Newsgroups: rec.puzzles
From: hoey@AIC.NRL.Navy.Mil (Dan Hoey)
Date: Tue, 17 May 1994 16:02:17 GMT
Subject: Re: Geometric Construction Puzzle

mpra...@magnus.acs.ohio-state.edu (Manish S Prabhu) writes:

> Given a straight line, a point and a circle (on the same side of
> the line as the point), construct a circle that touches the line,
> the circle and passes through the point.

What makes this problem difficult is that circles are harder to work
with than lines (they aren't as straight).  But there is a powerful
tool, called _inversion in a circle_, that allows us to turn circles
into lines.  Besides that, it's worth looking at inversion just
because it's pretty.  Let P+oo be the plane together with a point oo
at infinity.  Given any circle C in the plane with center c and radius
R, the inversion in C is defined to be a mapping V from P+oo to itself
as follows: V(c)=oo, V(oo)=c, and otherwise p and V(p) are points on
the same ray from c for which d(c,p) d(c,V(p)) = R^2.

It is immediate that V is a bijection from P+oo to itself and that V
is its own inverse.  What makes inversion useful is that every line or
circle in P+oo is mapped to a line or circle in P+oo.  (This can be
verified easily by calculation; I don't know a geometric proof).  The
image of any line or circle containing the center c must be a line,
since finite circles do not contain V(c)=oo.  The mapping preserves
tangency of lines and circles.

Furthermore, it is easy to construct the image under V of a point,
line, or circle.  If a point p is inside the circle of inversion
C, this may be done by raising the perpendicular from cp at p to
intersect C at a point b, then raising the perpendicular to cb at b to
intersect the ray cp at a point q; that V(p)=q can be seen from the
similar right triangles cpb and cbq.  If p is outside C, construct
a circle of diameter cp to intersect C at a point b, then drop a
perpendicular from b to cp (since cbp is a right angle, this is the
previous case taken in reverse).  Mapping of a circle or line can be
constructed by taking a diameter (of the circle) or perpendicular (to
the line) through c, mapping the endpoint(s), and constructing the
line or circle on the resulting diameter or perpendicular.  Note that
the center of a circle is _not_ in general mapped to the center of the
image of the circle.

So to solve the given construction problem.  I will describe the case
of the given point and the constructed circle exterior to the given
circle; similar methods work for other cases.  We first construct a
distance H, which is half the distance from the point to the circle.
Construct a circle C1 about the given point with radius H; construct a
circle C2 concentric with the given circle and with radius greater by
H than that of the given circle; construct a line L3 parallel to the
given line and H units away in the direction of the given circle and
point.  We will now construct a circle tangent to C1, C2, and L3.

Note that C1 and C2, as constructed, are tangent.  If L3 intersects
the point of tangency, then consider the point of tangency to be a
zero-radius circle tangent to the three of them.  If not, take an
arbitrary circle C0 about the point of tangency, and invert C1, C2,
and L3 in C0.  C1 and C2, since they are tangent at the center of C0,
will be mapped to parallel lines L1', L2'; L3 will be mapped to a
circle C3'.  Let K be half the distance between L1' and L2'.
Construct the circle concentric with C3' and with a radius greater by
K; construct the the parallel line equidistant from L1' and L2'.  The
circles of radius K with centers at the intersection of the line and
circle will be tangent to L1', L2', and C3'.  Inverting these circles
with respect to C0 yields circles tangent to C1, C2, and L3.

We now have at least one circle tangent to C1, C2, and L3.  Increasing
the radius of such a circle by H will yield a circle, containing the
given point and tangent to the given line and circle, Q.E.C.

I hope there is a shorter construction; perhaps the inversion can be
skipped entirely.  But I've mostly bothered writing this so that more
people can know about inversion in a circle, a useful and beautiful
method.  Interested readers might want to try proving that inversion
preserves not only tangency, but perpendicularity of lines and
circles.  Also, consider finding those circles and lines that are
invariant under inversion.  And find a geometric interpretation for
inversion, preferably one that illustrates the preservation of lines
circles, and perpendicularity.

am...@FreeNet.Carleton.CA (Brice Wightman) answered:

> 1. Draw a line parallel to the given line (call this the "parallel")
> 2. Drop the perpendicular from the given point to the given line.
> 3. Draw a radius of the circle and produce outside of the circumference.
> 4. With centre at the intersection of the circumference and the
> radius and length equal to the perpendicular previously constructed
> strike an arc on the radius produced.
> 5. With centre at the centre of the given circle and radius equal to
> the distance from the centre to the intersection constructed in (4)
> draw a circle.
> 6. With centre at the intersection of the circle constructed in (5)
> and the "parallel" and radius equal to the length of the
> perpendicular, draw a circle.  This is the required circle.

It is unfortunate for a construction to be given without any hint of
why the author considers it a solution to the problem; it is most
unfortunate when such a construction is incorrect, as Brice's is.
(For instance, the parallel need not intersect the circle in step 6,
and even if it does the final circle need not be tangent to, nor even
intersect, the given line.)  Due to the lack of explanation, I find it
impossible to figure out whether the error is due to minor problems
with a correct answer, or to a complete failure to understand the
techniques and requirements of geometric construction.

Dan Hoey
Hoey@AIC.NRL.Navy.Mil