 21 Dec 2003 22:09:30 -0600 (CST) Received: (from rusin@localhost) by vesuvius.math.niu.edu (8.9.3/8.9.3) id WAA21390; Sun, 21 Dec 2003 22:09:31 -0600 (CST) Date: Sun, 21 Dec 2003 22:09:31 -0600 (CST) Message-Id: <200312220409.WAA21390@vesuvius.math.niu.edu> From: rusin@math.niu.edu To: jscholes@kalva.demon.co.uk Subject: Putnam 1942 B5 Cc: rusin@math.niu.edu X-Spam-Status: No, hits=-1.5 required=5.0 tests=BAYES_20,NO_REAL_NAME version=2.55 X-Spam-Level: X-Spam-Checker-Version: SpamAssassin 2.55 (1.174.2.19-2003-05-19-exp) Status: O Content-Length: 2229 1942 B5 B5. Let f(x) = x/(1 + x^6 sin^2x). Sketch the curve y = f(x) and show that \int_0^\infty f(x) dx exists. [For a sketch, I would simply: sketch y=sin(x); square y-coordinates to get the graph of y=sin(x)^2; sketch x^6, and then use this to "stretch" the graph of y=sin(x)^2 vertically; add 1 ; take reciprocals; multiply by x (probably noting the asymptotic approach to the graphs of y=x and y=1/x^5). In truth, a hand sketch is likely to vastly underestimate the vertical growth!] Answer: Clearly the denominator is continuous and at least equal to 1, so f is continuous and so the integral over any finite interval exists; in fact it is less than the integral of x, over any subinterval of [0, \infty). Also the integrand is positive for x>0, so the integrals over [1,N] increase with N. Therefore we need only show that those integrals remain bounded (and thus have a limit as N --> \infty). For any x > 0 we may we write x = n pi + z for some (unique) integer n >=0 and some z with |z| <= pi/2. Then sin(x) = (-1)^n sin(z) and so sin(x)^2 = sin(z)^2 >= z^2/(pi/2)^2. Also x >= (n - 1/2) pi, so x^6 > (n - 1/2)^6 pi^6. Thus the denominator of the integrand is at least 1 + C^2 z^2 where C = 2 (n-1/2)^3 pi^2. So on the interval [ (n-1/2)pi, (n+1/2)pi ] we may bound the integrand with x/(1 + C^2 z^2) = n pi/(1 + C^2 z^2) + z/(1 + C^2 z^2) where z = x - n pi. This upper bound is easily integrated: the integral of the second summand is zero over [-pi/2, pi/2] and the other summand has antiderivative n pi/C arctan( C z ) and so its integral over this interval is at most n pi/C (pi) = (1/2) n/(n-1/2)^3. So we have an upper bound for the integral over the n-th period of the sine function. This gives an upper bound for the integral over the right tail of the real line: 1/2 \sum n/(n + 1/2)^3. This converges e.g. by the Limit Comparison test (compare to the convert series \sum 1/n^2) (In fact, it sums to 3 zeta(2) - 7/2 zeta(3) which is less than 1 !) Note that the same reasoning applies if we replace the exponent "6" with any number greater than "4", but similar reasoning shows that the integral of x/(1 + x^4 sin(x)^2 ) is divergent. (I did not expect this!) From rusin@math.niu.edu Mon Dec 15 17:26:06 2003 Received: from vesuvius.math.niu.edu (vesuvius [131.156.3.93]) by sanford.math.niu.edu (8.12.10/8.12.10) with ESMTP id hBFNQ625028705; Mon, 15 Dec 2003 17:26:06 -0600 (CST) Received: (from rusin@localhost) by vesuvius.math.niu.edu (8.9.3/8.9.3) id RAA04370; Mon, 15 Dec 2003 17:26:07 -0600 (CST) Date: Mon, 15 Dec 2003 17:26:07 -0600 (CST) Message-Id: <200312152326.RAA04370@vesuvius.math.niu.edu> From: rusin@math.niu.edu To: jscholes@kalva.demon.co.uk Subject: Putnam 1948 A6 Cc: rusin@math.niu.edu X-Spam-Status: No, hits=1.1 required=5.0 tests=NO_REAL_NAME version=2.55 X-Spam-Level: * X-Spam-Checker-Version: SpamAssassin 2.55 (1.174.2.19-2003-05-19-exp) Status: O Content-Length: 2363 Another problem with an implicit smoothness assumption which we were apparently supposed to guess. At least I think so: I don't claim much expertise with physics; you might want to check that I haven't completely bungled the idea. This should wrap up 