Putnam 1997 (Problems and Solutions)

Transcription

1 Putna 997 (Probles and Solutions) A. A rectangle, HOMF, has sides HO =and OM =5. A triangle ABC has H as the intersection of the altitudes, O the center of the circuscribed circle, M the idpoint of BC, andf the foot of the altitude fro A. What is the length of BC? Solution. In the figure below, let G be such that MG =,letx = BF = GC, and let y = AH. Since O is the circucenter, we have y =5 2 +(+x) 2. The slope of the line AB is (5 + y) /x and the slope of HC is 5/ (22 + x). Since these lines eet at right angles, the product of the slopes is. Thus, 5+y 5 x 22+x =or y =22x + x 2. Hence y =5 2 +(+x) 2 = x + x 2 = y, or y 2 5y 50 = 0 or (y 0) (y +5) =0. Thus, y =0,x=3and BC = = 28.

2 A2. Players, 2, 3,..., n are seated around a table and each has a single penny. Player passes a penny to Player 2, who then passes two pennies to Player 3. Player 3 then passes one penny to player 4, who passes two pennies to Player 5, and so on, players alternately passing one penny or two to the next player who still has pennies. A player who runs out of pennies drops out of the gae and leaves the table. Find an infinite set of nubers n for which soe player ends up with all n pennies. Solution. Suppose that at soe point there are an odd nuber greater than of reaining players, say p,p 2,..., p 2k+ (k ), none in danger of dropping out on their next pass. Then there is an odd nuber of passes starting with p s pass to p 2 and ending with p 2k+ s pass to p. Thus, if p passes 2 coins he will receive 2 coins, and if he passes he will receive. The sae is true for any other player and the gae will not terinate. Thus, we ust avoid this situation, or we are stuck with a nonterinating gae. Suppose that to begin with there are n =2 (where for notational convenience, 3) players, say p,..., p 2. Note that p drops out and p 2 passes his two and drops out, p 3 passes and ends up with 2, p 4 drops out, but p 5 ends up with 2. Continuing, p 3,p 5,..., p 2 each have 2 coins and p 2 drops out when he passes 2 to p 3 who now has 4. There are now players and we are stuck if is odd. Thus, we need =2 for soe. Relabel the players q,..., q 2.Nowq has 4 coins, the rest have 2, andq starts this stage by passing to q 2.Sincethere areanevennuber(2 ) of players, q will eventually receive 2 (for a total of 5) andpass again to q 2 to begin the second round of the stage, but then q 2 drops out, as do q 4,q 6,..., q 2. When q 2 passes 2 to q and drops out, q will have 7 and the rest q 3,q 5,..., q 2 each have 4. At the beginning of this new stage, there are now players left which ust be even, or else =(i.e., we have a winner) or we are stuck. Continuing, we see that the gae terinates if and only if is a power of 2, since half of the players (those repeatedly passing 2) eventually drop out at each stage which necessarily ust have an even nuber of players if not a winner. Observe that all the players passing 2 in a stage, have the sae nuber of coins, and so they all drop out in the sae final round of the stage. Thus, a gae with n =2 players terinates if and only if =2 =2 k for soe k 0, or n =2 =2 2 k + =2 h +2, h =2, 3, 4,... will yield terinating gaes. Of course, n =2and n =4gaes also terinate. Incidentally, for n =2 h +2, the original player p 3 is the winner since each stage ends with p 3 receiving 2. 2

3 A3. Evaluate Z µx µ x3 2 + x5 2 4 x x x x dx 0 Solution. The first series is xe 2 x2. For the second series, we use the fact that for n =0,, 2,..., Z π 0 cos 2n θ dθ = = 3 5 (n ) (2n ) 2n π = n (2n) 2 π (2n)! (2n) 2 π. Then second series is (using R π 0 cos2n+ θ dθ =0), R π R π 0 cos2n θ dθ π x 2n π 0 = cosk θ dθ x k = (2n)! k! π n=0 Thus, the desired integral is Z Z π = π n=0 Z π 0 0 e x cos θ dθ. Z π Z Z π 0 (x cos θ) k dθ k! xe 2 x2 e x cos θ dθ dx = e 2 r2 e r cos θ rdrdθ 0 π 0 π 0 0 = Z Z e 2 (x2 +y 2 ) e x dxdy = Z Z π 0 π e 2 e 2 2x+) (x2 dx e 2 y2 dy 0 = Z Z π e 2 e 2 (x )2 dx e 2 y2 dy = π e ³ 2 2π 2π = e. 2 n=0 3

