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1 Problems Ted Eiseberg, Sectio Editor ********************************************************* This sectio of the Joural offers readers a opportuity to exchage iterestig mathematical problems ad solutios. Please sed them to Ted Eiseberg, Departmet of Mathematics, Be-Gurio Uiversity, Beer-Sheva, Israel or fax to: Questios cocerig proposals ad/or solutios ca be set to <eisebt@1.et>. Solutios to previously stated problems ca be see at < Solutios to the problems stated i this issue should be posted before May 15, 1 5: Proposed by Keeth Korbi, New York, NY Give positive itegers a, b, c, d) such that, a + b + c + d) a + b + c + d ) with a < b < c < d. Fid positive itegers x, y ad z such that x ab + ad + bd ab + ac + bc, y bc + bd + cd bc + ab + ac, z bc + bd + cd ac + ad + cd. 51: Proposed by Keeth Korbi, New York, NY Give covex cyclic ) quadrilateral ABCD with iteger legth sides where AB, BC, CD 1 ad with AB < BC < CD. The iradius, the circumradius, ad both diagoals have ratioal legths. Fid the possible dimesios of the quadrilateral. 5: Proposed by Neculai Staciu, Buzău, Romaia Solve i R, l x + ) x l y + y + 1 y x 1 x y+1) x y : Proposed by Pedro Patoja, Natal-RN, Brazil Evaluate, π/4 l 1 + si ) x si 4 x + cos 4 dx. x 1

2 54: Proposed by José Luis Díaz-Barrero, Barceloa, Spai Let f : R R be a o-costat fuctio such that, fx + y) fx) + fy) 1 + fx)fy) for all x, y R. Show that 1 < fx) < 1 for all x R. 55: Proposed by Ovidiu Furdui, Cluj-Napoca, Romaia Fid the sum, ) + + 1) 1 l 1 l + 1. Solutios 518: Proposed by Keeth Korbi, New York, NY Part I: A isosceles right triagle has perimeter P ad its Morley triagle has perimeter x. Fid these perimeters if P x + 1. Part II: A isosceles right triagle has area K ad its Morley triagle has area y. Fid these areas if K y + 1 Solutio by David E. Maes, Oeota, NY For part I, P + ) ad x ) For part II, K ) ad y Deote the isosceles right triagle by ABC with the right agle at vertex C ad sides a, b, c opposite the vertices A, B, C respectively. The a b ad c a, whece P + )a. The side legth s of the Morley triagle of ABC is give by s 8R si A si B si C where R is the circumradius of triagle ABC. The R abc a + b + c)b + c a)c + a b)a + b c) a + ) a ) a a ) a. Usig the idetity si z 1 cos z, oe calculates si A si B si 15 1 cos. 4

3 Therefore, ) ) a 1 s 8R si 15 si 8 4 ) a so that the perimeter x of the Morley triagle is give by x s + ) a ) a + 1 or The equatio P x + 1 implies a Hece, I part II, P + )a + ) ad x The equatio K y + 1 implies a 8 Hece, ) a ) K a ad [ y 4 s ] ) a ) a. 8 K ) ) a. ) a + 1; that is, a ). 19 ad y Also solved by Ekel Hyselaj, Uiversity of Techology, Sydey, Australia joitly with Elto Bojaxhiu, Kriftel, Germay; Paul M. Harms, North Newto, KS; Kee-Wai Lau, Hog Kog, Chia; Charles McCracke, Dayto, OH; Boris Rays, Brookly, NY; David Stoe ad Joh Hawkis joitly), Statesboro, GA, ad the proposer. 518: Proposed by Keeth Korbi, New York, NY A covex petago ABCDE, with iteger legth sides, is iscribed i a circle with diameter AE. Fid the miimum possible perimeter of this petago. Solutio by Kee-Wai Lau, Hog Kog, Chia We show that the miimum possible perimeter of this petago is 164.

