Cevian Nest: What Is It About?
A Mathematical Droodle


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Copyright © 1996-2017 Alexander Bogomolny

The applet may suggest the following statement:

Let ΔQ1Q2Q3 be the cevian triangle of point P with respect to ΔP1P2P3. Let ΔR1R2R3 be the cevian triangle of point Q with respect to ΔQ1Q2Q3. Then triangles P1P2P3 and R1R2R3 are perspective (from a point): the lines P1R1, P2R2, P3R3 are concurrent.

( ΔQ1Q2Q3 is called a cevian triangle triangle of point P with respect to ΔP1P2P3, if its vertices Q1, Q2, Q3 serve as the feet of the cevians P1Q1, P2Q2, P3Q3 in ΔP1P2P3 through point P.)

The configuration of three triangles inscribed into each other is known as the Cevian Nest.


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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 4 = 1 and 5 = 2. For k = 1, 2, 3, let Sk be the intersection of PkQk and Qk-1Qk+1 and Tk be the intersection of PkRk and Pk-1Pk+1. By Ceva's theorem,

P3Q1·P1Q2·P2Q3

Q1P2·Q2P3·Q3P1
 = 1

and

Q2S1·Q3S2·Q1S3

S1Q3·S2Q1·S3Q2
 = 1

By the invariance of cross-ratio of four collinear points under projection (from Pk),

Pk - 1Qk·TkPk + 1

QkPk + 1·Pk - 1Tk
 = 
Qk + 1Sk·RkQk - 1

SkQk - 1·Qk + 1Rk

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

 
T1P2·T2P3·T3P1

P3T1·P1T2·P2T3
 = 
R1Q3·R2Q1·R3Q2

Q2R1·Q3R2·Q1R3

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 P1T1, P2T2, P3T3 are concurrent. The second is equal to 1 only if the lines Q1R1, Q2R2, Q3R3 are concurrent.

Note: Vladimir Nikolin had a somewhat different perspective on Cevian nests.

Menelaus and Ceva

 62042013

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