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1 Problems Ted Esenberg, Secton Edtor ********************************************************* Ths secton of the Journal offers readers an opportunty to echange nterestng mathematcal problems and solutons. Proposals are always welcomed. Please observe the followng gudelnes when submttng proposals or solutons:. Proposals and solutons must be legble and should appear on separate sheets, each ndcatng the name and address of the sender. Drawngs must be sutable for reproducton. Proposals should be accompaned by solutons. An astersk (*) ndcates that nether the proposer nor the edtor has suppled a soluton.. Send submttals to: Ted Esenberg, Department of Mathematcs, Ben-Guron Unversty, Beer-Sheva, Israel or fa to: Questons concernng proposals and/or solutons can be sent e-mal to: <esen@math.bgu.ac.l> or to <esenbt@3.net>. Solutons to the problems stated n ths ssue should be posted before June 5, 8 54: Proposed by Kenneth Korbn, ew York, Y. Gven trangle ABC wth a =, b = 5, and wth equal cevans AD and BE. Fnd the permeter of the trangle f AE BD = CE CD. 55: Proposed by Kenneth Korbn, ew York, Y. Part I: Fnd the value of = ( 4 ) Arcsn Part II: Fnd the value of = ( 4 ) Arcsn : Proposed by John ord, Spokane, WA. Locate a pont (p, q) n the Cartesan plane wth ntegral values, such that for any lne through (p, q) epressed n the general form a + by = c, the coeffcents a, b, c form an arthmetc progresson. 57: Proposed by M.. Deshpande, agpur, Inda. Let ABC be a trangle such that each angle s less than 9. Show that a c sn B + tan A = b a sn C + tan B = c b sn A + tan C

2 where a = l(bc), b = l(ac), and c = l(ab). 58: Proposed by José Lus Díaz-Barrero, Barcelona, Span. Wrte the polynomal as a product of two polynomals wth nteger coeffcents. 59: Mchael Broznsky, Central Islp, Y. In a horse race wth horses the horse wth the number one on ts saddle s referred to as the number one horse, and so on for the other numbers. The outcome of the race showed the number one horse dd not fnsh frst, the number two horse dd not fnsh second, the number three horse dd not fnsh thrd and the number four horse dd not fnsh fourth. However, the number fve horse dd fnsh ffth. How many possble orders of fnsh are there for the ten horses assumng no tes? Solutons 4996: Proposed by Kenneth Korbn, ew York, Y. Smplfy: ( ) ( ) )( + 3. Soluton by José Hernández Santago, (student, UTM, Oaaca, Méco.) ( ) ( ) )( + 3 = = ( ) + ( ) + 3 ( ) 3 ( ) ( 3 ) = ( ) ( 3 )(( ) + ) + )(( 3 + ) = ( 5 3 ) 3 = (3 )3 + 3 (5 3 ) 3 = = 5 Also solved by Bran D. Beasley, Clnton, SC; Mchael Broznsky, Central

