SOLUTIONS. Let a, b, and c be nonnegative real numbers such that a + b + c = 3. Prove. 4 ca + c2 a. 4 ab 1.

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1 44/ SOLUTIONS SOLUTIONS No problem is ever permaetly closed. The editor is always pleased to cosider for publicatio ew solutios or ew isights o past problems [ : 54, 543] Proposed by Hug Pham Kim, studet, Staford Uiversity, Palo Alto, CA, USA. that Let a, b, ad c be oegative real umbers such that a + b + c = 3. Prove a b 4 bc + b c 4 ca + c a 4 ab. Solutio by Oliver Geupel, Brühl, NRW, Germay. Sice for oegative real umbers such that a + b + c = 3 we have a b + b c + c a + abc 4 () (See the lemma i the solutio of Problem 3549 [:53].), the a b 4 4 bc + b c 4 ca + c a 4 ab = a b So it suffices to prove that bc 4 bc + + b c a b c 4 bc + b c a 4 ca + c a b 4 ab abc. Clearig deomiators, it becomes ca ab 4 ca + + c a 4 ab + 4 ab 4 bc + bc 4 ca + ca 4 ab. 3(ab + bc + ca) + abc(a b + b c + c a + abc) 64 8(a + b + c)abc 4(a b + b c + c a ). After applyig iequality () ad homogeizig, the iequality ca be writte i the form 3 9 (a + b + c) (ab + bc + ca) + 4 (a + b + c)abc (a + b + c)4 8(a + b + c)abc 4(a b + b c + c a ). Clearig deomiators agai, expadig ad adoptig the otatio [α, β, γ] = X sym a α b β c γ, Crux Mathematicorum, Vol. 38(), December

2 SOLUTIONS / 45 it becomes 6([3,, ] [4,, ]) + 33([,, ] [,, ]), which is true by Muirhead s theorem. This completes the proof. Notice that the equality holds for a = b = c =, or a =, b =, c =, or ay permutatios of these values. Also solved by GEORGE APOSTOLOPOULOS, Messologhi, Greece; PAOLO PERFETTI, Dipartimeto di Matematica, Uiversità degli studi di Tor Vergata Roma, Rome, Italy; ad the proposer. The proposer posted this problem i 8, o the Mathliks forum, see http: // www. artofproblemsolvig. com/ Forum/ viewtopic. php? t= Also, you ca fid a ice solutio by Vo Quoc Ba Ca o http: // cahag7. wordpress. com/ 9/ / / iequality-37-p-k-hug [ : 54, 543] Proposed by Nguye Thah Bih, Haoi, Vietam. For a arbitrary poit M i the plae of triagle ABC defie D, E, ad F to be the secod poits where the circumcircle meets the lies AM, BM, ad CM, respectively. If O, O, ad O 3 are the respective cetres of the circles BCM, CAM, ad ABM, prove that DO, EO, ad F O 3 are cocurret. Idetical solutios by Šefket Arslaagić, Uiversity of Sarajevo, Sarajevo, Bosia ad Herzegovia ad Salem Malikić, studet, Simo Fraser Uiversity, Buraby, BC, with the otatio modified by the editor. Editor s commet. To avoid the eed for umerous special cases, we shall make use of directed agles: the symbol P QR will represet the agle through which the lie QP must be rotated about Q to coicide with QR, where P QR < 8. Properties of directed agles are discussed i Roger Johso s Advaced Euclidea Geometry, Dover reprit (96). Our goal is to prove that ot oly A are DO, EO, F O 3 cocurret, but φ their commo poit lies o the circumcircle of ABC. Deote by S the secod poit where the lie DO itersects O S the circumcircle, ad let φ = MAC = O E DAC ad θ = CBM = CBE. Because A, C, D, E ad S are all o the same circle we have M O SE = DSE = DSC + CSE = DAC + CBE = φ + θ. () B θ D C Sice O ad O are circumcetres of triagles M BC ad M AC, respectively, we have MO C = MAC = φ O Copyright c Caadia Mathematical Society, 3

