SMT 2013 Team Test Solutions February 2, 2013

Transcription

1 1 Let f 1 (n) be the numbe of divisos that n has, and define f k (n) = f 1 (f k 1 (n)) Compute the smallest intege k such that f k ( ) = Answe: 4 Solution: We know that = Theefoe, f 1 ( ) = = 19 53, so has 4 3 = 64 divisos f 1 (64) = 7, and f 1 (7) = This means that f 4 ( ) =, so k = 4 In unit squae ABCD, diagonals AC and BD intesect at E Let M be the midpoint of CD, with AM intesecting BD at F and BM intesecting AC at G Find the aea of quadilateal MF EG Answe: 1 1 Solution: Let (ABC) denote the aea of polygon ABC Note that AF B MF D with AB/MD =, so we have DF = 1 3 BD This implies that (MF D) = 1 3 (MBD) = 1 3 ( 1 (CBD)) = By symmety, (MGC) = 1 as well Theefoe, we have (MF EG) = (CED) (MBD) (MGC) = = Nine people ae pacticing the tiangle dance, which is a dance that equies a goup of thee people Duing each ound of pactice, the nine people split off into thee goups of thee people each, and each goup pactices independently Two ounds of pactice ae diffeent if thee exists some peson who does not dance with the same pai in both ounds How many diffeent ounds of pactice can take place? Answe: 80 Solution 1: Given a pemutation of nine people, let us have the fist thee people be in one goup, the second thee people in anothe goup, and the last thee people in a thid goup We want to compute how many pemutations geneate the same goup Note that thee ae (3!) 3 ways to pemute people within each goup, and thee ae 3! ways to pemute the oveall goups, 9! so the answe is (3!) 4 = 80 Solution : Note that if thee people ae doing this, thee is tivially exactly one unique iteation If six people ae doing this, then abitaily label one peson Thee ae ( 5 ) goups that can be ceated with this peson, and then the othe thee people ae foced to be in a goup, so thee ae ( 5 ) iteations fo six people If nine people ae doing this, then abitaily label one peson Thee ae ( 8 ) goups that can be ceated with this peson, and then the othe six people can fom goups in ( 5 ) ways, so thee ae ( ( 5 ) 8 ) = 80 iteations fo nine people 4 Fo some positive integes a and b, (x a + abx a ) b (x 3 + 3bx + 37) a = x x 41 + Find the odeed pai (a, b) Answe: (7, 3) Solution: The fist tem is just (x a ) b (x 3 ) a = x ab+3a, so ab + 3a = 4 Using the binomial theoem, the second tem is ( ( ) b 1 (x a ) b 1 (abx a 1 ))(x 3 ) a + ( ( a 1) (x 3 ) a 1 (3bx ))(x a ) b = (ab + 3ab)x ab+3a 1, so ab + 3ab = 16 Factoing these two equations gives a(b + 3) = 4 and ab(b + 3) = 16 Dividing the second equation by the fist gives b = 3 Then, substituting that into the fist equation gives a = 7