1948. dave 1948 A6. [ Do either (1) or (2): ] (1) On each element ds of a closed plane curve there is a force 1/R ds, where R is the radius of curvature. The force is towards the center of curvature at each point. Show that the curve is in equilibrium. Answer: Obviously a central force applied to each point on a circular piece of string would collapse the string to a point; so "equilibrium" here evidently implies that the curve is to be rigid and thus may be treated as a single object to be moved (rather than a set of points to be deformed). Reference to "ds" suggests that the curve admits a piecewise-differentiable parameterization f : [0,L] --> R^2 in terms of arc length; since the curve is closed, f(0) = f(L). The radius of curvature is 1/|| f''(t) ||, so the force acting on an "element" ds has magnitude ||f''(t)|| ds and in the same direction as f''(t) (since (d/dt)(f'(t) . f'(t)) = 0 ), so the "element" of force dF is precisely f''(t) ds = f''(t) dt. The integral along a path from f(t0) to f(t1) where the curve is smooth is then f'(t1) - f'(t0), and in particular if the whole curve is smooth, the total force acting on the curve is f'(L) - f'(0), that is, it equals the change in the tangent direction where the two ends of the curve join together. Assuming the curve is also smooth at that point, the difference is zero: there is no net force, and so the curve (as a body) is under equilibrium. To see that smoothness is necessary, simply consider the boundary of a lune bounded by arcs of two circles of very different radii (in the limiting case, the boundary of a half-disk): on the portion where the arc is straighter, the force is smaller (and is applied to a shorter arc); on the other arc, the force is larger. On both arcs the net force is, by symmetry, directed along the line joining the centers of the two circles. There is then a net force in the direction pointing from the more curved arc towards the straighter one. (In the limiting case it is clear there is a nonzero force toward the straight edge and a force of zero toward the curved edge.) From rusin@math.niu.edu Fri Dec 12 13:41:12 2003 Received: from vesuvius.math.niu.edu (vesuvius [131.156.3.93]) by sanford.math.niu.edu (8.12.10/8.12.10) with ESMTP id hBCJfC25006195; Fri, 12 Dec 2003 13:41:12 -0600 (CST) Received: (from rusin@localhost) by vesuvius.math.niu.edu (8.9.3/8.9.3) id NAA26023; Fri, 12 Dec 2003 13:41:12 -0600 (CST) Date: Fri, 12 Dec 2003 13:41:12 -0600 (CST) Message-Id: <200312121941.NAA26023@vesuvius.math.niu.edu> From: rusin@math.niu.edu To: jscholes@kalva.demon.co.uk Subject: Putnam 1948B2 Cc: rusin@math.niu.edu X-Spam-Status: No, hits=-1.5 required=5.0 tests=BAYES_20,NO_REAL_NAME version=2.55 X-Spam-Level: X-Spam-Checker-Version: SpamAssassin 2.55 (1.174.2.19-2003-05-19-exp) Status: O Content-Length: 1849 You can tell I have work to do that I'm avoiding... Here's another answer for an old Putnam exam. dave 1948 B2. A circle radius r is tangent to the three coordinate planes (x = 0, y = 0, z = 0) in space. Find the locus of its center. Answer: How far is a point from a plane, measured along another plane? There is probably some simple way to answer this but I had to work it out. If the center is at x0,y0,z0 and the unit normal to the plane containing the circle is (A,B,C) then that plane is A (x-x0) + B (y-y0) + C (z-z0) = 0, which intersects the plane z=0 (say) along the line where A (x-x0) + B (y-y0) = C (z0). Parameterizing the line as the set of points (x,y,z) = (x0 - B t + A C z0/(1-C^2), y0 + A t + B C z0/(1-C^2), 0) we can find the point on the line nearest the center, and easily show that that point corresponds to t=0: (x,y,z) = (x0 + A C z0/(1-C^2), y0 + B C z0/(1-C^2), 0) and can