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(Maybe Very) Hard Geometric Inequality
Posted: Fri Feb 24, 2012 10:47 pm
by Corei13
Let $\triangle ABC$ is a triangle with circumradius $R$. Extend $AB$ to $R$, $BC$ to $P$ an $CA$ to $Q$ and let $D,E,F$ be the midpoints of $BC$, $CA$, $AB$ respectively.
Prove that,
\[\frac{PQ}{AB}+\frac{QR}{BC}+\frac{RP}{CA} \ge \frac{PD+QE+RF}{R}\]
Re: (Maybe Very) Hard Geometric Inequality
Posted: Tue Feb 28, 2012 3:40 pm
by sakibtanvir
Corei13 wrote:Let $\triangle ABC$ is a triangle with circumradius $R$. Extend $AB$ to $R$, $BC$ to $P$ an $CA$ to $Q$ and let $D,E,F$ be the midpoints of $BC$, $CA$, $AB$ respectively.
Prove that,
\[\frac{PQ}{AB}+\frac{QR}{BC}+\frac{RP}{CA} \ge \frac{PD+QE+RF}{R}\]
Re: (Maybe Very) Hard Geometric Inequality
Posted: Tue Feb 28, 2012 6:46 pm
by Phlembac Adib Hasan
I think one $R$ is length and the other is a point.