4 A4. Let G be a group with identity e and φ : G G be a function such that φ (g ) φ (g 2 ) φ (g 3 )=φ (h ) φ (h 2 ) φ (h 3 ) whenever g g 2 g 3 = e = h h 2 h 3. Prove that there exists an eleent a in G such that ψ (x) =aφ (x) is a hooorphis (that is, ψ (xy) =ψ (x) ψ (y) for all x and y in G). Solution. Since x xe = e 3, we have φ x φ (x) φ (e) =φ (e) 3 or φ x φ (x) =φ (e) 2 or φ x = φ (x) φ (e) 2. Since yy e = e 3,wehave φ (y) φ y φ (e) =φ (e) 3 or φ (y) φ y = φ (e) 2 or φ y = φ (e) 2 φ (y). Since x (xy) y = eee,we have φ x φ (xy) φ y = φ (e) 3, and so φ (xy) = φ x φ (e) 3 φ y = φ (x) φ (e) 2 φ (e) 3 φ (e) 2 φ (y) = φ (x) φ (e) φ (y). 4

5 A5. Let N n denote the nuber of ordered n-tuples of positive integers (a,a 2,...,a n ) such that /a +/a /a n =. Deterine whether N 0 is even or odd. Solution. Consider the involution of the set of solutions (a,a 2,a 3,a 4,a 5,a 6,a 7,a 8,a 9,a 0 ) 7 (a 2,a,a 4,a 3,a 6,a 5,a 8,a 7,a 0,a 9 ) The nuber of points that are not fixed is even. Thus, we need only to consider the solutions of the for (a,a,a 3,a 3,a 5,a 5,a 7,a 7,a 9,a 9 ). On this set of reaining solutions, consider the involution (a,a,a 3,a 3,a 5,a 5,a 7,a 7,a 9,a 9 ) 7 (a 3,a 3,a,a,a 7,a 7,a 5,a 5,a 9,a 9 ) The points that are fixed are of the for (a,a,a,a,a 5,a 5,a 5,a 5,a 9,a 9 ). By one ore involution we need only consider the nuber of solutions of the for (a,a,a,a,a,a,a,a,a 9,a 9 ). For these 8 a + 2 a 9 =,andsoa 9 = 2a a 8. The set of possible pairs (a,a 9 ) is {(9, 8), (0, 0), (2, 6), (6, 4), (24, 3)}. Thus,N 0 is odd. 5

6 A6. For a positive integer n and any real nuber c, define x k recursively by x 0 =0, x =, and for k 0, x k+2 = cx k+ (n k)x k. k + Fix n andthentakec to be the largest value for which x n+ =0. Find x k in ters of n and k, k n. Solution. The recurrence relation is as one ight find while solving a differential equation using power series. We have (k +)x k+2 = cx k+ (n k)x k x k+2 (k +)t k = c x k+ t k n x k t k + kx k t k x k+2 (k +)t k = c x k+ t k (n ) x k t k + t 2 k= (k ) x k t k 2 Let f(t) = P x k+t k =+ P x k+2t k+ = P k= x kt k. Then the above says Hence f 0 (t) = cf(t) (n ) tf(t)+t 2 f 0 (t) f 0 (t) c (n )t = f(t) t 2 = B +t + A t = 2 (n +c) +t 2 (n c). t f(t) =( t) 2 (n c) ( + t) 2 (n +c). Now f(t) will be a polynoial of degree n if 2 (n +c) and 2 (n c) are nonnegative integers, say j and n j (i.e., j =0,..., n ). Since f(t) = P x k+t k,wewouldthenhavex n+ =0. Since j = 2 (n +c), c =2j n +. Thus, soe possible values for c for which x n+ (n, c) =0are the n values n +, n +3,..., 2(n ) n +=n. However, it is clear that x k (n, c) is a polynoial of degree at ost k in c. Indeed, x 2 (n, c) =c, and fro x k+2 = cxk+ (n k)xk k+ the result follows fro induction. Thus, the values the above values are the only values of c for which x n+ (n, c) =0. The largest of these values is c = n. Then f(t) =( t) 2 (n c) ( + t) 2 (n +c) =(+t) n, 6

