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Geometry ProblemLevel: Mathematics Education, High School, Honors Geometry, College.Problem submitted by Charles T.Click the figure below to see the complete problem 860.
∵ BE//AC and DM(B)//BA∴ ABDM(B) is a parallelogram⇒ DM(A) = M(A)M(B) = 1/2 AB = AM(C)⇒ AM(A)DM(C) is a parallelogram⇒ AM(A) = DM(C)Also, BD = AM(B) = 1/2 AC = CM(B)⇒ BDCM(B) is a parallelogram⇒ BM(B)= DC***Let S(ABC) = area of ΔABC∵ M(A)M(C)//AC∴ S(CM(A)M(C)) = S(M(A)M(B)M(C)) = 1/4 S(ABC)and S(DM(A)M(C)) = S(BM(A)M(C)) = 1/4 S(ABC)∵ M(B)M(C)//BC∴ S(CDM(A)) = S(BM(A)M(B)) = S(BM(A)M(C)) = 1/4 S(ABC)∴ S(CDM(C)) = 3/4 S(ABC)
By affine transformation,(all properties are preserved) map triangleABC into an equiliateral traingle. It is then easy to observe that Mc-A-D-Ma , is a parrallelogram and B-D-C-Mb is a rectangle. Hence the first two statements are proved.Therefore, triangle C-D-Mc has lengths which are equal to the length of medians of ABC which is in a ratio of sqrt(3)/2 : 1. So, the area would be 3/4 : 1.q.e.d.
http://img39.imageshack.us/img39/4396/problem860.bmp1.note that BMbCD is a parallelogram => BMb=DCSince AMc=MaMb=DMa => AMcDMa is a parallelogram => AMa=DMc2. Area of ABC= area of parallelogram BMbCDSince CMbMcMa is a parallelogram => CMb=MaMb and MaF=1/2. CMbSo MbH/McG=2/3Area of Triangle(ABC)/Area of parallelogram(MbCDB)=1/2. (McG/MbH)=3/4So Area CDMc=3/4 Area ABC
Area CDMc= AreaDBMc+AreaACMc=1/2 Area DBAC Area BDMc= Area BMcMa=1/2 Area BDMaMc=1/4Area ABCAreaACMc=1/2 Area ABC So Area CDMc=1/2 Area ABC+1/4Area ABC=3/4 Area ABC
Let is (ABC)=S and E,Z,H midpoints the sides AC,BC,ABABDE ,BDCE , BDZH is parallelograms (HZC) =(HBC)/2=(W/2)/2=S/4(HDZ)=(HZB)=(BHC)2=(S/2)/2=S/4(ZDC)=(BZE)=(BEC)/2=(S/2)/2=S/4Therefore, (HCD)=(HZC)+ (HDZ)+ (ZDC)=3S/4Plan : http://img18.imageshack.us/img18/2807/geogebra.png
Demonstratia relatieiArea CDMc= AreaDBMc+AreaACMc=1/2 Area DBAC In trapezul ABDC daca notam cu P intersectia dreptelor DMc cu AC,din Mc mijlocul AB =>Area CDMc= AreaCPMc,Area BDMc= AreaAPMc si din AreaCPMc= AreaAPMc+AreaACMc =>Area CDMc= AreaDBMc+AreaACMc=1/2 Area DBAC
Jacob Ha - I don't understand this step: S(BM(A)M(B)) = S(BM(A)M(C)) How do you show that? I can't see the two as being congruent. Thank you!
MaMb = MaD = AMc and MaD//AMc hence AMcDMa is a //ogramSo AMa = DMcExtend McMa to meet DC at P.MaP = MbC/2 = b/4If S(ABC) = S S(MaMcD) = S/4S(MaDP) = S/8Further DP = PCSo S(CDMc) = 2(S/8 + S/4) = 3/4 SSumith PeirisMoratuwaSri Lanja