4 Suppose that O is the ceter of the circle. Let AB p, BC q, CD r, DE s, AE d, AOB α, BOC β, COD γ, DOE δ, where p, q, r, s, d are positive itegers ad α, β, γ, δ are positive umbers such that α + β + γ + δ π. We have si α p d d, cos α p, si β q d d d, cos β q, d si γ r d d, cos γ r, ad si δ s d d. Sice si δ cosα + β + γ), so d s d p d q d r ) ) ) d p qr d q rp d r pq. It is ot hard to see that if at least oe of the d p d q d r is irratioal, the d s is also irratioal. Hece we seek the primitive Pythagorea triples with d m +, p { m, m } such that d p d q ) ) ) d r d p qr d q rp d r pq d is a positive iteger. We ow fid with a computer that the miimum perimeter is 164, give by d, p, q, s) 65, 16, 5, 5, ) ad other combiatios. Commet by Editor. David Stoe ad Joh Hawkis of Statesboro, GA approached the problem i the above maer. They ra a MATLAB program checkig all iteger combiatios for p, q, r, s, ad d 5, ad usig the data from this program they showed that they, like Kee-Wai, had actually foud the smallest solutio. They the wet o to list some additioal petagos satisfyig the requiremets of the problem, but with larger perimeters. I have substituted Kee-Wai s otatio ito David 4

5 ad Joh s matrix.) d p q r s perimeter David ad Joh wet o to state that they did t kow if a aalytical proof exists as opposed to a computer oe) that shows that the miimum perimeter is give by d, p, q, s) 65, 16, 5, 5, ). They the made the followig commets. There is iterestig territory for further exploratio of these iteger-valued, covex, petagos iscribed i a semi-circle Korbi petagos?). We could defie p, q, r, s, d) to be primitive if there is o commo factor ad hope to fid a parametric-type formula for geeratig the primitive oes. Based o our few examples, it appears that i a primitive petago, s must be the o-prime hypoteuse of a primitive Pythagorea triple, that d is the product of primes cogruet to 1mod 4) ad p, q, r ad s must be legs of some right triagle havig d as its hypoteuse. It also seems that d must be expressible i more tha oe way as the sum of squares. It looks like may primitive petagos exist. For example with d 5 5 1, there are several primitive petagos ad oe multiple of our miimal example). Two of these primitive petagos eve have the same perimeter! Also solved by David Stoe ad Joh Hawkis of Statesboro, GA, the proposer. 5184: Proposed by Neculai Staciu, Buzău, Romaia 5

6 If x, y ad z are positive real umbers, the prove that x + y)y + z)z + x) x + y + z)xy + yz + zx) 8 9. Solutio 1 by Pedro Patoja, Natal-RN, Brazil If x, y, z are positive real umbers the, x + y)y + z)z + x) xy + yz + zx)x + y + z) xyz. So, x + y)y + z)z + x) xy + yz + zx)x + y + z) 1 xyz xy + yz + zx)x + y + z). By the AM-GM iequality, x + y + z xyz ad xy + yz + zx xyz), which implies that xy + yz + zx)x + y + z) 9xyz. Ad this implies that xyz xy + yz + zx)x + y + z) 1 9. So, 1 xyz xy + yz + zx)x + y + z) Therefore, x + y)y + z)z + x) xy + yz + zx)x + y + z) 8 9. Solutio by Bruo Salgueiro Faego, Viveiro Spai 9x + y)y + z)z + x) 8x + y + z)xy + yz + zx) 9xy + yz + zx + y )z + x) 8x + y + z)xy + yz + zx) 9 xy + yz + zx)z + x) + y z + x) 8z + y + z)xy + yz + zx) 9 x + y + z)xy + yz + zx) xy + yz + zx)y + y z + x) 8x + y + z)xy + yz + zx) 9 x + y + z)xy + yz + zx) xy y z xyz + y z + xy ) 8x + y + z)xy + yz + zx) 6

