PUTNAM PREP FOR THE FIRST THREE PRACTICE SESSIONS V11

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1 PUTNAM PREP FOR THE FIRST THREE PRACTICE SESSIONS V11 Contents 1. Basic Information about Practices 2 2. Number Theory October 5, A Few Interesting Problems Basic Num. Th. Techniques/Theorems/Terms Putnam & Beyond Num. Th. Prob. for those with Less Time Soln. to Num. Th. Problems at the Beginning 4 3. Differential Equations October 12, A Few Interesting Problems Basic Diff. Eq. Techniques/Theorems/Terms Putnam & Beyond Diff. Eq. Prob. for those with Less Time Soln. to Diff. Eq. Problems at the Beginning 9 4. Inequalities October 19, A Few Interesting Problems Basic Ineq. Techniques/Theorems/Terms Putnam & Beyond Ineq. Prob. for those with Less Time Soln. to Ineq. Problems at the Beginning 16 1

2 2 1. Basic Information about Practices You are MOST WELCOME to take the Putnam exam on Saturday December 3 without coming to our Wednesday afternoon 4:45 practice sessions. Indeed when we took the Putnam exam, most people (including us) took the exam that way just a six hour funfest with some elementary problems, and a chance for some recognition that might be appreciated in job and grad school applications. Skill at Putnam problems sometimes is useful on research problems and in real life, but the problems are different and we encourage you to view Putnam problem solving as potentially a part, not the core of your mathematical development. However, for those interested, we offer a weekly practice session from 4:45 to 6 pm. In recent years, the format of these has generally been about forty-five minutes of individual problem solving followed by half an hour of discussion and presentation of solutions. Most of our sessions have a topic theme. Frequently these line up with portions of the recent book Putnam & Beyond by Razvan Gelca and Titu Andreescu. Cornell students have free access to electronic copies of this - see our website putnam for details. Indeed there are excellent support books for Putnam preparation nowadays. A collection are on reserve in the Math Library and/or are mentioned at our website. But in many cases, working through these books does take more time than many people have. Even trying all the problems in one subsection of Putnam and Beyond often takes a long time. If you are looking for a shorter path through some of the chapters, we suggest here a few examples and problems you might starting by concentrating on. In the following pre-session notes for our first three sessions, we also include a few favorite problems (with solutions at the end of each section.) They ve been chosen to illustrate why we think you may find it mathematically enriching and interesting to work at Putnam type problems.

3 PUTNAM PREP FOR THE FIRST THREE PRACTICE SESSIONS V Number Theory October 5, A Few Interesting Problems. (Solutions are two pages later.) Problem 1: (Putnam 1985 A4) Define a sequence {a i } by a 1 = 3 and a i+1 = 3 a i for i 1. Which integers between and 99 inclusive occur as the last two digits in the decimal expansion of infinitely many a i? Problem 2: For how( many ) integers n between 49 and 1 (inclusive) is the n binomial coefficient odd? 49 Problem 3: Show that the equation x 2 + y 2 + z 2 = 2xyz has no nontrivial solutions. Problem 4: Find all integral solutions of x 2 3y 2 = Basic Num. Th. Techniques/Theorems/Terms. Modular arithmetic, Chinese Remainder Theorem, Little Fermat, Euler, Wilson, totient, Euclidean algorithm, unique factorization, linear diophantine equations, Pythagorean triplets, Pell, infinite descent Putnam & Beyond Num. Th. Prob. for those with Less Time. Putnam & Beyond Readings: Chapter 5, pages (Fermat s Infinite Descent Principle). A basic technique. The first example is an easy one, but the reference to the maximum modulus principle may be unfamiliar to those who haven t studied complex analysis. (Complex analytic functions always attain their maximum modulus on the boundary of their domains.) 76 is easy. 77 a bit harder. 78 is cute but not easy for all (The Greatest Integer Function). Discussion is amazing but may not be quick (Factorization and Divisibility). Examples 1 and 2 on page 253 are quick. 724 is quick. As is with the right observation (Prime Numbers). The first example is interesting but elaborate with an apparent typo in the centered equation just before the end. The example at the bottom of page 255 is good and quick. Polignac s formula on page 257 is quite intuitive and easy to carry with you

