End-of-Year Contest. ERHS Math Club. May 5, 2009

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1 Ed-of-Year Cotest ERHS Math Club May 5, 009 Problem 1: There are 9 cois. Oe is fake ad weighs a little less tha the others. Fid the fake coi by weighigs. Solutio: Separate the 9 cois ito 3 groups (A, B, ad C) with 3 cois i each group. Weigh groups A ad B agaist each other. If A ad B are equal, the the last weighig will ivolve oly the 3 cois i group C sice we kow that there is o fake coi i either A or B to lesse the weight of oe of these groups that we weighed. Otherwise, the ext weighig will ivolve the 3 cois i the group that weighed less (either A or B); the group that weighs less must have a fake coi that lesses the overall weight. I this group of three, pick cois to weigh agaist each other. If the are equal, the fake coi must be the oe we did ot weigh. If ot, the fake coi has to be the oe that weighed less. Proposed by Mayowa Omokawaye Problem : Imagie a hypothetical society with a alphabet that has 6 letters, if you ca. Bob seds Je a message that uses oly letters of the alphabet. Before he set it, he coded it usig a mooalphabetic cipher. What is the probability that a radomly chose key will decrypt the message with o icorrect letters i the ew plaitext? I a mooalphabetic substitutio cipher oe letter represets aother letter i decryptio. Also i this cipher oe letter ALWAYS represets the same letter i plaitext. The key is the list of correspodig letters from cipher text to plai text. Solutio: We kow that there is oly oe correct key for of the letters, but the coded letters correspodig to the other four ca be arraged i ay way possible sice they are ot used i the message. There are! ways to arrage the four uused letters to a key. There are 6! possible! keys. Therefore, the aswer is: 6! 1

2 Proposed by Mayowa Omokawaye Problem 3: Make 10 usig a 1 ad a 0. You ca use Additio, subtractio, divisio, ad multiplicatio Factorials Expoets Decimals But o cocateatio, so you ca t just put the 1 ext to the 0 ad call it te! (10 poits) First Solutio: 0!.1 = 10 Secod Solutio: (by Holde Lee).1 0! = 10 Proposed by Lius Hamilto Problem : Prove that if poits) c 1+b Solutio: Replace b with ta(arcta(b)) to get 1 ad c 0 the si x + b cos x = c has a solutio for x. (10 si x + ta(arcta(b)) cos x = c Now multiply both sides by cos(arcta(b)) to get cos(arcta(b)) si x + si(arcta(b)) cos x = c cos(arcta(b)) si(arcta(b) + x) = c cos(arcta(b)) Readig the values off a 1 b 1 + b right triagle gives that cos(arcta(b)) = 1 1+b, so the equatio becomes si(arcta(b) + x) = c 1 + b c 1+b 1, so it must have a arcsi. Therefore we ca take the arcsi of both sides, givig c arcta(b) + x = arcsi( ) 1 + b Fially, subtractig arcta(b) from both sides gives c x = arcsi( ) arcta(b) 1 + b

