Putnam n +1 (1 P (O n 1)) 2n +1. 2n +1 (2 (n 1) + 1) 2n +1 = (2 (n 1) + 1) P (O n 1)+1. (2(n 2)+1)P (On 2 )+1 2(n 1)+1. O n (n 1) + n 1 2n +1

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1 Putnam 00 A. Consider a set S and a binary operation, i.e., for each a, b S, a b S. Assume (a b) a b for all a, b S. Provethata (b a) b for all a, b S. Solution. Wehaveb ((b a) b) (b a) a (b a). A. You have coins C,C,...,C n.foreachk, C k is biased so that, when tossed, it has probability /(k +) of falling heads. If the n coins are tossed, what is the probability that the number of heads is odd? Express the answer as a rational function of n. Solution. Let O n be the event that n tosses result in an odd number of heads, and let E n be the event that n tosses result in an even number of heads. We have P (O )/3 and P (O n ) P (C n lands on T ) P (O n )+P (C n lands on H) P (E n ) P (O n )+ n + n + ( P (O n )) n + P (O n )+ n + n + n n + P (O n )+ n + ( (n ) + ) P (O n )+ n + ( (n ) + ) ³ ((n )+)P (On )+ (n )+ + n + ( (n ) + ) P (O n )+ n + ( (n i)+)p (O n i)+i n + ( (n (n )) + ) P O n (n ) + n n n n + n n +. A3. For each integer m, consider the polynomial P m (x) x 4 (m +4)x +(m ). For what values of m is P m (x) the product of two non-constant polynomials with integer coefficients? Solution. Wefirst find the roots r,r,r 3,r 4 of P m (x). Then P m (x) (x r )(x r )(x r 3 )(x r 4 ).

2 Thus, one of the nonconstant polynomial factors would be of form x r for some root r, or (x r )(x r ) for some roots r and r. The quadratic formula yields r ³(m +4)± p (m +4) 4(m ) m +± p (m +) (m ) (m +)± m ³ m ±. So n m o {r,r,r 3,r 4 } +, m, m +, m If one of the nonconstant polynomial factors is of form x r, then the only possibility is m and r 0. This possibility is realized, since P (x) x 4 8x x x 8. If one of the nonconstant polynomial factors is of form (x r )(x r ), then note that (x r )(x r )x (r + r ) x + r r in which case r + r and r r must be integers. The possibilities for r + r are 0, ± m, and ± (not an integer). We have ½ ª m +, m r + r 0 {r,r } ª m, m + ½ m + m + r r m m +, which is an integer only if m is n for some integer n 0. Wethenhave n m o {r,r,r 3,r 4 } +, m, m +, m n o n +, n, n +, n and ³ ³ ³ ³ x n + x n ³ ³ ³ ³ x n x n + x n 4n x n +4n P n (x). In the remaining case, r + r ± m {r,r } n ± m +, ± m o r r m

3 and m must either be (as before) or the square of an integer n 0. Ifm n,then n {r,r,r 3,r 4 } n +,n, n +, n o, and indeed ³ ³ x n + ³ ³ x n ³ ³ x n + ³ ³ x n x xn + n x +xn + n x 4 (n +4)x +(n ) P n (x). Thus, m must be of the form n or n for an integer n 0. A4. Triangle ABC has an area. Points E,F,G lie, respectively, on sides BC, CA, AB such that AE bisects BF at point R, BF bisects CG at point S, andcg bisects AE at point T. Find the area of the triangle RST. Solution. Chooser, s, t, sothat CE CB r, AF AC s, BG BA t Let (UVW) denote the area of a triangle with vertices U, V, W.Then (AEB) (AEF ), since the triangles have the same base AE and since AE bisects FB thesamealtitudeas well. Also (ABE) BE EC (ABC) BC r. BC BC Similarly, (ACE) CE CB (ACB) (CEF) CF CA (CEA). Thus, Hence, (CEF) CF CA (CEA)CF CE (ACE) CF CA CA CB (ACB) ( s) r (ACB) ( s) r. (ABC) (AEB)+(AEF )+(CEF) ( r)+( r)+( s) r r rs r ( + s) 3

