Answers 335the mid-point on BH, and make CL equal to BK. BC is said to be divided in
medial section, and we have found KL, the side of the pentagon. Now (see
A 1
Esecond diagram) lay KM and LN equal to KL, so that M and N may lie on
BA and CD respectively. Fold PQ and lay MO and NO equal to KM and
LN. Then KMONL is the pentagon required. For this solution I am indebted
to a little book, Geometrical Exercises in Paper Folding, by T. Sundara Row
(Madras, 1893). *
- FOLDING AN OCTAGON
By folding the edge CD over AB
we can crease the middle points E
and G. In a similar way we can find
the points F and H, and then crease
the square EHGF. Now fold CH on
EH and EC on EH, and the point
where the creases cross will be I. Pro-
ceed in the same way at the other
three corners, and the regular octa-
gon, as shown, will be marked out by
the creases and may be at once cut
out.
l* Currently available in a Dover paperback repnnt -M. G.)