1 THE ACADEMY CORNER No. 16 Bruce Shawyer All communications about this column should be sent to Bruce Shawyer, Department of Mathematics and Statistics, Memorial University ofNewfoundland,St.John’s,Newfoundland,Canada. A1C 5S7 Thismonth, we presentsome more ofthesolutionsto a universityen- trance scholarshipexaminationpaperfromthe1940’s,whichappearedinthe April 1997issueofCRUX withMAYHEM[1997: 129]. 3 4. (a) Supposethat a = 0andc = 0,andthat ax +bx+c hasafactor 2 6 6 2 2 of the form x +px+1. Showthat a c = ab. (cid:0) 3 3 2 (b) In this case, prove that ax + bx + c and cx + bx + a have a common quadratic factor. Solution. (a) The given conditions on the coe(cid:14)cients show that x = 0 is not a solution. Also, the other factor mustbe linear, and, onexamining coe(cid:14)cients, mustbe ofthe form ax+c. Therefore 3 2 3 2 ax +bx+c = (x +px+1)(ax+c) = ax +(c+ap)x +(a+cp)x+c: Comparingcoe(cid:14)cients gives c+ap = 0; a+cp = b: 2 c c Thus, p = a, so that a a = b, givingthe required result. (cid:0) (cid:0) 1 (b) Since x = 0, we write y = , and examine the corresponding x 6 equations iny to getthispart. 5. Prove that allthe circles inthefamilyde(cid:12)nedbythe equation 2 2 2 2 x +y a(t +2)x 2aty 3a = 0 (cid:0) (cid:0) (cid:0) (a (cid:12)xed, t variable) touch a (cid:12)xedstraight line. 2 Solution. We re-write the equation as 2 2 2 2 t 2 t x a 1+ +(y at) = a 2+ : (cid:0) 2 (cid:0) 2 (cid:18) (cid:18) (cid:19)(cid:19) (cid:18) (cid:18) (cid:19)(cid:19) Thisshowsthatthelinex = aisatangentline. (They{coordinate is (cid:0) 2at.) 6. Find the equation of the locus of a point P which moves so that the 2 2 2 tangents from P to the circle x + y = r cut o(cid:11) a line segment of length2r on the linex = r. Solution. 2 2 2 From the circle, x +y = a , we (cid:12)nd the points of intersection with the line(y q)= m(x p). (cid:0) (cid:0) 2 2 2 Substitutingy = mx mp+q, we get x +(mx mp+q) = a , (cid:0) (cid:0) leadingto the two solutions: 2 2 2 2 m(mp q) a (m +1) m p +q(2mp q) x = (cid:0) (cid:6) (cid:0) (cid:0) : 2 m +1 p 2 2 For the line to be a tangent, the discriminant, (cid:1) = a (m + 1) 2 2 (cid:0) m p +q(2mp q),must bezero. Sowe solve(cid:1) = 0, to get (cid:0) 2 2 2 pq a p +q a m = (cid:0) : p2 a2 (cid:6) p2 a2 p (cid:0) (cid:0) So, we have that the equations of the tangent lines from (p;q) to the 2 2 2 circle x +y = a are: 2 2 2 pq a a p q y = (cid:0) (cid:0) x (cid:0)a2 p2 (cid:6) p p2 a2 ! (cid:0) (cid:0) 2 2 2 2 2 a p q pq=(a p )+ (cid:0) (cid:0) p+q: (cid:0) (cid:0) (cid:0) (cid:6)p p2 a2 ! (cid:0) Settingx = a inthese, andsubtracting the two valuesleadsto 2 2 2 2 (a+p) = p +q a : (cid:0) Thusthe locusofP isgivenby 2 y = 2a(a+x): Thisisaparabola openingto therightwithcentral axis,thex{axisand nose at x = a. There are, of course, three points that should be (cid:0) excluded: ( a;0), (a;2a),and (a; 2a). (cid:0) (cid:0) 3 7. IfthetangentsatA, B,C,to thecircumcircle oftriangle ABC meet 4 the opposite sides at D, E, F, respectively, prove that D, E, F, are collinear. Solution. E F P C B D Q R A For notational ease, we write PB = PC = (cid:11), QC = QA = (cid:12) and RB = RA = (cid:13). ApplyMenelaus’Theoremfortriangle PQRwithtransversalsBDC, 4 EAC and BAF PB RD QC QC RD 1 = = ; (cid:0) BR DQ CP BR DQ PE RA QC PE RA 1 = = ; (cid:0) ER AQ CP ER CP PB RA QF PB QF 