Putnam Solutions, 2007

Transcription

1 Putnam Solutions, 7 These solutions are based on discussions among our contestants (and our friends from Carlton College) Saturday afternoon and evening, with revisions as various errors were caught by careful reviewers! 7 A Find all values of α for which the curves y = αx + αx + /4 and x = αy + αy + /4 are tangent to each other. Solution. The reflection (x, y) (y, x) interchanges the curves. If one curve lies entirely on one side of the diagonal line y = x, then it cannot intersect its reflection, which lies on the other side. Therefore, if these curves intersect at all, they will have a point of intersection on the diagonal. This is enough to show that they cannot have a common tangent at off-diagonal points, because a common tangent would be an intersection of multiplicity at least two and, in conjunction with its mirror image, would imply the two curves intersect in at least five points (counting multiplicity), an impossibility for two polynomials of degree two. Therefore, we can restrict our search for mutually tangent points to those of the form (t, t). The slope of the first curve at (t, t) is /(dx/dy) = /(αt + α) and the slope of the second curve at (t, t) is dy/dx(t) = αt + α. For the two curves to be tangent to each other at (t, t), we need these slopes to be the same. Whence Writing u = t + ½ gives t = αt + αt + /4 and /(αt + α) = αt + α, or = (αt + α). = α (3/6)α + 4αu 4(αu) and u = (/α). There are two cases: u and α have the same or different signs; that is, either u = /(α) or u = -/(α). () Same sign: = α (3/6)α + ; solutions are (3 ± 5)/ = {3/, /3}. () Different sign: = α (3/6)α 3; solutions are (3 6)/. Page /

2 Comment. The figures illustrate the four solutions. 3 Reflection Alpha=-.96 Diagonal Reflection Alpha=3.6 Diagonal Reflection Alpha=.667 Diagonal Reflection Alpha=.5 Diagonal A Find the least possible area of a convex set in the plane that intersects both branches of the hyperbola xy = and both branches of the hyperbola xy = -. Solution. The answer is 4. Let X = (x, /x), Y = (y, -/y), Z = (-z, -/z), and W = (-w, /w) be four of the points of intersection, one per branch, with x, y, z, and w all positive. The convex hull of these four points lies inside the convex set and therefore cannot have greater area. We can further reduce the area if (-w,/w) (-z,-/z) (y,-/y) (x,/x) Page /

3 we can decrease the area of any of the triangles formed by three of these vertices, such as XYW. That triangle has base YW. The value of x making the height minimum occurs where the slope of the tangent, -/x, is parallel to the base; that is, -/x = (-/y /w) / (y + w). This together with three similar formulas obtained from contemplating the other three triangles implies x = y = z = w. The areas in all these cases are 4. Because there are no other critical points and nothing is achieved by letting the variables approach zero or infinity, this is the global minimum area. 7 A3 Let k be a positive integer. Suppose that the integers,, 3,, 3k + are written down in random order. What is the probability that at no time during this process, the sum of the integers that have been written up to that time is a positive integer divisible by 3? Your answer should be in closed form, but may include factorials. Solution. The probability is (k+)!k! / [(3k+)(k)!]. Because divisibility by three is all that matters, we work modulo three. Evidently the positions of the zeros do not change any of the divisibility properties. We ask, then, for the equivalent probability that a random sequence of k zeros, k+ ones, and k twos has no zero partial sums (modulo 3). A sequence satisfying the conditions cannot begin with a zero. This leaves (k+)/(3k+) of the sequences to consider, but for them, the positions of the zeros do not matter. If a partial sum is ever, the next value in the sequence cannot be, and if a partial sum is ever, the next value cannot be. A straightforward induction shows the only possible sequence (ignoring the zeros) is,,,,,,,. There are (k+)!k! such sequences out of all (k+)! sequences and they are equally probable. Multiplying by (k+)/(3k+) produces the answer. 7 A4 A repunit is a positive integer whose digits in base are all ones. Find all polynomials f with real coefficients such that if n is a repunit, then so is f(n). Solution. The polynomials are all those in the form f(x) = ( e ( + 9x) d ) / 9, d =,, 3, ; e =,,,. The repunits are the integers ( k )/9, k =,, 3,. This inspires us to define g(x) = + 9f((x )/9). The condition on f is equivalent to saying that for each positive integer k, there exists a positive integer m such that g( k ) = m. As k grows arbitrarily large, the leading term of g dominates, say g d x d, showing that m = kd + log (g d ), implying g(x) = e x d for integers d and e. The stated solution is obtained from f(x) = (g( + 9x) )/9. The restriction e is needed to make f() an integer (because it s a repunit), then an easy check shows all such f actually work. Page 3/