3 ad we are doe. Proposed by Lius Hamilto Problem 5: A house is made up of a array of rooms. Each room has a light ad a switch. The state of the buildig is determied by which lights are o ad which lights are off. Flippig the switch i a room does ot toggle the light i that room, but rather the lights i the rooms adjacet to it (rooms sharig a commo wall with it). Iitially all the lights are off. Fid, i terms of, a. The umber of states that ca be attaied by flippig some series of switches. ( poits) b. The umber of subsets S of switches with the followig property: If all lights are iitially off, ad exactly those switches i S are flipped, the all lights will be off agai. Iclude the empty set i your cout. ( poits) c. Prove your aswer to (a) ad (b). (7 poits) Solutio: a. ( 1) b. c. Number the rows ad colums 1,,... from top to bottom ad left to right. Represet the rooms as ordered pairs (r, c) where r, c deote the row ad colum of the room, respectively. For a subset A of rooms, let s(a) deote the operatio of togglig all switches i rooms i A, ad let s(l) deote the operatio of togglig the switch i room l. Note that togglig a switch a eve/ odd umber of times has the same effect as togglig it 0, 1 time, respectively, because togglig a switch twice has o effect. Furthermore, the order i which switches are performed does t matter. Thus all reachable states ca be reached by a operatio s(a), for some subset A of switches. First Approach: (i) Let T be the set of rooms i the first 1 rows. We show if A, B are two distict subsets of T, the s(a) s(b). Suppose by way of cotradictio that s(a) = s(b). Order the lights (lexicographically) as follows: we say that (r 1, c 1 ) (r, c ) ((r 1, c 1 ) comes before (r, c )) if ad oly if r 1 < r OR r 1 = r ad c 1 < c. Note for ay (r, c) T, the last light chaged by s(r, c) is light (r + 1, c). s(a) = s(b) implies that ( deotes compositio) s(a) s(a B) = s(b) s(a B) ad sice o each side the switches i A B are switched twice, (A \ (A B) = A \ B), s(a \ B) = s(a \ B) Let l = (r, c) be the last switch i oe of A \ B or B \ A. Suppose without loss of geerality that it is i A \ B. The o the left side, light (r + 1, c) is toggled. However, sice o the right side the last switch toggled is before l, the last light toggled is before (r + 1, c) ad (r + 1, c) is ot toggled, cotradictio. Thus s(a) s(b). 3

4 (ii) We show that for every light (, c) there exists a set A of switches i T such that s(, c) s(a) is the idetity trasformatio. Ideed, switchig all lights (x, y) (icludig (, c)) satisfyig (1) x + y + c (mod ) () c x + y + c (3) c x y c has o et effect: All lights (x, y) with x + y + c (mod ) are either toggled times (if () ad (3) is satisfied), times (if x + y = ± (c + 1) or x y = ±(c 1)), or 0 times (otherwise). All lights (x, y) with x + y + c (mod ) are ot toggled. Thus for every subset A of rooms, we ca fid a set of switches cotaied i rooms i T that has the same effect as s(a), ad thus a set B T such that s(a) = s(b). I light of (i), we coclude the umber of iequivalet operatios possible, ad hece the umber of attaiable states, is T = ( 1). (That is, all attaiable positios ca be obtaied by oly togglig switches i the first 1 rows.) Let F be the set of all sets of switches satisfyig (b), ad R a set of subsets of rooms with the largest size such that s(l) is distict for every distict l R. The for every subset of rooms A, there exists a uique B R such that s(a) = s(b). The we have s((a \ B) (B \ A)) is the idetity trasformatio, so (A \ B) (B \ A) F. Thus every set A ca be associated with a pair (B, C) such that B R, C F, B C = A. Sice the total umber of subsets is, = R F = T F F = Secod Approach: (by Lius Hamilto) We prove part (b) first. There are ways to flip switches i oly the bottom row of the house. Every combiatio gives a specific o/off patter o the bottom row. Now give a row R of lights i some state, there is oly 1 way to flip some switches i the row above that turs all lights i R off, amely, to toggle the switches i the rooms right above rooms where lights are o. (Note togglig switches i rooms or more rows above R has o effect o lights i R.) By part (ii) of the previous solutio, for every oe of subsets of rooms i the bottom row, there exists 1 subset of rooms cotaiig them such that if all switches i those rooms are toggled, all lights will be off agai. Thus such a set is uique for each subset of rooms i the bottom row. Fiish as i the first solutio but this time F is kow, ad we are solvig for R. Remark: The set G of possible chages caused by flippig some sequece of switches forms a abelia group. That is, for ay a, b, c G (a) Commutativity a b = b a (b) Associativity (a b) c = a (b c) (c) Existece of idetity a I = a where I = s(φ) (d) Existece of iverse a a 1 = I where a 1 = a. We say that G is geerated by {s(t) t T } because every elemet i G ca be attaied by compositio of elemets of T.