3 46/ SOLUTIONS ad CO M = CBM = θ. Moreover, the radii O M = O C ad O M = O C, so that the quadrilateral MO CO is a kite, implyig that O O C = MO C = φ ad CO O = CO M = θ. () Cosequetly, i O CO we have O CO = φ + θ. (3) We deote by O the poit where ES itersects CO ad will prove that i fact, O = O. From () ad (3) we have O CO = O CO = φ + θ = O SE = O SO. Cosequetly the poits O, C, O, S are cocyclic, whece (with the help of ()) O O C = O SC = DSC = DAC = φ = O O C. Because there is exactly oe lie through O, amely O O, that makes a directed agle of φ with the lie CO, we coclude that O = O ; that is, EO passes through S, as claimed. Iterchagig the roles of B ad C, oe shows similarly that F O 3 also passes through S. Also solved by GEORGE APOSTOLOPOULOS, Messologhi, Greece; MICHEL BATAILLE, Roue, Frace ( solutios); ad the proposer. Apostolopoulos provided a eat computatio, usig complex umbers to show that the poit where DO itersects the circumcircle is represeted by a expressio that is symmetric i A, B, ad C [ : 54, 543] Proposed by Michel Bataille, Roue, Frace. Give k 4, Š, let {a } = be the sequece defied by a =, a = ad the recursio a + = a + + ka. Evaluate = a. Solutio by AN-aduud Problem Solvig Group, Ulaabaatar, Mogolia. Usig roots of the characteristic equatio, x x k =, of the give recursio, we have + + 4k + 4k a = +, k ( /4, ). Crux Mathematicorum, Vol. 38(), December

4 SOLUTIONS / 47 Therefore, = a = X = = = + x + = + 4k + 4k + ( x) := S(x), Œ where x = + +4k. Sice = x = = Z x t dt = Z x = t dt = Z x dt = l( x) t ad the = ( x) = l x, S x X (x) = + ( x) = = = l( x) + x l x = (l x l( x)). Hece S(x) = l x l( x) + C. P Sice lim x + l x l( x) = ad S() = we have the fial result a = π 6 l + + 4k = = l + 4k. = π π 6, the C = 6, ad Also solved by AN-ANDUUD Problem Solvig Group, Ulaabaatar, Mogolia; GEORGE APOSTOLOPOULOS, Messologhi, Greece; ŠEFKET ARSLANAGIĆ, Uiversity of Sarajevo, Sarajevo, Bosia ad Herzegovia; RADOUAN BOUKHARFANE, Polytechique de Motréal, Motréal, PQ; PAUL BRACKEN, Uiversity of Texas, Ediburg, TX, USA; CHIP CURTIS, Missouri Souther State Uiversity, Jopli, MO, USA; OLIVER GEUPEL, Brühl, NRW, Germay; ANASTASIOS KOTRONIS, Athes, Greece; SALEM MALIKIĆ, studet, Simo Fraser Uiversity, Buraby, BC; PAOLO PERFETTI, Dipartimeto di Matematica, Uiversità degli studi di Tor Vergata Roma, Rome, Italy; M. A. PRASAD, Idia; JOEL SCHLOSBERG, Bayside, NY, USA; ALBERT STADLER, Herrliberg, Switzerlad; HAOHAO WANG ad YANPING XIA, Southeast Missouri State Uiversity, Cape Girardeau, MO, USA; ad the proposer. Copyright c Caadia Mathematical Society, 3

