Geometry Problem. Post your solution in the comment box below.

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

Details: Click on the figure below.

## Saturday, December 17, 2016

### Geometry Problem 1296: Heron's Formula: Area, Three Medians

Subscribe to:
Post Comments (Atom)

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