5 Proposed by Holde Lee Problem 6: > 1 buies sit o a umber lie such that the maximum distace betwee ay two of them is d. Each step, buies are selected. The buy at the left, say A, jumps some distace x > 0 to the right, while the buy at the right, say B, jumps the same distace x to the left, such that A is still to the left of (or occupies the same locatio as) B. (Buies may jump over each other.) After a fiite umber of steps, let S be the sum of the distaces traveled by all buies. a. Fid the miimum L(, d), as a fuctio of ad d, such that we always have S L(, d). (8 poits) b. Prove your aswer to part (a). (7 poits) Solutio: a. L(, d) = d, eve ( 1)d, odd b. Give the buies flags labeled 1,,... from left to right. Whe a buy passes aother buy durig a jump, they switch flags. I this way, the flags are always i icreasig order from left to right. Furthermore, each step ca be decomposed ito substeps, where i a substep two flags move the same way as buies do durig a step. (The times whe flags are switched divide a step ito substeps.) S is the sum of the distaces traveled by the flags. L(, d) Now ote that, if it is fiite, is a fuctio oly of. This is sice if S = S 1 is attaiable d for buies ad maximum distace d 1 0, the S = S 1 d1 is attaiable for buies ad maximum distace d by multiplyig all distaces (distaces betwee buies, ad jumps) by d 1 d ad vice versa. Hece d 1 L(, d 1 ) = L(, d ) L(, d 1) = L(, d ). d d 1 d L(, d) We fid the maximum of. Suppose the iitial ad fial positios of flag i are x i ad d y i, respectively. Note that i each step, the sum of the positios of the buies remais ivariat. Suppose flag i moves distace l i, r i to the left ad right, respectively, ad let d i = l i r i. Let d j = mi i (d i ) ad d k = max i (d i ). The d k d j = (x k x j ) (y k y j ) x k x j d (1) d 1 + d d = x i y i = 0 () i=1 i=1 d l 1 + l l = r 1 + r r (3) Now every time a flag labeled with a umber i N moves to the left, a flag to the left of it, that is, a flag with umber j < i N must move to the right. Hece l 1 + l l N r 1 + r r N 1 () 5

6 for 1 N. I particular, l 1 = 0. () ad () imply S = l 1 + l l + r 1 + r r = (l 1 + l l ) (d 1 + d d ) = [(l 1 + l l ) (d 1 + d d )] i ( d j ) = = ( i=1 ( d j ) i=j ( j)d j ) Let e 1, e,... e be the permutatio of d 1, d,... d i odecreasig order. The by the Rearragemet Iequality, S ( ( j)e j ) (5) The from (1), We fid the maximum of subject to the coditios S d = ( ( j)e j) ( ( j)e j) (6) d e e 1 ( ( j)e j) e e 1 (7) (a) e 1 + e e = 0 (b) e 1 e,... e 1 e We ca fid f 1,... f such that e i = f i f f for each 1 i. Thus it suffices to fid the maximum of uder the sigle coditio T = ( ( j)(f j e 1+...e )) f f 1 = +1 (j )f j f f i (c) f 1 f,... f 1 f Note that whe i + 1 the coefficiet of e i i the umerator is opositive, so T will ot decrease if e i is replaced by e 1. For i > + 1, the coefficiet is positive, so T will icrease if e i is replaced 6