2 5 A polygonal pism is made fom a flexible mateial such that the two bases ae egula n -gons (n > 1) of the same size The pism is bent to join the two bases togethe without twisting, giving a figue with n faces The pism is then epeatedly twisted so that each edge of one base becomes aligned with each edge of the othe base exactly once Fo example, when n =, the untwisted and one-twist cases ae shown below; in both diagams, each edge of one base is to be aligned with the edge of the othe base with the same label a b b a c d d n =, untwisted c a b a d c d c n =, one twist b Fo an abitay n, what is the sum of the numbe of faces ove all of these configuations (including the non-twisted case)? Answe: (n + ) n 1 Solution: Thee ae n cases, which can be consideed based on thei divisibility by powes of Suppose we twist the pism by x of a full otation whee x = k y and y is odd Note that n unde this twist, each of the oiginal n faces is linked to evey x th face Since y is odd, we see that the fist multiple of x divisible by n is n k x Thus, each face in the new figue is made up of n k faces fom the oiginal (untwisted) fom and thee ae n = k sides fo this figue n k Next, we must conside how many otations have x of the fom k y fo a fixed value of k The numbe of integes less than o equal to n that ae divisible by k and not k+1 is n n k k+1 Fo k < n, this equals n k 1 Finally, we add in the untwisted case, with n faces Thus, the total numbe of sides is n 1 k n k 1 + n = n n 1 + n = (n + ) n 1 k=0 6 How many distinct sets of 5 distinct positive integes A satisfy the popety that fo any positive intege x 9, a subset of A sums to x? Answe: 4 Solution: Let a 1,, a 5 denote the 5 elements of A in inceasing ode Let S denote 5 i=1 a i Fist, note that 1, A because thee ae no othe ways to obtain 1 and Hence, we only have to think about the othe thee elements of A Note that thee ae 5 1 = 31 non-empty subsets of A, so at most two of those subsets ae not useful to the subset-sum constaint, eithe by having sum geate than 9 o being edundant with anothe subset We condition on the value of S This sum is clealy at least 9, and must be at most 31, since S, S 1, and S ae all achievable subset-sums, so we equie S 9 If S = 31, then A \ {1} has sum 30, so each subset has a distinct sum Theefoe, 4 A because the only othe way to get 4 would be 1 + 3, but 3 A would imply thee wee two diffeent ways to get 3, namely 1 + and 3 Similaly, since the subsets of {1,, 4} can sum to any positive intege less than 8, 8 A 16 A fo the same eason fo the set {1,, 4, 8}, and so A is completely detemined

3 If S = 30, we may have exactly one pai of subsets with the same sum Hence, we still get 4 A because 3 A would imply a 1 + a = a 3 and a 1 + a + a 5 = a 3 + a 5 Similaly, 8 A Finally, since we know all elements of A must sum to 30, we choose a 5 = 15 If S = 9, then we still have 4 A because thee will be moe than two edundant pais of subsets if 3 S In geneal, we cannot have x + y = z fo x, y, z A because thee would be too many edundant sets Hence, a 4 7 It can be at most 8, since othewise thee would be no way to achieve a sum of 8, so thee ae two cases fo a 4 Each choice of a 4 detemines a 5 by the condition on S We can veify that a 4 = 7, a 5 = 15 woks because {1,, 4} can geneate all sums 7, so {1,, 4, 7} can geneate all sums 14 Adding 15 clealy yields all sums 9 The othe case can be checked tivially Hence, in total thee ae 4 viable sets 7 Find all eal values of u such that the cuves y = x + u and y = x u intesect in exactly one point Answe: ( 1, 0) 1 4 Solution: Thee ae two possibilities: eithe the cuves y = x + u and x = y + u intesect in exactly one point, o they intesect in two points but one of the points occus on the banch y = x u Case 1: the two cuves ae symmetic about y = x, so they must touch that line at exactly one point and not coss it Theefoe, x = x + u, so x x + u = 0 This has exactly one solution if the disciminant, ( 1) + 4(1)(u) = 1 + 4u, equals 0, so u = 1 4 Case : y = x + u intesects the x-axis at ± u, while y = x u stats at x = u and goes up fom thee In ode fo these to intesect in exactly one point, we must have u < u, o u > u (note that u must be positive in ode fo any intesection points of y = x + u and x = y + u to occu outside the fist quadant) Hence we have u(u + 1) < 0, o u ( 1, 0) 8 Rational Man and Iational Man both buy new cas, and they decide to dive aound two acetacks fom time t = 0 to time t = Rational Man dives along the path paametized by x = cos(t) y = sin(t) and Iational Man dives along the path paametized by x = cos y = sin t t Find the lagest eal numbe d such that at any time t, the distance between Rational Man and Iational Man is not less than d Answe: Solution: We can un-paametize this equations easily to see that Rational Man is taveling along the cicle x + y = 1

