# Orthic Semiperimeter

There are many many formulas for the area of a triangle. Among the most useful is $S = pr$, where $S$ is the area, $p$ the semiperimeter, and $r$ the inradius of a triangle. Among the little known is the one Ross Honsberger discovered in the end of the 19th century publication Mathematical Visitor:

$S = qR$,

where, as before, $S$ is the area of a triangle, $R$ its circumradius, and $q$ the semiperimeter of its orthic triangle.

The insight into this formula comes from Fejér's solutions to the Fagnano's problem.

A. Bogomolny, 17 January 2015, Created with GeoGebra

### Proof

Let $AD$, $BE$, and $CF$ be the altitudes in $\Delta ABC$. Triangle $DEF$ is known as the orthic triangle of $\Delta ABC$. Let $E'$ and $E''$ be the reflections of $E$ in $BC$ and $AB$, respectively. Due to the mirror property of the altitudes, $E'DFE''$ is a straight line and $\Delta E'BE''$ is isosceles $(BE' = BE = BE'')$. Also, by the construction, $\angle E'BE'' = 2\angle ABC$.

Most importantly,

\begin{align} E'E'' &= E'D + DF + FE'' \\ &= DE + DF + EF \\ &= 2q. \end{align}

It thus follows that, for $G$ being the foot of the altitude from $B$ in $\Delta E'BE''$, $q = E'G = BE'\cdot\mbox{sin}\angle ABC = BE\cdot\mbox{sin}\angle ABC$. So $q = BE\cdot\mbox{sin}\angle ABC$, but, by the Law of Sines, $\displaystyle 2R = \frac{AC}{\mbox{sin}\angle ABC}$. Also, as is well known, $\displaystyle S = \frac{AC\cdot BE}{2}$. Now, combining the three identities gives the required result:

\begin{align} q &= BE\cdot\mbox{sin}\angle ABC \\ &= \frac{2S}{AC}\cdot \frac{AC}{2R} \\ &= \frac{S}{R}. \end{align}

### References

1. R. Honsberger, Mathematical Diamonds, MAA, 2003, p 19