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Let-ABCD-be-a-convex-quadrilateral-and-let-k-gt-0-be-a-real-number-Find-the-locus-of-points-M-in-its-interior-such-that-MAB-2-MCD-k-




Question Number 16066 by Tinkutara last updated on 21/Jun/17
Let ABCD be a convex quadrilateral  and let k > 0 be a real number. Find  the locus of points M in its interior  such that  [MAB] + 2[MCD] = k.
LetABCDbeaconvexquadrilateralandletk>0bearealnumber.FindthelocusofpointsMinitsinteriorsuchthat[MAB]+2[MCD]=k.
Answered by mrW1 last updated on 28/Jun/17
Let AB=a, CD=b  ⇒[MAB]=(1/2)ah_1   ⇒[MCD]=(1/2)bh_2   with  h_1 =distance from M to side AB  h_2 =distance from M to side CD    [MAB] + 2[MCD] = k=constant  ⇒(1/2)(ah_1 +2bh_2 )=k  ⇒(1/2)(ah_1 +b′h_2 )=k with b′=2b
LetAB=a,CD=b[MAB]=12ah1[MCD]=12bh2withh1=distancefromMtosideABh2=distancefromMtosideCD[MAB]+2[MCD]=k=constant12(ah1+2bh2)=k12(ah1+bh2)=kwithb=2b
Commented by mrW1 last updated on 28/Jun/17
Commented by mrW1 last updated on 29/Jun/17
Case 1: AB//CD  h_1 +h_2 =h=constant  (1/2)(ah_1 +b′h_2 )=k  ⇒(1/2)[ah_1 +b′(h−h_1 )]=k  ⇒(a−b′)h_1 +b′h=2k    if a=b′:  ⇒ b′h=2k ⇒ k=((b′h)/2)=((ah)/2)  i.e. k must be ((ah)/2), otherwise there is  no solution. In this situation M can  be everywhere in interior of ABCD.    if a≠b′:  (a−b′)h_1 +b′h=2k  ⇒h_1 =((2k−b′h)/(a−b′))=((k/((b′h)/2))−1)×((1/((a/(b′))−1)))×h=constant  ⇒the locus of M is a straight line  parallel to AB and CD. But there  is a solution only if the value of k is  between ((ah)/2) and ((b′h)/2), see diagram above.  When k=((ah)/2), M lies on side CD.  When k=((b′h)/2), M lies on side AB.
Case1:AB//CDh1+h2=h=constant12(ah1+bh2)=k12[ah1+b(hh1)]=k(ab)h1+bh=2kifa=b:bh=2kk=bh2=ah2i.e.kmustbeah2,otherwisethereisnosolution.InthissituationMcanbeeverywhereininteriorofABCD.ifab:(ab)h1+bh=2kh1=2kbhab=(kbh21)×(1ab1)×h=constantthelocusofMisastraightlineparalleltoABandCD.Butthereisasolutiononlyifthevalueofkisbetweenah2andbh2,seediagramabove.Whenk=ah2,MliesonsideCD.Whenk=bh2,MliesonsideAB.
Commented by mrW1 last updated on 29/Jun/17
Case 2: AB and CD meets at O  ∠AOD=α    (1/2)(ah_1 +b′h_2 )=k  ⇒ ah_1 +b′h_2 =2k  that means point M has to fulfill that  ah_1 +b′h_2 =constant which is 2k.    We make OA′=AB=a and  OD′=2CD=2b=b′.    We get A′D′.  Every point on A′D′ fulfills that   ah_1 +b′h_2 =constant.  Every point on any straight line // A′D′  fulfills also this condition.    Each line parallel to A′D′ stands for  a certain value of k from 0 to +∞.    We have following special lines // A′D′:    A1 corresponding to k_1 =2A_(ΔACD)   D2 corresponding to k_2 =A_(ΔADB)   C3 corresponding to k_3 =A_(ΔACB)   B4 corresponding to k_4 =2A_(ΔCBD)     So we have:  if k<k_1  ⇒ no solution for point M    if k_1 ≤k≤k_2  ⇒ Locus of M is a part of straight line from AD to AB    if k_2 ≤k≤k_3  ⇒ Locus of M is a straight line EF // A′D′    if k_3 ≤k≤k_4  ⇒ Locus of M is a part of straight line from BC to AB    if k>k_4  ⇒ no solution for point M
Case2:ABandCDmeetsatOAOD=α12(ah1+bh2)=kah1+bh2=2kthatmeanspointMhastofulfillthatah1+bh2=constantwhichis2k.WemakeOA=AB=aandOD=2CD=2b=b.WegetAD.EverypointonADfulfillsthatah1+bh2=constant.Everypointonanystraightline//ADfulfillsalsothiscondition.EachlineparalleltoADstandsforacertainvalueofkfrom0to+.Wehavefollowingspeciallines//AD:A1correspondingtok1=2AΔACDD2correspondingtok2=AΔADBC3correspondingtok3=AΔACBB4correspondingtok4=2AΔCBDSowehave:ifk<k1nosolutionforpointMifk1kk2LocusofMisapartofstraightlinefromADtoABifk2kk3LocusofMisastraightlineEF//ADifk3kk4LocusofMisapartofstraightlinefromBCtoABifk>k4nosolutionforpointM
Commented by Tinkutara last updated on 29/Jun/17
Thanks Sir! Very much.
