THE UNIVERSITY OF TORONTO UNDERGRADUATE MATHEMATICS COMPETITION. In Memory of Robert Barrington Leigh. March 9, 2014

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1 THE UNIVERSITY OF TORONTO UNDERGRADUATE MATHEMATICS COMPETITION I Memory of Robert Barrigto Leigh March 9, 4 Time: 3 hours No aids or calculators permitted. The gradig is desiged to ecourage oly the stroger studets to attempt more tha five problems. Each solutio is graded out of. If the sum of the scores for the solutios to the five best problems does ot exceed 3, this sum will be the fial grade. If the sum of these scores does exceed 3, the all solutios will be graded for credit.. The permaet, per A, of a matric A a i,j ), is equal to the sum of all possible products of the form a,σ) a,σ) a,σ), where σ rus over all the permutatios o the set {,,, }. This is similar to the defiitio of determiat, but there is o sig factor.) Show that, for ay matrix A a i,j ) with positive real terms, per A! i,j a i,j.. For a positive iteger N writte i base umeratio, N deotes the iteger with the digits of N writte i reverse order. There are pairs of itegers A, B) for which A, A, B, B are all distict ad A B B A. For example, a) Determie a pair A, B) as described above for which both A ad B have two digits, ad all four digits ivolved are distict. b) Are there ay pairs A, B) as described above for which A has two ad B has three digits? 3. Let be a positive iteger. A fiite sequece {a, a,, a } of positive itegers a i is said to be tight if ad oly if a < a < < a, all ) differeces aj a i with i < j are distict, ad a is as small as possible. a) Determie a tight sequece for 5. b) Prove that there is a polyomial p) of degree ot exceedig 3 such that a p) for every tight sequece {a i } with etries. 4. Let fx) be a cotiuous realvalued fuctio o [, ] for which fx)dx ad xfx)dx. a) Give a example of such a fuctio. b) Prove that there is a otrivial ope iterval I cotaied i, ) for which fx) > 4 for x I. Please tur over page for additioal problems.

2 5. Let be a positive iteger. Prove that ) { } ) : odd, ). 6. Let fx) x 6 x 4 + x 3 x +. a) Prove that fx) has o positive real roots. b) Determie a ozero polyomial gx) of miimum degree for which all the coefficiets of fx)gx) are oegative ratioal umbers. c) Determie a polyomial hx) of miimum degree for which all the coefficiets of fx)hx) are positive ratioal umbers. 7. Suppose that x, x,, x are real umbers. For i, defie y i max x, x,, x i ). Prove that Whe does equality occur? y 4x 4 y i x i+ x i ). i 8. The hyperbola with equatio x y has two braches, as does the hyperbola with equatio y x. Choose oe poit from each of the four braches of the locus of x y ) such that area of the quadrilateral with these four vertices is miimized. 9. Let {a } ad {b } be positive real sequeces such that a u > ad Prove that ad b a b a ) v >. ) b a ) u log v.. Does there exist a cotiuous realvalued fuctio defied o R for which ffx)) x for all x R?

3 Solutios. The permaet, per A, of a matric A a i,j ), is equal to the sum of all possible products of the form a,σ) a,σ) a,σ), where σ rus over all the permutatios o the set {,,, }. This is similar to the defiitio of determiat, but there is o sig factor.) Show that, for ay matrix A a i,j ) with positive real terms, per A! i,j a i,j. Solutio. By the Arithmetic-Geometric Meas Iequality, the sum defiig the permaet havig! terms) is ot less that! times the product of all these terms raised to the power /!. Each term i the defiitio of the permaet has factors, so that product of all the terms has! factors. There are etries a i,j ; each appears the same umber! )! times. The result follows.. For positive iteger N writte i base umeratio, N deotes the iteger with the digits of N writte i reverse order. There are pairs of itegers A, B) for which A, A, B, B are all distict ad A B B A. For example, a) Determie a pair A, B) as described above for which both A ad B have two digits, ad all four digits ivolved are distict. b) Are there ay pairs A, B) as described above for which A has two ad B has three digits? Solutio. a) Let A a+b ad B u+v where a, b, u, v 9. The coditio AB B A leads to 99au bv), so that a ecessary coditio for the property to hold is au bv. There are may possibilities, icludig a, b; u, v), 3; 6, ). Thus, for example, [The possibilities iclude A, B), 4),, 63), 3, 6),, 84), 4, 8), 3, 64), 4, 63), 4, 84), 3, 96), 6, 93), 34, 86), 46, 96), 48, 63).] b) Let A a + b ad B u + v + w, where abuw ad a, b, u, v, w 9. The coditio AB B A leads to au bw) + [av w) + bu v)]. Therefore av w) + bu v) must be a multiple of. If the two terms of this sum differ i sig, the the absolute value of the sum caot exceed If the two terms of the sum have the same sig, the the sigs of v w ad u v must be the same, ad the absolute value of the sum caot exceed av w) + bu v) 9 v w) + u v) 9 u w 8. Therefore av w) + bu v), so that au bw. If v w ad u v do ot vaish, they must have opposite sigs. These coditios are ecessary ad sufficiet for a example. Trial ad error leads to a, b; u, v, w), ;, 3, ). Thus, we obtai I fact, we ca replace these umbers by multiples that ivolve o carries to obtai the examples s, 3t) where s 4 ad t 3. Other examples of pairs are 3, 68), 8, 45). Commet. At a higher level, we have that

4 3. Let be a positive iteger. A fiite sequece {a, a,, a } of positive itegers a i is said to be tight if ad oly if a < a < < a, all ) differeces aj a i with i < j are distict, ad a is as small as possible. a) Determie a tight sequece for 5. b) Prove that there is a polyomial p) of degree ot exceedig 3 such that a p) for every tight sequece {a i } with etries. Solutio. a) A sequece havig all differeces differet remais so if the same iteger is added or subtracted from each term. Therefore a for ay tight sequece. Sice there are 5 ) differeces for a tight sequece with five etries, a 5. We show that a 5 caot occur. Suppose, if possible, that a ad a 5. Sice each of the differeces from to, iclusive, must occur, ad sice the largest differece is the sum of the four differeces of adjacet etries, the differeces of adjacet etries must be,, 3, 4 i some order. The differece caot be ext to either of the differeces or 3, for the there would be a differece betwee oadjacet etries would be 3 or 4, respectively. Thus is the differece of either the first two or the last two etries ad the adjacet differece would be 4. But the a 3 a a 5 a 3 5, which is ot allowed. Sice there is o other possibility, a 5. Tryig a 5, we fid that the sequece {,, 5,, } is tight. b) The strategy is to costruct a sequece that satisfies the differece coditio such that the differeces betwee pairs of terms have differet cogruece classes with respect to some modulus, depedet oly o the umber of terms betwee them. Thus, we mae the differece betwee adjacet terms cogruet to, betwee terms separated by oe etry cogruet to, etc.. To this ed, let a ad a i+ a i + + i ) ) for i. The, I particular, a i i + i )i ) ). a + ) ) ). Whe i < j, we have that modulo ). Furthermore, a j a i a j a j ) + a j a j ) + + a i+ a i ) j a j+ a i+ ) a j a i ) a j+ a j ) a i+ a i ) j i) ) >. It follows that all the differeces are distict. Ay tight sequece with terms must have its largest term o greater tha p) ). Commet. This upper boud is udoubtedly too geerous ad it may be that there is a upper boud quadratic i. 4. Let fx) be a cotiuous realvalued fuctio o [, ] for which fx)dx ad xfx)dx. a) Give a example of such a fuctio. b) Prove that there is a otrivial ope iterval I cotaied i, ) for which fx) > 4 for x I. 4

5 Solutio. a) Tryig a fuctio of the form cx ) yields the examples fx) x 6. Other examples are fx) π cosπx) ad fx) π siπx). b) If there existed o such iterval I, the fx) 4 for all x [, ]. The < 4 fx)x /))dx /) x)dx + 4 fx))/) x)dx + fx)x /))dx x /))dx 4 /8) + 4 /8). Note that the iequality is strict. Equality would require that fx) 4 o, ) ad fx) 4 o, ), a impossibility for a cotiuous fuctio. Note. B. Yaghai draws attetio to Problem 5 of Chapter appearig o page 84 of Michael Spiva, Calculus W.A. Bejami, 967), where it is required to show that, if g is twice differetiable o [, ] with g) g ) g ) ad g), the g c) 4 for some c [, ]. Settig gt) t x t)fx)dx, whereupo g t) t fx)dx ad g t) ft), yields the desired result. [This is based o problem #45 i the America Mathematical Mothly.] 5. Let be a positive iteger. Prove that ) { } ) : odd,. Solutio. The values of each of the sums for 5 are,, 5/6, /3, 8/5. The equality of the secod ad third sums is straightforward to establish. Let S deote the leftmost sum. Sice ) ), we have that Usig the fact that S [ )]. [ )] [ )] + ) + + ) for, it ca be show that S ad that S + S + + [ )], for. Therefore S For each positive iteger, let T odd ). 5