5 48/ SOLUTIONS Proposed by Pham Va Thua, Haoi Uiversity of Sciece, Haoi, Vietam. Let x, y, ad z be oegative real umbers such that x + y + z =. Prove that r x + y r + y + z r + x + z 6. [Editor s ote: Both AN-aduud Problem Solvig Group, Ulaabaatar, Mogolia; ad Albert Stadler, Herrliberg, Switzerlad, poited out that the same problem by the same proposer had appeared as Example of the article Square it published i Vol., No. 5 (8) of Mathematical Excalibur ( hk/excalibur/v_5.pdf). The solutio give there, which used the Cauchy- Schwarz Iequality, is similar to most of the submitted solutios we received.] Solved by ARKADY ALT, Sa Jose, CA, USA; GEORGE APOSTOLOPOULOS, Messologhi, Greece; ŠEFKET ARSLANAGIĆ, Uiversity of Sarajevo, Sarajevo, Bosia ad Herzegovia; MICHEL BATAILLE, Roue, Frace; PAUL BRACKEN, Uiversity of Texas, Ediburg, TX, USA; KEE-WAI LAU, Hog Kog, Chia; SALEM MALIKIĆ, studet, Simo Fraser Uiversity, Buraby, BC; PAOLO PERFETTI, Dipartimeto di Matematica, Uiversità degli studi di Tor Vergata Roma, Rome, Italy; M. A. PRASAD, Idia; HAOHAO WANG ad JERZY WOJDYLO, Southeast Missouri State Uiversity, Cape Girardeau, Missouri, USA; ad the proposer [ : 54, 544] Proposed by Michel Bataille, Roue, Frace. Let a be a positive real umber. f : (, ) (, ) such that for all positive x,y. (x + a)f(x + y) = af(yf(x)) I. Solutio by Oliver Geupel, Brühl, NRW, Germay. Fid all strictly mootoe fuctios a x+a It is straightforward to check that f(x) = is a solutio. We prove that it is uique. Suppose that f meets the requiremets of the problem. Sice f is mootoe, it has a limit L (possibly ifiite) from the right at. For each x >, we have that lim f(x + y) = a y + x + a lim y + f(yf(x)) = a x + a L. Thus the limit L must be a positive real umber. Sice a mootoe fuctio has at most coutably may discotiuities, f(x) = (al)/(x + a) o a dese subset of (, ) ad so f(x) must be decreasig. For ay x > ad each ɛ >, we ca choose positive umbers x ad x such that x ɛ < x < x < x < x + ɛ ad a x + a L = f(x ) f(x) f(x ) = a x + a L. Crux Mathematicorum, Vol. 38(), December

6 SOLUTIONS / 49 Therefore, we must have that f(x) = al x + a for all x >. Substitutig this expressio ito the fuctioal equatio reveals that L must be. Note that we did ot require the mootoicity to be strict. II. Solutio by M. A. Prasad, Idia. Defie g(x) = f(ax) for x >. We get that (x + )g(x + y) = g(yg(x)) for x, y >. For sufficietly large values of y, we may write (x + )g(x + y) = g( + (yg(x) )) = g((yg(x) )g()) = (yg(x) )g() g g(x + ) g(x + ) = (x + )g x + + yg(x) g(x + ) g(). Sice the fuctio is strictly mootoe, it is oe-oe, so that x + y = x + + yg(x) g(x + ) g(). For each x, this holds for ifiitely may values of y, so that, equatig terms idepedet of y ad coefficiets of y, we get that x = x + g() (x + )g(x + ) = g(). g(x + ) ad g(x + ) = g(x)g(). Elimiatig g(x+) from these equatios yields that (x+)g(x) =, from which f(x) ca be obtaied. Also solved by GEORGE APOSTOLOPOULOS, Messologhi, Greece; ad the proposer [ : 54, 544] Proposed by Nguye Thah Bih, Haoi, Vietam. Let the icircle of triagle ABC touch the sides BC at D, CA at E, ad AB at F. Costruct by ruler ad compass the three mutually taget circles that are iterally taget to the icircle, oe at D, oe at E, ad oe at F. Solutio by George Apostolopoulos, Messologhi, Greece. Copyright c Caadia Mathematical Society, 3