4 Similarly, s ( + t) and t ( + r). Thus, r ( + s),s( + t),t( + r) Now By symmetry, r +s + +t and t ( + r) +t +t t + +t t ( + t ++t) +t +t t + t 0 ³ ³ t 5 or 5+ (negative) r s t ³ 5. Note that (RST) (ABC) ( (ABR)+ (BCS)+ (CAT)). Also, (ABR) BR BF (ABF ) (ABF ) AF AC (ABC) s (ABC) t, and similarly (BCS) (CAT) t. Thus, (RST ) ³ t + t + t A5. Prove that there are unique positive integers a, n such that a n+ (a +) n 00. Solution. Suppose a n+ (a +) n 00. Noticethat 00 a n+ (a +) n + 0moda Thus a is a factor of Notethata 6,sincethen n+ 3 n 00, 4

5 which is impossible since 00 is divisible by 3 unlike n+. More generally since a n+ (a +) n 00 0mod3 a mod3 is impossible, as is a 0mod3.Thus,a mod3,and ( ) n (a +) n a n+ mod3 n even. Recall that a n+ +has a factor of a +when n is even. Thus, a +is a factor of a n+ + (a +) n as is a itself. Only two factors differ by, namely 3 and 4. Thus, a 3and the smallest value for n is. Wehave It remains to show that no even n> satisfies 3 n+ 4 n 00 If n>, 4 n 0mod8, and 00 mod8.thus, 3 n+ mod8 However, 3 5mod8 and 3 5 mod8. Hence, 3 n+ 5mod8 if n is even Thus, n> and n even is not possible. A6. Can an arc of a parabola inside a circle of radius have a length greater than 4? Solution. The answer is yes. Consider the arc of the parabola y cx inside the circle x +(y ),wherec>. x +( cx ) x + 4 c x 4 cx 0 x 4 c x + c 0 x 0or x ± c c The length of the parabolic arc from x 0to x c c is then c c Z c c 0 Z c 0 +c x dx (using u cx) +u du (using the substitution z sinhu) ³ arcsinh u + u +u 0 c 5

6 We have arcsinh c + q c + c c ³ arcsinh c + c p +4(c ) c c arcsinh c r + 7 4c + 3 4c r 7 4c + 3 r 4c > 74c 7 > 4c c, for c sufficiently large. Thus, it suffices to prove that c arcsinh c > c for c sufficiently large, but this is so since arcsinh x increases without limit as x. Indeed, d arcsinh x arcsinh (x) dx x lim lim x log x x lim. x x +x Note: Mathematica yields a maximal length of for c B. Let n be an even positive integer. Write the numbers,,...,n in the squares of an n n grid so that the k-throw,fromlefttoright,is (k )n +, (k )n +,...,(k )n + n. Color the squares of the grid so that half of the squares in each row and in each column are red and the other half are black (a checkerboard coloring is one possibility). Prove that for each coloring, the sum of the numbers on the red squares is equal to the sum of the numbers on the black squares. Solution. Instead of writing the numbers,,...,n in the squares, write 0,,,...,n. Since half the squares are red and half are black, the assertion to be proven holds iff it holds with this new numbering. Now rewrite each number in base n in the form d d,where we take d 0in the first row. Note that d is the same in each row, while d is the same in each column. Since half of the squares in each row and in each column are red and the other half are black, the sum of the d s for the red squares is the same as the sum of the d s for the black squares in any row, and hence the sum of all red d s and all black d s is the same. Considering columns, the same equality holds for the sums of d s for red and d s for black squares. Thus, adding base n yields the same sum for red and black squares. B. Find all pairs of real numbers (x, y) satisfying the system of equations x + y (x +3y )(3x + y ) x y (y4 x 4 ). 6

