2011 Problems with Solutions

Transcription

1 1. Argumet duplicatio The Uiversity of Wester Australia SCHOOL OF MATHEMATICS AND STATISTICS BLAKERS MATHEMATICS COMPETITION 2011 Problems with Solutios Determie all real polyomials P (x) such that P (2x) = P (x) P (x) for all x R. Solutio. [By Aaro Mayard ad Phillip Meg, both 3rd year, UWA] Let P (x) be a real polyomial such that P (2x) = P (x) P (x) for all x R. The zero polyomial is oe such polyomial P (x). Suppose that P (x) is ot the zero polyomial ad let be the degree of P (x). The the degree of P (2x) is. If is equal to 0 or 1, the P (x) P (x) will be the zero polyomial, ad so will ot be equal to P (2x). If 2, the the degree of P (x) P (x) will be ( 1) + ( 2) = 2 3. Therefore, P (2x) = P (x) P (x) implies that = 2 3 ad hece = 3. So, P (x) is a cubic polyomial of the form a + bx + cx 2 + dx 3, with d 0. Therefore, ad so which implies that P (2x) = a + 2bx + 4cx 2 + 8dx 3 P (x) P (x) = (b + 2cx + 3dx 2 )(2c + 6dx) = 2bc + (4c 2 + 6bd)x + 18cdx d 2 x 3 a + 2bx + 4cx 2 + 8dx 3 = 2bc + (4c 2 + 6bd)x + 18cdx d 2 x 3 a = 2bc 2b = 4c 2 + 6bd 4c = 18cd 8d = 18d 2. Solvig this set of simultaeous equatios, first we fid d = 4 9 (recall d 0), from which we see that successively c = 0, b = 0 ad a = 0, givig us the solutio P (x) = 4 9 x3. Therefore, the oly real polyomials P (x) such that P (2x) = P (x) P (x) for all x R are P (x) = 0 ad P (x) = 4 9 x3. 2. So tyred A car has 4 tyres, ad i its boot are stored 3 spare tyres km. Each tyre ca be used for What is the maximum distace that the car ca be drive? (The tyres ca be iterchaged as may times as you wat.)

2 Solutio. [By Aaro Mayard ad Phillip Meg, both 3rd year, UWA] There are 7 tyres, each of which ca travel km, givig a total of tyre km. Sice 4 tyres eed to be attached to the car for it to travel, the maximum distace the car ca travel is at most = km. The table below shows oe way the car may be drive for km: Distace car has travelled First km Next km Next km Next km Next km Tyre 1 Tyre 2 Tyre 3 Tyre 4 Tyre 5 Tyre 6 Tyre 7 Thus we have show that the maximum distace that the car ca travel is bouded above by km, ad that km is achievable. Therefore, the maximum distace that the car ca travel is km. 3. Iequality Prove that (!) 2 > for all 3. Solutio. [By Tuo Li, 2d year, UWA] We ca orgaise the factors of (!) 2 i the followig way: 1 (!) 2 = ( 1) (( 1) 2) (( 2) 3) (1 ) = ( i)(i + 1). Hece the iequality (!) 2 > ca be rewritte as Moreover we have equality i = 0 or i = 1. Therefore we have 1 ( i)(i + 1) > 1. (1) ( i)(i + 1) For each iteger i [0, 1], 1, sice ( i)(i + 1) i( i 1) = 1 + ad i, i 1 0. ( i)(i + 1) 1 ( i)(i + 1) = 1 exactly whe i( i 1) = 0, that is, whe = 1. Sice 3, the product has at least 3 factors; so at least oe of the factors (say for i = 1) is strictly greater tha 1. So, for 3 the iequality (1) is strict. Hece, (!) 2 > for all 3. 2

3 4. Black or white cube A cube is assembled from cubes all of whose faces are white. We pait all of the faces of the large cube black, ad the disassemble it. A blidfolded ma reassembles the large cube from the 27 little cubes. What is the probability that all the faces of the reassembled cube are completely black? Solutio. [By Aaro Mayard, 3rd year, UWA] Of the 27 little cubes, oe is completely white, six are black o oe face, twelve are black o two (edge-sharig) faces ad eight are black o three (corer-sharig) faces. Whe the large cube is reassembled, each of the little cubes has 27 possible positios ad 24 possible orietatios. Therefore, the total umber of possible large cubes is 27! Suppose that all of the faces of the reassembled large cube are completely black. There is 1! possible way to positio the white cube, 6! possible ways to positio the cubes with oe black face, 12! for the cubes with two black faces, ad 8! for the cubes with three black faces. There are also 24 ways to oriet the white cube, 4 ways to oriet each of the cubes with oe black face, 2 ways to oriet each of the cubes with two black faces, ad 3 ways to oriet each of the cubes with three black faces. Therefore, the total umber of possible large, completely black cubes is 1! 6! 12! 8! Therefore, the probability that the reassembled cube is completely black is 6! 12! 8! ! 12! 8! 27! = 27! = This is approximately ; i other words, a very small probability. 5. Touchig circles Let circles K 1 ad K 2 be touchig at poit P, with the smaller circle K 1 iside K 2. Let lie l be taget to K 1 at A ad itersect K 2 at poits B ad C. Show that P A is the (iterior) agle bisector of BP C. Solutio. [By Tuo Li, 2d year, UWA] We make extesive use of the followig stadard theorem. Theorem 1 (Alterate Segmet Theorem). If P Q is a chord of a circle K, X is a poit exteral to K such that XP is the taget lie to K through P, ad A is aother poit o K that is o the opposite side of P Q to X, the XP Q = P AQ. A Proof. Let O be the cetre of K. The P AQ = 1 2 P OQ, agles at circumferece ad cetre stadig o same arc P Q OP Q + XP Q = 90 ad OP Q = OQP 180 = OP Q + OQP + P OQ = 2 OP Q + P OQ 90 = OP Q P OQ P XP Q = 1 2 P OQ = P AQ 3 O X Q