4 7 A5 Suppose that a finite group has exactly n elements of order p, where p is a prime. Prove that either n =, or p divides n +. Solution. There are + kp Sylow p-subgroups and, because they are all conjugate and every element of order p is contained in some Sylow p-subroup, they partition the n elements of order p into + kp equal-size collections. The number of elements of order p in any p-group is always one less than a power of p, implying n + ( + kp)(-) + modulo p. Comment. OK, it seems a bit like cheating to invoke the Sylow theorems. Let s provide a more detailed analysis. Let the group be G. We do not need to assume it is finite, but we will suppose the subset A G of order-p elements has a finite number n > members. We are going to count A modulo p. Pick any a A and define P to be the subgroup of G generated by a, so that P = {ε, a, a,, a p- }. G acts on A (that is, permutes its elements) via conjugation, because (gbg - ) p = gb p g - = gg - = ε for any g in G and b in A. As a permutation of A, the generator a consists of a collection of disjoint cycles, all of a size dividing p. Because p is prime, all cycles must have length p or. This implies the orbits of the P-action on A each have just one element (fixed points) or exactly p elements. Modulo p, the size of A equals the number of fixed points. A fixed point is an element of A that commutes with P. Suppose b and c are fixed points. Then a(bc) = (ab)c = (ba)c = b(ac) = b(ca) = (bc)a implies bc commutes with a and therefore with P. Thus, the fixed points generate a subgroup Z of G in which P is central (and therefore normal). We can now proceed inductively. Let n be the smallest integer for which the proposition fails. The canonical projection from Z to Z/P is well-defined as a group homomorphism, because P is normal in Z. It is a p-to-one mapping which sends elements in Z of order p to elements whose order is either p or. Because all elements of P map to the element of order in Z/P, and P has a nontrivial element (a), the number of elements of order p in Z/P is less than n. Consequently the proposition is true in Z/P: either there are no elements of order p or else the number of elements of order n, plus one, is a multiple of p. In either case the number of elements of order n or one in Z (namely, the number of fixed points plus one) is still a multiple of p. Therefore n + is also a multiple of p, showing our initial assumption about n is false: there can be no integers for which the proposition fails, QED. Here is an example to help make this less abstract. Let G be the group of permutations of {,,3,4}. It contains nine elements of order : A = {(), (3), (4), (3), (4), (34), ()(34), (3)(4), (4)(3)}. (This is cycle notation: () designates the permutation Page 4/

5 , and implicitly 3 3 and 4 4.) Let a = (), so that P = {ε, ()} is acting on A. Conjugation by () is easy to compute: it exchanges the roles of and in the permutation. For instance, a(3)a - = (3). The action of P on A is completely determined by the action of () itself, and looks like this: () () (3) (3) (3) (4) (4) (4) (34) (34) ()(34) ()(34) (3)(4) (3)(4) (3)(4). Thus the fixed points are {(), (34), ()(34)}. The group they generate, Z, also includes the identity. The canonical map from Z to Z/P sends () and ε to εp and sends both (34) and ()(34) to (34)P. Within Z/P there is one element of order along with the identity of order. The inverse image of each of these in Z consists of two elements, for a total of four elements, including ε. 7 A6 A triangulation of a polygon P is a finite collection of triangles whose union is P, and such that the intersection of any two triangles is either empty, or a shared vertex, or a shared side. Moreover, each side of P is a side of exactly one triangle in. Say that is admissible if every internal vertex is shared by 6 or more triangles. Prove that there is an integer M n, depending only on n, such that any admissible triangulation of a polygon P with n sides has at most M n triangles. Solution. Here is a sketch. The strategy is to decompose admissible triangulations into smaller ones. There are two ways to do this. If a triangulation includes a diagonal of the polygon, cut it along the diagonal. Otherwise, remove the polygon and all edges adjacent to the polygon s vertices, yielding a triangulation of the derived polygon. This process eventually terminates in a triangulation with no diagonals and no interior vertices, of which there are only three: the complete graphs on one, two, and three vertices. We can prove that the process of derivation (a) decreases the number of vertices on the polygon by at least six and (b) removes a number of faces equal to the sum of the original polygon vertices and the new polygon vertices. This is enough to show inductively that the number of faces of an admissible triangulation never exceeds some crude bound such as M n = (n ). The hard part is showing (a). All the necessary information is encoded by the sequence of degrees of vertices of P within its triangulation. Specifically, for any boundary vertex v of P, let its deficiency d(v) = max(, 6 deg(v)). This is the minimum number of additional edges needed to bring the degree of v up to 6. Page 5/

