Fagnano's Problem: What is it?
A Mathematical Droodle
Izvolsky's solution
What if applet does not run? 
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Copyright © 19962017 Alexander Bogomolny
The applet attempts to illustrate a relatively novel solution to Fagnano's problem. The solution by N. A. Izvolsky has been published in Russian in 1937 in the first series of collections Mathematics Education (n 10) and later included into S. I. Zeitel's New Triangle Geometry (UchPedGiz, 1962), also in Russian.
What if applet does not run? 
Let O be the circumcenter of ΔABC and points P, Q, R located on sides BC, AC, and AB respectively. Joining O to P, Q, R yields three quadrilaterals: OQAR, ORBP, OPCQ. Let's start with the latter. The quadrilateral OPCQ consists of two triangles OCP and OCQ with common base OC. If h and g denote altitudes to the base in the two triangles, we can write

where r is the circumradius of ΔABC. Similarly we obtain

Adding the three we see that

Hence
(1)  Perimeter(PQR) ≥ 2·Area(ABC)/r. 
Now, by Nagel's theorem, the sides of the orthic triangle are perpendicular to the radiusvectors from O to vertices of ΔABC. Which means that, for the orthic triangle, (1) turns in an identity. Thus the orthic triangle solves Fagnano's problem. That the solution is unique follows from the fact that for any inscribed triangle PQR (1) becomes an identity if and only if its sides are perpendicular to the said radiusvectors. The sides of such a triangles are then parallel to the sides of the orthic triangle. But this is impossible for an inscribed triangle different from orthic.
 Schwarz's solution
 Fejér's solution
 Samelson's solution
 Izvolsky's solution
 Fagnano's Problem in Reverse
Activities Contact Front page Contents Store Geometry
Copyright © 19962017 Alexander Bogomolny
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