check that this point is of distance |z0|/sqrt(1-C^2) from the center. By symmetry, the distance from the center to the other planes, as measured along the plane of the circle, is similarly |x0|/sqrt(1-A^2) and |y0|/sqrt(1-B^2) If only the point (x0,y0,z0) is given, we can then determine the coefficients in the equation describing the relevant plane: we must have |A| = sqrt(1 - (x0/r)^2), |B| = sqrt(1 - (y0/r)^2), |C| = sqrt(1 - (z0/r)^2). Since we assumed (A,B,C) was a unit vector, this requires (x0^2+y0^2+z0^2)= 2 r^2. Apart from that, the only restriction is that we be able to compute values for A, B, C --- that is, we need each of x0, y0, z0 to be at most r in absolute value. So the correct locus seems to be that portion of the sphere of radius sqrt(2) r obtained by removing the "triangular" segment of the sphere in each quadrant, bounded by the planes |x0|=r, |y0|=r, |z0|=r. From rusin@math.niu.edu Thu Dec 11 15:24:09 2003 Received: from vesuvius.math.niu.edu (vesuvius [131.156.3.93]) by sanford.math.niu.edu (8.12.10/8.12.10) with ESMTP id hBBLO825028444; Thu, 11 Dec 2003 15:24:08 -0600 (CST) Received: (from rusin@localhost) by vesuvius.math.niu.edu (8.9.3/8.9.3) id PAA23474; Thu, 11 Dec 2003 15:24:09 -0600 (CST) Date: Thu, 11 Dec 2003 15:24:09 -0600 (CST) Message-Id: <200312112124.PAA23474@vesuvius.math.niu.edu> From: rusin@math.niu.edu To: jscholes@kalva.demon.co.uk Subject: Putnam 1948 B4 Cc: rusin@math.niu.edu X-Spam-Status: No, hits=0.2 required=5.0 tests=BAYES_30,NO_REAL_NAME version=2.55 X-Spam-Level: X-Spam-Checker-Version: SpamAssassin 2.55 (1.174.2.19-2003-05-19-exp) Status: O Content-Length: 2408 [Determine the eigenvalues of the operator on C^0( (0,1) ) given by integrating with kernel min(x,y) . Hmm -- a functional analyst probably knows in an instant how to put this into a larger context...] Adapt as you see fit; this answer probably spends too much time on explicit integration! dave 1948 B4. R is the reals. For what \lambda can we find a continuous function f : (0, 1) ---> R, not identically zero, such that \int_0^1 min(x, y) f(y) dy = \lambda f(x) for all x \in (0, 1)? Ans: The left side of the desired inequality is F(x) = \int_0^x y f(y) dy + \int_x^1 x f(y) dy. We may pull the factor x outside the second integral; then it is clear by the fundamental theorem of calculus that this F is always differentiable on (0,1) and F'(x) = x f(x) + \int_x^1 f(y) dy - x f(x) = \int_x^1 f(y) dy. Again FTC makes this F' differentiable and guarantees that F''(x) = - f(x) . Now the question is, for which real values of lambda is there a function f with F(x) = lambda f(x). Note that if lambda were zero, this would require F(x) = 0 identically, so F''(x) = 0 too, so f(x)=0, which was forbidden. So lambda is nonzero, and thus f(x) = F(x)/lambda is necessarily twice differentiable, with f''(x) = (-1/lambda) f(x). If lambda > 0 this means f(x) = A cos(x/sqrt(lambda)) + B sin(x/sqrt(lambda)) for some real constants A and B; if lambda < 0 we have instead a similar form, f(x) = A exp(x/sqrt(lambda)) + B exp(-x/sqrt(lambda)). It remains only to discover the combinations of A,B, and lambda which make the integral F(x) equal to lambda f(x). In either case, we can compute the integral easily: when lambda>0 we get F(x) = lambda f(x) - A lambda + x sqrt(lambda) (A sin(1/sqrt(lambda)) - B cos(1/sqrt(lambda))) so this function f meets the desired condition (F=lambda f) iff A = 0 and cos(1/sqrt(lambda))=0, i.e. lambda = (2/((2n+1)\pi))^2 for some integer n. Similarly when lambda < 0 we find F(x) = lambda f(x) + ... where the remaining expression is a sum of constant multiples of exp(x/sqrt(lambda)), exp(-x/sqrt(lambda)), x, and constants. Since those functions are linearly