Thread: Circumradius 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

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)...

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

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

now we just have to show that abc/4s >= 2s/p ,which reduces to abc>= 8(p-a)(p-b)(p-c). now we know that a^2-(b-c)^2<a^2.so similarly we could write
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

Not standard way...

please show the standard way,it will be very helpful

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