1 = = : (cid:0) BR AQ FP AQ FP These leadto (cid:13) +(cid:12)+QD BR (cid:13) = = ; (1) QD (cid:0)QC (cid:0)(cid:12) (cid:11)+(cid:13) +ER CP (cid:11) = = ; (2) ER (cid:0)RA (cid:0)(cid:13) QF AQ (cid:12) = = : (3) FQ+(cid:11)+(cid:12) (cid:0)PB (cid:0)(cid:11) Now consider the product of ratios to prove D, E and F collinear by the converse ofMenelaus: PE RD QF : ER DQ FP Thisisequalto (cid:11)+(cid:12) +ER (cid:13) +(cid:12) +QD QF : ER QD FQ+(cid:11)+(cid:12) Using (1), (2) and (3) above, we get a value of 1, and the result is (cid:0) proved. 4 THE OLYMPIAD CORNER No. 187 R.E. Woodrow AllcommunicationsaboutthiscolumnshouldbesenttoProfessorR.E. Woodrow,DepartmentofMathematicsandStatistics,UniversityofCalgary, Calgary, Alberta, Canada. T2N 1N4. Anotheryearhaspassed|Howthetimeseemsto(cid:13)y. Iwouldparticu- larlyliketothankJoanneLongworthforherexcellenthelpinpullingtogether A the information for thecolumn and herwork inproducing agood LTEX copy fortheEditor-in-Chief{usuallyunderenormoustimepressurebecauseIam behindschedule. TheCornercouldnotexistwithoutitsreaderswhosupplymewithgood Olympiadmaterials andsendintheirinterestingsolutionsandcomments to problems in the Corner. I hope we have missedno one in the followinglist of contributors. Miguel AmengualCovas JoelKamnitzer BobPrielipp Se(cid:20)fketArslanagic(cid:19) DeepeeKhosla ToshioSeimiya MansurBoase DerekKisman MichaelSelby ChristopherBradley MurrayKlamkin ZunShan Sabin Cautis MarcinKuczma D.J. Smeenk AdrianChan AndyLiu DarylTingley Byung-Kuy Chun BeatrizMargolis PanosTsaoussoglou Mihaela Enachescu VedulaMurty Ravi Vakil GeorgeEvagelopoulos RichardNowakowski DanVelleman ShawnGodin ColinPercival StanWagon JoanneJuszun(cid:19)ska EdwardWang Thank youall, andallthe bestfor 1998! th The(cid:12)rstOlympiadwegiveforthenewyearisthe17 Austrian-Polish Mathematics Competition, written in Poland, June 29{July 1, 1994. My th thanksgotoRichardNowakowski,CanadianTeamLeadertothe35 IMOin HongKongandto WaltherJanous,Ursulinengymnasium,Innsbruck,Austria for sendingacopy to me. 17th AUSTRIAN{POLISH MATHEMATICS COMPETITION Poland, June 29{July 1, 1994 1. Thefunctionf : R Rsatis(cid:12)esforallx Rthe conditions ! 2 f(x+19) f(x)+19 and f(x+94) f(x)+94: (cid:20) (cid:21) 5 Showthat f(x+1) = f(x)+1for allx R. 2 2. Thesequence (an)isde(cid:12)nedbythe formulas 1 2an a0 = and an+1 = for n 0; 2 2 1+an (cid:21) and thesequence (cn)isde(cid:12)nedbythe formulas 2 c0 = 4 and cn+1 = cn 2cn +2 for n 0: (cid:0) (cid:21) Prove that 2c0c1:::cn(cid:0)1 an = for all n 1: cn (cid:21) 3. Arectangularbuildingconsistsoftworowsof15squarerooms(situ- ated likethecellsintwoneighbouringrowsofachessboard). Eachroom has three doors which lead to one, two or all the three adjacent rooms. (Doors leading outside the buildingare ignored.) The doors are distributed insuch awaythat onecanpassfromanyroom toanyother onewithoutleavingthe building. How many distributionsof thedoors (inthe wallsbetween the 30 rooms) can befoundsoasto satisfythe givenconditions? 4. Let n 2be a (cid:12)xed natural number andlet P0 be a (cid:12)xedvertex of (cid:21) theregular(n+1)-gon. TheremainingverticesarelabelledP1;P2;::: ;Pn, in any order. To each side of the (n + 1)-gon assign a natural number as follows: ifthe endpointsofthesideare labelledPi andPj,theni j isthe (cid:0) number assigned. Let S be the sumofallthe n+1numbers thusassigned. (Obviously,Sdependsontheorderinwhichtheverticeshavebeenlabelled.) (a) What istheleast valueof S available(for (cid:12)xedn)? (b) How many di(cid:11)erent labellingsyieldthisminimumvalueofS? 