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. . . . . . In this unfinished page, the Pappus Chain of circles in an arbelos is presented. The page needs fixing in a number of ways, some of which were pointed out by Bill Nixon:
I am having trouble following the text. The labels
on the drawing are missing. Where is point O? You
mention a perpendicular from point A, where is it? You said construct a segment OA and its midpoint m, I don't see a point labeled m. Please review the text and drawing and correct
any omission so that the text and the drawing
match.
Point O is the center of the large circle. Point O1 is the center of the left inner circle, and O2 is the center of the right inner circle.
The following is from NRICH, posted by a person named "arbelos":
The constructions I use to invert a point are:
To invert a point,A through a circle C with centre O when A is outside the
circle:
i) construct segment OA and its midpoint,m
ii) construct circle,C', centre m, radius mO
iii) P is the intersection of C and C'
iv) construct the perpendicular from P to OA
v) let A-1 be the point of intersection of the perpendicular and OA
To invert a point,A through a circle C with centre O when A is inside the circle:
i) construct the ray OA and then the perpendicular from A
ii) let P be the point of intersection of the perpendicular and C
iii) construct the perpendicular from OP passing through P
iv) let A-1 be the point of intersection of the perpendicular from OP through P and the ray OA
In both of the above cases, A-1 is the inverse of A wrt C.
Having constructed the arbelos, construct perpendiculars at B and C and then the
series of circles c2, c3 etc as shown in the figure below.
Construct the circle c1, centre A radius AP.
Inversion of the points X,Y,Z (on c2) through c1 gives points which lie on the
circumference of the incircle (which is then easily constructed - although there
are easier methods (ie no inversion) to construct the incircle). Inversion of c3
through c1 gives the next circle in the chain etc.
If you have the time, I'd appreciate any thoughts you have on the following
construction which I found by 'messing' around with GSP.
I'm trying to analyse why it works and am hoping that there isn't any inversion
'hidden' in the method:
1. Construct the incircle as shown, giving points of tangency (B, C and D)with
the 3 semicircles of the arbolos:
2. Construct perpendicular at A. Construct the tangent at B which cuts the perpendicular at E. Mark off the perpendicular in divisions equal to AE (giving E', E'' etc). Construct tangents from E', E'' etc to semicicle AF. The points of tangency (G,H,I etc), as shown below, being points on successive circles in Pappus chain.
It's interesting to note, that as part of the construction, a pencil of
circles appears with axis the midpoint of A and the centre of semicircle AF.
3. As in 2 above, points of tangency of successive circles in Pappus chain with
semicircle FL are found as shown below:
4. Points of tangency between circles in the chain lie on a circle as shown below. Points N, R and G lie on the next circle which can now be constructed etc
Last post! Have found the following with above construction and using similar triangles, Pythagoras and some trig. I'm sure the distances could be shown to be correct by some other method eg inversion, and complete proof.
From Mathworld, Pappus Chain, Arbelos, Archimedes Circles, Bankoff Circle, Steiner Chain
From Cut-the-knot, Inversion of Pappus Chain In Arbelos, Steiner/Pappus Sangaku, Gothic Arc, Ellipse In Arbelos
Pappus-related things you might have been looking for when you found this page.
Pappus Theorem, which is more about points and lines, and is a special case of Pascal's Theorem, in which a hexagon is inscribed in a conic. I link to it here, because you may have been looking for it when you found this page.
Pappus' Centroid Theorem in solid geometry
The webmaster and author of this Math Help site is Graeme McRae.