6 The T ad T + T odd + ) ) + ) odd ) + odd ) + ) +. Therefore the recursios for S ad T agree whe ad satisfy the same equatios. Thus the result holds. 6. Let fx) x 6 x 4 + x 3 x +. a) Prove that fx) has o positive real roots. b) Determie a ozero polyomial gx) of miimum degree for which all the coefficiets of fx)gx) are oegative ratioal umbers. c) Determie a polyomial hx) of miimum degree for which all the coefficiets of fx)hx) are positive ratioal umbers. Solutio. a) Note that fx) x )x 4 ) + x 3 x ) x + ) + x 3 from which we see that fx) > for all x >. Alteratively, fx) x 6 + x 3 + x 3 x 4 ) + x ) x 4 x ) + x x ) + x 3 + from which we see that there are o roots i either of the itervals [, ] ad [, ). b) It is straightforward to see that multiplyig fx) by a liear polyomial will ot achieve the goal. We have that fx)x + bx + c) x 8 + bx 7 + c )x 6 + b)x 5 + b c )x 4 + c b)x 3 + c)x + bx + c from which we see that taig c ad b will yield the desired polyomial gx). Examples of suitable gx) are x + x + ad x + ). c) From b), we ote that o quadratic polyomial will serve. However, taig hx) x 3 +x +x+, we fid that fx)hx) x 9 + x 8 + x 7 + x 6 + x 5 + x 4 + x + x + x +. Commet. It is ow that for a give real polyomial fx), there exists a polyomial gx) for which all the coefficiets of fx)gx) are oegative resp. positive) if ad oly if all the roots of fx) are positive resp. oegative). Aother example is fx) x 3 x +. Sice fx) x 3 + x) xx ) +, we see that there are o roots i [, ] ad [, ). No liear polyomial x + c will do for gx). Sice fx)x + bc + c) x 5 + bx 4 + c )x 3 + b)x + b c)x + c, we require that c b ad b. The oly possibility is that gx) x + x +. For strictly positive coefficiets i the product, set gx) ax 3 + bx + cx + d. The fx)gx) ax 6 + bx 5 + c a)x 4 + a + d b)x 3 + b c)x + c d)x + d. 6

7 For g to be suitable, we require that c > a, a + d > b ad b > c > d. We ca tae gx) 3x 3 + 5x + 4x + 3. [This example is due to Horst Bruotte i Düsseldorf, Germay.] 7. Suppose that x, x,, x are real umbers. For i, defie y i max x, x,, x i ). Prove that Whe does equality occur? y 4x 4 y i x i+ x i ). i Solutio. We ca simplify the problem. Suppose that for some i < j, y i y i+ y j < y j+. The y i x i+ x i ) + y i+ x i+ x i+ ) + + y j x j+ x j ) y i x j+ x i ), so that all of x i+,, x j do ot figure i the terms of the right side. Therefore, with o loss of geerality, we ca assume that x < x < < x so that y i x i for i, ad y is equal to the maximum of x ad x. The right side of the iequality is equal to 4x 4[x x x ) + x x x ) + + x x x )] x + x x ) + x x ) + + x x ) + x x + x x ) + x x ) + + x x ) + x + x x ). Therefore, the right side is greater tha or equal to both of x ad x. The desired result follows. Equality occurs if ad oly if x for each i. Commet. It is atural to try a proof by iductio. Whe, the right side is equal to 4x 4x x + 4x x + x + x x ) which is clearly greater tha or equal to either of x ad x. Suppose that the iequality holds for m. The 4x m+ 4 [ m y i x i+ x i ) 4x m+ 4x m + 4x m 4 i 4x m+ 4x m + y m 4y m x m+ x m ) K m i y i x i+ x i ) Sice y m+ is the greater of x m+ ad y m, there are two cases to cosider. Whe y m+ x m+, the, sice x m y m y m+ ad y m+ y m y m, However, whe y m+ y m, the K y m+ y m ) + 4x m y m x m ) y m+ y m ) y m+. K y m + 4x m+ x m )x m+ + x m y m ) ad it is ot clear whether the secod term is always oegative. 7 ] 4y m x m+ x m )