5. Solvethe equation 1 3 (x+y)(y+z)(z+x)+(x+y+z) = 1 xyz 2 (cid:0) inintegers. 6. Let n > 1 be an odd positive integer. Assume that the integers x1;x2;::: ;xn 0satisfythe system ofequations (cid:21) 2 2 (x2 x1) +2(x2 +x1)+1 = n (cid:0) 2 2 (x3 x2) +2(x3 +x2)+1 = n (cid:0) ...................................... 2 2 (x1 xn) +2(x1 +xn)+1 = n : (cid:0) Show that either x1 = xn or there exists j with 1 j n 1 such that (cid:20) (cid:20) (cid:0) xj = xj+1. 6 7. Determine all two-digit (in decimal notation) natural numbers n = (ab)10 = 10a+b(a 1)withtheproperty thatforeveryintegerxthe a b (cid:21) di(cid:11)erence x x isdivisibleby n. (cid:0) 8. Consider the functional equation f(x;y) = a f(x;z)+ b f(y;z) withrealconstants a, b. Forevery pairofreal numbersa, bgivethe general 2 formoffunctionsf : R Rsatisfyingthegivenequationforallx;y;z R. ! 2 9. On the plane there are given four distinct points A, B, C, D lying (inthisorder) ona lineg,at distancesAB = a, BC = b,CD = c. (a) Construct, whenever possible, a point P, not on g, such that the angles\APB,\BPC, \CPD are equal. (b)Prove thatapointP withtheproperty asaboveexistsifandonlyif the followinginequalityholds: (a+b)(b+c) < 4ac. As a second source of problem pleasure for winter evenings (that is if youarehavingwinterthisJanuary!),wegivetheSecondRoundoftheIranian National Mathematical Olympiad. My thanks go to Richard Nowakowski, th CanadianTeamLeadertothe35 IMOinHongKongforcollectingtheprob- lems andforwardingthem to me. IRANIAN NATIONAL MATHEMATICAL OLYMPIAD February 6, 1994 Second Round 1. Supposethatpisaprime numberandisgreater than3. Prove that p p 7 6 1isdivisibleby 43. (cid:0) (cid:0) 2. Let ABC be an acute angled triangle with sides and area equal to a, b, c and S respectively. Show that the necessary and su(cid:14)cient condition for existence of a point P inside the triangle ABC such that the distance between P and the vertices of ABC be equal to x, y and z respectively is that there be a triangle with sides a, y, z and area S1, a triangle with sidesb,z, xandarea S2 andatriangle withsidesc, x, y andarea S3 where S1 +S2 +S3 = S. 3. Let n and r be natural numbers. Find the smallest natural num- ber m satisfyingthis condition: For each partition of the set 1;2;::: ;m f g into r subsets A1;A2;::: ;Ar there exist two numbers a and b in some Ai a 1 (1 i r)suchthat 1 < 1+ . b n (cid:20) (cid:20) (cid:20) 4. G is a graph with n vertices A1;A2;::: ;An such that for each pairofnonadjacentvertices Ai andAj there existsanother vertexAk thatis adjacent to both Ai andAj. (a) Findthe minimumnumberofedgesof suchagraph. 7 (b) If n = 6 and A1;A2;A3;A4;A5;A6 form a cycle of length 6, (cid:12)nd the number of edges that must be added to this cycle such that the above condition holds. 5. Show that if D1 and D2 are two skew lines, then there are in(cid:12)- nitelymanystraightlinessuchthattheirpointshaveequaldistancefromD1 and D2. 