Fagnano's Problem in Reverse

Reflect ΔABC 5 times in its sides as if rolling it each time over a side. Note that, on the fifth reflection, side AB becomes parallel to its original position with the same orientation. Connect point P on the original side AB to its image in the fifth reflection. Line PP is parallel (and equal) to lines AA and BB. Imagine folding the reflections starting with the last triangle over lines (sides) of reflection and mark positions of the line PP (light gray below). Let P be such that line PP is entirely located in the stripe formed by the six triangles. Prove that line PP folds into a hexagon inscribed into (the original) ΔABC.


This applet requires Sun's Java VM 2 which your browser may perceive as a popup. Which it is not. If you want to see the applet work, visit Sun's website at http://www.java.com/en/download/index.jsp, download and install Java VM and enjoy the applet.


What if applet does not run?

Acknowledgement

The problem was suggested to me by Hans Samelson, Professor Emeritus of Mathematics, Stanford University, and is strongly reminiscent of H. A. Schwarz's solution to Fagnano's problem.

Explanation

Fagnano's problem

|Activities| |Contact| |Front page| |Contents| |Geometry| |Store|

Copyright © 1996-2017 Alexander Bogomolny


This applet requires Sun's Java VM 2 which your browser may perceive as a popup. Which it is not. If you want to see the applet work, visit Sun's website at http://www.java.com/en/download/index.jsp, download and install Java VM and enjoy the applet.


What if applet does not run?

The line PP folds into an inscribed hexagon simply because its two ends are the final reflections of each other in a series of five reflections, which are repeated backwards by the folding procedure.

Note that line PP is always parallel to a side of the orthic triangle inscribed into ΔABC. (And this is true for all 6 triangles involved.) This is a consequence of the mirror property of the orthic triangle.

The hexagon inscribed in a triangle has also been discussed in the context of the Tucker and Lemoine circles.

|Activities| |Contact| |Front page| |Contents| |Geometry| |Store|

Copyright © 1996-2017 Alexander Bogomolny

 62028354

Search by google: