Sunday, February 24, 2013

Problem 860: Triangle, Three Medians, Parallel, Parallelogram, Area, Congruence

Geometry Problem
Level: Mathematics Education, High School, Honors Geometry, College.

Problem submitted by Charles T.
Click the figure below to see the complete problem 860.

Online Geometry Problem 860: Triangle, Three Medians, Parallel, Parallelogram, Area, Congruence


  1. ∵ 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)

  2. 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.

    1.note that BMbCD is a parallelogram => BMb=DC
    Since AMc=MaMb=DMa => AMcDMa is a parallelogram => AMa=DMc
    2. Area of ABC= area of parallelogram BMbCD
    Since CMbMcMa is a parallelogram => CMb=MaMb and MaF=1/2. CMb
    So MbH/McG=2/3
    Area of Triangle(ABC)/Area of parallelogram(MbCDB)=1/2. (McG/MbH)=3/4
    So Area CDMc=3/4 Area ABC

  4. Area CDMc= AreaDBMc+AreaACMc=1/2 Area DBAC
    Area BDMc= Area BMcMa=1/2 Area BDMaMc=1/4Area ABC
    AreaACMc=1/2 Area ABC
    So Area CDMc=1/2 Area ABC+1/4Area ABC=3/4 Area ABC

  5. Let is (ABC)=S and E,Z,H midpoints the sides AC,BC,AB
    ABDE ,BDCE , BDZH is parallelograms
    (HZC) =(HBC)/2=(W/2)/2=S/4
    Therefore, (HCD)=(HZC)+ (HDZ)+ (ZDC)=3S/4

    Plan :

  6. Demonstratia relatiei
    Area 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

  7. 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!

  8. MaMb = MaD = AMc and MaD//AMc hence AMcDMa is a //ogram

    So AMa = DMc

    Extend McMa to meet DC at P.
    MaP = MbC/2 = b/4

    If S(ABC) = S
    S(MaMcD) = S/4
    S(MaDP) = S/8

    Further DP = PC
    So S(CDMc) = 2(S/8 + S/4) = 3/4 S

    Sumith Peiris
    Sri Lanja