6. f(x) and g(x) are polynomials with real coe(cid:14)cients such that for f(x) f(x) in(cid:12)nitely many rational values x, is rational. Prove that can be g(x) g(x) written asthe ratio oftwo polynomialswithrational coe(cid:14)cients. Nowweturntoreaders’solutionsandcommentsforproblemsposedin th theCorner. Firstsolutionstoaproblemofthe25 UnitedStatesofAmerica Mathematical Olympiad[1996: 203{204]. 1. Prove that the average ofthe numbers (cid:14) nsinn ; n = 2;4;6;::: ;180 (cid:14) iscot1 . Solutionby D.J.Smeenk, Zaltbommel,theNetherlands. Let (cid:14) (cid:14) (cid:14) (cid:14) x = 2sin2 +4sin4 + +90sin90 + +178sin178 (cid:14) (cid:1)(cid:1)(cid:1) (cid:14) (cid:1)(cid:1)(cid:1) = (2+178)sin2 +(4+176)sin4 + (cid:14) (cid:14) (cid:14) (cid:1)(cid:1)(cid:1) (cid:14) = 180(sin2 +sin4 + +sin88 )+90sin90 : (cid:1)(cid:1)(cid:1) Then x (cid:14) (cid:14) (cid:14) x(cid:22) = = 2sin2 +2sin4 + +2sin88 +1 90 (cid:1)(cid:1)(cid:1) (cid:14) (cid:14) (cid:14) (cid:14) (cid:14) (cid:14) (cid:14) (cid:14) x(cid:22)sin1 = 2sin2 sin1 +2sin4 sin1 + +2sin88 sin1 +sin1 : (cid:1)(cid:1)(cid:1) Now (cid:14) (cid:14) (cid:14) (cid:14) 2sin2 sin1 = cos1 cos3 (cid:14) (cid:14) (cid:14)(cid:0) (cid:14) 2sin4 sin1 = cos3 cos5 (cid:0) (cid:14) (cid:14) (cid:1)(cid:1)(cid:1) (cid:14) (cid:14) 2sin88 sin1 = cos87 cos89 : (cid:0) Hence (cid:14) (cid:14) (cid:14) (cid:14) x(cid:22)sin1 = cos1 cos89 +sin1 (cid:14) (cid:0) = cos1 : (cid:14) Thusx(cid:22) = cot1 , asrequired. 8 Next we give a comment on a problem, and one solution to another from the1994Italian Mathematical Olympiad. 2. [1996:204]ItalianMathematicalOlympiad1994. Findallintegersolutionsofthe equation 2 3 y = x +16: Comment byMiguelAmengual Covas,Cala Figuera,Mallorca,Spain. Theequationhasnosolutionsinintegersdi(cid:11)erentfromx = 0,y = 4. (cid:6) For a proof see Theorem 20, page 102of W. Sierpinski’sElementary Theory ofNumbers. 4. [1996: 204]ItalianMathematicalOlympiad1994 Letr bealineintheplaneandletABC beatrianglecontained inone 0 0 0 of the half-planesdetermined by r. Let A , B , C be the points symmetric 0 to A, B, C with respect to r; draw the line through A parallel to BC, the 0 0 linethroughB paralleltoAC andthelinethroughC paralleltoAB. Show that thesethree lineshave acommon point. Solutionby MiguelAmengual Covas,Cala Figuera,Mallorca,Spain. qC There is no need that ABC be 0 4 A contained in one of the half planes q q B r determined byr. Aq qB0 If the coordinates of A are (x1;y1), thecoordinatesofBare(x2;y2)and q 0 C thoseofC are(x3;y3)inaCartesian coordinate system with r as x-axis, 0 then A has coordinates (x1; y1), 0 (cid:0) 0 B hascoordinates (x2; y2)andC (cid:0) hascoordinates (x3; y3). (cid:0) We have: 0 The equationof thelinethroughA parallel to BC is (y2 y3)x (x2 x3)y x1(y2 y3) y1(x2 x3) = 0: (cid:0) (cid:0) (cid:0) (cid:0) (cid:0) (cid:0) (cid:0) 0 The equationof thelinethroughB parallel to CA is (y3 y1)x (x3 x1)y x2(y3 y1) y2(x2 x1) = 0: (cid:0) (cid:0) (cid:0) (cid:0) (cid:0) (cid:0) (cid:0) 0 The equationof thelinethroughC parallel to AB is (y1 y2)x (x1 x2)y x3(y1 y2) y3(x1 x2) = 0: (cid:0) (cid:0) (cid:0) (cid:0) (cid:0) (cid:0) (cid:0) Since the three equations when added together vanish identically, the linesrepresented bythem meet ina point. 