7 Solution. By adding and subtracting the two given equations, we obtain the equivalent set of equations /x x 4 +0x y +5y 4 /y 5x 4 +0x y + y 4. Multiplying the first by x and the second by y, weget Adding these yields Subtracting yields Thus, x 5 +0x 3 y +5xy 4 5x 4 y +0x y 3 + y 5 3x 5 5x 4 y +0x 3 y +0x y 3 +5xy 4 + y 5 (x + y) 5. x 5 5x 4 y +0x 3 y 0x y 3 +5xy 4 y 5 (x y) 5 x + y 3 /5 x y, and so x (3 /5 +)/ and y (3 /5 )/. B3. For any positive integer n, lethni denote the closest integer to n.evaluate n hni + hni n. Solution. Since (m /) m m +/4 and (m +/) m + m +/4, wehave that hni m if and only if m m +/4 n m + m +/4 However, since n is an integer, this is equivalent to m m + n m + m 7

8 Hence n hni + hni n ( m + m ) m m X m hnim mx +m nm m+ n hni + hni n mx +m m nm m+ m + m n ( m + m ) (m m+) m X m ( m + m ) (m m+) ( m ) ( m + m ) m ³ m +m m m m(m ) (m+)m m X m(m ) +3. m n0 n ( m + m ) (m m) m m m m(m ) m ³ m +m m m m3 m(m ) B4. LetS denote the set of rational numbers different from {, 0, }. Define f : S S by f(x) x /x. Prove or disprove that \ f (n) (S), n where f (n) denotes f composed with itself n times. Solution. Suppose that n is reduced to lowest terms; i.e., n and m have no common m prime factors. Then f( n m ) n m m n n m (n + m)(n m) mn mn We claim that n m is reduced to lowest terms: If a prime p divides mn then it divides m mn or it divides n. If p divides m, then it cannot divide n, and hence it cannot divide n ± m. If p divides n, then it cannot divide m, and hence it cannot divide n ± m. Ineithercase,p cannot divide (n + m)(n m), and the claim holds. Definethedegreeof n (reduced to lowest terms) by deg n m m m + n. Thus, deg ³f( n ) n m + mn 3+ mn, m since n m is smallest (i.e., 3) for {n, m} {, }. Hence, deg ³f( n ) ³ n ) deg 3+ mn ( m + n ) m m ++ mn ( m + n ) +( m ) ( n ) 8

9 Hence the degree of any fraction in f (n) (S) is at least n. Since any rational number has finite degree, T n f (n) (S). B5. Let a and b be real numbers in the interval (0, /), andletg be a continuous realvalued function such that g(g(x)) ag(x)+bx for all real x. Prove that g(x) cx for some constant c. Solution. Pickx 0 arbitrary, and define x n recursively by x n+ g(x n ) for n 0. x n+ g(x n+ )g(g(x n )) ag(x n )+bx n ax n+ + bx n () We can findsolutionsofthisrelationoftheformx n r n. To determine r, notethat r n+ ar n+ + br n 0r n+ ar n+ br n r n r ar b ³ r 0or r r ± : a ± a +4b. The general solution of () is then x n c + r n + + c r n. If we can show that c 0(resp. c + 0) for all choices of x 0,thenc + x 0 (resp. c 0) and and we are done. Note that g (x 0 )x 0 r + (resp. g (x 0 )x 0 r ) r + <r < 0 <r + < + q , 4 so that c + r n + dominates for large positive n, while c r n dominates for large negative n. However, x n is not defined for negative n unless g exists. We prove that g : R R is - and onto as follows. Note that g (u) g (v) ag(u)+ bu g(g(u)) g(g(v)) ag(v)+ bv ag(u)+ bv bu bv u v. Thus, g is -, and hence must be strictly increasing or strictly decreasing. If g is strictly increasing and lim x + g (x) L + < +, then we have the contradiction g(l + ) lim g(g(x)) lim (ag(x)+bx). x + x + Thus, g is strictly increasing lim x + g (x) +. If lim x g (x) L >, then we have the contradiction g(l ) lim g(g(x)) lim (ag(x)+bx). x x 9

