3rd Bay Area Mathematical Olympiad

Transcription

1 2001 3rd Bay Area Mathematcal Olympad February 27, 2001 Problems and Solutons 1 Each vertex of a regular 17-gon s colored red, blue, or green n such a way that no two adacent vertces have the same color. Call a trangle multcolored f ts vertces are colored red, blue, and green, n some order. Prove that the 17-gon can be cut along nonntersectng dagonals to form at least two multcolored trangles. (A dagonal of a polygon s a a lne segment connectng two nonadacent vertces. Dagonals are called nonntersectng f each par of them ether ntersect n a vertex or do not ntersect at all.) Soluton: (One of many smlar solutons.) Denote the colors by 1, 2, 3. Notce that all three colors must be used. Ths s true because f only two colors were used, the vertex colorng would be of the type whch s mpossble, snce 17 s odd (ths would make two adacent vertces the same color). Hence there are three consecutve vertces colored (wthout loss of generalty) 123, respectvely. (For otherwse, f 3 consectve vertces had coloraton of the form aba, the next three vertces [overlappng two vertces] would have to be colored bab, etc., forcng the pattern ababab...whch s not possble wth an odd number of vertces. Thus we have 4 cases dependng on the colors of the vertces adacent to ths segment: 21231, 21232, 31231, It s easy to see that the desred constructon can be done n each case; the fgure below llustrates ths

2 Remark: You may wsh to try a more general problem: show that there exsts a trangulaton of the 17-gon for whch all trangles are multcolored. (A trangulaton s a dssecton of an entre polygon nto trangles, where each trangle s formed from vertces and dagonals (or sdes) of the orgnal polygon, and no trangles overlap [besdes sharng sdes or vertces]. There are many ways to trangulate a gven n-gon, but all of them use (n 2) trangles.) 2 Let JHIZ be a rectangle, and let A and C be ponts on sdes ZI and ZJ, respectvely. The perpendcular from A to CH ntersects lne HI n X, and the perpendcular from C to AH ntersects lne HJ n Y. Prove that X, Y and Z are collnear (le on the same lne). Soluton: Observe that XAI = XHC = HCJ. Hence XAI HCJ, and thus XI/HJ = AI/CJ. Lkewse, YJ/HI = CJ/AI. Puttng these together yelds XI/HJ = HI/YJ, and hence XI ZI = ZJ XZI ZY J. YJ Snce JZI = 90, ths mmedately mples YZX = 180, and X, Y, Z are collnear. X Z A I Y C J B H Remark: It turns out that ths problem s equvalent to the Poncelet Branchon Theorem: Let A, B and C be three ponts on a rectangular hyperbola (a hyperbola wth perpendcular asymptotes.) Then the orthocenter of ABC also les on the hyperbola. D E A B H C

3 Proof : In the proectve plane, let D and E be the two ponts of ntersecton of the lne at nfnty l wth the two asymptotes of the hyperbola. Note that snce s a conc, D and E are also the ponts of ntersecton of wth l. We apply : Converse of Pascal s theorem: If the three pars of opposte sdes n a hexagon ntersect n collnear ponts, then the hexagon s nscrbed n a conc. Note that through any 5 ponts n general poston (no 3 of them collnear), there passes a unque conc. In our stuaton, the ponts A, B, C, D and E le on a hyperbola, (and hence they are n general poston). To show that the orthocenter H of ABC also les on ths hyperbola, t suffces to verfy that the ponts X, Y and Z are collnear, where X = AB HD, Y = BC HE, and Z = AE CD. Thus, the problem s equvalent to Let J H I Z be a rectangle, and let A and C be ponts on sdes Z I and Z J, respectvely. The perpendcular from A to C H ntersects lne H I n X, and the perpendcular from C to AH ntersects lne H J n Y. Then X, Y and Z are collnear. But ths was our problem #2! 3 Let f (n) be a functon satsfyng the followng three condtons for all postve ntegers n: (a) f (n) s a postve nteger, (b) f (n + 1) > f (n), (c) f ( f (n)) = 3n. Fnd f (2001). We wll present two solutons. The frst one was presented by the problem proposer, and was the method used essentally by all the students but one. The second soluton, whch won the BAMO 2001 Brllancy Prze, s due to Andrew Dudzk of Lynbrook HS. Andrew s soluton, (whch we only sketch) s not shorter, nor does t use any ngenous mathematcal trcks. But t s superor to the other solutons, because t llumnates the nature of f (n) n a very smple way. Soluton 1: We wll show that f (2001) must equal We start by provng a lemma whch gves us some of the values of f (n). Lemma: For n = 0, 1, 2,..., (a) f (3 n ) = 2 3 n ; and (b) f (2 3 n ) = 3 n+1. Proof : We use nducton. For n = 0, note that f (1) 1, otherwse 3 = f ( f (1)) = f (1) = 1, whch s mpossble. Snce f (k) s a postve nteger for all postve ntegers k, we conclude that f (1) >1. Snce f (n + 1) > f (n), f s ncreasng. Thus 1 < f (1) < f ( f (1)) = 3or f (1) = 2. Hence f (2) = f ( f (1)) = 3. Suppose that for some postve nteger n 1, f ( 3 n) = 2 3 n and f ( 2 3 n) = 3 n+1. Then, and f (3 n+1 ) = f ( f (2 3 n ) ) = 2 3 n+1, f (2 3 n+1 ) = f ( f (3 n+1 ) ) = 3 n+2,