7 by e. Let a = f 1 f f = e 1. Usig (7), the maximum for T is = (( 1) )( a) (( 1) )(a) a = whe is eve. Whe is odd, the maximum occurs i (7) whe e 1 =... = e +1 e 1 =... = e +1 = a +1 1 ; hece the maximum is (( 1) = [( 3 3 = [ 3( 1) = = 1 )( + 1 ) ( 3 3 ] )( a) (( ) )(a 1 ) a + a +1 1 )( 1 )( )] 1 Note we used the formula for arithmetic series. We have just proved that Now we show this boud is optimal. S d =, eve 1, odd = a ad (8) Lemma: Give ay arragemet of flags (where flags are i icreasig order from left to right) if we repeatedly move pairs of flags (1, ), (, 3),..., ( 1, ), each time movig the flags i the i=1 pair to the same locatio, i that order, the the locatio of all flags will coverge to x = x i. Proof: If the positio of all flags coverges to x the we have x = x x x = x. Let x i (t) be the positio of the ith flag after t repetitios. The x (t) x i (t) is decreasig, so it coverges to some D 0. It suffices to cosider D 0. Note that after the (t+1)th step of the algorithm, every poit betwee x 1 (t) ad x (t) has bee jumped over (or o) by oe buy. Ideed, after movig pairs (1, ),..., (i 1, i) to the same locatio, flag i has ot moved right. The whe (i, i + 1) are moved together, all poits betwee x i (t) ad x i+1 (t) have bee jumped over (or o). So the total distace traveled by the buies i the (t + 1)th step is at least x (t) x 1 (t) D, ad the total distace traveled by the buies is ubouded if the step is repeated, a cotradictio. Now if is eve, the take x 1 =... = x ad x +1 =... = x = x 1 + d, ad if is odd, take x 1 =... = x +1, x +3 =... = x = x 1 + d, ad carry out the algorithm described i 7

8 the lemma. Let d i (t), S(t) deote d i, S after t repetitios. As t, all positios approach x so d 1 (t),..., d (t) d, d +1 (t),..., d (t) d for eve ad d 1(t),..., d +1 (t) d 1, d +3 (t),..., d (t) d +1 for odd. We have equality i () always because the oly time a flag i would move right is if flag i + 1 moves left (1 i < ). Puttig d i = lim t d i (t) i the iequalities, they must still hold. Equality i (5) holds because d i is odecreasig i i. Fially the maximum i (8) is attaied because the stated equality coditio is satisfied. Thus ad this shows the boud is optimal. lim S(t) = t, eve 1, odd Remark: Usig the ideas of the proof, we ca show that the maximum S for a give iitial coditio ca be approached by ay sequece of steps where all buies coverge to the same spot, ad o buy ever jumps over aother buy. Coversely, if a buy jumps over aother buy at some time, the it is impossible to get arbitrarily close to the maximum. Proposed by Holde Lee (9) Problem 7: Cosider + 1 circular pacakes whose origial radii a 0, a 1, a,..., a obey the recursio a i+1 = 1 a i ad the largest pacake has uit origial radius (i.e. a 0 = 1). A pacake has the property where its radius a i icreases by a factor of 1 + r j whe compressed by a pacake of radius r j directly above. If all pacakes are stacked i the order of decreasig radii (i.e. a 0, a 1, a,..., a ), the what is the radius r 0 of the bottom pacake (the largest). Solutio: From this poit γ = 1 for simplificatio. Settig up the recursive expressios It follows from equatio (11) that r i = (1 + r i+1 ) a i (10) a i+1 = γa i (11) a i = γ i a 0 = γ i (1) Cosiderig the stack of pacakes, the radius r of the top pacake must be equal to its origial radius a. Therefore, by equatio (10) ad equatio (1), r 1 = (1 + r ) a 1 = (1 + γ ) γ 1 = γ 1 + γ 1 Proceedig with r ad r 3 to idetify a patter yields, r = γ + γ 3 + γ 3 3 8

9 r 3 = γ 3 + γ 5 + γ γ 6 By ispectio, the geeralized expressio is (for 0 i ), r i = r i = = i i r i = i γ (j+1) j k=0 (i k) γ (j+1)+ j k=0 k i j k=0 1 γ (j+1)+ 1 j(j+1) i(j+1) i γ ( i+ j )(j+1) At this poit, fidig the radius of the bottom pacake is straightforward by the use of equatio (13) r 0 = r = γ ( + )(j+1) j Or Gradig rubric: r 0 = = γ j (j+1) ( ) 1 1 j(j+1) For the correct expressio for a i as i equatio (11)- pts For a attempt to derive r i - pts For the correct expressio for r i as i equatio (13)- pts For the correct fial aswer- pts Proposed by Aki Morriso (13) 9