4 with a peiod of π, while Iational Man is tavelling along the ellipse (x 1) 16 + y 4 = 1 with a peiod of π Now, we claim that d is equal to the smallest distance between a point on the given cicle and a point on the given ellipse This is because fo any numbe [0, 1), we can find a positive intege multiple of whose factional pat is abitaily close to, using a Pigeonhole agument Moe pecisely, fo any n N, we conside,,, n Now divide the egion between 0 and 1 into n equally-spaced intevals Fo a given [0, 1), find the inteval it falls into Eithe one of ou n multiples of falls into this inteval (and thus is at most 1 n fom ), o none of them do, in which case two numbes fall into the same inteval, and thus thei diffeence has factional pat of magnitude less than 1 n Now, it is clea that we can take a multiple of this numbe that is within 1 n of Thee is a slight complication if this numbe is negative, but we simply appoximate 1 instead of and then multiply by 1 to get a positive numbe with factional pat close to Applying this fact to ou poblem, we conside Rational Man s position at any time t This is the same as his position at time t + πn fo all n N Now, if Iational Man assumes some position at time t, then he also assumes it at time t + πm fo all m N By the fact poven above, we can always choose an m such that t + πm is abitaily close to t + πn fo some n N (divide though by π to make this cleae) Since the two dives can get abitaily close to any pai of points on thei espective paths, d must simply be the shotest distance between these two paths Now we make the obsevation that given a cicle of adius centeed at O and a point P outside this cicle, the shotest distance fom P to the cicle is along the line that passes though O This is evident by applying the Tiangle Inequality to tiangle OP Q, whee Q is any point on the cicle that is not on the line OP Hence, minimizing distance between the ellipse and the cicle is equivalent to minimizing distance between the ellipse and the cente of the cicle, ie the oigin Hence, we set out to minimize subject to the constaint so we ae minimizing (x 1) 16 x + y + y 4 = 1, x (x 1) = 1 4 (3x + x + 15) This attains its minimum value at x = 1 3, so the minimum squaed distance fom the oigin is We want one less than the distance to the oigin as ou final answe, so epot 3 9 Chales is playing a vaiant of Sudoku To each lattice point (x, y) whee 1 x, y < 100, he assigns an intege between 1 and 100, inclusive These integes satisfy the popety that in any ow whee y = k, the 99 values ae distinct and ae neve equal to k; similaly fo any column whee x = k Now, Chales andomly selects one of his lattice points with pobability

5 popotional to the intege value he assigned to it Compute the expected value of x + y fo the chosen point (x, y) Answe: Solution: We claim that when 100 is eplaced by n, the answe is n n 3n = 3n n+ 3n By symmety and lineaity of expectation, we need only compute the expected value of y, then multiply by two Fist, each i = 1,, n 1 is seen n times (once in each ow except fo ow i), while n is pesent in evey ow Hence, the sum of all values is n 1 n(n 1) + (n ) i=1 Meanwhile, the sum of values in ow i has weight Hence, the desied expectation is n 1 i=1 i = n(n 1) + 1 n(n 1)(n ) = n(n 1) (1 + 1 ) (n ) = 1 n (n 1) 1 n(n + 1) i n(n + 1) i = (n(n + 1) i) n i = n + 1 n 1 (n 1) n(n 1) Multiplying by two gives the final answe = n + 1 i=1 i n 1 3n = 3n n + 6n n 1 n (n 1) 10 A unit cicle is centeed at the oigin and a tangent line to the cicle is constucted in the fist quadant such that it makes an angle 5π/6 with the y axis A seies of cicles centeed on the x axis ae constucted such that each cicle is both tangent to the pevious cicle and the oiginal tangent line Find the total aea of the seies of cicles ( ) ( ) Answe: π(+ 3) 8 = π(7+4 3) 3 8 = π = π Solution: Let α = 5π/6 Fist, notice that because the tangent line has constant slope, the intesection point on evey cicle must occu at the same angle with espect to the cicle s cente Likewise, the line between the intesection points on two cicles must coincide with the tangent line Let cicle 1 have adius R and cente at (X, 0) and cicle have adius It follows that cicle has a cente at (X + R +, 0) Thus the two points of intesection with the tangent line ae (X + R cos α, R sin α) and (X + R + + cos α, sin α) The line between these must have slope cot α, so sin α R sin α X + R + + cos α (X + R cos α) = cot α = = R tan α/ i=1 i

