## Conway’s Circle Theorem as special case of Side Divider Theorem The prolific and enigmatic mathematician John Conway, at the age of 82, tragically passed away on 11 April 2020 due to COVID-19 complications. He worked in the theory of finite groups, knot theory, number theory, combinatorial game theory and coding theory. He also participated in online chatrooms about mathematics, and contributed to several branches of recreational mathematics, probably the most well-known being his invention of the Game of Life.

One of the mathematical gems discovered by Conway around 2002 is the following elementary geometry theorem: Given any triangle ABC, extend its sides at each vertex by the length of the side opposite each vertex. Then PQRSTU is cyclic.

As it turns out, Conway's Circle is a special case of a more general theorem, namely, the Side Divider theorem from De Villiers (1994, 1996, 2007), and more recently by Braude (2021), and is presented below.

Side Divider Theorem
Given any △ABC with an arbitrary point P on line AB. Construct BQ = BP, CR = CQ, AS = AR, BT = BS, CU = CT. Then AU = AP, and PQRSTU is cyclic.
1) In the dynamic sketch below, click on the buttons in the top left numbered 1-5 to show the construction steps. What do you observe?
2) Click on the 'Show CircumCircle PQRSTU' button.
3) Drag P along line AB to move outside segment AB, as well as the vertices of △ABC, to explore more.

#### .sketch_canvas { border: medium solid lightgray; display: inline-block; } Conway’s Circle Theorem as special case of Side Divider Theorem

Construction Tip: If you'd like to construct your own dynamic sketch for the Side Divider Theorem, do not use circles to construct BQ = BP, CR = CQ, etc. as the way in which intersections with circles are defined in software like Sketchpad and GeoGebra will prevent the sketch from working correctly when P is dragged outside ABC. Rather use rotations in the same direction at each vertex to map BP onto BQ, etc. In other words, basically using 'directed angles' (De Villiers, 2020).

Side Divider Theorem Challenge
1) Can you explain why (prove that) AU = AP?
2) Can you explain why (prove that) PQRSTU is cyclic?
3) Can you find a formula for the radius of circle PQRSTU in terms of AP, AB, AC, the semi-perimeter of ABC, and the radius of the incircle?
4) Check your answer in 3) above by clicking on the 'Show Formula Radius Circle PQRSTU' button.
5) Explain why when dragging P past A on line AB the displayed formula no longer appears to work correctly.

Connection to Tucker Circle
Click on the 'Show △XYZ' button on the middle right to show a (variable) triangle XYZ for which the circumcircle of PQRSTU is the Tucker Circle of △XYZ.
Acknowledgement: Thank you to Dào Thanh Oai from Vietnam who pointed out this interesting connection to the Tucker Circle to me at the beginning of July 2023 in the Romantics of Geometry group on Facebook.

Conway’s Circle Theorem as Special Case
6) Reduce the size of △ABC if necessary, then drag P outside AB, on the extension of AB, until BP = AC (= b).
7) If you have difficulty with 6) above, click on the 'Link to Conway Special Case' button to move to a new dynamic sketch, where P has already been dragged so that BP = AC (= b). What do you notice?
8) Though it should be visually obvious that this new sketch in 7) corresponds to the conditions of Conway's Circle Theorem, can you explain why (prove that) is the case?
9) Next click on the 'Show Circumcircle PQRSTU' button to check your answer. What do you notice?
10) Can you explain (prove) your observations?
11) What does the earlier formula for the radius of circle PQRSTU in 4) above become in the case of Conway's circle?

Angle Divider Theorem
As discussed in De Villiers (1994, 1996, 2007) there exists an interesting ’side-angle’ dual of the Side Divider Theorem above. This dual for any △ABC can be formulated as follows (and you can navigate to its dynamic illustration by clicking on the 'Link to Angle Divider Theorem' button): Construct any angle divider (line) AX of ∠A. Rotate line AB around B by the (directed) ∠XAB, and label its intersection with AX as P. Note that this construction obviously implies that ∠ABP = ∠PAB. In a similar way, construct angle divider CQ of ∠C so that ∠BCQ = ∠QBC, angle divider AR of ∠A so that ∠CAR = ∠RCA, angle divider BS of ∠B so that ∠ABS = ∠SAB and angle divider CT of ∠C so that ∠BCT = ∠TBC. If U is the intersection of lines CT and AP, then ∠UCA = ∠CAU, and PQRSTU is a tangential (circumscribed) hexagon.
Explore
1) In the dynamic sketch, click on the buttons in the top left numbered 1-6 to show the construction steps. What do you observe?
2) Click on the 'Show Incircle PQRSTU' button.
3) Drag X so that the angle divider line AX moves outside segment ∠A, as well as the vertices of △ABC, to explore more.
4) Can you drag X so that the incircle (or excircle) of PQRSTU coincides with the circumcircle of △ABC? If so, when does this happen, and can you explain why?

Angle Divider Theorem Challenge
1) Can you explain why (prove that) ∠UCA = ∠CAU?
2) Can you explain why (prove that) PQRSTU is tangential (circumscribed)?

Published Paper (2023)
My paper Conway’s Circle Theorem as a special case of a more general Side Divider Theorem with proofs of the above has been published in the June 2023 issue of the Learning & Teaching Mathematics (no. 34) journal of AMESA. All Rights are Reserved.

References
1) Braude, E. (2021). Conway’s Circle Theorem: A Short Proof, Enabling Generalization to Polygons.
2) De Villiers, M. (1994, 1996, 2007). Some Adventures in Euclidean Geometry. Morrisville, NC: Lulu Press, pp. 58-61; p. 67; 195-198; 202.
(The 1994 & 1996 editions were published by the University of Durban-Westville (now University of KwaZulu-Natal), South Africa).
3) De Villiers, M. (2020). The Value of using Signed Quantities in Geometry. Learning & Teaching Mathematics, No. 29, pp. 30-34.