4 as desred. Ths completes the nducton, and establshes the lemma. Contnung wth our soluton, there are 3 n 1 ntegers m such that 3 n < m < 2 3 n and there are 3 n 1 ntegers m such that f ( 3 n) = 2 3 n < m < 3 n+1 = f ( 2 3 n). Snce f s an ncreasng functon, f ( 3 n + m ) = 2 3 n + m, for 0 m 3 n. Therefore f ( 2 3 n + m ) = f ( f ( 3 n + m )) = 3 ( 3 n + m ) for 0 m 3 n. Hence f (2001) = f ( ) = 3 ( ) = Soluton 2: (Sketch) Andrew Dudzk s nsght was to recognze that f (n) deals n a very smple way, wth the base-3 representaton of n. Let n = a 1 a n a t be the base-3 dgts of n. For example, f n = 50 n base 10, we would wrte n = 1212 n base 3 and hence a 1 = 1, a 2 = 2, a 3 = 1, a 4 = 2. Dudzk proved the followng: 1. If a 1 = 1, then f (n) = 2a 2 a 3 a t. 2. If a 1 = 2, then f (n) = 1a 2 a 3 a t 0 These two statements can be proven easly wth nducton; we leave ths an an exercse for the reader. Note that Dudzk s formulas allow us to mmedately compute f (2001). In base-3 notaton, 2001 s equal to Thus, by formula #2, f ( ) = = , and ths equals 3816 n base-10 notaton. 4 A kngdom conssts of 12 ctes located on a one-way crcular road. A magcan comes on the 13th of every month to cast spells. He starts at the cty whch was the 5th down the road from the one that he started at durng the last month (for example, f the ctes are numbered 1 12 clockwse, and the drecton of travel s clockwse, and he started at cty #9 last month, he wll start at cty #2 ths month). At each cty that he vsts, the magcan casts a spell f the cty s not already under the spell, and then moves on to the next cty. If he arrves at a cty whch s already under the spell, then he removes the spell from ths cty, and leaves the kngdom untl the next month. Last Thanksgvng the captal cty was free of the spell. Prove that t wll be free of the spell ths Thanksgvng as well. Soluton 1 (sketch): Number the ctes from 0 to 11. Encode the current state of affars by startng at cty 11, wrtng 1 f the cty s not under a spell, and a 0 f t s under a spell. Do the same thng for each cty, from cty 10 down to cty 0. Interpret the resultng lst as the bnary expanson of an nteger X between 0 and m = Some thought (usng modular arthmetc modulo m, and the way that addton works n base 2) shows that the rules can be nterpreted as sayng that f the magcan starts at cty k then 2 k s added to Y, where Y encodes the state as above. After 12 months the magcan starts once at all ctes, and ths has the net effect of addng = = m to the orgnal X whch obvously leaves the state unchanged.