6 Clealy, α < π/ so tan α/ < 1 Thus the adii obey a geometic seies: n=0 ( π[tan n α/] = cos 4 α ) sec α Plugging in α = 5π/6 gives π( + 3) What is the smallest positive intege with exactly 768 divisos? You answe may be witten in its pime factoization Answe: = Solution: Note that 768 = 8 3 We can immediately uppe bound the answe to It may be possible to incease exponents on small pimes and discad lage pimes to educe the answe Thee ae a few cases to conside (a) 5 is the lagest powe of 5 that divides the answe Theefoe, one of and 3 must contibute the facto of 3 to the numbe of divisos We have two subcases to conside at this point: i contibutes the facto of 3 We initially set 3 We can destoy 19 and 3 by using 5 and 3 3 ii 3 contibutes the facto of 3 We must use 3, and theefoe the powe of should be 7, destoying 19 and 3 also (b) 5 is the lagest powe of 5 that divides the answe We must theefoe use at least 3 and 3 3 Note that the vey fist subcase geneates the smallest poduct, so the answe is theefoe Suppose Robin and Eddy walk along a cicula path with adius in the same diection Robin makes a evolution aound the cicula path evey 3 minutes and Eddy makes a evolution evey minute Jack stands still at a distance R > fom the cente of the cicula path At time t = 0, Robin and Eddy ae at the same point on the path, and Jack, Robin, Eddy, and the cente of the path ae collinea When is the next time the thee people (but not necessaily the cente of the path) ae collinea? Answe: t = 3 π accos ( Solution: Define ) + +8R 4R ω 1 = π 3 ω = π Let (x 1 (t), y 1 (t)) be the location of Robin and (x (t), y (t)) be the location of Eddy at time t Let the cente of the path be the oigin and Jack s location be (R, 0) Then we have x 1 = cos(ω 1 t) y 1 = sin(ω 1 t) x = cos(ω t) y = sin(ω t)

7 The thee people ae collinea if and only if the slopes of the lines connecting any two people ae the same, ie y 1 x 1 R = y x R Coss multiplication and factoing gives us Plugging in gives us x 1 y x y 1 = R(y y 1 ) [cos(ω 1 t) sin(ω t) sin(ω 1 t) cos(ω t)] = R [sin(ω t) sin(ω 1 t)] This comes out to R sin(ω t ω 1 t) = sin(ω t) sin(ω 1 t) Replacing 3ω 1 = ω (fom the values given in the poblem), we get Using sum-to-poduct identities, we get Expanding the left side gives us R sin(ω 1t) = sin(3ω 1 t) sin(ω 1 t) R sin(ω 1t) = sin(ω 1 t) cos(ω 1 t) R sin(ω 1t) cos(ω 1 t) = sin(ω 1 t) cos(ω 1 t) So eithe the fist time the thee people ae collinea is when t = π o when Double angle identity of cosine gives us R cos(ω 1t) = cos(ω 1 t) R cos(ω 1t) = cos (ω 1 t) 1 This is a quadatic in cosine Multiplying both sides by R and solving the quadatic gives us cos(ω 1 t) = ± + 8R 4R The positive oot is the only collinea time that occus when Robin is still in the fist quadant Theefoe, it is the ealiest time < R implies cos(ω 1 t) = + + 8R 4R < 1, so it is in the ange of the cosine function Hence, the answe is ( t = 3 π accos + ) + 8R 4R

