Cevian Nest: What Is It About?
A Mathematical Droodle
What if applet does not run? 
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Copyright © 19962018 Alexander Bogomolny
The applet may suggest the following statement:
Let ΔQ_{1}Q_{2}Q_{3} be the cevian triangle of point P with respect to ΔP_{1}P_{2}P_{3}. Let ΔR_{1}R_{2}R_{3} be the cevian triangle of point Q with respect to ΔQ_{1}Q_{2}Q_{3}. Then triangles P_{1}P_{2}P_{3} and R_{1}R_{2}R_{3} are perspective (from a point): the lines P_{1}R_{1}, P_{2}R_{2}, P_{3}R_{3} are concurrent.
( ΔQ_{1}Q_{2}Q_{3} is called a cevian triangle triangle of point P with respect to ΔP_{1}P_{2}P_{3}, if its vertices Q_{1}, Q_{2}, Q_{3} serve as the feet of the cevians P_{1}Q_{1}, P_{2}Q_{2}, P_{3}Q_{3} in ΔP_{1}P_{2}P_{3} through point P.)
The configuration of three triangles inscribed into each other is known as the Cevian Nest.
What if applet does not run? 
The problem has been proposed by H. Gülicher (#1581, Mathematics Magazine Vol. 72, No. 4, October 1999). A solution by Daniele Donini has been published a year later (Mathematics Magazine Vol. 73, No. 4, October 2000, p. 325).
All subscripts below are interpreted cyclically modulo 3, so that
 =  1 
and
 =  1 
By the invariance of crossratio of four collinear points under projection (from P_{k}),
 = 

for k = 1, 2, 3. The product of the three equalities simplifies to
 = 

These two expressions are therefore either both equal to 1 or both are not. By Ceva's theorem, the first is equal to one only if the lines P_{1}T_{1}, P_{2}T_{2}, P_{3}T_{3} are concurrent. The second is equal to 1 only if the lines Q_{1}R_{1}, Q_{2}R_{2}, Q_{3}R_{3} are concurrent.
Note: Vladimir Nikolin had a somewhat different perspective on Cevian nests.
Activities Contact Front page Contents Geometry
Copyright © 19962018 Alexander Bogomolny