independent, we can set their coefficients to zero, which means that first A, then B, must be zero. So the solution is: the set of numbers of the form lambda = (2/((2n+1)\pi))^2 for some (positive) integer n. From rusin@math.niu.edu Fri Dec 12 08:58:15 2003 Received: from vesuvius.math.niu.edu (vesuvius [131.156.3.93]) by sanford.math.niu.edu (8.12.10/8.12.10) with ESMTP id hBCEwF25003865; Fri, 12 Dec 2003 08:58:15 -0600 (CST) Received: (from rusin@localhost) by vesuvius.math.niu.edu (8.9.3/8.9.3) id IAA25338; Fri, 12 Dec 2003 08:58:16 -0600 (CST) Date: Fri, 12 Dec 2003 08:58:16 -0600 (CST) Message-Id: <200312121458.IAA25338@vesuvius.math.niu.edu> From: rusin@math.niu.edu To: jscholes@kalva.demon.co.uk Subject: 1948 B6(1) Putnam Cc: rusin@math.niu.edu X-Spam-Status: No, hits=1.1 required=5.0 tests=NO_REAL_NAME version=2.55 X-Spam-Level: * X-Spam-Checker-Version: SpamAssassin 2.55 (1.174.2.19-2003-05-19-exp) Status: O Content-Length: 900 This one is very nice. B6. Do either (1) or (2): (1) Take the origin O of the complex plane to be the vertex of a cube, so that OA, OB, OC are edges of the cube. Let the feet of the perpendiculars from A, B, C to the complex plane be the complex numbers u, v, w. Show that u^2 + v^2 + w^2 = 0. Answer: If the length of a side is r, then we may write A = r(a,b,c), B=r(d,e,f), C=r(g,h,i) where (a,b,c), (d,e,f), (g,h,i) are mutually perpendicular unit vectors. Therefore the matrix with these entries in its columns is an orthogonal matrix. It follows that the ROWS (a,d,g) and (b,e,h) (as well as (c,f,i) ) are perpendicular unit vectors, that is, a^2 + d^2 + g^2 = 1, b^2 + e^2 + h^2 = 1, a b + d e + g h = 0 The complex numbers u,v,w are a+bi, d+ei, g+hi, so u^2 + v^2 + w^2 = (a^2+2abi-b^2)+(d^2+2dei-e^2)+(g^2+2ghi-h^2) = 1 - 1 + 2i(0) = 0. From rusin@math.niu.edu Sun Dec 21 18:21:07 2003 Received: from vesuvius.math.niu.edu (vesuvius [131.156.3.93]) by sanford.math.niu.edu (8.12.10/8.12.10) with ESMTP id hBM0L725015384; Sun, 21 Dec 2003 18:21:07 -0600 (CST) Received: (from rusin@localhost) by vesuvius.math.niu.edu (8.9.3/8.9.3) id SAA20654; Sun, 21 Dec 2003 18:21:09 -0600 (CST) Date: Sun, 21 Dec 2003 18:21:09 -0600 (CST) Message-Id: <200312220021.SAA20654@vesuvius.math.niu.edu> From: rusin@math.niu.edu To: jscholes@kalva.demon.co.uk Subject: Putnam 1949 A1 Cc: rusin@math.niu.edu X-Spam-Status: No, hits=0.2 required=5.0 tests=BAYES_30,NO_REAL_NAME version=2.55 X-Spam-Level: X-Spam-Checker-Version: SpamAssassin 2.55 (1.174.2.19-2003-05-19-exp) Status: O Content-Length: 8358 John, Many of these early questions amount to a bundle of computation. I've been trying to verify the calculations with Maple, but you really ought to verify the logic since I can't do that with a machine! dave 1949 A1. Do either (1) or (2) (1) Let L be the line through (0, -a, a) parallel to the x-axis, M the line through (a, 0, -a) parallel to the y-axis, and N the line through (-a, a, 0) parallel to the z-axis. Find the equation of S, the surface formed from the union of all lines K which intersect each of L, M and N. Answer: It's (most of) the surface defined by the equation xy + yz + zx + a^2 = 0 . The points on L are those whose coordinates have the form (a t1, -a, a) for some real number t1 ; similarly we may express the points on M and N. So a line K meeting all three of these must pass through points of each of these three forms. The three points must be collinear, which happens iff the vectors joining pairs of points are parallel, i.e. (a t1 - a, -a - a t2, 2a) must be a multiple of (2a, a t2 - a, -a - a t3). The multiplier is obviously (t1-1)/2 . (Here and below I assume a is nonzero.) So the collinearity condition is that -a - a t2 = (t1-1)/2 (a t2 - a) and 