000RM.dvi

1114 Lewis Carroll’s unused geometry pillow problem

Consider thedirectedangle∠BA′C. This is

∠BA′C=∠BA′W+∠WA′C

=2∠YA′W+2∠WA′Z

=2∠YA′Z

=− 2 ∠YAZ

sinceA′YAZis a rhombus. This means that∠BA′C=−^2 ∠BAC. The
reflection ofA′in the sideBCis therefore the pointQon the perpen-
dicular bisector such that∠BQC=2∠BAC, which is necessarily the
circumcenterOof triangleABC. We therefore conclude thatA′is the
reflection of the circumcenterOin the sideBC, and the reflection line
is the perpendicular bisector of the lineAA′.

O

D

H

C

A

B

N

A′

LetDbe the midpointBCandHthe orthocenter of triangleABC.
In a standard proof of the Euler line theorem, it is established thatAH=
2 OD,^1 and that the midpoint ofOHis the nine-point center of triangle
ABC. This means thatAH =OA′, andAOA′His a parallelogram. It
follows that the midpoint ofAA′is the same as that ofOH, the nine-
point centerNof triangleABC. The Lewis Carroll paper-folding line
is the perpendicular toANatN.

(^1) AH=2·OD=2RcosA, whereRis the circumradius of triangleABC.