# Problem 4160 from the Crux Mathematicorum

### Problem

### Solution

As we know, $D,E,F$ lie on the circumcircle $(ABC).$ Furthermore, we have angle distribution as shown below.

$[\Delta ABC]=2R^2\sin(2\alpha)\sin(2\beta)\sin(2\gamma),\\ [\Delta DEF]=2R^2\sin(\beta+\gamma)\sin(\gamma+\alpha)\sin(\alpha+\beta)=2R^2\cos\alpha\cos\beta\cos\gamma.$

It is also well known that

$r=4R\sin\alpha\sin\beta\sin\gamma.$

Combining everything,

$\displaystyle \begin{align} \frac{[\Delta DEF]}{[\Delta ABC]}&=\frac{2R^2\cos\alpha\cos\beta\cos\gamma}{2R^2\sin(2\alpha)\sin(2\beta)\sin(2\gamma)}\\ &=\frac{1}{8\sin\alpha\sin\beta\sin\gamma}=\frac{R}{2r}. \end{align}$

### Acknowledgment

This is problem 4160 from the Crux Mathematicorum (Vol. 42(6), June 2016.) I am grateful to Leo Guigiuc who co-authored this problem with Marian Cucoanes. Leo has kindly pointed me to the problem and sent me their solution.

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