4 4 733 is fairly easy. 744 good and systematic. 747 and 748 are interesting Polignac problems (Modular Arithmetic). Example 1 is quick and basic. Example 2 is interesting and may open some eyes; but will take more thought. 75, 751, and 752 relatively quick. 753 cute and quick with the right observation (Fermat s Little Theorem). A very basic theorem. The proof is a little exotic but illustrates a powerful technique. Example 1 page 262 relatively quick but tricky. Example 2 harder to come up with. 76 cute and quick. 761 quick - was on a recent practice Putnam. 762 first part easy and interesting. The application is harder, but quick (Wilson s Theorem). A basic theorem with a quick proof. The example is quite nice and quick. 769 is quick and cute. 77 is slick and quick. 771 is nice and famous. 772 is a good application (Euler s Totient Function). Quite a fundamental function. The proposition is another basic result. Proofs based on φ(mn) = φ(m)φ(n) when gcd(m, n) = 1 may be more accessible. Euler s theorem comes up a lot and two good proofs are given. The first example is a good and quick application (The Chinese Remainder Theorem). Again fundamental. The proof is given in an interesting constructive form; can also be viewed as a byproduct of techniques for solving linear diophantine equations. The example at the bottom of page 269 is amazing. 785 is a basic application. 786 is cute and in the end an easy Chinese Remainder Theorem Apllication. 787 is a good quick application Soln. to Num. Th. Problems at the Beginning. Soln. to Problem 1: We are interested in a n mod 1. Note all of the a n satisfy gcd(a n, 1) = 1.

5 PUTNAM PREP FOR THE FIRST THREE PRACTICE SESSIONS V11 5 Little Fermat says a p 1 1 mod p for a prime number p and a not divisible by p. More generally, Euler s theorem says if gcd(a, m) = 1, a φ(m) 1 mod m where φ(m) is the number of residue classes mod m which are coprime to m. Also note φ(ab) = φ(a)φ(b) when gcd(a, b) = 1 and φ(p k ) = p k p k 1 for p a prime. So a i+1 3 a i mod 1 is determined by a i mod φ(1) or a i mod 4 since φ(1) = φ( ) = φ(5 2 ) φ(2 2 ) = (5 2 5)(2 2 2) = 4. Then a i 3 a i 1 mod 4 is determined by a i 1 mod φ(4) or a i 1 mod 16 since φ(4) = φ(2 3 5) = φ(2 3 ) φ(5) = ( )(5 1) = 16. Continuing a i 1 3 a i 2 mod 16 is determined by a i 2 mod φ(16) or a i 2 mod 8 since φ(16) = = 8. Continuing a i 2 3 a i 3 mod 8 is determined by a i 3 mod φ(8) or a i 3 mod 4 since φ(8) = = 4. Continuing a i 3 3 a i 4 mod 4 is determined by a i 3 mod φ(4) or a i 3 mod 2 since φ(4) = = 2. All a i 4 for i 4 are odd, so working backwards, a i 3 3 mod 4 a i mod 8 3 mod 8 a i mod mod 16 a i 3 11 mod 4 ((3 4 ) 2 )(3 3 ) mod 4 (3 3 ) mod 4 27 mod 4 a i mod 1 ((3 6 ) 4 )(3 3 ) mod 1 (729 4 )3 3 mod mod 1 and mod 1 29( 13) 3 mod 1 29 (169 13) mod 1 So all but the first few terms end in ( 2197) mod 1 87 mod 1. Soln. to Problem 2: First note that (x + y) p = x p + y p (mod p) ( ) p for a prime number p since (mod p) when < a < p. So also a (x + y) pk = x pk + y pk (mod p). We can use this to establish ( ) ( ) b bi = Π a a i (mod p) where b = Σb i p i and a = Σa i p i give the base p expansions of a and b..