6 Let P be the derived polygon. Consider how to build up the original polygon in stages by erecting edges externally on P. For the new triangulation to be admissible, there must be at least d(v) edges attached to each vertex v. Proceeding around P, let v and w be two successive vertices. One new face is formed by going from v to a new vertex, then to w, then back along the edge of P from w to v. Every successive pair of vertices of P thereby determines a unique new vertex (and the deficiency of such a vertex is ). Furthermore, at any vertex v where d(v) exceeds, at least d(v) additional edges must be erected and therefore d(v) additional vertices must be created (and their deficiencies equal 3). Let D(P) be the sum of d(v) as v ranges over the vertices of P. Whence, the outer (original) polygon in the derivation has at least D(P) more vertices than P does. (This argument breaks down when P is a single point, but that does not ruin the induction, as we shall show.) Furthermore, there is a natural one-to-one correspondence between the new faces and the edges of both P and the new outer polygon. Therefore, all we need to show is that D(P) 6. This is an easy induction, because D(P) new vertices of deficiency 3 are created at each stage and in the base cases of,, and 3 vertices, we compute D(P) = 4, 6, and 6, respectively. It looks like things fail for the case of a single vertex, but we can fix this by considering all triangulations whose derived polygon is a single point: clearly these consist of k 6 triangles sharing that point, forming a k-vertex polygon. At each vertex v of that polygon d(v) = 3, implying D(P) = k 6. Thus the induction does go through as planned. Finally, it is straightforward to check that when two triangulations are sewn together along an outer edge (the reverse of splitting a triangulation along a diagonal of P), D(P) must increase. 7 B Let f be a polynomial with positive integer coefficients. Prove that if n is a positive integer, then f(n) divides f(f(n) + ) if and only if n =. Solution. Because f has positive integer coefficients, f(n) for all positive integral n. This means it makes sense to compute modulo f(n), where for any n we obtain f(f(n) + ) f() modulo f(n). When n = it is immediate that f() modulo f(n). Going in the other direction, this result also shows that if f(n) divides f(f(n) + ) that is, f(f(n) + ) modulo f(n) then f(n) divides f(). Because f(n) and f() are both positive, f() f(n). This would mean f(x) f() has a (real) zero in the interval [, ), which is impossible for any polynomial all of whose coefficients are positive, QED. Comment. In an example in The Art and Craft of Problem Solving (section 7.5), Paul Zeitz shows that if f(x) is a polynomial with integer coefficients, then for all integers n, f(f(n) + n) is a multiple of f(n). Page 6/