ThanksSir!Verymuch.
Commented by mrW1 last updated on 28/Jun/17
Commented by mrW1 last updated on 28/Jun/17
Here is how to determine the line EF  for a certain value of k.    Area of ΔOA′D′=A_0   A_0 =ab′sin α    A′D′ stands for k=k_0 =A_0   EF stands for k    ((OE)/(OA′))=(√(k/A_0 ))  ⇒OE=(√((k/A_0 ) ))a
HereishowtodeterminethelineEFforacertainvalueofk.AreaofΔOAD=A0A0=absinαADstandsfork=k0=A0EFstandsforkOEOA=kA0OE=kA0a
Answered by Tinkutara last updated on 29/Jun/17
Combining Problems 3 and 5 and using  the construction above, we deduce  that the area of XY ′M is constant.  (Figure). It results that the locus is  (according to the value of k) either a  segment KL parallel to XY ′, or a  point, or the empty set.
CombiningProblems3and5andusingtheconstructionabove,wededucethattheareaofXYMisconstant.(Figure).Itresultsthatthelocusis(accordingtothevalueofk)eitherasegmentKLparalleltoXY,orapoint,ortheemptyset.
Commented by Tinkutara last updated on 29/Jun/17
Commented by Tinkutara last updated on 29/Jun/17
This is my book′s solution. Here  Problem 3 is Q. 16064 and 5 is Q. 16066  problem. (This Question)
Thisismybookssolution.HereProblem3isQ.16064and5isQ.16066problem.(ThisQuestion)
Commented by mrW1 last updated on 30/Jun/17
In princip the answer is the same as  mine. The locus of M is KL which is  parallel to XY′ (but the figure shows  KL//XY, this is wrong)  And my answer is much more exact,  it tells you the different situations for  the locus of M depending on the ranges for  k: k_1  till t_4 . My answer tells also how  to determine the position of locus of  M for a certain value of k.  The answer of book mentions only  that the solution could be empty set,  but not when. My answer tells exactly  when, i.e. if k<k_1  or k>k_4  and gives  the values of k_1  and k_4  as well as other  special k−values k_2  and k_3  when the  locus of M is a single point (k=k_1 ,k=k_4 ), part of a  segment from AB to CD (k_1 <k<k_2 , k_3 <k<k_4  )or a whole  segment from AB to CD (k_2 ≤k≤k_3 ).    But the core of both solutions is the  same: the locus of M is a straight line  parallel to XY′ (=A′D′ by me) with  TX=AB (=OA′ by me) and  TY′=2CD (=OD′ by me)    Certainly my answer treats also the  case if AB//CD.
Inprinciptheansweristhesameasmine.ThelocusofMisKLwhichisparalleltoXY(butthefigureshowsKL//XY,thisiswrong)Andmyanswerismuchmoreexact,ittellsyouthedifferentsituationsforthelocusofMdependingontherangesfork:k1tillt4.MyanswertellsalsohowtodeterminethepositionoflocusofMforacertainvalueofk.Theanswerofbookmentionsonlythatthesolutioncouldbeemptyset,butnotwhen.Myanswertellsexactlywhen,i.e.ifk<k1ork>k4andgivesthevaluesofk1andk4aswellasotherspecialkvaluesk2andk3whenthelocusofMisasinglepoint(k=k1,k=k4),partofasegmentfromABtoCD(k1<k<k2,k3<k<k4)orawholesegmentfromABtoCD(k2kk3).Butthecoreofbothsolutionsisthesame:thelocusofMisastraightlineparalleltoXY(=ADbyme)withTX=AB(=OAbyme)andTY=2CD(=ODbyme)CertainlymyanswertreatsalsothecaseifAB//CD.
Commented by Tinkutara last updated on 30/Jun/17
Thanks Sir! I knew your solution was  correct. But I posted the solution in my  book to show that it has a good  collection of questions but not solutions.
ThanksSir!Iknewyoursolutionwascorrect.ButIpostedthesolutioninmybooktoshowthatithasagoodcollectionofquestionsbutnotsolutions.
Commented by mrW1 last updated on 30/Jun/17
They are really interesting questions.
Theyarereallyinterestingquestions.

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