THE COMPASSES PUZZLE.
(
Patchwork Puzzles)
It is curious how an added condition or restriction will sometimes
convert an absurdly easy puzzle into an interesting and perhaps
difficult one. I remember buying in the street many years ago a little
mechanical puzzle that had a tremendous sale at the time. It consisted
of a medal with holes in it, and the puzzle was to work a ring with a
gap in it from hole to hole until it was finally detached. As I was
walking along the street I very soon acquired the trick of taking off
the ring with one hand while holding the puzzle in my pocket. A friend
to whom I showed the little feat set about accomplishing it himself, and
when I met him some days afterwards he exhibited his proficiency in the
art. But he was a little taken aback when I then took the puzzle from
him and, while simply holding the medal between the finger and thumb of
one hand, by a series of little shakes and jerks caused the ring,
without my even touching it, to fall off upon the floor. The following
little poser will probably prove a rather tough nut for a great many
readers, simply on account of the restricted conditions:--
Show how to find exactly the middle of any straight line by means of the
compasses only. You are not allowed to use any ruler, pencil, or other
article--only the compasses; and no trick or dodge, such as folding the
paper, will be permitted. You must simply use the compasses in the
ordinary legitimate way.
Answer:
Let AB in the following diagram be the given straight line. With the
centres A and B and radius AB describe the two circles. Mark off DE and
EF equal to AD. With the centres A and F and radius DF describe arcs
intersecting at G. With the centres A and B and distance BG describe
arcs GHK and N. Make HK equal to AB and HL equal to HB. Then with
centres K and L and radius AB describe arcs intersecting at I. Make BM
equal to BI. Finally, with the centre M and radius MB cut the line in C,
and the point C is the required middle of the line AB. For greater
exactitude you can mark off R from A (as you did M from B), and from R
describe another arc at C. This also solves the problem, to find a point
midway between two given points without the straight line.
I will put the young geometer in the way of a rigid proof. First prove
that twice the square of the line AB equals the square of the distance
BG, from which it follows that HABN are the four corners of a square. To
prove that I is the centre of this square, draw a line from H to P
through QIB and continue the arc HK to P. Then, conceiving the necessary
lines to be drawn, the angle HKP, being in a semicircle, is a right
angle. Let fall the perpendicular KQ, and by similar triangles, and from
the fact that HKI is an isosceles triangle by the construction, it can
be proved that HI is half of HB. We can similarly prove that C is the
centre of the square of which AIB are three corners.
I am aware that this is not the simplest possible solution.