7 9x + y + z)xy + yz + zx) 9xyz 8x + y + z)xy + yz + zx) 8x + y + z)xy + yz + zx) + x + y + z)xy + yz + zx) 8xyz xyz 8x + y + z)xy + yz + zx) 8x + y + z)xy + yz + zx) 8xyz + x + y + z)xy + yz + zx) xyz 8x + y + z)xy + yz + zx) 8x + y + z)xy + yz + zx) 8xyz + xyz xyyzzx xyz 8x + y + z)xy + yz + zx) 8x + y + z)xy + yz + zx) 8xyz + 9xyz xyz 8x + y + z)xy + yz + zx) 8x + y + z)xy + yz + zx) 8x + y + z)xy + yz + zx) 1, which is equivalet to the proposed iequality, where the AM GM iequality has bee applied. We ote that equality holds if, ad oly if, x y z. Solutio by Paul M. Harms, North Newto, KS Let y tx ad z rx where x, r, ad t are positive teal umbers. The the iequality to be proved becomes 1 + t)t + r)1 + r) 1 + r + t)r + rt + t) 8 9. The followig iequalities are equivalet to the above iequality t)t + r)1 + r) 81 + r + t)r + rt + t) 9t + t + rt + t + r + rt + r t) 8r + rt + t + r + t + rt + r t r + t 6rt + t + r + rt + r t. To prove the last iequality we write the left side of this iequality as follows: t rt + r ) + rt t + 1) + tr r + 1) t r) + rt 1) + tr 1). Sice all three terms are o-egative, the above iequalities are correct. Thus the iequality i the problem has bee proved. Solutio 4 by Ekel Hyselaj, Uiversity of Techology, Sydey, Australia ad Elto Bojaxhiu, Kriftel, Germay 7

8 Normalizig the LHS oe ca assume that x + y + z 1 ad so our iequality will become equivalet to provig x + y)y + z)z + x) xy + yz + zx) 8 9 subject to x + y + z 1 ad the fact that x, y ad z are positive real umbers. We will use the Lagrage Multiplier method to fid the miimum of the fuctio x + y)y + z)z + x) fx, y, z) subject to gx, y, z) x + y + z 1 ad the fact that xy + yz + zx) x, y ad z are positive real umbers. Doig easy maipulatios we have that fx, y, z) < f x, f y, f z > ad < xy + z)xy + yz + zx xy + yz + zx), yz + x)xy + yz + zx zx + y)xy + yz + zx xy + yz + zx), xy + yz + zx) > gx, y, z) < g x, g y, g z >< 1, 1, 1 > applyig the Lagrage Multiplier method, the extremes will be the solutios of the system of the equatios { fx, y, z) λ gx, y, z) which is equivalet to gx, y, z) 1 xy + z)xy + yz + zx) xy + yz + zx) yz + x)xy + yz + zx) xy + yz + zx) λ λ zx + y)xy + yz + zx) xy + yz + zx) λ x + y + z 1 where λ is a real umber. Solvig this easy system of equatios we have that the solutios will be 1 x, y, z, λ) {1, 1, 1, ), 1, 1, 1, ), 1, 1, 1, ),, 1, 1, 8 } 9) Usig the fact that x, y ad z are positive real umbers, the oly poit of iterest will be 1 x, y, z), 1, 1 ) ad the value of the fuctio at that poit will be ) ) 1 f, 1 ), ) + 1 )

9 1 Gettig the value of the fuctio fx, y, z) at aother poit, let say x, y, z), 1 4, 1 4) we have ) ) 1 f, 1 4 4), ) + 1 ) > we have that the extreme poit x, y, z), 1, 1 is a miimum ad this is the ed ) of the proof. Solutio 5 by Kee-Wai Lau, Hog Kog, Chia It ca be checked readily that x + y+)y + z)z + x) x + y + z)xy + yz + zx) xy z) + yz x) + zx y) + 8 9x + y + z)xy + yz + zx) 9, ad the iequality of the problem follows. Solutio 6 by Adrea Fachii, Catú Italy Let p x + y + z, q xy + yz + zx ad r xyz. The the give iequality becomes I.e., pq r pq 8 9. pq 9r, that we ca prove easily usig the AM-GM iequality, pq x + y + z)xy + yz + zx) xyz x y z 9r. So the proposed iequality is proved. Commet by Albert Stadler of Herrliberg, Switzerlad I tried to geeralize this problem to variables ad cojectured the followig statemet: Let x 1, x,..., x be real positive umbers,. The i1 x i + x i+1 ) x 1 x x x i i1 1 x i1 i For this says that x 1 + x ) x 1 + x ) which is true. For we have the statemet of problem For 4 this says that, with the assumptio that x +1 x 1. x 1 + x )x + x )x + x 4 )x 4 + x 1 ) x 1 + x + x + x 4 )x 1 x x + x 1 x x 4 + x 1 x x 4 + x x x 4 ) 4 4, which is equivalet to x 1 x x x 4 ), ad this is obviously true. However it turs out that the statemet is false for 5 as is evideced by the couterexample x 1, x, x, x 4, x 5 ) 8,, 1,, 8). 9