4 Now let us cosider the give problem. Let l itersect the taget lie to K 1 ad K 2 (through P ) at X. Without loss of geerality, take C to be farther tha B from X. Let BP itersect K 1 at D. Let P C itersect K 1 at E. By Theorem 1, DP A = BAD, (chord AD, taget BA, circle K 1 ) (2) EP A = CAE, (chord AE, taget CA, circle K 1 ) (3) DAP = XP B, (chord P D, taget XP, circle K 1 ) (4) BCP = XP B, (chord P B, taget XP, circle K 2 ) (5) AEP = BAP, (chord AP, taget BA, circle K 1 ) (6) So we have BAD + DAP = BAP = AEP, by (6) = CAE + ACE, AEP is exterior to EAC BAD = CAE, sice DAP = BCP, by (4) ad (5), ad BCP = ACE (same agle) DP A = EP A by (2) ad (3) P A is the agle bisector of DP E = BP C. P X B D A K 2 K 1 l E C 6. Itegral boxes A vector ṽ = (x, y, z) R 3 is itegral if each compoet is a iteger. Prove that if ũ, ṽ ad w are mutually orthogoal itegral vectors with the same legth L, the L is a iteger. Solutio. [By Mitchell Misich, 3rd year, UWA] Give three mutually orthogoal itegral vectors ũ, ṽ, with the same legth L 0, they defie a parallelepiped that is actually a cube. (Note w that we ca igore the trivial case L = 0, sice 0 Z, ad so there is othig to prove i this case.) The volume of the cube defied by ũ, ṽ, is L w 3 = ũ (ṽ ) ad so L w 3 is a iteger. Furthermore, L 2 = ũ ũ ad so L 2 is a iteger. Therefore, L = L3 L 2 is ratioal, ad sice L2 is a iteger, L must also be a iteger. Theorem. If N Q ad N 2 Z the N Z. Proof. Sice N Q, N = a for some a, b Z with b 0 ad a, b coprime, i.e. gcd(a, b) = 1. b Sice N 2 Z, we have b 2 divides a 2 ad hece for ay prime divisor p of b, we have p 2 a 2 = p a 2 = a a = p a or p a, by Euclid s Lemma = p a. Thus ay prime divisor p of b is also a divisor of a, so that gcd(a, b) p. So we have a cotradictio, uless b has o prime divisors. Therefore, b = 1 ad N Z. 4

5 7. Coutig digits From a positive iteger (i decimal form), we form aother oe α() as follows: write the umber of eve digits of the the umber of odd digits of, the the total umber of digits of. For istace α( ) = ad α(7777) = 044 = 44. Is there a umber k such that, for ay, α i () = k, for i sufficietly large? Solutio. [Ispired by Karl Beidatsch, 3rd year, Curti] The aswer is: Yes, with umber k = 123 beig always ultimately reached. Call a sequece,, α(), α 2 (),... a α-sequece, ad for brevity let us write it i the form α α() α α 2 () α. We first show that if has 1, 2 or 3 digits, the α i () = 123 for i sufficietly large. (Note that wheever a α-sequece reaches a umber that has bee see previously, we ca stop, sice the sequece thereafter will cotiue i the maer of the sequece previously aalysed; we idicate this by a statemet of form proceed as above.) If has 1 digit which is odd, the: α 11 α 22 If has 1 digit which is eve, the: α 101 α 123. α 202 α 303 α 123. If has 2 digits, both odd, the: α 22 α proceed as for 1 odd digit. If has 2 digits, both eve, the: α 202 α proceed as for 1 odd digit. If has 2 digits, oe odd, oe eve, the: α 112 α 123. If has 3 digits, all odd, the: α 33 If has 3 digits, two odd, oe eve, the: α 123. α 22 If has 3 digits, oe odd, two eve, the: α 213 α 123. α proceed as for 1 odd digit. If has 3 digits, all eve, the: α 303 α proceed as for 1 odd digit. I particular, 123 α 123. We ow prove the statemet: If has l digits, l 4, the α() has strictly less tha l digits. Suppose l itself has m digits, that is, 10 m 1 l < 10 m. The the umber of odd digits of, umber of eve digits of, ad total umber of digits of, are each at most m. Hece α() has at most 3m digits. For m 2, we have 3m < 10 m 1, sice 10 m 1 = (1 + 9) m (m 1), by Berouilli s Iequality = 1 + 9m m , sice m 2 > 3m; so 3m < l ad α() has strictly fewer digits tha. For m = 1, we have 3m = 3 < l sice l 4 ad α() has strictly fewer digits tha. I coclusio, if has three digits or fewer, α i () = 123 for i sufficietly large (i fact, i 5); if has more tha three digits, successive applicatios of α give umbers with progressively fewer digits, util a umber is reached with three or fewer digits, ad the we ca apply the previous statemet (so that, we at least have i log 10 () + 5). 5

