Re: Another Euclidean geometry problem

William Elliot wrote :
On Mon, 3 Nov 2008, [ISO-8859-1] José Carlos Santos wrote:

I would to like to know how to do this with compass and ruler alone (note
that I do not know whether or not it can be done): given two non-parallel
straight lines _r_ and _s_ and a circle _c_ which encloses the intersection
point of _r_ and _s_, to find the points P (there are two of them) such that
the distances from P to _r_, from P to _s_, and from P to _c_ are all equal.

The locus of points that are equidistance from r and s are along
the bisectors of angles rs.

If center c = rs (intersection rs), there can be four such points.
Draw a cord from cr to cs. Bisect the angle between cord and c.

? which angle ?

The point you want is the intersection of that bisector and the
bisector of angle rs.


I don't think so.
First of all your point is equidisatnt from _r_ and _s_, but not at
same distance of circle _c_ : it is at same distance from a *chord*
of _c_.
Also it is not stated that the center of circle is intersect point
of _r_ and _s_, just that this intersection point is *inside* the
circle.

The problem is in fact to find a circle which is tangent to both _r_,
_s_ and _c_, which is the well known Appolonius problem :
one of the 10 cases from 3 (points,lines,circles).

This one is not too hard :
There are 8 solutions. 4 inside the given circle, and 4 outside.
Consider one of the inside solutions.
Let R the radius of _c_ and increase the radius of the searched circle
by R, decreasing the radius of given circle to 0, and shifting the
two lines by R.
This will results into the same center P of unknown circle.
Now the problem becomes : given two lines _r'_ and _s'_ and a point c
(center of _c_), to construct a circle tangent to _r'_ and _s'_ and
going through c.
Let I = intersection of _r'_ and _s'_
Draw any circle (k) centered on the angle bisector of _r'_ and _s'_
and tangent to _r'_ and _s'_.
Line Ic intersects this circle in A and A'.
Then do a dilation of circle (k) from center I such that A (or A')
comes into c. You get the two solutions.

Regards.







--
Philippe Ch., mail : chephip+news@xxxxxxx
site : http://mathafou.free.fr/ (recreational mathematics)


.

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