2 = (t1-1)/2 (-1-t3) i.e. -4 = (t1+1)(t2-1) and -4 = (t3+1)(t1-1) Note that these equations show that t1 cannot equal 1 nor -1. Otherwise, we may solve these equations for t2 and t3. That is, we discover that there is a unique line K passing through (a t1, -a, a) and meeting both line M and line N, for every real t1 except 1 and -1. Writing the points on the line in the form (a t1, -a, a) (1+u)/2 + (a, a t2, -a) (1-u)/2 , we then find that the set of points on K may be written P := a ( (1-u+t1+t1 u)/2, (u-t1 u-2)/(t1+1), u ) Setting these three coordinates equal to x,y,z respectively, we may solve for t1 and u in terms of x and z, then substitute into the expression for y and simplify to get xy + yz + zx + a^2 = 0. It is probably true that the problem posers expected nothing more than this, but this procedure is a bit vague and indeed the correct locus of points is not quite equal to the surface described implicitly by this single equation. So in the remainder of our answer to (a) we address these technicalities. So let's do this a bit more carefully. The surface we are to find is precisely the set of points in R^3 parameterized as P, with t1 different from 1 and -1. Since the value t1=1 is excluded, this parameterization misses the point a(1, -1, u) . (More precisely, one can check that the ONLY pairs (t1,u) which make the first two coordinates of P be a and -a respectively are those with t1=1.) Apart from the points on this line, the surface we want is precisely the set of points in R^3 parameterized as P. On the one hand, a simple calculation shows that each of these points P satisfies the equation xy + yz + zx + a^2 = 0 . But on the other hand, not every point satisfying this equation will be in the image of our parameterization P . Indeed, suppose (x,y,z) is a point whose coordinates satisfy xy + yz + zx + a^2 = 0 . If this is to be in the image of the parameterization, we clearly require u to be z/a. Then using the first coordinate of P we see that in order for (x,y,z) to be in the parameterization, we require that (z+a) t1 = (2x+z-a) As long as z is different from -a, this specifies the unique possible choice of t1 which will force P to yield our initial coordinates (x,y,z) ; this value of t1 isn't 1 (unless x=a) nor -1 (unless x = -z). So from the coordinates x and z we have (uniquely) determined the necessary values of t1 and u in general. We may then check that indeed the second coordinate of P will simplify to (x, -(x z + a^2)/(x+z) , z ), and so as long as (x,y,z) lies on the surface xy + yz + zx + a^2 = 0, this will be the same as (x,y,z). (If z IS equal to -a, then this analysis does not dictate what t1 must be, but we can compute the appropriate value of t1 by setting the second coordinate of P equal to y, as long as y is different from a.) So our conclusion is that the whole implicit surface lies in the image of the parameterization EXCEPT FOR the points where (i) x= a, y=-a (those are the points requiring t1 = 1 ) (ii) x=-a, z= a (those are the points requiring t1 = -1 ) (iii) y= a, z=-a (those are the points where no t1 may be computed ) These points form the three lines which are reflections through the origin of our original three lines. (Our original three lines ARE themselves part of the parameterized surface, as can be checked straight from the definition of that surface.) This is a surface of revolution around the line x=y=z (it's a hyperboloid of one sheet -- minus the three lines of course.) It is a pretty picture to draw it and the cube with vertices (+-a, +-a, +-a) : two of the vertices lie on the axis of revolution, and the other six are the intersections of one of the original lines and one of the deleted lines. This Putnam problem