7 and f(t) = P x k+t k x k+ = n k or 0 k =0 x k = k = n k k =2,..., n 0 k>n. 7

8 B. Let {x} denote the distance between the real nuber x and the nearest integer. For each positive integer n, evaluate F n = 6n = in({ 6n }, { 3n }). (Here in(a, b) denotes the iniu of a and b.) Solution. Wehave { 6n } = ½ 6n 3n 6n 3n 6n { 3n 3n/2 3n } = 3n 3n/2 3n 3n 3n 3n 3 2 = 9n 2 2 9n 3n 2 6n Now in({ 6n }, { 6n 3n/2 3n })= in 6n, 3n 3n/2 3n in 6n, 3n 3n 3n 3 2 = 9n 2 in 6n, 2 9n 3n 2 6n Using 6n 3n 2n, 6n 3n 4n 6n 2 3n 6n,. we have in({ 6n }, { 3n })= 6n 3n/2 6n 3n/2 2n 3n 2n 3n 3n 3n 4n 6n 4n 9 2 n 9 6n 2 n 6n 6n 2n = 3n 2n 3n 3n 3n 4n 6n 4n 6n 8

9 Hence, F n = = 6n = 2n = 6n in({ 6n }, { 3n }) 3n 6n + = =2n à 2n 2 = ³ 3n 3n =2n n =3n 4n =3n ³ 6n 3n + 6n =4n =4n! ³ 6n +(3n 2n +) (4n 3n +)+(6n 4n +) = 6n 2 (2n)(2n ) (5n ) n +(7n ) n 2 (0n ) 2n +2n = n. 9

10 B2. Let f be a twice-differentiable real-valued function satisfying f(x)+f 00 (x) = xg(x)f 0 (x), where g(x) 0 for all real x. Provethat f(x) is bounded. Solution. Multiplying by f 0 (x), we get 2 d f(x) 2 + f 0 (x) 2 = f(x)f 0 (x)+f 0 (x)f 00 (x) = xg(x)f 0 (x) 2. dx Thus, f(x) 2 + f 0 (x) 2 is nondecreasing for x 0 and nonincreasing for x 0. Hence f(x) 2 f(x) 2 + f 0 (x) 2 f(0) 2 + f 0 (0) 2. 0

11 B3. For each positive integer n, write the su P n = / in the for p n/q n, where p n and q n are relatively prie positive integers. Deterine all n such that 5 does not divide q n. Solution. WeaywritethesuS n := P n = / in the for S n = = n / K 5 k bn/5 k c 5- bn/5c 5-2 / + bn/5 c 5 2 K / = 5- / k F bn/5 k c, where 5 K is the largest power of 5 not greater than n and M F M :=. Consider a su of reciprocals of integers n,..., n k not divisible by 5, say = n 2...n k + n n 3...n k n n 2...n k = N n n 2 n k n n 2...n k D with D = n n 2...n k. Let n i betheassociateofn i ; i.e., n i n i od(5). Note that D (n + n n k ) = n n 2...n k (n + n n k ) n 2...n k + n n 3...n k n n 2...n k = N (od 5) Thus, D (n + n n k ) N (od 5) Since 5 - D we have 5 N n + n n k 0(od5) Note that =, 2=3, 3=2, 4=4 Working od 5, wehave 5- =, +2 =, +2+3 =, = = = = = 0, etc..