10 Thus, under the assumptions, g strictly increasing implies that g : R R is - and onto. If g is strictly decreasing and lim x + g (x) L + >, thenwehavethecontradiction g (L + ) lim g(g(x)) lim (ag(x)+bx) +. x + x + If lim x g (x) L < +, then we have the contradiction g(l ) lim g(g(x)) lim (ag(x)+bx). x x Hence in either case, g : R R is - and onto (i.e., as x goes from to +, g (x) strictly increases from to +, org (x) strictly decreases from + to ). At any rate, we may define x n g (x n ) for n 0. Suppose that c + 60and c 60. We will arrive at a contradiction. Note that x n c + r n + + c r n c r n + c + c r+ r n. ³ Since r+ r >, c + r+ c r n < for n sufficiently large and negative, say n<n<0. Then for n<n<0, sign(x n )c ( ) n alternates and x n > c r n as n. Since x n g (x n ), there are arbitrarily large x>0 for which g (x) < 0, andsog must be decreasing. We also have x n c + r+ n + c r n c + r+ n + c n r. c + r + ³ Since r r + <, c r c + r + n < for n sufficiently large, say n>n>0. Thus,ifc + > 0, then for n>n>0, g (x n )x n c +r n + and so there are arbitrarily large values of x for which g (x) > 0. This cannot happen if g is decreasing, as we have just shown is the case. If c + < 0, then g (x n )x n c +r n + and there are arbitrarily large negative values of x for which g (x) is negative. This is also impossible for g decreasing. Hence, for each x 0,wehaveg (x 0 )r + x 0 or g (x 0 )r x 0. Since g is continuous, both of the sets {x 0 (0, ) :g (x 0 ) r + x 0 0} and {x 0 (0, ) :g (x 0 ) r x 0 0} are closed in (0, ). Since (0, ) is connected either g (x 0 ) r + x 0 for all x 0 > 0, or g (x 0 )r x 0 for all x 0 > 0. Similarly, this holds for x 0 < 0. We must eliminate the possibility g (x) r ± x for x>0 and g (x 0 )r x for x<0, but we know this since g is increasing or decreasing. 0

11 B6. Assume that (a n ) n is an increasing sequence of positive real numbers such that lim a n /n 0. Must there exist infinitely many positive integers n such that a n i +a n+i < a n for i,,...,n? Solution. Here, increasing must mean strictly increasing, since otherwise a constant sequence provides an immediate counter-example. Actually, non-decreasing and not eventually constant is enough. The condition a n i + a n+i < a n means that the line connecting (n i, a n i ) and (n + i, a n+i ) in R has midpoint n, (a n i + a n+i ) which lies below (n, a n ). Equivalently, this means that there is a line L n through (n, a n ) such that (n i, a n i ) and (n + i, a n+i ) are below L n. Thus, it suffices to show that there are infinitely many n such that there is a line L n through (n, a n ) such that all of the points (n i, a n i ) and (n + i, a n+i ) for i,,...,n lie below L n. The sequence a i a ª i is (eventually) positive for i and has limit 0. Thus, it has a positive maximum, say m which is attained for some largest i >. Thus, the line L i through (i,a i ) with slope m : a i a i > 0 has the property that all points (i, a i ) with i<i are on or below this line, while all points (i, a i ) for i>i are strictly below (since i is largest). The ratios a i a i a i i for i>i are then all less than m and lim i a i i i i 0. Hence m a i a i i i achieves a positive maximum, say ε. The line, say L i, with slope m ε through (i,a i ) then lies above (i, a i ) for all i>i. The points (i, a i ) for i<i also lie below this line since they are on or below the line with slope m through (i,a i ).Wenow repeat the same argument, starting with i instead of to produce a largest i,suchthat the maximum of a i a i i i for i>i is achieved for i i. This maximum, say m i,islessthan m ε and so the line with slope m i through (i,a i ) will have the property that all points (i, a i ) will lie on or below this line. Again, there is some ε > 0, such that the line with slope m i ε through (i,a i ) will lie above all other points (i, a i ),i6 i. In this way, we produce infinitely many points (i k,a ik ) suchthereisalinethrough(i k,a ik ) which lies above all points (i, a i ),i6 i k.