“費(fèi)馬點(diǎn)”與中考數(shù)學(xué)試題

“費(fèi)馬點(diǎn)”與中考數(shù)學(xué)試題

Fermat'sPointandMiddleSchoolMathTestQuestionsFermat,aFrenchamateurmathematician,wasknownasthe"KingofAmateurMathematics"andwasoneoftheinventorsofanalyticgeometry.Fermat'sPointisthepointwiththesmallestsumofdistancestothethreeverticesofatriangle.Fermat'sconclusionisthatforatrianglewithanglesnotexceeding120degrees,Fermat'sPointisthepointwheretheanglesbetweeneachsideareall120degrees.Foratrianglewithoneangleexceeding120degrees,Fermat'sPointisthevertexofthatangle.Nowlet'sbrieflyexplainhowtofindpointPsothatthesumofitsdistancestothethreeverticesoftriangleABCisthesmallest,whichistheso-calledFermatproblem.AsshowninFigure1,rotatetriangleAPCcounterclockwisearoundpointAby6degreestoobtaintriangleAP'C',andconnectPP'.Then,triangleAPP'isanequilateraltrianglewithAP=PP',andP'C'=PC.Therefore,PA+PB+PC=PP'+PB+P'C'.PointC'canbeseenasthefixedpointobtainedbyrotatingsegmentACcounterclockwisearoundpointAby6degrees,andBC'isafixedlength.Therefore,whenpointsB,P,P',andC'arecollinear,PA+PB+PCisthesmallest.Atthistime,∠BPA=180°-∠APP'=180°-60°=120°,∠APC=∠AP'C'=180°-∠AP'P=180°-60°=120°,and∠BPC=360°-∠BPA-∠APC=360°-120°-120°=120°.Therefore,wheneachinternalangleoftriangleABCislessthan120degrees,thepointPwearelookingforhasanglesof120degreesoneachsideofthetriangle.Wecandraw120-degreearcsonsidesABandBC,andtheintersectionofthetwoarcsinsidethetriangleispointP.Whenoneinternalangleisgreaterthanorequalto120degrees,thepointPwearelookingforisthevertexoftheobtuseangle.Fermat'sproblemtellsusthatthereissuchapointwiththesmallestsumofdistancestothreefixedpoints,andthesolutionistouserotationaltransformation.Thisarticleprovidesexamplesofhow"Fermat'sPoint"hasappearedinrecentmiddleschoolmathtestpapersforstudentstolearnandreference.Forexample1(2008GuangdongMiddleSchoolEntranceExam),givenasquareABCDandamovingpointEinsidethesquare,theminimumsumofthedistancesfromEtopointsA,B,andCis2+6.Whatisthelengthofthesideofthesquare?Analysis:ConnectACandwefindthatthesumofthedistancesfrompointEtopointsA,B,andCisactuallythesumofthedistancesfromEtothethreeverticesoftriangleABC.ThisisactuallyavariationofFermat'sproblem,justwithadifferentbackground.AsshowninFigure2,connectACandrotatetriangleAECclockwisearoundpointCby60degreestoobtaintriangleGFC.ConnectEF,BG,andAG,anditcanbeseenthattrianglesEFCandAGCarebothequilateraltriangles,soEF=CE.Also,FG=AE.Therefore,AE+BE+CE=BE+EF+FG(Figure4).PointBandpointGarefixedpoints(GisobtainedbyrotatingpointAclockwisearoundpointCby60degrees).Therefore,segmentBGistheminimumsumofdistancesfrompointEtopointsA,B,andC,andatthistime,pointsEandFarebothonBG(Figure3).Assumingthelengthofthesideofthesquareisa,thenBO=CO=26.GC=2a,GO=a,BG=BO+GO=3a.因?yàn)辄c(diǎn)E到A、B、C三點(diǎn)的距離之和的最小值為2+6=8，所以a+a=8/3，解得a=4/3.在平面直角坐標(biāo)系xOy中，△ABC三個(gè)頂點(diǎn)的坐標(biāo)分別為A(-6,0)，B(6,0)，C(0,4/3).延長(zhǎng)AC到點(diǎn)D，使CD=DE∥AB交BC的延長(zhǎng)線于點(diǎn)E.(1)求D點(diǎn)的坐標(biāo)：由于CD=DE，所以D的坐標(biāo)為(0,16/3).(2)作C點(diǎn)關(guān)于直線DE的對(duì)稱點(diǎn)F，分別連接DF、EF，若過(guò)B點(diǎn)的直線y=kx+b將四邊形CDFE分成周長(zhǎng)相等的兩個(gè)四邊形，確定此直線的解析式：直線BM的解析式為y=-3x+63.(3)設(shè)G為y軸上一點(diǎn)，點(diǎn)P從直線y=kx+b與y軸的交點(diǎn)出發(fā)，先沿y軸到達(dá)G點(diǎn)，再沿GA到達(dá)A點(diǎn)，若P點(diǎn)在y軸上運(yùn)動(dòng)的速度是它在直線GA上運(yùn)動(dòng)速度的2倍，試確定G點(diǎn)的位置，使P點(diǎn)按照上述要求到達(dá)A點(diǎn)所用的時(shí)間最短.解法：使P點(diǎn)到達(dá)A點(diǎn)所用的時(shí)間最短，就是使MQ+AQ/2最小，即使G點(diǎn)到A、B、M三點(diǎn)的距離和最小.把△MQB繞點(diǎn)B順時(shí)針旋轉(zhuǎn)60°，得到△M'Q'B，連接QQ'、MM'，可知△QQ'B、△MM'B都是等邊三角形，則QQ'=BQ.又M'Q'=MQ，所以MQ+AQ/2=BQ+M'Q'+AQ/2=BM'+OM+AQ/2=63-3b/2+4/3+√(b^2+63^2)/2.因此，要使MQ+AQ/2最小，就要使√(b^2+63^2)最小，即b=0.所以G的坐標(biāo)為(0,0).可得角BMO=3度，因此QK=MQ。為了使MQ+AQ最小，只需使AQ+QK最小。根據(jù)“垂線段最短”原理，當(dāng)點(diǎn)A、Q、K在一條直線上時(shí)，AQ+QK最小，且此時(shí)的QK垂直于BM，因此點(diǎn)Q即為所求的點(diǎn)G。過(guò)點(diǎn)A作AH⊥BM于點(diǎn)H，則AH與y軸的交點(diǎn)為所求的點(diǎn)G。由OB=6，OM=63，可得角OBM=60度，因此角BAH=30度。在直角三角形OAG中，OG=AO·tanBAH=23，因此點(diǎn)G的坐標(biāo)為（，23）（點(diǎn)G為線段OC的中點(diǎn)）。如果點(diǎn)P為△ABC所在平面上一點(diǎn)，且∠APB=∠BPC=∠CPA=120度，則點(diǎn)P叫做△ABC的費(fèi)馬點(diǎn)。在銳角△ABC的費(fèi)馬點(diǎn)P中，如果∠ABC=60度，PA=3，PC=4，則PB的值為23。如圖8，在銳角△ABC的外側(cè)作等邊△ACB′，并連接BB′，則BB′過(guò)△ABC的費(fèi)馬點(diǎn)P，且BB′=PA+PB+PC。將△ACP繞點(diǎn)C順時(shí)針旋轉(zhuǎn)60度到△B′CE，連接PE，則△EPC為正三角形。因?yàn)椤螧′EC=∠APC=120度，∠PEC=60度，所以∠B′EC+∠PEC=180度，即P、E、B′三點(diǎn)在同一直線上。因?yàn)椤螧PC=120度，∠CPE=60度，所以∠BPC+∠CPE=180度，即B、P、E三點(diǎn)在同一直線上。因此，B、P、E、B′四點(diǎn)在同一直線上，即BB′過(guò)△ABC