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

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