8 13 A boad has, 4, and 6 witten on it A peson epeatedly selects (not necessaily distinct) values fo x, y, and z fom the boad, and wites down xyz + xy + yz + zx + x + y + z if and only if that numbe is not yet on the boad and is also less than o equal to 013 This peson epeats this pocess until no moe numbes can be witten How many numbes will be witten at the end of this pocess? Answe: Solution: We claim that N can be witten on the boad if and only if N + 1 has a pime factoization of the fom 3 a 5 b 7 c, whee a + b + c is odd It emains to actually pove this Note that if we wite N = xyz + xy + yz + zx + x + y + z, then we have that N + 1 = (x + 1)(y + 1)(z + 1) Note that the oiginal numbes,, 4, and 6, ae each less than the pimes 3, 5, and 7, espectively Theefoe, we ensue that the only pimes which can divide any valid N + 1 ae 3, 5, and 7 Futhemoe, these numbes each have exponents summing to 1, an odd intege, so theefoe since we multiply thee integes with an odd sum of exponents, we ensue that all numbes which emain have an odd sum of exponents It emains to compute all numbes of the fom 3 a 5 b 7 c, whee each numbe is less than o equal to 013 and the sum of the exponents is odd Thee ae such numbes 14 You have a mete long sting You choose a point along the sting unifomly at andom and make a cut You discad the shote section If you still have 05 metes o moe of sting, you epeat You stop once you have less than 05 metes of sting On aveage, how many cuts will you make befoe stopping? Answe: 8 4 log Solution: Let f(x) be the aveage numbe of cuts you make if you stat with x metes of sting Fo x [0, 1 ), we have f(x) = 0 To calculate f(x) fo x 1, say you make 1 cut that bings the length of the sting to y Then you aveage a total of 1 + f(y) cuts Note that y is distibuted unifomly at andom fom x to x So we aveage 1 + f(y) ove y [ x, x], which gives us f(x) = x x x/ f(y) dy + 1 Now we have an initial condition and ecuence It can be easily veified that 0, if x [0, 1 ) f(x) = 4x 1, if x [ 1, 1) 5x x log x, if x [1, ] satisfies the initial condition and ecuence Theefoe the answe is f() = 8 4 log (Note that technically we should also pove that the above f(x) is the unique solution to the ecuence We could just look at the ecuence and see that it is completely obvious that it defines a unique function If you would like a igoous poof, ead below whee we find the solution by solving two ODEs Sufficiently nice ODEs have unique solutions so the solution is unique) How did we discove this solution to the ecuence? Fist, let F (x) = x 0 f(t) dt In tems of F, the ecuence is F (x) = (F (x) F (x/)) + 1 x

