# 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.

### 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

- R. Honsberger,
*Mathematical Diamonds*, MAA, 2003, p 19

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