3 Islp, Y; John Boncek, Montgomery, AL; Else M. Campbell, Donne T. Baley, and Charles Dmnne (jontly), San Angelo, TX; José Lus Díaz-Barrero, Barcelona, Span; Paul M. Harms, orth ewton, KS;. J. Kuenz, Oshkosh, WI; Kee-Wa Lau, Hong Kong, Chna; Carl Lbs, Kngston, RI; R. P. Sealy, Sackvlle, ew Brunswck, Canada; Davd Stone and John Hawkns (jontly), Statesboro, GA, and the proposer. 4997: Proposed by Kenneth Korbn, ew York, Y. Three dfferent trangles wth nteger-length sdes all have the same permeter P and all have the same area K. Fnd the dmensons of these trangles f K = 4. Soluton by Donne Baley, Else Campbell, and Charles Dmnne (jontly), San Angelo, TX. Let a, b, c be the sdes of the trangle and, for convenence, assume that a b c. By Heron s Formula, ( ) ( ) ( ) ( ) (4) a + b + c a + b c a b + c a + b + c = () Snce a, b, c are postve ntegers, t s easly demonstrated that the quanttes (a + b c), (a b + c), and ( a + b + c) are all odd or all even. By (), t s clear that n ths case, they are all even. Therefore, there are postve ntegers, y, z such that a + b c =, a b + c = y, and a + b + c = z. Then, a = + y, b = + z, c = y + z, a + b + c = ( + y + z), and a b c mples that y z. Wth ths substtuton, () becomes (4) = yz ( + y + z) () Snce y z < + y + z, () mples that 4 < yz ( + y + z) = (4) and hence, 4 =, where m denotes the greatest nteger m. Therefore, the possble values of are,, 3, 4, 5, 6, 7, 8, 9,,, 4, 5, 6, 8, (snce must also be a factor of (4) ). Further, for each, () mples that.e., y 3 < yz ( + y + z) = (4), y 3 (4), (and y s a factor of (4) ). Once we have assgned values to and y, () becomes z ( + y + z) = (4) y, (3) whch s a quadratc equaton n z. If (3) yelds an ntegral soluton y, we have found a vable soluton for, y, z and hence, for a, b, c also. By fndng all such solutons, we

4 can fnd all Heronan trangles (trangles wth ntegral sdes and ntegral area) whose area s 4. Then, we must fnd three of these wth the same permeter to complete our soluton. The followng two cases llustrate the typcal steps encountered n ths approach. Case. If = and y = 8, (3) becomes z + 9z 98 =. Snce ths has no ntegral solutons, ths case does not lead to feasble values for a, b, c. Case. If = and y = 4, (3) becomes z + 6z 3675 =, whch has z = 49 as ts only postve ntegral soluton. These values of, y, z yeld a = 6, b = 5, c = 73, and P = 5. The results of our approach are summarzed n the followng table. y z a b c P ow, t s obvous that the last three entres consttute the soluton of ths problem. Also solved by Bran D. Beasley, Clnton, SC; Paul M. Harms, orth ewton, KS; Davd Stone and John Hawkns (jontly), Statesboro, GA, and the proposer. 4998: Proposed by Jyot P. Shwalkar & M.. Deshpande, agpur, Inda. Let A = [a,j ], =,, and j =,,, be a trangular array satsfyng the followng condtons: ) a, = L() for all ) a, = for all 3) a,j = a,j + a,j + a,j a,j for j ( ). If T () = a,j for all, then fnd a closed form for T (), where L() are the Lucas j= numbers, L() =, L() = 3, and L() = L( ) + L( ) for 3.

5 Soluton by Paul M. Harms, orth ewton, KS. ote that a,j s not n the trangular array when j =, so we set a, =. From Lucas numbers a, = a, + a, for >. For >, Therefore we have T () = a, + a, + + a, + = (a, + a, ) + (a, + a, + a, a, ) + +(a, + a, + a, a, ) +. (a, + a, + a, a, ) = ( ) + + a, ( ). ote that n T () each term of row ( ) appears twce and subtracts out and each term of row ( ) ecept for the last term ( ), s added to tself. The term ( ) appears once. If we wrte the last term,, of T () as = ( ) +, then T () = T ( ) +. The values of the row sums are: T () = T () = 5 T (3) = (5) + T (4) = ((5) ( + ) + = ) (5) + + T (5) = [(5) + ] + + = 3 (5) + + +, and n general T () = (5) + ( ) = (5) + ( ) = (6) = (3) for. Also solved by Carl Lbs, Kngston, RI;. J. Kuenz, Oshkosh, WI; R. P. Sealy, Sackvlle, ew Brunswck, Canada; Davd Stone and John Hawkns (jontly), Statesboro, GA, and the proposers. 4999: Proposed by Isabel Díaz-Irberr and José Lus Díaz-Barrero, Barcelona, Span. Fnd all real trplets (, y, z) such that + y + z = +y + y+z + z+ = 6 9 Soluton by Davd E. Manes, Oneonta, Y. The only real soluton s (, y, z) = ( 3, 3, ). ote that these values do satsfy each of 3 the equatons. By the Arthmetc-Geometrc Mean Inequalty, 6 9 = +y + y+z + z+

