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Saturday, December 17, 2016
Geometry Problem 1296: Heron's Formula: Area, Three Medians
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Problem 1296
ReplyDeleteLet D is medium BC and E medium AB.Then the points E,D and H are collinear.
Let (ABC) area triangle ABC then (BGH)=(BGD)+(GDH)+(BDH)=(DEC)/2+(ADG)+(EDC)=
(ABC)/4+(ADC)/2+(BEC)/2=(ABC)/4+(ABC)/4+(ABC)/4=3(ABC)/4.So (ABC)=4/3*(BGH)=
4/3*sqrt(m(m-d)(m-e)(m-f)).
APOSTOLIS MANOLOUDIS 4 HIGH SCHOOL KORYDALLOS PIRAEUS GREECE
name M point GH meet BC. S(BGM)=1/2 S(ABC) - (1/4)(1/2)S(ABC)
ReplyDelete=> S(BGM)= 3/8S(ABC) => S(BGH)=3/4 S(ABC)
S(ABC)=4/3 S(BGH)
https://goo.gl/photos/nraBV6ncgemgkNL47
ReplyDeleteDefine points M, N, E per attached sketch
Per observation we have parallelograms GBEC, ANEM
So NE=AM=d
CE=GB=e
Per the result of problem 860 we have S(ABC)= 4/3. S(NCE)
Apply Heron’s formula in triangle NCE
S(NCE)= sqrt(m. (m-e)(m-f)(m-d))
Where m= half perimeter of triangle NCE
So S(ABC)=4/3. sqrt(m. (m-e)(m-f)(m-d))
Let U, V and W be the midpoints of AB, BC and VC respectively and let X be the centroid of ∆ ABC. Let UV and GW meet at H.
ReplyDeleteEasily CGVH is a parellogram and since VH = GC = AG =b/2 and VH//AG,
AGHV is also a parallelogram and hence AV = d = GH.
Similarly CH = & // to BU, hence BHCU is a parallelogram and so CU = BH.
Therefore BGC is a ∆ with sides d,e,f.
Now S(BGW) = ¾ S(BCG) = ½ S(BGH)
So ¾ X S(ABC) / 2 = ½ S (BGH)
Therefore S(ABC) = 4/3 S(BGH) and the result follows since ∆BGH has sides d,e,f
Sumith Peiris
Moratuwa
Sri Lanka