6 6 This is seen by observing (x + y) b (x + y) Σ ib i p i Π i ( x pi + y pi) b i Π i Σ k ( bi k ) ( x kpi y (b i k)p i) (mod p) and comparing the coeficients of x a y b a. Since 49 = = and we are interested in n 1, we first do the problem for n 127. For the binomial coefficient to be odd, the 7 or less bit number n in base 2 must look like ( or 1)11( or 1)( or 1)( or 1)1 2 a total of 2 4 = 16 possibilities. Since > 1, the eight possibilities whose 64 column is 1 are too big, and we have the remaining eight possibilites where the binomial coefficient is odd. Soln. to Problem 3: First, let (x, y, z) be a nontrivial solution of ( ) x 2 + y 2 + z 2 = 2xyz. Mod 4, squares are or 1 and 2xyz is unless all of x, y, z are odd. So any nontrivial solution consists entirely of even integers. Writing x = 2 x, y = 2ȳ, z = 2 z, we have a nontrivial solution ( x, ȳ, z) to x 2 + y 2 + z 2 = 4xyz. But then the same argument modulo 4 shows that x, ȳ, z must all be even and we can repeat. Motivated by this, we use Fermat s method of descent to show there are no nontrivial solutions. Let ( x, ỹ, z) be a minimal nontrivial solution to ( ) x 2 + y 2 + z 2 = 2 k xyz. for some positive integer k where by minimal we mean having the smallest positive value of x among all nontrivial solutions to ( ) regardless of the positive value of k. The above argument modulo 4 shows that all of x, ỹ, z are even. But then ( x 2, ỹ 2, z 2 ) would solve ( ) with k increased by 1 contradicting the assumption that ( x, ỹ, z) was a solution minimal in x value. So no nontrivial solutions to ( ) exist. Soln. to Problem 4: This equation is called Pell s equation and will work similarly if 3 is replaced by any other square-free positive integer. Since (x, y) a solution implies ( x, y) is also (and there are no solutions with x = ), we just need to think about solutions with x >.

7 PUTNAM PREP FOR THE FIRST THREE PRACTICE SESSIONS V11 7 The equation may be rewritten as ( ( ) x y ) ( 3 x + y ) 3 = 1 or zz = 1 where z = x + y 3 (with x, y Z) and z = x y 3. (Note that the product of two numbers of the form x + y 3 with x, y integers can be simplified into a number of the same form.) Let S = {x + y 3 : x, y Z, x >, and (x, y) solves ( )}. Then ab = ab means z 1, z 2 S z 1 z 2 S. Also z S 1 z S as well since zz = 1 and 1 z = z for elements of S. Thus the product and quotient of elements of S are also in S. When an element of S x + y 3 is greater than 1, x y 3 < 1 since their product is 1. Thus the set of solutions S with x > is partitioned into two subsets S + = {x + y 3 S : x > and y } S = {x y 3 S : x > and y > }. S + consists of those elements of S bigger than or equal to 1; S those elements between and 1. Now let u be the smallest element of S + bigger than 1. We claim all solutions of ( ) with x > (interpreted as elements of S) are whole number powers u d of u. For if u = x + y 3 is any other positive solution with x >, since u > 1, there is a unique integer k so that u k < u < u k+1. Multiplication by u k gives a solution with 1 < uu k < u. But uu k being a solution bigger than 1 forces it to be in S + contradicting the assumption that u was the smallest element bigger than 1. So no additional solution u exists. In our case of x 2 3y 2 = 1, u = by inspection. Hence all solutions of ( ) interpreted as numbers z = x + y 3 are simply ±u k for k Z.