10 Also solved by Daiel Lopez Aguayo, Istitute of Mathematics UNAM) Morelia, Mexico; Dioe Bailey, Elsie Campbell ad Charles Dimiie, Sa Agelo, TX; Scott H. Brow, Motgomery, AL; Michael Brozisky, Cetral Islip, NY; David E. Maes, Oeota NY; Ágel Plaza, Uiversity of Las Palmas de Gra Caaria, Spai; Paolo Perfetti two solutios), Departmet of Mathematics, Uiversity Tor Vergata, Rome, Italy; David Stoe ad Joh Hawkis joitly), Statesboro, GA, ad the proposer. 5185: Proposed by José Luis Díaz-Barrero, Barceloa, Spai Calculate, without usig a computer, the value of [ )] si arcta + arcta + arcta + arcta + arcta + arcta. ) 5) 7) 11) 1) 11 Solutio 1 by Adrea Fachii, Catú, Italy Kowig that the argumet of the product of complex umbers is the sum of the argumets of the factors, we ca see that θ arcta +arcta +arcta +arcta ) 5) 7) is the argumet of the followig multiplicatio 1 1 +arcta +arcta 11) 1) + i) 5 + i) 7 + i) 11 + i) 1 + i) i) ) multiplyig i the usual way, we obtai the pure imagiary umber 696i, so θ π ) π ad the fially we have si 1. Solutio by Aastasios Kotrois, Athes, Greece The followig idetities are well kow: arcta a + arcta b π arcta a + arcta 1 a π arcta a + b 1 ab, ab < 1 arcta a + b 1 ab + π, ab > 1 a > arcta a + b 1 ab π, ab > 1 a <, a >, a < 1 Applyig these formulas to the pair arcta ) ) arcta 1 7), arcta 1 11 ad to arcta 1, arcta, ad the repeatig to 5) 1 1), we obtai that 1

11 ) )) si arcta + arcta + arcta + arcta + arcta + arcta ) 5) 7) 11 1) 11 si arcta si π 1. ) 11 + arcta 111 )) Also solved by Bria D. Beasley, Clito, SC; Dioe Bailey, Elsie Campbell ad Charles Dimiie, Sa Agelo, TX; Michael C. Faleski two solutios), Uiversity Ceter, MI; Paul M. Harms, North Newto, KS; Ekel Hyselaj, Uiversity of Techology, Sydey, Australia joitly with Elto Bojaxhiu, Kriftel, Germay; Keeth Korbi, New York, NY; David E. Maes, Oeota NY; Kee-Wai Lau, Hog Kog, Chia; Ágel Plaza, Uiversity of Las Palmas de Gra Caaria, Spai; Paolo Perfetti, Departmet of Mathematics, Uiversity Tor Vergata, Rome, Italy; Boris Rays two solutios), Brookly, NY; Neculai Staciu, Buzău, Romaia with Titu Zvoaru, Comăesti, Romaia; David Stoe ad Joh Hawkis joitly), Statesboro, GA; Albert Stadler, Herrilberg, Switzerlad, ad the proposer. 5186: Proposed by Joh Nord, Spokae, WA k Fid k so that b ) a x + b dx 1 a b ) a x + b dx. Solutio by Ágel Plaza, Departmet of Mathematics, Uiversity of Las Palmas de Gra Caaria, Spai It is clear that if b, the the equatio holds for every k. Assumig that the parameter b, it may be removed, ad the problem the becomes to fid k so that k ) a x dx 1 a ) a x dx. a a We may assume that a >. Itegratig we obtai: ) a x ] k 1 ) a x ] a ) a k 1 1 a a a. Ad, therefore k a 1 1 ) if is eve, while k a 1 ± 1 ) if is odd. Commet by David Stoe ad Joh Hawkis of Statesboro, GA. Whe a ad b are positive, we have the usual area iterpretatio of our result. The problem asks us to determie how far alog we should move to capture half of the area from to a. ) b I this case, the graph of y x a) has y itercept, b), ad drops off to its a a x itercept a, ), so the itegral b ) a x + b dx actually represets the area uder the curve. 11