9 Fo x [ 1, 1), the initial condition tells us that F (x/) = 0, so the ecuence simplifies to F (x) = F (x) + 1 x Also notice that we have an initial condition F ( 1 ) = 0 Since multiplying by x is the same as diffeentiation fo x, we might guess that a degee polynomial solves this diffeential equation If we plug in a geneal degee polynomial, we find that we ae coect and that the solution is F (x) = x x fo x [ 1, 1) Now that we know F (x) on x [ 1, 1), we can plug that into ou oiginal ecuence fo F to get the following diffeential equation, valid fo x [1, ]: F (x) = F (x) x + x This time we could ty anothe degee polynomial, but it won t wok Specifically, if we ty out F (x) = ax + bx we get ax + b = (a 1)x + b + Thee is no choice of a, b that satisfies this We need to somehow get a tem involving x on one side without getting it on the othe side in ode to balance the x s on each side Notice that x log x will give us an x when we diffeentiate but not when we multiply by x So that might wok And indeed it does We can plug in F (x) = ax + bx + cx log x, solve fo the coefficients (keeping in mind the initial condition F (1) = 1 that we get fom ou pevious expession fo F ), and get F (x) = 3x x log x x Now we know F (x) on all of [0, ], so we can diffeentiate it to get f(x) The esult is exactly the expession fo f(x) that we have above 15 Suppose we climb a mountain that is a cone with adius 100 and height 4 We stat at the bottom of the mountain (on the peimete of the base of the cone), and ou destination is the opposite side of the mountain, halfway up (height z = ) Ou climbing speed stats at v 0 = but gets slowe at a ate invesely popotional to the distance to the mountain top (so at height z the speed v is (h z)v 0 /h) Find the minimum time needed to get to the destination Answe: 504 log + 500π Solution 1: Fo ease of notation, let 0 = 100 and h = 4 Begin by flattening the cone into a secto of a cicle with adius R = 0 + h The poblem then is equivalent to finding the optimal path fom the pola point (, θ) = (R, 0) to the point ( R, 0 R π) on the flattened cone We can find an optimal path by constucting a new distance metic that measues elapsed time by consideing standad Euclidean distance along with a facto that accounts fo velocity Obseve that any point on the secto with adius (distance along the cone s suface to the cente) and height (on the cone) z satisfies h z h = R by simila tiangles Theefoe, the speed at adius on the secto is R v 0

10 Let the optimal path cuve be given by γ(θ) = ((θ), θ) We wish to optimize the integal that gives the total time spent along the cuve γ We can measue length by the standad pola aclength fomula, and we can measue speed using the fomula above Hence, we can measue time by looking at distance divided by speed: 0R π θ=0 0R π distance d speed = + dθ 0 R v 0 = R 0 R π 1 + v 0 0 ( 1 d ) dθ dθ We now wish to find a coodinate tansfomation in which this path is a staight line, so that the minimum time will just be the Euclidean distance between the endpoints We can do this by choosing a new coodinate so that 1 d dθ = d dθ By integating, it is easily seen that one such substitution is log =, which esults in the endpoints ( (log R, 0) and log R, ) 0 R π, so ou integal is ( log R log R ) + 0 R π = (log ) + 0 R π Multiplying by the constant tems we factoed out of this integal ealie, ou final answe (and minimum time) is R v0 (log ) + 0 v0 π = 504(log ) + 500π Solution : Rathe than thinking of the velocity as deceasing as we climb up the mountain, we can imagine the mountain is gowing lage as we climb up it In fact, no matte how fa you ve climbed up the mountain, it will appea as you have made no pogess Thus, the mountain can be thought of as an infinitely tall cylinde As a sanity check, note that at height z, the cicumfeence of the cone will be h z h z h π, and ou velocity is h v 0, so it will take π v 0 units of time to go aound the cicumfeence This is independent of ou height z, confiming ou intuition that the mountain should be teated as a cylinde Ou poblem then becomes detemining whee ou oiginal destination on the cone coesponds to on this cylinde As checked above, the adius of the cylinde is the same as the adius of the base of the cone So all that s left to detemine is the height on the cylinde that coesponds to a height of z = on the cone We do so by evaluating the integal s s s xdx whee s denotes the length of the slant of the cone To justify this integal, we note that when ou distance to the top of the mountain is s s q, ou speed is q s times as long to tavel This integal evaluates to s log take s s q of ou oiginal speed, so a small distance dx will Now, tavelling on a cylinde is the same as tavelling on the plane, so we uncul the cylinde If we denote ou initial position as (0, 0), then ou destination has coodinates (π, s log ) The distance to ou destination can now be calculated by Pythagoas: π + s (log ) The time

11 π +s (log ) then to get to ou destination is v 0 We ae given that = 100 and v 0 = The slant is easily calculated s = + h = Plugging these values in, we see that the time equied to get to ou destination is 504(log ) + 500π