12 Writing F M = NM D M, we have N M,,, 0, 0(od5)if M, 2, 3, 4, 5(od5), respectively. Note that if M 5, thenf M = F M and N M = N M 0(od5). Recall Consider the last ter S n = K 5 k F bn/5 k c ( ) 5 K F bn/5 K c in ( ). We have n/5 K =, 2, 3 or 4. If n/5 K =, 2,or 3, then the nuerator of F bn/5k c is not divisible by 5. In this case, S n ust have a factor of at least 5 K in its denoinator. Thus, K =0in this case and n =, 2,or 3. Henceforth, suppose that n/5 K =4. In this case, we have F 4 =+/2+ /3 +/4 = 25 2 and 5 F K bn/5k c = 2 5. If K =0,thenS K 2 4 = F 4 = 25 2 and n =4has the desired property. If K, the preceding ter in ( ) is 5 F K bn/5 K c,and n/5 K =4 n/5 K =4, 2 5, 22 5, 23 5, 24 5 n/5 K =20, 2,..., 24. If n/5 K =2, 22, 23, then 5 F K bn/5 K c has nuerator not divisible by 5 and the denoinator has a factor of 5 K. As all other denoinators in the su have lower powers of 5 (in particular the last ter is 2 5 ), in this K 2 case S n will have denoinator divisible by 5 unless K. Thus, in the case n/5 K =2, 22, 23, we have n = bnc = n/5 K =2, 22, 23. Now suppose that n/5 K =20, 24. We have F 20 = and F 24 = As the nuerators of F 20 and F 24 are divisible by 25, theter 5 F K bn/5 K c has denoinator with a power of 5 of at least K 3. The su of the two ters in ( ) before and after this ter is 5 K 2 F bn/5 K 2 c + 5 K F bn/5 K c = 5 K 2 µ F bn/5 K 2 c + 2 Since 2 3 and the nuerator of F bn/5 K 2 c can be only be congruent to ± or 0, the denoinator of this su has a factor of at least 5 K 2, and all of the other ters have denoinators with lower powers of 5. Thus, we ust have K = or 2. For K =,n=20, 24 and for K =2,we have bn/5c =20, 24 n = 00 04, Thus, in view of all of the above, the only values for n with the desired property are 4, 20 24, 00 04, and

13 B4. Let a,n denote the coefficient of x n in the expansion of ( + x + x 2 ). Prove that for all k 0, b 2k 3 c 0 ( ) i a k i,i. i=0 Solution. Notethatsincethedegreeof( + x + x 2 ) is 2, wehavea,n =0 for n>2. Hence a k i,i =0for i>2(k i) or 3i >2k. Hence the upper liit on the su only serves to eliinate ters which would be 0 or undefined. Let us define a,n =0for n<0 or <0. Thusinthesuaboveandallsus below, we ay take the index to run over all integers. a +,n x n = (+x + x 2 ) + =(+x + x 2 )( + x + x 2 ) n = (+x + x 2 ) k a,k x k = k a,k (x k + x k+ + x k+2 ) = n a,n x n + n a,n x n + n a,n 2 x n = n (a,n + a,n + a,n 2 ) x n. Thus, a +,n = a,n + a,n + a,n 2. Let s k := P b 2k 3 c i=0 ( )i a k i,i = P i ( )i a k i,i.then s k+ : = ( ) i a k+ i,i = ( ) i (a k i,i + a k i,i + a k i,i 2 ) i i = ( ) i a k i,i ( ) i a (k ) (i ),i +( ) i 2 a (k 2) (i 2),i 2 i We have = s k s k + s k 2 s 2 = i ( ) i a 2 i,i =0,s = i ( ) i a i,i =0,s 0 = i ( ) i a 0 i,i = a 0,0 = Thus, s = s 0 s + s 2 = s 2 = s s 0 + s =0 s 3 = s 2 s + s 0 =0 s 4 = s 3 s 2 + s = s 5 = s 4 s 3 + s 2 = 3

14 Assue that s 4k =,s 4k+ =,s 4k+2 =0,s 4k+3 =0. Then s 4(k+) = s 4k+4 = s 4k+3 s 4k+2 + s 4k+ =0 0+=, s 4(k+)+ = s 4k+5 = s 4k+4 s 4k+3 + s 4k+2 = 0+0=, s 4(k+)+2 = s 4k+6 = s 4k+5 s 4k+4 + s 4k+3 = +0=0, and s 4(k+)+3 = s 4k+7 = s 4k+6 s 4k+5 + s 4k+4 =0 +=0. Thus, s k is 0 or for all k 0, and in particular 0 s k. 4