6 8. Polyomials Does there exist a polyomial p(x, y) with real coefficiets such that p(m, ) is a o-egative iteger (i.e. i Z 0 ), if m, Z 0 ad such that p : (Z 0 ) 2 Z 0 is a bijectio? Note. A polyomial p(x, y), i.e. oe i two variables, becomes a polyomial i oe variable by settig either variable to a costat. For example, take p(x, y) = 2x 3 + xy 2 + 7xy + y 2. Solutio. [Ispired by Aaro Mayard, 3rd year, Saul Freedma, 1st year, UWA] À la Cator s zig-zag argumet, we umber the pairs of o-egative itegers (x, y) i the followig way: x y More precisely we defie p, recursively, as follows:. p(0, 0) = 0, p(x, 0) = p(0, x 1) + 1, if x 1, p(x, y) = p(x + 1, y 1) + 1, if y 1. Observe that every o-egative iteger appears exactly oce i the body of the table. Hece p is a bijectio as required. We wat to show that p is a polyomial with real coefficiets. We claim that p(x, 0) = 1 2x(x + 1). This is trivially true for x = 0. Assume it is true for x, ad let us deduce it for (x + 1). We have p(x + 1, 0) = p(0, x) + 1 Hece the claim follows by iductio. Now = p(1, x 1) + 2. = p(x, 0) + x + 1 = 1 2x(x + 1) + (x + 1) = 1 2 (x + 1)(x + 2). p(x, y) = p(x + 1, y 1) + 1 = p(x + 2, y 2) + 2. = p(x + y, y y) + y = p(x + y, 0) + y which is a polyomial with real coefficiets. = 1 2 (x + y)(x + y + 1) + y = 1 2 x y2 + xy x y, 6

7 9. A map of a square Let f : R 2 R be a map such that f(a) + f(b) + f(c) + f(d) = 0 wheever a, b, c, d are the 4 vertices of a square. Is it true that f(x) = 0 for all x R 2? Solutio. [By Tuo Li, 2d year, UWA] Choose a arbitrary poit X R 2. Costruct a square ABCD, with X as its cetre. Let E, F, G, H be the midpoits of AD, AB, BC, CD, respectively. Sice AF XE, DEXH, BGXF ad CHXG are squares, we have: f(a) + f(f ) + f(x) + f(e) = 0 (7) f(d) + f(e) + f(x) + f(h) = 0 (8) f(b) + f(g) + f(x) + f(f ) = 0 (9) f(c) + f(h) + f(x) + f(g) = 0 (10) A E D F X H B G C Addig (7), (8), (9), (10), we have ( f(a) + f(b) + F (C) + f(d) ) + 2 ( f(e) + f(f ) + F (G) + f(h) ) + 4f(X) = 0. (11) But ABCD ad EF GH are also squares; hece f(a) + f(b) + f(c) + f(d) = 0 ad f(e) + f(f ) + f(g) + f(h) = 0, ad so (11) reduces to 4f(X) = 0 f(x) = 0. Sice x = X was arbitrarily chose, it follows that: Yes, it is true that f(x) = 0 for all x R 2. 7

8 10. A prime degree polyomial A polyomial of degree 2011 with real coefficiets is such that P () = {0, 1, 2,..., 2011}. What is the value of P (2012)? + 1 for all itegers Solutio. We are give a polyomial P (x) over R (i.e. with real coefficiets) of degree Let Q(x) = (x + 1)P (x) x. The Q(x) is a polyomial of degree 2012 over R ad for {0, 1, 2,..., 2011}, Q() = ( + 1) + 1 = 0, so that Q(x) has zeros 0, 1, 2,..., Sice Q(x) is of degree 2012 ad has 2012 distict zeros, it ca have o other zeros. Hece, for some real costat k. Now, 2011 Q(x) = k (x i) 1 = Q( 1) 2011 = k ( 1 i) 2011 = k ( 1) 2012 (i + 1) = k 2012! k = ! Q(x) = ! 2011 (x i) 2013 P (2012) 2012 = Q(2012) = (2012 i) 2012! = ! 2012! = 1 P (2012) = = 1. 8