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.
Sunday, February 24, 2013
Problem 860: Triangle, Three Medians, Parallel, Parallelogram, Area, Congruence
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∵ BE//AC and DM(B)//BA
ReplyDelete∴ 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.
ReplyDeleteTherefore, 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.bmp
ReplyDelete1.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
Area CDMc= AreaDBMc+AreaACMc=1/2 Area DBAC
ReplyDeleteArea 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
Let is (ABC)=S and E,Z,H midpoints the sides AC,BC,AB
ReplyDeleteABDE ,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/4
Therefore, (HCD)=(HZC)+ (HDZ)+ (ZDC)=3S/4
Plan : http://img18.imageshack.us/img18/2807/geogebra.png
Demonstratia relatiei
ReplyDeleteArea 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))
ReplyDeleteHow 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 //ogram
ReplyDeleteSo 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
Moratuwa
Sri Lanja
From Mid-point theorem, MbMa=AB/2 and // to AB => ABDMb //gm and BD=AMb=MbC=McMa , DMb=AB
ReplyDeleteMcMa cuts the //gm ABDMb into two congruent //gms => AMa=McD --------(1)
Since BD=MbC and BD//MbC => BDCMb is //gm and BMb=DC------------(2)
Let 2S be the area of triangle ABC and denote the intersection of McD and BC as F
S(CBMc)=S and from (1), it can be derived that F is the midpoint of the //gm BDMaMc
=> S(BMcF)=S/4
=> S(CMcF)=3S/4
Since DMc=AMa and F is the mid-point=> BC is a median of CDMc
=> S(CDMc)=6S/4=3(2S)/4=3/4 Area of ABC ------------(3)