may be generalized a bit: Given four "generic" lines in R^3, there are exactly two lines which pass through all four. (This indented statement is invariant under linear transformations, and under the action of that group a "generic" set of lines is equivalent to a set of lines parallel to those in the Putnam question, but meeting the planes at points (0,-b,1), (1,0,-c), (-a,1,0) -- only slightly more general a problem, and leading to the corresponding surface (c+1)xy + (a+1)yz + (b+1)zx + (bc-1)x + (ca-1)y + (ab-1)x + (abc+1) = 0. ) For generalizations to higher dimensions, see http://www.math-atlas.org/96/skewer ================================================================ (2) Let S be the surface xy + yz + zx = 0. Which planes cut S in circles? In parabolas? Answer: This surface is a cone (a cone which fits neatly inside the hyperboloid of part (1) ! ), as can be seen by changing to an orthonormal basis consisting of v1=(1,1,1)/sqrt(3), v2=(1,0,-1)/sqrt(2), and v3=(1,-2,1)/sqrt(6) (These are polar vectors for the given quadratic form.) In the associated new coordinate scheme the surface S is X^2 - Y^2/2 - Z^2/2 = 0. Now we can describe geometrically which planes are which; it's not clear whether an algebraic description was [also] required. The intersection of a cone with a plane is a circle iff the plane is perpendicular to the central axis (and does not pass through the central point of the cone); in this case, the axis is the set of multiples of v1 and so these are the planes of the form x + y + z = c for any nonzero constant c. (This is a one-parameter family of planes.) The intersection of a cone with a plane is a parabola iff the plane is perpendicular to a normal vector of the cone (and is not tangent to the cone). At any point (x0,y0,z0) on the cone except the origin, the normal vector is parallel to the gradient of the function F(x,y,z)=(xy+yz+zx) whose level surface F^{-1}(0) is S, that is, the normal vector is parallel to (y0+z0, z0+x0, x0+y0). The corresponding planes are those described by equations (y0+z0)x + (z0+x0)y + (x0+y0)z = c where c can be any constant except the value causing tangency: c = (y0+z0)x0 + (z0+x0)y0 + (x0+y0)z0 = 2(x0y0+y0z0+z0x0) = 0. Note that points P and lambda P have parallel normal vectors and so they lead to identical families of parallel planes cutting the surface in parabolas. Thus although the cone is 2-dimensional, there is only a one-dimensional family of such normal directions, and thus only a 2-dimensional family of planes cutting the cone in a parabola. We can list these planes irredundantly by using only points (x0,y0,z0) lying (say) on the plane x + y + z = 3; all other points are multiples of these. There we may write the plane's equation equally well as (3-x0)x + (3-y0)y + (3-z0)z = c We may even trace out these points (x0,y0,z0) on the intersection of the cone and this plane, if we like; the points have x0 = 1 + sqrt(3) cos(theta) + sin(theta) y0 = 1 - 2 sin(theta) z0 = 1 - sqrt(3) cos(theta) + sin(theta) and so as theta and c vary we find a 2-parameter family of planes. (Of course a generic plane cuts the cone either in a hyperbola or an ellipse.) From rusin@math.niu.edu Mon Dec 8 16:50:13 2003 Received: from vesuvius.math.niu.edu (vesuvius [131.156.3.93]) by sanford.math.niu.edu (8.12.10/8.12.10) with ESMTP id hB8MoD25000847; Mon, 8 Dec 2003 16:50:13 -0600 (CST) Received: (from rusin@localhost) by vesuvius.math.niu.edu (8.9.3/8.9.3) id QAA13834; Mon, 8 Dec 2003 16:50:15 -0600 (CST) Date: Mon, 8 Dec 2003 16:50:15 -0600 (CST) Message-Id: <200312082250.QAA13834@vesuvius.math.niu.edu> From: rusin@math.niu.edu To: jscholes@kalva.demon.co.uk Subject: 1949 Putnam Cc: rusin@math.niu.edu X-Spam-Status: No, hits=1.1 