8 8 3. Differential Equations October 12, A Few Interesting Problems. (Solutions are two pages later.) Problem 1: (Putnam 21 A3) Suppose that the function h : R 2 R has continuous partial derivatives and satisfies the equation h(x, y) = a h (x, y) + b h(x, y) x y for some constants a, b. Prove that if there is a constant M such that h(x, y) M for all (x, y) R 2, then h is identically zero. Problem 2: For the differential equation dx dt = x3 t with initial condition x() = x, show that there is a value c so that for both x > c and x < c, the solutions for t have vertical asymptotes. Problem 3: (Putnam 25 B3) Let P (x 1,..., x n ) denote a polynomial with real coefficients in the variables x 1,..., x n, and suppose that ( ) x 2 P (x 1,..., x n ) = (identically) n and that x 2 1 Show that P = identically. x x 2 n divides P (x 1,..., x n ). Problem 4: Let x = f(t) and x = g(t) be linearly independent solutions (, ) of the ordinary differential equation t d2 x dt 2 + cdx dt + x = for some constant c. Show that between any two zeroes of f(t), there is a zero of g(t) Basic Diff. Eq. Techniques/Theorems/Terms. List to be added. Putnam & Beyond Readings: Chapter 3, Sections , pages Putnam & Beyond Diff. Eq. Prob. for those with Less Time (ODE s of First Order). Idea of separable equations quite basic and likely familiar. First example on page 192 easy but worthwhile. Might also be phrased as Let y(x) = x f(t) dt; then we have a sep. eqn... 1+t 2 The example on page 193 is quite nice and easy to follow, but perhaps only if you ve already studied Green s Theorem and potentials in multivariable calculus.

9 PUTNAM PREP FOR THE FIRST THREE PRACTICE SESSIONS V is quite easy and cute. 555 is quick and elementary. 556 is not easy, but illustrates an interesting extension of a well known idea. 558 Interesting. There is a typo in the definition of g in the solution, but after correction, the solution is quite nice to follow and different from that posted in the solutions to the 25 Putnam. 559 Cute (ODE s of Higher Order). This section is easy reading. The example on page 195 is a nice use of complex form. The example on page 196 is easy to follow and has a couple of good general tricks. 56 is quick with the right idea. 561 quick and easy. 562 also quite easy. 563 uses a good systematic approach. 565 cute (Problems Solved with Techniques of ODE s). First example is easy to follow though you may be tempted by a diferent approach. Second example idea is clear but details will take a while. 568 Nice one to try can be quick. 569 Another enlightening potentially quick problem. 57 Also relatively quick but not necessarily easy Soln. to Diff. Eq. Problems at the Beginning. Soln. to Problem 1: The method of lines is a way of solving linear first order partial differential equations of the form ( ) c(x, y)z(x, y) = a(x, y) z + b(x, y) z x y as well as some more general first order possibilities.

10 1 It is based on the observation that if (x(t), y(t)) solve the system of ODE s dx dt = dy dt = then along these solution curves (which are called characteristics), the original equation says d z(x(t), y(t)) = c(x, y)z(x(t), y(t)). dt So a solution to this first order ODE specifies z values along a characteristic. Consequently an initial condition of z along some curve in the xy plane transverse to the characteristics will allow us to determine z in a neighborhood of the initial value curve. Let (x, y ) be any point in R 2. Here the characteristics through (x, y ) are lines (x(t), y(t)) = (x + at, y + bt) and our equation says d dt (z(x + at, y + bt)) = z(x + at, y + bt), the equation of exponential growth. So z(x + at, y + bt) = z(x, y )e t and we see that z remains bounded along the characteristic only if z(x, y ) = in which case z is identically zero along the characteristic. Since (x, y ) was arbitrary, this shows h(x, y ) must be for all (x, y ). Soln. to Problem 2: Caution: This solution hopefully conveys some good ideas but may be somewhat amateurishly written. It s also been suggested that it would work out more easily using fences x 3 t = ±1 instead of x 3 t = ±x 2. First note that a solution to an ODE like dx dt = x 2 with initial condition x(t ) = x > develops vertical asymptotes in forward time because of explicit integration: x dx t x x 2 dx = dt t giving x = 1 (t t ) 1 x. Similarly dx dt = x2 with x(t ) = x <. Hubbard & West have a wonderful way of describing how one can compare solutions of differential equations to other curves. (You can read more about this in Volume I of their Differential Equations book, on reserve in the Math Library.) A curve given by x = α(t) is a (strong) lower fence for the ODE dx dt = f(t, α(t)) on an interval if dα dt < f(t, α(t)). It is immediate that no solution of the ODE can cross a lower fence from above as t increases. Similarly (strong) upper fences ( dβ dt below as t increases. a b > f(t, β(t))) cannot be crossed from

