Let r, R, and s be the inradius, circumradius, and semiperimeter of triangle ABC.
If /A ≥ 90°, prove that

    r     a sin(A)
   --- ≤ ----------
    R        2s

Hi friends!

The initial relation can be be write : 2*r*s < a*R*sin(A)

But using relation :

 - r*s=S (S is the triangle's area)

 - a=2*R*sin(A) (sin theorem)

.. the initial relation can be write like this :

2*S < a^2/2 and 4*S < a^2 and after S = a*h(a)/2 , where h(a) is the altitude from A,  

h(a) < a/2 

To prove this relation i find an intermediate between h(a) and a/2. This intermediate is m(a), mean the median from A.

So i will prove now that h(a) < m(a) < a/2

a) For the first part is obvious that h(a) < m(a)

b) For the second part i use the expresion of the median  m(a)^2 =  (2*(b^2+c^2)-a^2) / 4 and after little calculus i must prove that  a^2 > b^2+ c^2.

But the cos theorem give us that a^2 = b^2 + c^2 - 2*b*c*cos(A) and because A is greater than 90, cos(A) < 0 and a^2 > b^2+c^2.

So h(a) < m(a) < a/2 and the (nice) problem is done!

 

Edited on March 28, 2008, 5:47 am

Posted by Chesca Ciprian on 2008-03-27 16:32:22