9 Its coordinates are found,bysolvingbetween anytwo, to be 2 2 2 (x1+x2x3)(y2 y3)+(x2+x3x1)(y3 y1)+(x3 +x1x2)(y1 y2) (cid:0) (cid:0) (cid:0) ; x1(y2 y3)+x2(y3 y1)+x3(y1 y2) (cid:18) (cid:0) (cid:0) (cid:0) 2 2 2 (y1 +y2y3)(x2 x3)+(y2 +y3y1)(x3 x1)+(y3 +y1y2)(x1 x2) (cid:0) (cid:0) (cid:0) : x1(y2 y3)+x2(y3 y1)+x3(y1 y2) (cid:0) (cid:0) (cid:0) (cid:19) Now we jump to the October 1996 number of the Corner, and two th solutionstoproblemsofthe10 IBEROAMERICANMathematicalOlympiad [1996: 251{252]. 1. (Brazil). Determine all the possiblevalues of the sum of the digits of the perfect squares. Solution by Mansur Boase, student, St. Paul’s School, London, Eng- land. The squarescan onlybe 0, 1,4 or 7 mod 9. Thus the sum of the digits of a perfect square cannot be 2, 3, 5, 6 or 8 mod 9, sincethe number itselfwouldthenbe 2,3, 5, 6or 8 mod9. We shall show that the sum of the digits of a perfect square can take every valueof theform 0, 1, 4or 7 mod9. m 2 2m m (10 1) = 10 2 10 +1 = 99 9 8 0 0 1; m 1: (cid:0) (cid:0) (cid:1) (cid:1)(cid:1)(cid:1) (cid:1)(cid:1)(cid:1) (cid:21) m(cid:0)1 m(cid:0)1 The sum of the digitsis 9m, givingall the|va{lzues}grea|te{rzt}han or equal to 9 congruent to 0 mod 9 m 2 2m m (10 2) = 10 4 10 +4 = 99 9 6 0 0 4; m 1: (cid:0) (cid:0) (cid:1) (cid:1)(cid:1)(cid:1) (cid:1)(cid:1)(cid:1) (cid:21) m(cid:0)1 m(cid:0)1 The sumofthe digitsis9m+1,whichgiv|es{azllv}alue|sg{rzea}ter thanor equal to 10congruent to 1 mod 9. m 2 2m m (10 3) = 10 6 10 +9 = 99 9 4 0 0 9; m 1: (cid:0) (cid:0) (cid:1) (cid:1)(cid:1)(cid:1) (cid:1)(cid:1)(cid:1) (cid:21) m(cid:0)1 m(cid:0)1 Thesumofthedigitsis9m+4,whichtake|sev{ezry}valu|eg{zrea}terthanorequal to 13whichiscongruent to 4 mod9 m 2 2m m+1 (10 5) = 10 10 +25 = 9 900 0 25: (cid:0) (cid:0) (cid:1)(cid:1)(cid:1) (cid:1)(cid:1)(cid:1) m(cid:0)1 m(cid:0)1 The sum of the digits is 9(m 1)+7 = 9m |2,{zfro}m| w{zhich} we get every (cid:0) (cid:0) value greater thanor equalto 7 congruent to 7 mod 9. We havetaken care ofalltheintegers apart from 0, 1,4, whichare the 2 2 2 sumsofthe digitsof0 , 1 , and2 respectively. 10 5. (Spain). The inscribed circumference in the triangle ABC is tan- gent to BC, CA and AB at D, E and F, respectively. Suppose that this circumference meets AD againat its mid-pointX; that is,AX = XD. The lines XB and XC meet the inscribed circumference again at Y and Z, re- spectively. Showthat EY = FZ. Solutionby ToshioSeimiya, Kawasaki, Japan. A Since \BFY = \BXF and \FBY = \XBF we have BFY 4 and BXF are similar, sothat 4 FY : FX = BF : BX: (1) X Similarlywe get E F DY : DX = BD : BX: (2) AsBF = BD, wehavefrom (1)and Y Z (2) that B D C FY : FX = DY : DX: Since AX = DX we get FY : FX = DY : AX: (3) Since X, F,Y, D are concyclic wehave \FYD = \AXF: (4) Thuswe getfrom (3) and(4) that FYD issimilarto FXA. Hence \YFD = \XFA =4\XDF so that FY4XD. Similarly we k have EZ XD. ThusFY EZ. k k Therefore FYZE isan isoscelestrapezoid andthenEY = FZ. WenowturntotwosolutionstotheMaxiFinale1994oftheOlympiade Mathe(cid:19)matiqueBelge[1996: 253{254]. 1. Unpentagone planconvexe adeuxanglesdroits nonadjacents. Les deux co^te(cid:19)s adjacents au premier angle droit out des longueurs e(cid:19)gales. Les deuxco^te(cid:19)sadjacentsausecondangledroitontdeslongueurse(cid:19)gales. Enrem- plac(cid:24)ant par leur point milieu les deux sommets du pentagone situe(cid:19)ssur un seulco^te(cid:19) decesanglesdroits,nousformonsunquadrilate(cid:18)re. Cequadrilate(cid:18)re admet-il ne(cid:19)cessairement unangledroit?