11 PUTNAM PREP FOR THE FIRST THREE PRACTICE SESSIONS V11 11 In this problem, since cubic curves t = x 3 + ax 2 ( sidewise, if we put t on the horizontal axis) have nearly horizontal tangents as t, x get large - say t t 1 ), it is clear that t = x 3 x 2 is a lower fence for t large and t = x 3 + x 2 is an upper fence. t = x 3 and t = x 3 ± x 2 Also, because of the local existence of solutions to ODE s (with smooth coefficients), the only issue in developing vertical asymptotes is whether or not solutions are defined for all forward time. Since t = x 3 x 2 is a lower fence, any solution with t > t 1 to the given ODE with x(t ) above the lower fence (i.e. x(t ) greater than the positive solution to x 3 x 2 = t ) will have a solution to dx dt = x 2 as a lower fence as well; since that lower fence has a vertical asymptote, such a solution will also not be defined for all time and so must develop a vertical asymptote. Similarly a solution to the ODE with x(t ) below the upper fence (for t > t 1 ) AND x(t ) < will have a solution to dx dt = x 2 as an upper fence, and is again forced by the upper fence to stop existing in finite time, and so develop a vertical asymptote. (Note as well that x = is an upper fence here.) So at this stage, the only possible candidates for solutions without asymptotes are (1) Solutions which for some value of t > t 1 are below the upper fence t = x 3 + x 2 but still in the first quadrant. (2) Solutions remaining for large t below the lower fence t = x 3 x 2 and above the upper fence t = x 3 + x 2. For the first case, suppose a solution for large t > t 1 is below the upper fence t = x 3 + x 2 but with x(t ). Then using the solutions to dx dt = x2 as a lower fence, we easily see that such a solution, if defined for all t > t would either cross the t-axis (in which case we ve already shown a vertical

12 12 asymptote) or else approach as t. In the latter case, we can use the easy upper fence of a line of slope 1 to show that once the solution gets below x = 1 for t large, this simple fence forces it below the t axis where we ve already shown it must develop a vertical asymptote. So all that remains is showing that there cannot be two distinct solutions x 1 (t) < x 2 (t) defined for all t which remain above the upper fence and below the lower fence. (This is the part that may be easier using fences x 3 t = ±1.) This can be ruled out by observing that the vertical distance between the two fences t = x 3 ± x 2 is bounded by a constant for t large while (by the mean value theorem on the x variable) the difference between two solutions x 2 (t) and x 1 (t) is bounded below by positive exponential growth. In explicit terms for this case: since x 1 < x 2. dx 2 dt = x3 2 t dx 1 dt = x3 1 t d(x 2 x 1 ) = dt (x 2 x 1 )(x x 1 x 2 + x 2 1) 3x 2 1(x 2 x 1 ) So since at most one solution can stay between the fences for t large, there can be at most one initial condition for which no vertical asymptote develops. Soln. to Problem 3: Put Q = x x2 n. Since Q is homogeneous, P is divisible by Q if and only if each of the homogeneous components of P is divisible by Q. It is thus sufficient to solve the problem in case P itself is homogeneous, say of degree d. (A function f(x 1,..., x n )) is said to be homogeneous of degree d if f(tx 1,..., tx n )) = t d f(x 1,..., x n )). Differentiating this equation and setting t = 1 establishes Euler s relation Σx i f = df.) x i Suppose that we have a factorization P = Q m R for some m >, where R is homogeneous of degree d and not divisible by Q. Write 2 as shorthand for ; Easy calculation giving the formula x 2 x 1 2 n 2 (fg) = g 2 f + f 2 g + 2 f g and Euler s relation n i=1 x i R x i = dr.