7 43/ SOLUTIONS C D T c I a O T b B E I b T a I c F A We will use otatio A to represet a circle with cetre at A. Let A, B, ad C be the circles with radii AE, BF, ad CD, respectively. First, we will costruct a small circle O, exterally taget to all of A, B, ad C. (This is ot a difficult but rather cumbersome costructio ad it ca be foud for example o the website: http: // oz. thu. edu. tw/ ~ g9754/ soddycircles. html.) If T a, T b, T c deote the correspodig taget poits, as o the diagram, the the circle I c iscribed i ABO is the circumcircle of F T a T b. Similarly, let I a ad I b be the iscribed circles i BCO ad CAO, respectively. We claim that the circles I a, I b, I c satisfy coditios of the problem. Firstly, they are taget to the lies AB, BC, ad CA at poits E, D, ad F, respectively, which meas that they are taget iterally to the icircle of ABC at the correspodig poits E, D, ad F. Secodly, I a ad I b are both taget to the lie CO at the commo poit T c, hece they are exterally taget to each other. Sice similar coclusio refers to the remaiig pairs of circles, the claim holds. Also solved by MICHEL BATAILLE, Roue, Frace; OLIVER GEUPEL, Brühl, NRW, Germay; ad the proposer [ : 54, 544] Proposed by Michel Bataille, Roue, Frace. For positive iteger, prove that ta π ta π ta 3π 6 7 is a iteger ad fid the highest power of 7 dividig this iteger. Solutio by Roy Barbara, Lebaese Uiversity, Faar, Lebao, expaded slightly by the editor. We prove more geerally that f() = ta π 7Š + ta π 7 is a iteger for all N. c. Let a = ta π 7Š, b = ta π 7 Š Š + ta 3π 7 Š Š ad c = ta 3π. 7 The f() = a + b + Crux Mathematicorum, Vol. 38(), December

8 SOLUTIONS / 43 It is well kow (see []) that ta π 7, ta π 7, ad ta 3π 7 are zeros of the polyomial x 6 x x 7. Therefore a, b, ad c are three (distict) roots of x 3 x + 35x 7. It follows that a + b + c =, ab + bc + ca = 35 ad abc = 7. Hece ad a + b + c = (a + b + c) (ab + bc + ca) = 7 = 37, a 3 +b 3 +c 3 = (a+b+c)(a +b +c ab bc ca+3abc = (37 35)+ = 777. Sice a +3 + b +3 + c +3 = (a + b + c)(a + + b + + c + ) we get (ab + bc + ca)(a + + b + + c + ) + abc(a + b + c ) f( + 3) = f( + ) 35f( + ) + 7f() () ad sice f() =, f() = 37, ad f(3) = 777 are all itegers, it follows from () that f() is a iteger for all N. We ow show that 7 f(3): that is, the highest power of 7 dividig f(3) is. Sice f(), f(), ad f(3) are all multiples of 7, we have 7 f(k) for k =,, ad 3. Suppose that 7 k f(3k ), 7 k f(3k ) ad 7 k f(3k) for all k =,,..., for some. The replacig i () by 3, 3, ad 3, respectively, we get f(3 + ) = f(3) 35f(3 ) + 7f(3 ) () f(3 + ) = f(3 + ) 35f(3) + 7f(3 ) (3) f(3 + 3) = f(3 + ) 35f(3 + ) + 7f(3). (4) Usig () (4) we obtai successively that 7 + f(3 + ), 7 + f(3 + ), ad 7 + f(3 + 3). That is, 7 + f(3( + )). Hece by iductio we coclude that 7 f(3) for all N. To complete the proof, it remais to show that 7 + f(3). To this ed, we use iductio to prove that f(3) 7 3 (mod 7). (5) Sice f(3) = 777 = 7 ad 3 (mod 7), (5) is true for =. Suppose (5) holds for some. We let f(3) = 7 q where q 3 (mod 7). We also set f(3+) = 7 + r, f(3+) = 7 + s, ad f(3+3) = 7 + t, where r, s, ad t are itegers. The by (4) we obtai 7 + t = (7 + s) 35(7 + r)+7(7 q) which, upo dividig by 7 +, yields t = 7(3s) 7(5r) + q q 3 (mod 7). That is f(3+3) (mod 7) ad the iductio is complete. Copyright c Caadia Mathematical Society, 3