required=5.0 tests=NO_REAL_NAME version=2.55 X-Spam-Level: * X-Spam-Checker-Version: SpamAssassin 2.55 (1.174.2.19-2003-05-19-exp) Status: O Content-Length: 1977 With the 2003 Putnam behind us, I was looking over various Putnam sites. Yours is impressively complete, but I see you must have grown tired or something, leaving a few holes here and there. Picking one at random I thought I'd offer to patch a gap. 1949 B2. Do either (1) or (2) (1) Prove that \sum_2^\infty cos(ln ln n) / ln n diverges. (2) Let k, a, b, c be real numbers such that a, k > 0 and b^2 < ac. Show that \int_U (k + ax^2 + 2bxy + cy^2)^{-2} dx dy = \pi/( k sqrt(ac - b^2)), where U is the entire plane. In (1) we must show that the sequence of partial sums fails to converge. But fail it must because it's not even a Cauchy sequence. For any integer k we can let n1 = ceiling( exp(exp( 2 pi k - pi/4 )) ) and n2 = floor( exp(exp( 2 pi k + pi/4 )) ) . Then the difference between the n1-th and n2-th partial sums is not at all small: in this range, ln ln n is strictly between 2 pi k - pi/4 and 2 pi k + pi/4, so the cosine is at least 1/sqrt(2). Also 1/ln n is at least 1/ln n2 . On the other hand, the number of such summands is n2-n1 > (1/2) n2 (surely one of the most dramatic inequalities I've ever written!...) so this difference between the partial sums is more than constant.n2 / ln(n2), which not only fails to go to zero as k --> infinity, but in fact grows very big, very fast! (2) The given hypotheses make ax^2 + 2bxy + cy^2 be a positive definite quadratic form; we may apply a rotation to assume that b = 0, and then scale the two axes to assume a = c = 1. (The latter change increases the integral by a factor of sqrt(ac). ) We then need only evaluate \int_U 1/(k + x^2+y^2)^2 = \int_0^\infty \int_0^{2\pi} 1/(k + r^2)^2 r d\theta dr = \pi \int_0^\infty 1/(k + r^2)^2 2r dr = \pi (k + r^2)^(-1) |_0^\infty = \pi/k . I note that Robin Chapman used this kind of result in his solution to one of the 2003 problems! Are you interested in polishing off the remaining problems? I could give some of them a go. dave From jscholes@kalva.demon.co.uk Wed Dec 10 15:57:22 2003 Received: from anchor-post-34.mail.demon.net (anchor-post-34.mail.demon.net [194.217.242.92]) by sanford.math.niu.edu (8.12.10/8.12.10) with ESMTP id hBALvM25018956 for ; Wed, 10 Dec 2003 15:57:22 -0600 (CST) Received: from kalva.demon.co.uk ([158.152.30.114]) by anchor-post-34.mail.demon.net with esmtp (Exim 3.35 #1) id 1AUCKd-0002Km-0Y for rusin@math.niu.edu; Wed, 10 Dec 2003 21:57:23 +0000 Mime-Version: 1.0 X-Sender: kalva@pop3.demon.co.uk Message-Id: In-Reply-To: <200312082250.QAA13834@vesuvius.math.niu.edu> References: <200312082250.QAA13834@vesuvius.math.niu.edu> Date: Wed, 10 Dec 2003 21:51:48 +0000 To: rusin@math.niu.edu From: J Scholes Subject: Re: 1949 Putnam Content-Type: text/plain; charset="us-ascii" ; format="flowed" X-Spam-Status: No, hits=-1.3 required=5.0 tests=BAYES_30,IN_REP_TO,REFERENCES version=2.55 X-Spam-Level: X-Spam-Checker-Version: SpamAssassin 2.55 (1.174.2.19-2003-05-19-exp) Status: O Content-Length: 638 Many thanks for the solution you sent. It looks correct to me. I will get it up soon. Thanks also for the reminder on this year's Putnams. >With the 2003 Putnam behind us, I was looking over various Putnam sites. >Yours is impressively complete Thanks. >, but I see you must have grown tired or >something, leaving a few holes here and there. Yes. You are right! I did grow exceedingly tired of early Putnams. Any further help you could give on solutions for the missing 1940s would be much appreciated. >Picking one at random >I thought I'd offer to patch a gap. > 1949 B2. Do either (1) or (2) -- Kind regards John Scholes