7 7 B Suppose that f :[, ] has a continuous derivative and that f ( x) dx =. Prove that for every α (, ), α f ( x) dx max f ' ( x). 8 x Solution. Define F(x) = x f ( t) dt and let a be a value in (, ) at which F is maximized. The Fundamental Theorem of Calculus implies F is differentiable with derivative Ff; its value at a a critical point must be zero. If F(a) = there is nothing left to show, because then F is identically zero, so we infer f(a) =. For any x, Taylor s Theorem asserts there is a number c between a and x for which Let x = and x = in turn: F(x) = F(a) + (x a) f (a) + ½(c a) f '(a) = F(a) + ½(c a) f '(a). = F() = F(a) + ½(c a) f '(a) for c in (, a) and = F() = F(a) + ½(c a) f '(a) for c in (a, ). Taking the values of c that maximize (c a) in the closures of these intervals (namely, and ) gives F(a) ½(a) f '(a) and F(a) ½( a) f '(a), respectively. In at least one of the two expressions a cannot exceed ½, yielding QED. 7 B3 F(x) = F (x) F (a) = F(a) ½(½) f '(a) /8 max [ < x < ] f '(x), Let x = and for n, let x n+ = 3x n + x n 5. In particular, x = 5, x = 6, x 3 = 36, and x 4 = 7. Find a closed-form expression for x 7. ( a means the largest integer a.) Solution. An answer is x 7 = (a 7 + b 7 )/ + (a 7 b 7 )/ 5, where α = and β = 3 5 (the conjugate of α). More generally, we claim x n = (α n + β n )/ + (α n β n )/ 5 = f(n). The proof is by induction, with the base case n = an easy calculation. Assume then that x n = (α n + β n )/ + (α n β n )/ 5 for a particular n. By definition, Page 7/

8 x n+ = 3x n + x n 5 = αx n = α[(α n + β n )/ + (α n β n )/ 5] = (α n+ + β n+ )/ + (α n+ β n+ )/ 5 + αβ n / αβ n / 5 β n+ / + β n+ / 5 = f(n+) + β n (α β)(½ / 5) = f(n+) + β n (α β)(½ / 5). The induction will be complete if we can show the second term is always zero; that is, < β n (α β)((½ / 5) < for all n. However, because < β <, QED. β n ( 5 ) < ( 5 ) <, and < β n (α β)(½ / 5) = β n ( 5)(½ / 5) = β n ( 5 ) <, Comment. x n+ = 6x n 4x n- for all positive n. This follows from our formula for x n. Comment. One can readily discover the formula for x n by making a small table of powers of α and comparing them to known values of x n. To prepare for the calculations, note that (3 + 5)(a + b 5) = 3a + 5b + (3b + a) 5, so that the coefficients (a, b) of the n th power determine the coefficients (3a+5b, 3b+a) of the n+ st power. n a b x n The pattern x n = a + b is clear. To obtain a closed formula, we need formulas that extract the coefficients a and b of α n. However, whenever α n = a + b 5, β n = a b 5, implying whence a + b = f(n). a = (α n + β n )/ and b = (α n β n )/( 5), Comment. Engel problem 4.7 (the last in the book) proves that for every positive integer n, divides + (3 + 5) n. The solution introduces β, uses the vanishing size of β n, and exploits the initial comment concerning the recursive relationship. Reinforcing this is a similar problem (4.66) concerning powers of 3 +. Page 8/

9 7 B4 Let n be a positive integer. Find the number of pairs P, Q of polynomials with real coefficients such that and deg(p) > deg(q). (P(X)) + (Q(X)) = X n + Solution. There are n+ solutions. Let ζ be a primitive 4n th root of unity (ζ n = -). Use it to factor the right side and, simultaneously, factor the left side over the complex numbers: P(X) + Q(X) n n j+ j+ = ( R( X ))( R( X )) = ( X ) ( X ζ ) j= R(X) = ±(P(X) + iq(x)). ζ where Because [X] has unique factorization, ±R(X) is determined by choosing a subset S of {,,, n-} and selecting terms from the first product corresponding to S and terms from the second product corresponding to the complement of S, thereby ensuring the remaining terms are their complex conjugates. Moreover, ±R(X) completely determines P and Q as its real and imaginary parts, respectively. The leading term is X n, which is real and corresponds to P, so indeed P does have the higher degree. The number of subsets is n and, independently of that, there are two ways to choose the sign of R(X), giving * n = n+ solutions. Comment. As a check, and to show how this works in a concrete case, a straightforward calculation gives the + solutions (P(X), Q(X)) = (X, ), (X, -), (-X, ), and (-X, -) when n =. They correspond to the factorization X + = (X i)(x + i). The case n = is less trivial. Four of the + = 8 solutions come directly from the factorization j= X 4 + = (X ζ)(x ζ 3 )(X ζ 7 )(X ζ 5 ) with ζ = ( + i)/ (ζ 7 is the conjugate of ζ and ζ 5 is the conjugate of ζ 3 ): (X ζ)(x ζ 3 ) = X + i( X), P(X) = X, Q(X) = X, (X ζ)(x ζ 5 ) = X + i(-), P(X) = X, Q(X) = -, (X ζ 3 )(X ζ 7 ) = X + i(), P(X) = X, Q(X) =, (X ζ 7 )(X ζ 5 ) = X + i(- X), P(X) = X, Q(X) = - X. The other four solutions come from negating these. Page 9/

