## 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.

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.
q.e.d.

3. http://img39.imageshack.us/img39/4396/problem860.bmp
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
(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

6. Demonstratia relatiei
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