15 B5. Prove that for n 2, o n o n (odn) Solution. Let t := 2 and let t n := 2 t n. We are to show t n =2 t n t n (od n). We firstchecktheresultinthecasen =2 k. Note that t n 2 2 k 2 tn 2 tn 2 od 2 k t n t n od 2 k Thus, for the case n =2 k, it suffices to check that t 2 k 2 2 k. This is true if k =. If it is true for k, then t 2 k+ 2 >t 2k =2 t 2 k k 2 k+, and so t 2k 2 2 k by induction. We handle the case n =2 k d where d is odd, by using induction on d. The case d =has been done. First note that since (d, 2) =, 2 ϕ(d) (odd). Thus, we have the key observation that t 2 t (od ϕ (d)) 2 t t 2 (odd) 2 t 2 2 t (od d) t t (od d). Now t n t n od 2 k d t n t n od 2 k and t n t n (od d) Since n 2 k, we have already shown t n t n od 2 k. Thus, we need only show that t n t n (od d) and by the key observation it suffices to show that t n 2 t n (od ϕ (d)). We have d n 0 := ϕ (d) =2 k0 d 0, where d 0 is odd. Since d 0 <d,by induction we have t n 0 t n 0 (od n 0 ) or t n 0 t n 0 (od ϕ (d)). However, we need t n 2 t n (od ϕ (d)). Since n d n 0,we have n 2 n 0. Hence, if we could show the stronger result that not only t n t n (od n) but also t n t n t n+ t n+2 (od n), 5

16 (ad infinitu), then we would have t n0 t n 0 t n t n t n+ (od ϕ (d)) The stronger result t n t n t n+ t n+2 (od n) is true in the base case n =2 k. Then assuing it holds for d 0 <dinductively, we have t n 0 t n 0 t n t n t n+ (od ϕ (d)) in which case by the key observation, We already know that Thus, t n 0 t n t n t n+ (od d). t n t n+ od 2 k. t n t n+ t n+2 od 2 k d, and we have proven this stronger result by induction. 6

17 B6. The dissection of the triangle shown below (into four congruent right triangles siilar to the original) has diaeter 5/2. Find the least diaeter of a dissection of this triangle into four parts. (The diaeter of a dissection is the least upper bound of the distances between pairs of points belonging to the sae part.) Solution. Lower bounds on the least diaeter of a dissection of the triangle into 4 parts can be obtained by selecting 5 points and taking the sallest distance between pairs. Indeed, any dissection into four parts ust have one region containing 2 of the 5 points and that region has diaeter at least the distance between the two points. Thus, optiistically, we seek 5 points so that the iniu distance between pairs is as large as possible. Then we attept to find a dissection where no region has diaeter larger than this iniu distance. Let A =(0, 4),B =(0, 0),C =(3, 0). The point on the segent AC which is distance t fro (3, 0) is 5 t (0, 4) + 5 t (3, 0) = t, 4 5 t. The square of the distance fro this point to the point (0, 4 t) on AB (at distance t fro A) is µ3 35 t, 45 t 2 (0, 4 t) =25 8t t2 Setting this equal to t 2 yields 25 8t t2 = t 2 t = 25 3, 5 Thus, A is distance 25 3 fro D:= 0, 27 (0, ) which is at distance 25 3 fro E:= , = 24 3, 3 20 (.8462,.5385). It is easy to verify that 25 3 is the iniu distance between pairs of the five points A,B,C,D,E. 25 Thus, 3 is a lower bound on the diaeter of any dissection of ABC into 4 regions. We try to show that this lower bound is realized. Let F = 3 2, 0 and G = 3 2,h where h>0 is chosen so that BG = CG = µ 2 µ h 2 = 26 3 We find h = Thus,G= 3 2, (.5,.2034). Let H = 0, and let I = , = 5 3, 3 32 (.538, 2.465) be the point on AC at distance 25 3 fro A. Then BG = 25 3, CG =

18 Now, the triangle ADI and the two quadrilaterals BFGH and FGEC clearly have diaeter For the pentagon HGEID, we need to check that HE 3, but the height of E is less than that of the idpoint of DH since µ 3 < Thus, all 4 regions have diaeter equal to the lower bound