An Inequality for Cevians And The Ratio of Circumradius and Inradius

Source

An Inequality for Cevians And The Ratio of Circumradius and Inradius

Problem

An Inequality for Cevians And The Ratio of Circumradius and Inradius - problem

Solution

By the Law of Cosines, in $\Delta XYC,$

$\displaystyle \begin{align} XY^2 &= XC^2+YC^2-2\cdot XC\cdot YC\cdot\cos C\\ &\ge 2\cdot XC\cdot YC\cdot(1-\cos C)=4\cdot XC\cdot YC\cdot\sin^2\frac{C}{2}. \end{align}$

It follows that $\displaystyle XY\ge 2\sqrt{XC\cdot YC}\cdot\sin\frac{C}{2}.\,$ Similarly, $\displaystyle YZ\ge 2\sqrt{YA\cdot ZA}\cdot\sin\frac{A}{2}\,$ and $\displaystyle ZX\ge 2\sqrt{XB\cdot ZB}\cdot\sin\frac{B}{2}.\,$

Multiplying the three we get

$\displaystyle \small{XY\cdot YZ\cdot ZX\ge 8\sqrt{XC\cdot XB\cdot YC\cdot YA\cdot ZA\cdot ZB}\cdot\sin\frac{A}{2}\cdot\sin\frac{B}{2}\cdot\sin\frac{C}{2}}.$

Thus,

$\displaystyle\small{\frac{XB}{XY}\cdot\frac{YC}{YZ}\cdot \frac{ZA}{ZV}\le\sqrt{\frac{XB}{XC}\cdot \frac{YC}{YA}\cdot \frac{ZA}{ZB}}\cdot \frac{\displaystyle 1}{8\cdot\sin\frac{A}{2}\sin\frac{B}{2}\sin\frac{C}{2}}=\frac{R}{2r}}$

because, by Ceva's theorem, $g=\displaystyle \frac{XB}{XC}\cdot \frac{YC}{YA}\cdot \frac{ZA}ZB.$

Acknowledgment

The problem by Titu Andreescu has been posted by Miguel Ochoa Sanchez at the Peru Geometrico facebook group and commented with a solution (above) by Marian Dinca. I am grateful to Leo Giugiuc for bringing this problem to my attention.

 

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