prove that for any triangle ,R>= 2r,where R and r are the circumradius and inradius of the triangle respectively.

post in different proofs (Hi)

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- July 7th 2010, 08:35 PMearthboyCircumradius and inradius.
prove that for any triangle ,R>= 2r,where R and r are the circumradius and inradius of the triangle respectively.

post in different proofs (Hi) - July 7th 2010, 09:12 PMAlso sprach Zarathustra
Not standard way...

Let ABC be some triangle,with AB=a, AC=b, BC=c,and with area S.

Let, p=(a+b+c)/3, hence, r=S/p.

We, also know that: R=abc/4S

I'ii leave like that, maybe someone would like to proceed somehow(if possible)... - July 7th 2010, 11:28 PMearthboy
THANKS!(Bow)

oh yes!i wish to complete your proof:Quote:

Let ABC be some triangle,with AB=a, AC=b, BC=c,and with area S.

Let, p=(a+b+c)/2, hence, r=S/p.

We, also know that: R=abc/4S

abc>=sqrt( a^2-(b-c)^2)sqrt(b^2-(c-a)^2)sqrt(c^2-(a-b)^2)

=(a+b-c)(b+c-a)(c+a-b)=8(p-a)(p-b)(p-c) Q.E.D

Quote:

Not standard way...

- July 8th 2010, 08:20 AMearthboy
hey!

post in other proofs.is there a proof with the help of the eulers theorem: (SI)^2=R^2-2Rr,Where S,I,R,R are the circumcenter,incenter,circumradius,inradius respectively(Talking)