9 43/ SOLUTIONS Refereces [] mathworld.wolfram.com/trigoometryaglespi7.html. Also solved by AN-ANDUUD Problem Solvig Group, Ulaabaatar, Mogolia; GEORGE APOSTOLOPOULOS, Messologhi, Greece; OLIVER GEUPEL, Brühl, NRW, Germay; JOEL SCHLOSBERG, Bayside, NY, USA; ALBERT STADLER, Herrliberg, Switzerlad; HAOHAO WANG ad JERZY WOJDYLO, Southeast Missouri State Uiversity, Cape Girardeau, Missouri, USA; ad the proposer [ : 54, 544] Proposed by Ovidiu Furdui, Campia Turzii, Cluj, Romaia. Fid the value of lim Z + x dx. I. Solutio by Albert Stadler, Herrliberg, Switzerlad. The Let I = Z / Z + x dx, ad J = / = Z / dx I Z / + x dx. dx, ad Z / x dx J Z / x dx. Sice Z / x dx = Z / x dx = Š /, + we fid that lim I = lim J =, so that the required limit is equal to. II. Solutio by George Apostolopoulos, Messologhi, Greece; Moubiool Omarjee, Lycée Heri IV, Paris, Frace; ad M. A. Prasad, Idia(idepedetly). Sice + x < + x for < x <. The Z p + x + dx < Z ( + x )dx = + +. O the other had, whe < c <, we have that Z + x dx = > Z c Z c + x dx + dx + Z c Z c x dx = c + + x dx + ( c+ ). Crux Mathematicorum, Vol. 38(), December

10 SOLUTIONS / 433 Therefore the required limit is ot less tha c + for each c (, ). It follows that the limit is. Also solved by PAUL BRACKEN, Uiversity of Texas, Ediburg, TX, USA; KEE-WAI LAU, Hog Kog, Chia; PAOLO PERFETTI, Dipartimeto di Matematica, Uiversità degli studi di Tor Vergata Roma, Rome, Italy; ad the proposer. I the secod solutio, Omarjee avoided the extra parameter by takig the specific value c = /. Perfetti ad the proposer made a variable trasformatio y = x that redered the itegral as the sum of two itegrals over [, ] ad [, ], each of which teded to. Lau made a trasformatio y = / x. Two icorrect solutios were received [ : 54, 544] Proposed by Mehmet Şahi, Akara, Turkey. Let ABC deote a triagle, I its iceter, ad ρ a, ρ b, ad ρ c the iradii of IBC, ICA, ad IAB, respectively. Prove that ta(75 ) ρ a ρ b ρ c a + b + c. I. Solutio by George Apostolopoulos, Messologhi, Greece. Let α = B + C, β = C + A ad γ = A + B. The α + β + γ = 8, a = r(ta β + ta γ), b = r(ta γ + ta α), c = r(ta α + ta β), IB = r sec β, IC = r sec γ ad [IBC] = ar ad [IBC] = ar, so that = cos β + cos γ + ρ a r r si α with aalogous expressios for /ρ b ad /ρ c. Observe that ta 75 = 3+ ad ta α ta β ta γ = ta α+ta β+ta γ 3 ta 6 = 3 3 (from the covexity of the taget fuctio). The Thus, it is required to show that 3 ta α ta β ta γ + X 8 ta 75 a + b + c = r ta α ta β ta γ. cyclic cos β + cos γ cos α ta β ta γ This iequality holds for the two terms respectively, that for the secod term relyig o a applicatio of the Arithmetic-Geometric Meas iequality. II. Solutio by Oliver Geupel, Brühl, NRW, Germay. We ote that ta 75 = ta( ) = + 3. Let r, R ad s deote the iradius, circumradius ad semiperimeter, respectively, of triagle ABC. Let h I, h A, h B ad [IAB] deote the altitudes ad area of triagle IAB. We have that h I = r, h A = c si(b/) ad h B = c si(a/). Hece AB + AI + BI = ρ a [IAB] = h I + h B + h A = r + c si(a/) + c si(b/). Copyright c Caadia Mathematical Society, 3