13 PUTNAM PREP FOR THE FIRST THREE PRACTICE SESSIONS V11 13 lead to = 2 P = 2mnQ m 1 R + Q m 2 R + 2 n i=1 = Q m 2 R + (2mn + 4md)Q m 1 R. m 1 R 2mx i Q x i Since m >, this forces R to be divisible by Q, contradiction. Soln. to Problem 4: For two solutions x = f(t) and x = g(t) of a homogeneous second order linear ODE d 2 x dt 2 + p(t)dx dt + q(t)x = like this one with p(t) = c t, one can form their Wronskian W (t) defined by W (t) = f(t) f (t) g(t) g (t) = fg f g. It is easy to differentiate this formula for W with respect to time obtaining W = fg f g = f( pg qg) ( pf qf)g = pw. So the Wronskian satisfies the separable equation 1 dw = p(t) W dt ( ) t and so on any interval [a, b], W (t) = W exp a p(t) dt where W is its value at t = a. In particular the Wronskian does not change signs on the interval. Apply this now to the situation in the problem where f(a) = f(b) =. First note that if the Wronskian were at t = a, then (f(a), f (a)) and (g(a), g (a)) would be proportional, and by the uniqueness of solutions theorem, so would f and g be, contradicting linear independence. Similarly at t = b. So W when t = a implying f (a)g(a). Since W has the same sign at t = b, f (b)g(b) as well, and with the same sign as f (a)g(a). Since f is never zero on (a, b), it is of constant sign there and the nonzero numbers f (a) and f (b) must be of opposite sign. But then g(a) and g(b) are also of opposite sign, and so g has a zero on (a, b) by the intermediate value theorem.

14 14 4. Inequalities October 19, A Few Interesting Problems. (Solutions are two pages later.) Problem 1: Show that for non-negative x,y, and z, one has the bound x 2 y 3 + x 2 z 3 + y 2 x 3 + y 2 z 3 + z 2 x 3 + z 2 y 3 x 4 y + x 4 z + y 4 x + y 4 z + z 4 x + z 4 y. Problem 2: On an equilateral triangle with area A, the product of any two sides is equal to 4A 3. Show that this is the extreme case in the sense that for a triangle of area A, there must exist two sides whose product is at least as large as 4A 3. Problem 3: (Putnam 24 A6) Suppose that f(x, y) is a continuous realvalued function on the unit square x 1, y 1. Show that 1 ( ( 1 2 f(x, y)dx) dy + f(x, y)dy) dx ( 1 1 f(x, y)dx dy) Problem 4: Show that for non-negative x,y, and z, (f(x, y)) 2 dx dy. xyz 1 = (1 + x)(1 + y)(1 + z) Basic Ineq. Techniques/Theorems/Terms. Squares, Cauchy-Schwartz, Holder, triangle, AM-GM, Minkowski, Chebyshev, convex functions, Jensen, increasing functions, rearrangement, power means Putnam & Beyond Ineq. Prob. for those with Less Time. Putnam & Beyond Readings: Chapter 2, Section 2.1 pages 25-45; Section pages ; Section pages ; (Algebraic Identities). The example at the top of page 27 makes a good combination with problems 9, 91, and 92. These emphasize how an algebraic identity can be used to solve equations and produce inequalities. 82 is quick and programmatic. 81 is similar in spirit to the first example of the section, but simpler. 83 is a nice use of the Lagrange identity expressing (a 2 + b 2 )(c 2 + d 2 ) as the sum of two squares.

15 PUTNAM PREP FOR THE FIRST THREE PRACTICE SESSIONS V (x 2 ). The two examples on page 29 show the virtues of writing expressions as almost the sum of squares. 96 is quite quick. 95 and 94 are relatively direct. 97 and 98 are hard but interesting. 11 is a nice combining the equations example. 12 is also very nice (Cauchy Schwartz). There are many nice proofs of C-S. The one in the text might be more naturally motivated as x x y y 2 The top example on page 34 is a nice quick application of C-S instead of Lagrange multipliers or another maxmin technique. If you ve run into the concept already, both the applications on page 34 can also be viewed as C-S applied to a different inner product ; e.g. one where < (x, y), (x, y) >= 2x 2 + 3y 2 + 4z 2 14 is a nice quick application. 13 is also quite simple. 19 and 111 are nice and fairly simple 18 is amazing and not too hard. 17 is a nice bound (Triangle Inequality). The example on page 37 illustrates the important application of estimating the location of zeroes of polynomials. 113 is quick. Many of the others may be tough for most people (Arithmetic Mean - Geometric Mean). Two nice arguments for this inequality are presented. The first example is quick and neat. 121 is quick by other methods but also has a cute AM-GM approach. 122 also has many quick solutions including a natural AM-GM one. 123 very quick after a nice programmatic observation. 125 programmatic and fairly quick