12 For eve, the graph bottoms out at a, ) ad stays above the x-axis ad we see that k 1 1 ) a is the magical spot where we halve the area. +1 For odd, the graph slices through the x axis at a, ) ad is symmetric about this x itercept, so we have two values of k. The first k 1 a a, actually marks the +1 spot where half of the area from to a is achieved, while the secod k a + a +1 a marks the spot where the et area oce agai equals half of the area from to a. Note that the geometrical iterpretatio is more complicated whe a ad/or b is egative, but the same values of k provide the correct area iterpretatio. Also solved by Daiel Lopez Aguayo, Istitute of Mathematics UNAM) Morelia, Mexico; Bria D. Beasley, Clito, SC; Michael C. Faleski, Uiversity Ceter, MI; Paul M. Harms, North Newto, KS; Ekel Hyselaj, Uiversity of Techology, Sydey, Australia joitly with Elto Bojaxhiu, Kriftel, Germay; Kee-Wai Lau, Hog Kog, Chia; David E. Maes, Oeota NY; Paolo Perfetti, Departmet of Mathematics, Uiversity Tor Vergata, Rome, Italy; Boris Rays, Brookly, NY; Neculai Staciu, Buzău, Romaia; David Stoe ad Joh Hawkis joilty), Statesboro, GA ad the proposer. 5187: Proposed by Ovidiu Furdui, Cluj-Napoca, Romaia Let f : [, 1], ) be a cotiuous fuctio. Fid the value of f 1 f lim ) + ) + + f ). Solutio by Paolo Perfetti, Departmet of Mathematics, Uiversity Tor Vergata, Rome, Italy The aswer is e Proof. 1 lfx))dx f k ) exp{ 1 l[f k )]} l f k ) ) + O 1 ). We observe that beig the fuctio cotiuous, it is bouded above ad below ad the lower boud is positive by the positivity of the fuctio amely < m fx) M for ay x [a, b]. This allowed us to write O1/ ) i the last term regardless the presece of the fuctio fx). Thus, f 1 f ) + )... f k ) ad use l1 + x) x + Ox )) k1 lf k )) + O 1 ). 1 + p + O 1 ) 1 + p + O 1 )) exp{ l1 + p + O 1 )} exp{p + O p ) + O 1 ) + Op )}. 1

13 The quatity p is clearly the Riema sum of 1 lfx))dx ad the the last expoetial is bouded below ad above sice lm) 1 lfx))dx lm). We obtai exp{p + O p ) + O 1 ) + Op )} e 1 lfx))dx. Also solved by Kee-Wai Lau, Hog Kog, Chia; Neculai Staciu, Buzău, Romaia; Albert Stadler, Herrilberg, Switzerlad, ad the proposer. Addedum The ame of Bria D. Beasley of Clito, SC was iadvertetly left off the list of havig solved problem Sorry Bria, mea culpa. Also, Albert Stadler of Herrilberg, Switzerlad oticed two typos i the February, 1 issue of the colum. I his solutio to 5176, the fourth lie i the first equatio array lists the term y, but it should be y. Ad i his solutio to 5178, the last lie should have bee x + y + z ad ot x + y + z. Agai, sorry, mea culpa. 1

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Problems Ted Eiseberg, Sectio Editor ********************************************************* This sectio of the Joural offers readers a opportuity to exchage iterestig mathematical problems ad solutios.