11 434/ SOLUTIONS Similar idetities hold for /ρ b ad /ρ c. We fid that (a + b + c) + + ρa ρ b ρ c 3(a + b + c) = + (a + b + c) r a si(b/) + b si(c/) + c si(a/) + (a + b + c) a si(c/) + b si(a/) +. c si(b/) By the Arithmetic-Geometric Meas iequality, we have that 3(a + b + c) r 6s = (s a)(s b)(s c)/s 6s3/ (s/3) 3/ = 6 33/ = 8 3. Applyig the Cauchy-Schwarz iequality, the Arithmetic-Geometric Meas iequality ad Euler s iequality R r i successio, we obtai that (a + b + c) a si(b/) + b si(c/) + c si(a/) si(a/) + si(b/) + si(c/) Ž r 9 4R si(a/) si(b/) si(c/) = 9 3 r 8. 3 Aalogously, (a + b + c) a si(c/) + b si(a/) + c si(b/) 8. Cosequetly, (a + b + c) + + 8( + 3) = 8 ta 75. ρa ρ b ρ c Equality holds if ad oly if triagle ABC is equilateral. Also solved by ŠEFKET ARSLANAGIĆ, Uiversity of Sarajevo, Sarajevo, Bosia ad Herzegovia; MICHEL BATAILLE, Roue, Frace; CHIP CURTIS, Missouri Souther State Uiversity, Jopli, MO, USA; SALEM MALIKIĆ, studet, Simo Fraser Uiversity, Buraby, BC; NECULAI STANCIU, George Emil Palade Secodary School, Buzău, Romaia ad TITU ZVONARU, Comáeşti, Romaia (joit); ad the proposer. Crux Mathematicorum, Vol. 38(), December

12 SOLUTIONS / [ : 54, 544] Proposed by Michel Bataille, Roue, Frace. Let ABC be a triagle ad a = BC, b = CA, c = AB. Give that ap A + cp B + bp C = cp A + bp B + ap C = bp A + ap B + cp C for some poit P, show that ABC is equilateral. Solutio by Chip Curtis, Missouri Souther State Uiversity, Jopli, MO, USA. The problem is missig a ecessary restrictio; the correct statemet (exteded to three-dimesioal space by the editors): Let l be the lie i Euclidea space that is that is perpedicular to the plae of the give triagle ABC ad passes through its circumceter; if there exists a poit ot o l that satisfies the give equatios, the the triagle must be equilateral. Set x = P A, y = P B, ad z = P C, ad assume the labels have bee chose so that z. The give system of equatios the becomes a(x z )+b(z y )+c(y x ) = ad a(z y )+b(y x )+c(x z ) =, or, equivaletly, ax by + cy cx = (a b)z () ay + by bx + cx = (c a)z () Case. If a b, c, the we ca multiply the first equatio by c a, the secod by a b, ad subtract to obtai (c a)(ax by + cy cx ) (a b)( ay + by bx + cx ) =, or, after regroupig, (x y )(bc + ca + ab a b c ) =. Because a b, c the AM-GM iequality implies that bc+ca+ab a b c. Hece, x = y, ad (from ()) z = ax by + cy cx a b = x (a b) a b = x. Cosequetly, P A = P B = P C; that is, P is equidistat from the three vertices so that it lies o the lie l. Observe that whe P A = P B = P C, each of the three expressios i the problem statemet are equal to P A (a + b + c). Case. If a = b the equatios () ad () become (a c)(x y ) = ad (c a)(x z ) =. If i additio, x = y = z, the agai, P lies o l. Otherwise, c = a = b ad ABC is equilateral, as claimed. Also solved by MARIAN DINCĂ, Bucharest, Romaia; OLIVER GEUPEL, Brühl, NRW, Germay; M. A. PRASAD, Idia; ad the proposer. The solutio submitted by the proposer clearly idicates that he had iteded to exclude the circumcetre as a possible positio for P (as i the restatemet of the problem). Amog the other three submissios, two dealt with specialized iterpretatios, ad oe simply provided the couterexample. Copyright c Caadia Mathematical Society, 3

SOLUTIONS. The edges of the tetrahedron are denoted a, b, c, d, e, f. Prove or disprove that 4 6 GD 1 9. a + 1 b + 1 c + 1 d + 1 e + 1 f.

SOLUTIONS. The edges of the tetrahedron are denoted a, b, c, d, e, f. Prove or disprove that 4 6 GD 1 9. a + 1 b + 1 c + 1 d + 1 e + 1 f. 8/ SOLUTIONS SOLUTIONS No problem is ever permaetly closed. The editor is always pleased to cosider for publicatio ew solutios or ew isights o past problems. 49. [995 : 58; 996 : 83-84] Proposed by Ja