5 Soluton 2: We shall denote the state of the kngdom by a 12-tuple of 0s and 1s, wth 0 and 1 meanng free of spell and under spell, respectvely. For example, suppose that the ntal state s (0, 1, 0, 0, 0, 0, 1, 0, 0, 0, 0, 0). Ths means that cty #1 s free of the spell, cty #2 s under the spell, cty #3 s free, etc. Next, defne S, for = 1, 2,...,12 to be the transformaton of the kngdom after the magcan makes a vst that starts at cty. For example, S 1 (0, 1, 0, 0, 0, 0, 1, 0, 0, 0, 0, 0) = (1, 0, 0, 0, 0, 0, 1, 0, 0, 0, 0, 0), snce she puts a spell on cty #1, then comes to cty #2 whch s already under the spell, frees t, and stops. Lkewse, S 8 (0, 1, 0, 0, 0, 0, 1, 0, 0, 0, 0, 0) = (1, 0, 0, 0, 0, 0, 1, 1, 1, 1, 1, 1), because she puts ctes #8 12 under spell, then contnues around the crcle to #1, puts t under spell, then fnally stops after freeng cty #2. One more example: S 3 (0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0) = (1, 1, 0, 1, 1, 1, 1, 1, 1, 1, 1, 1), because she puts every cty under the spell, startng wth #3, untl she comes to a cty that s already under the spell, and ths happens when she comes to cty #3 for the second tme. The key to ths problem s the remarkable fact that the S transformatons commute;.e. S S = S S for all, {1, 2, 3,...,12}. Let us assume ths fact, to see how t quckly solves the problem. Wthout loss of generalty, suppose the magcan starts a vst at cty #1. Then durng the course of the year, her subsequent vsts wll start at the followng ctes, n ths order: 6, 11, 4, 9, 2, 7, 12, 5, 10, 3, 8. In other words, durng the year she wll make 12 vsts, and each vst wll start at a dfferent cty. 1 In other words, the kngdom wll be transformed, n order, by S 1, S 6, S 11,... But snce these transformatons are commutatve, we can rearrange the order that we perform the transformatons. So, durng the course of the year, the kngdom wll be transformed by S 1, S 2, S 3,...,S 12, where we may choose the order n any way that we lke. Consder the state of the kngdom last Thanksgvng. Suppose ctes a, b, c,...are spell-free (where cty a s the captal) and ctes A, B,...are under the spell. Now we shall perform the transformatons, n ths order: S A, S B,...: Each of these are sngle-cty transformatons that merely free the startng cty and then stop. In other words, after these transformatons have been performed, all the ctes whch were under the spell are now free, and hence at ths pont, all ctes n the kngdom are free. (It may be that no ctes were under the spell n the frst place, so ths step may not take place.) S a : Startng wth a, all ctes get put under the spell (snce they were prevously all free), untl the magcan returns to a, frees t, and stops. Now all ctes are under the spell, except for the captal cty a. The remanng transformatons S b, S c...: Snce all ctes except a are under the spell, each of these are agan sngle-cty transformatons whch end up freeng ctes b, c,... Thus, after the 12 vsts, cty a s stll free. In fact, we have shown more, namely that the state of each of the 12 ctes of the kngdom wll repeat every 12 months. For example, f cty #5 was free last Thanksgvng, t wll be free ths Thanksgvng; f cty #7 was under the spell last Thanksgvng, t wll also be under the spell ths Thanksgvng, etc. 1 The sophstcated reader wll observe that ths s a consequence of the fact that 5 s relatvely prme to 12.

6 It remans to prove the key fact that the S transformatons are commutatve. Specfcally, we wll show that S (S (x 1, x 2,...,x 12 )) = S (S (x 1, x 2,...,x 12 )) for any dstnct, and for any 12-tuple (x 1, x 2,...,x 12 ). We can do ths by examnng several cases whch depend on the 12-tuple. 1. All the coordnates are zero. In other words, all towns are currently free of the spell. It s easy to check that performng S on ths wll put make all x k = 1 except x = 0. Then performng S wll change x to 0. The net result s that all values are 1 except for x = x = 0. Clearly, we wll have the same result f we perform S frst, and then S. 2. Exactly one of the coordnates s 1. (a) Suppose x = 1 and all others are zero. Performng S makes x = 0 and hence now all values are zero. Then S results n all values equallng 1 except x = 0. Conversely, performng S frst wll put all ctes from to 1 under the spell (travellng clockwse) untl the magcan reaches cty whch she then frees of the spell and stops. So now there s an arc of 1 s from x to x 1 and all other values are 0. Performng S on ths turns the values of the arc from to 1 nto 1 s, untl she reaches cty, and then she frees ths cty and stops. The net result agan s that all values are 1 except x = 0. The fgure below llustrates ths. We assume clockwse travel and denote free/spell by whte/black, respectvely. Start After S After S Start After S After S (b) Suppose x = 1 and all others are zero. Ths case s exactly the same as 2(a) above; ust nterchange and. (c) Suppose that x k = 1 and all others are zero, wth k and k. The fgure below llustrates the case where k les on the clockwse arc between and. The case where k les between and has a smlar pcture.