10 7 B5 Let k be a positive integer. Prove that there exist polynomials P (n), P (n),, P k- (n) (which may depend on k) such that for any integer n, n k k = k n n P ( n) + P ( n) + + Pk ( n) k k. ( a means the largest integer a.) Solution. Fix k and for each n write n in the form n = n/k k + r with r < k. Let x = n/k. We seek polynomials for which x k = P (kx + r) + xp (kx + r) + + x k- P k- (kx + r) for all positive integers x and all r with r < k. We will look for a solution where the degrees of the polynomials do not exceed k-, so that we can write k P i (z) = j= p z j. ij Expand the powers of z = kx + r using the Binomial Theorem and arrange by ascending powers of x. The constant terms (zeroth power) all come from P (kx + r) and equal p + p r + p r + + p k- r k-. These must be zero for r =,,, k=. It is well-known that there is a unique solution to this system of equations (using the theory of circulants or Vandermonde determinants), implying P is completely determined. Suppose inductively that P, P,, P j- have been determined. Examine the coefficients of x j. Because we are effectively working modulo x j+, there are no contributions from P j+,, P k-. There are k simultaneous equations for the coefficients of P j of the form p j + p j r + + p j,k- r k- = terms from P s of lower index (+ when j = k). Because this system is nonsingular, it has a unique solution, thereby determining P j, and thence by induction all the P s are uniquely determined, QED. k Really clever solution. Let f(x, y) = y i x, rewrite this in the form i= k k k n x = P ( y) + P ( y) x + + Pk ( y) x, and note that f (, n) k = for integral n. Page /

11 7 B6 For each positive integer n, let f(n) be the number of ways to make n! cents using an unordered collection of coins, each worth k! cents for some k, k n. Prove that for some constant C, independent of n, n e n / Cn n / 4 n / + Cn n / 4 f ( n) n e. Solution. Here is just a sketch, because solving this one seems to require some mathematical sophistication: generating functions, complex analysis, and asymptotic analysis. The generating function to partition arbitrary integers into pieces of size,, 6,..., through n! is F(t) where /F(t) = ( t)( t )( t 6 )... ( t n! ). (The coefficient of t n! is f(n).) The biggest singularity in its partial fraction expansion comes from the (-t) n term, so we expect this to dominate the growth. Its coefficient equals Lim [t ] (-t) n F(t) = /( Lim [t ] (+t)(+t+...+t 5 )...(+t+...+t [n!-] ) ) = /(*6*4*... * n!). Let's call this function /g(n). Expanding the (-t) n term with the Binomial Theorem shows it contributes a value of Comb(-n!, n) /g(n) to f(n). By several applications of Stirling s Formula, the natural log of this value is asymptotically log((n!+n )!) log(n!!) log(n!) log(g(n)) = n (n log(n) n) Sum of (k log(k) k), k from to n The usual techniques (compare to integrals) estimate the sum as (n / * log(n) n / 4) n / = n / * log(n) 3 n / 4. Subtracting and, as was implicit the foregoing, doing all calculations O(n log(n)), yields which is exactly what we want. log(f(n)) = n / * log(n) n / 4 + O(n log(n)), A gap remains: showing that the other n! singularities do not change the asymptotics. The two poles of order n- (that is, at /(+t) (n-) and /(-t) (n-) ) will contribute terms only /n! as large; the six poles of order n- will each contribute terms only /(n!(n!-)) as large, and generally the k! poles of order n+-k will each contribute terms only /(n!(n!-)...(n!-k+)) as large. Summing these factors and making a crude estimate shows the remaining poles contribute a truly inconsequential amount, so we re done. Page /