6 3 3 +y+z +y +z = /3 3 +y +z. Therefore, +y +z 4/3 so that + y + z 4/3 (). ote that 4 = ( + y + z) = + y + z + (y + yz + z), so that + y + z = 4 (y + yz + z). Substtutng n () yelds the nequalty y + yz + z 4 3. From ( y) + (y z) + (z ) wth equalty f and only f = y = z, one now obtans the nequaltes y + z y + yz + z 4 3. Hence. + y + z = y + yz + z = 4 3 ( y) + (y z) + (z ) =, and = y = z = 3. Also solved by Donne Baley, Else Campbell, Charles Dmnne and Karl Havlak (jontly), San Angelo, TX; Mchael Broznsky, Central Islp, Y; Kee-Wa Lau, Hong Kong, Chna; Paolo Perfett, Math Dept. U. of Rome, Italy; Bors Rays, Chesapeake, VA; Davd Stone and John Hawkns (jontly), Statesboro, GA, and the proposers. 5: Proposed by Rchard L. Francs, Cape Grardeau, MO. Of all the rght trangles nscrbed n the unt crcle, whch has the Morley trangle of greatest area? Soluton by Ken Korbn, ew York, Y. Gven ABC wth crcumradus R = and wth A + B = C = 9 o. The sde of the Morley trangle s gven by the formula = 8 R sn( A 3 ) sn(b 3 ) sn(c 3 ) = 8 sn( A 3 ) sn(b 3 ) = 4 sn( A 3 ) sn(b 3 ). wll have a mamum value f Then, A 3 = B 3 = 45o 3 = 5o. = 4 sn (5 o )

7 The area of ths Morley trangle s ( cos 3 o ) = 4 = cos 3 o = 3. ( 3) sn 6 o = 3 (7 4 3) = Comment by Davd Stone and John Hawkns: It may be the mamum, but t s pretty small! Also solved by Mchael Broznsky, Central Islp, Y; Kee-Wa Lau, Hong Kong, Chna; Davd Stone an John Hawkns (jontly), Statesboro, GA, and the proposer. 5: Proposed by Ovdu Furdu, Toledo, OH. Evaluate: ln ( ) d. Soluton by Kee-Wa Lau, Hong Kong, Chna. ( We show that ln )d = ln + π Denote the ntegral by I. Replacng by /, we obtan ) ln (( + )( + ) ln ( + ) I = d = d + = I + I + I 3, say. Integratng by parts, we obtan I = ln ( + )d( ) = + Replacng by /( ),we obtan I = ln( + ) ln( + ) d ln( + ) ( + ) d = ln( ) d = Replacng by / n I, we see that I = I = π. et note that 3 I 3 = ln(+) ln(+)d( ) = ln( + ) ( + ) d+ n= ln ( + ) d ( ) d. n = π 3. ln( + ) ( + ) d = J +J, say.

8 Replacng by /, then by and then by /, we have ln( + ) J = ( + ) d = ( )( + ) d Integratng by parts, we have, = ( )( + ) d = ( + ) d. + J = ln( ) + ln( + ) d = + ( ) n n= n = π 6 + π = π 4. We now evaluate J. Replacng + by and then by /, we have J = ( )( ) d = ( )( ) d = + π d = +K, say. 6 Replacng by K = ln( ) + d = ln( + ) + d + ln( ) ln( + ) d + ( ) = ln + ln + d. + By puttng y = +, we see that the last ntegral reduces to ln y dy = π + y. Hence, K = ln π, J = ln + π, I 3 = ln + 5π and fnally ( ) I = I + I + I 3 = π 3 + π 3 + ln + 5π = ln + π 6 as desred. Also solved by Paolo Perfett, Math. Dept., U. of Rome, Italy; Worapol Rattanapan (student at Montfort College (hgh school)), Chang Ma, Thaland, and the proposer.