7 k k k Start After S After S k k k Start After S After S 3. At least two coordnates equal 1. (a) Suppose x k = x l = 1, where k les on the clockwse arc from to and l les on the arc from to (t s possble that k or l may equal one or both of and.) Then the order of S and S wll not matter each transformaton alters non-overlappng arcs: S only affects the arc from to k, whle S affects the arc from to l. (b) One of the arcs descrbed n 3(a) contans only zeros. Wthout loss of generalty, suppose that the arc from to s all zeros (ncludng, ) and let k be the frst coordnate clockwse from such that x k = 1. The pcture below handles ths case. The ctes between k and are not shaded, but they could be black or whte; ther state s not relevant, snce they wll be unaffected by S or S. Start After S After S k k k Start After S After S k k k These cases encompass all possbltes, so we conclude that S and S commute for all 12-tuples and the proof s complete.

8 Remark: There s a thrd soluton method, whch examnes the number of tmes each cty s state s changed durng the course of the year. 5 For each postve nteger n, let a n be the number of permutatons τ of {1, 2,...,n} such that τ(τ(τ(x))) = x for x = 1, 2,...,n. The frst few values are a 1 = 1, a 2 = 1, a 3 = 3, a 4 = 9. Prove that dvdes a (A permutaton of {1, 2,...,n} s a rearrangement of the numbers {1, 2,...,n}, or equvalently, a one-to-one and onto functon from {1, 2,...,n} to {1, 2,...,n}. For example, one permutaton of {1, 2, 3} s the rearrangement {2, 1, 3}, whch s equvalent to the functon σ : {1, 2, 3} {1, 2, 3} defned by σ(1) = 2, σ(2) = 1, σ(3) = 3.) Soluton: Consder the permutatons τ of {1, 2,...,n} such that τ(τ(τ(x))) = x for x = 1, 2,...,n. Then for each x {1, 2,...,n}, there are only two possbltes. Ether x s a a fxed pont;.e., τ(x) = x, or else x s a member of a 3-cycle (xyz);.e. τ(x) = y,τ(y) = z,τ(z) = x, where x, y, z are 3 dstnct numbers. We can thus partton these permutatons nto two cases: Ether 1 s a fxed pont, or t s not. In the frst case, the remanng elements {2,...,n} can permuted n a n 1 ways. In the second case, 1 s part of a 3-cycle (1), where and, {2,...,n}. There are (n 1)(n 2) such 3-cycles (order counts). Then the remanng n 3 elements can be permuted n a n 3 ways. Hence we have (for n > 3) Defne b n to be the hghest power of 3 whch dvdes a n ;.e. a n = a n 1 + (n 1)(n 2)a n 3. (1) 3 b n a n. If n 0 (mod 3), then (n 1)(n 2) wll be a multple of 3, so equaton (1) yelds a n = a n 1 + 3ka n 3, where k s a postve nteger. Hence b n mn(b n 1, b n 3 + 1), n 0 (mod 3). (2) If n 0 (mod 3), then (n 1)(n 2) 1 (mod 3), but both (n 2)(n 3) and (n 3)(n 4) are multples of 3. Pluggng nto (1), we have a n a n 1 = (n 1)(n 2)a n 3 = (3u 1)a n 3, a n 1 a n 2 = (n 2)(n 3)a n 4 = 3va n 4, a n 2 a n 3 = (n 3)(n 4)a n 5 = 3wa n 5, for postve ntegers u,v,w. Addng these, we get a n a n 3 = (3u 1)a n 3 + 3va n 4 + 3wa n 5, so a n = 3(ua n 3 + va n 4 + wa n 5 ).

9 Ths mples that b n 1 + mn(b n 3, b n 4, b n 5 ), n 0 (mod 3), n > 5. (3) We know that b 1 = 0, b 2 = 0, b 3 = 1. Employng (2), we get b 4 1, b 5 1. Then (3) yelds b 6 1. Applyng (2) agan yelds b 7 1, b 8 1. But now when (3) s appled, we have b 9 2. It s evdent that contnung ths process, (2) and (3) wll ncrement b n by at least one as n ncreases by 6. In other words, Snce 2001 = , we are done. b 6k+3 k + 1. Remark: The above nequaltes are not the best. Drect computaton usng the recurrence (1) shows that a 2001 s a 3,830-dgt number and that a 2001.