8 [This problem is due to Mathias Beiglböc of the Uiversity of Viea, ad was commuicated to the cotest by Floria Herzig of the Uiversity of Toroto.] 8. The hyperbola with equatio x y has two braches, as does the hyperbola with equatio y x. Choose oe poit from each of the four braches of the locus of x y ) such that area of the quadrilateral with these four vertices is miimized. Solutio. Let N, W, S, E be arbitrary poits o the four braches that are symmetric respectively about the positive y axis orther), egative x axis wester), egative y axis souther) ad positive x axis easter). Cosider the diagoal NS of NW SE. If the secat through W that is parallel to NS passes through two distict poits of the wester brach, the the area [NW S] of triagle NSW ca be made smaller by replacig W by the poit of tagecy to the wester brach by a taget parallel to NS. Similarly, the area [NSE] ca be made smaller whe the taget at E is parallel to NS. We ca follow a similar argumet for the diagoal W E. Thus, for ay quadrilateral NW SE, we ca fid a quadrilateral of o larger area where the tagets to the hyberbolae at N ad S are parallel to W E ad the tagets to the hyperbolae at E ad W are parallel to NS. Thus we may assume that NW SE is a parallelogram. Observe that the slopes of the diagoals W E as well as the tagets to the orther ad souther braches rage over the ope iterval, ), ad that the slopes of the diagoals NS ad the tagets to the wester ad easter braches are similarly related. Each allowable taget slope occurs exactly twice, the tagets beig related by a 8 rotatio about the origi. Thus, i the miimizatio problem, we eed cosider oly parallelograms cetred at the origi with the foregoig taget ad diagoal relatioship. Let N a, b) be ay poit o the orther brach, where with o loss of geerality we may assume that a < b ad b a. The slope of the taget at N is a/b, so the poit E, located o the lie with this slope through the origi must be at b, a). Similarly, S a, b) ad W b, a). Thus, NW SE is i fact a rectagle whose sides are perpedicular to the asymptotes of the hyperbola ad whose area is NE ES )b a)) )b + a) b a ). Therefore, the required miimum area is. [This problem was cotributed by Robert McCa.] 9. Let {a } ad {b } be positive real sequeces such that ad Prove that ad a u > b a b a ) v >. ) b a ) u log v. Solutio. Observe that b a exp log b a ) exp v) exp. Suppose that v. The for sufficietly large values of, b a ad we have that [ ) a b a ) log + b )] a a ) b a a log b. a 8

9 Note that ad that t t log + t), so that as desired. ) b a b a a [ ] b a ) u log v If v, we eed to exercise more care, as b could equal a ifiitely ofte. I this case, the idices ca be partitioed ito two sets A ad B, where respectively b a for A ad b a for B; either of these sets could be ifiite or fiite. If ifiite, taig the it over A ad over B yields the result u log v, ad the problem is solved. [This problem was cotributed by the AN-aduud Problem Solvig Group i Ulaabataar, Mogolia.]. Does there exist a cotiuous realvalued fuctio defied o R for which ffx)) x for all x R? Solutio. The aswer is o. Sice ffx)) x defies a oe-oe fuctio, it follows that fx) itself is oe-oe. Therefore fx) is either strictly icreasig or strictly decreasig o R. But this is impossible, sice i either case, ffx)) would be icreasig. Solutio. The aswer is o. Suppose that f is such a fuctio ad that f) u. The fu) ff)) ad so u f) ffu)) u. Hece u ad f). Suppose that fv). The f) ffv)) v. Therefore fv) if ad oly if v. Let a ad b fa). The fb) a ad f a) b. Cosider the four umbers a, b, a, b repeated cyclically. The alterate oes have opposite sigs, so there must be two umbers of the same sig followed by oe of the opposite sig. With o loss of geerality, we ca assume that a ad b are positive while a ad b are egative. The fa) ad fb) have opposite sigs, so by the Itermediate Value Theorem, there must exist c betwee a ad b for which fc). But this is a cotradictio. Therefore, there is o cotiuous fuctio with the desired property. 9