16 (Sturm s Principle). The less familiar approach described in the latter parts of the section is quite powerful. 13, 135, and 136 are not too hard applications of this idea (Convex Functions). The section starts out with a good presentation of the basics. Holder s inequality is important and its prooof well illustrates a natural normalization approach. Another book on reserve, The Cauchy-Schwartz Master Class explains a ot more about this.... The treatment of convex sequences at the bottom of page 143 is less well lnown, but easy to follow. Jensen s inequality on page 146 is a powerful hammer at quickly establishing many inequalities. The generalized AM-GM inequalty is a good example (Inequalities involving Integrals). Example 1 on page 156 is easy to follow and learn. But likely very hard to do yourself. 473 Very doable. 474 Doable given the preparation. 475 Probably not easy and there s a small typo in the solution. But vewry nice! 476 Also very nice. 477 Also very nice, though perhaps easier to follow than to do. Cauchy-Schwartz, Minkowski, and Holder are all very basic. 478 and 479 are easy. 48 is quite nice and quick with the right idea. 482, 483, and 484 are also nice and not long Soln. to Ineq. Problems at the Beginning. Soln. to Problem 1: The arithmetic mean geometric mean inequality says x p y q px + qy for x, y when p + q = 1 and p, q. (It follows quickly from the convexity of ln x together with Jensen s inequality.) Motivated by we have (3, 2) = 2 3 (4, 1) + 1 (1, 4) 3

17 PUTNAM PREP FOR THE FIRST THREE PRACTICE SESSIONS V11 17 x 3 y 2 = ( x 4 y 1) 2 3 ( xy 4) ( x 4 y ) ( xy 4 ) (The fractions came from solving equations; their existence was guaranteed by betweeness on the line a + b = 5.) We get similar inequalities by interchanging x and y as well as by considering the other two pairs of variables {x, z} and {y, z}. Adding these six equations together gives the desired result. Soln. to Problem 2: Let T be a triangle with sides b, c, d and vertex angles B, C, D with the labeling angle B opposite side b,etc. Then A = bc sin D 2 bc = 2A sin D and its two permuted variations gives 1 3 (bc + bd + cd) = 2A ( 1 3 sin D + 1 sin C + 1 ). sin B But the function f(x) = 1 sin x = csc x is easily seen to be convex on (, π) based on a positive second derivative. So Jensen s inequality applied to 1 sin x when 1 3 (B + C + D) = π 3 says ( sin D + 1 sin C + 1 ) 1 sin B sin π = Since the average of the products of two sides is 4A 3, at least one such product must reach or exceed that value. Soln. to Problem 3: [f(x, y) + f(z, w) f(x, w) f(z, y)] 2 dx dy dz dw Expanding out and renaming variables, we find each of the terms in the requested inequality with an appropriate coefficient of ±4. Soln. to Problem 4: One approach is to homogenize and establish ((xyz) x)((xyz) y)((xyz) z) 8xyz. To do this we need lower bounds on (x + y + z)(xyz) 2 3 and (xy + xz + yz)(xyz) 1 3. As in the solution to problem 1 above, (3, 3, 3) = 1 3 (5, 2, 2) (2, 5, 2) + 1 (2, 2, 5) 3 (3, 3, 3) = 1 3 (4, 4, 1) (4, 1, 4) + 1 (1, 4, 4) 3

18 18 implies xyz = xyz = ) (x 5 1 ( ) y 3 z 3 x 2 1 ( ) y 3 z 3 x y 3 z 3 ) (x 4 1 ( ) y 3 z 3 x 4 1 ( ) y 3 z 3 x y 3 z ( ) (x + y + z)(xyz) ( (xy + xz + yz)(xyz) 1 3 ). So we do actually have ((xyz) x)((xyz) y)((xyz) z) = xyz + (x + y + z)(xyz) (xy + xz + yz)(xyz) xyz ( )xyz. from which xyz = 1 immediately gives the desired result.