The Parry circle containing X[2],X[15],X[16],X[23],X[110],X[111],X(352],X[353].
The Lester Circle containing
X[3],X[5],X[13],X[14],X[1117].
We shall show that these belong to families of circles. Each member of a family passes though at least four named centre and is orthogonal to a well-known circle. In total the families contain over 200 circles. Two examples involving centres early in ETC are found in §5. These are circles orthogonal to the circumcircle, and
(a) through X[1],X[2],X[23],X[36],
(b) through X[1],X[15],X[16],X[36].
The Parry circle appears in two families. This partly explains the fact that it passes through so many named centres.
We shall also show that the incircle contains many centres with fairly simple trilinears. This relies on recent work of Brisse. The same work also identifies points which, together with X[15] and X[16], define the unique circle orthogonal to the incircle, the circumcircle and the Brocard circle. See §9.
We shall need to refer to some known circles.
C the circumcircle,
B the Brocard circle,
N the nine-point circle,
I the incircle,
O the orthocentroidal circle.
The line through X[m] and X[n] will be denoted by l[m,n]. Some lines have more familiar names.
e the Euler line = l[2,3],
f the Fermat axis = l[13,14],
b the Brocard axis = l[3,6].
Theorem 1.
Suppose that A, B, C and D are distinct points and E is a line or circle.
(1) If A,B are inverse with respect to E, then any line or circle through A and B is orthogonal to E.
(2) If A,C are not inverse with respect to E, then there is a unique line or circle through A and C orthogonal to E.
(3) If A,B and C,D are inverse with respect to E, then there is a unique line or circle through A,B,C and D orthogonal to E.
The first two are easy exercises in inversive geometry. The third follows since, by (1), the lines or circles defined by A,B,C and A,C,D are orthogonal to E. By (2), these are identical.
ETC gives many examples of pairs inverse with respect to familiar central circles. By (3), two pairs for a given circle E give a central line or circle orthogonal to E.
We list pairs of inverse points for the circles mentioned above, together with the lines on which each pair lies. These lists are drawn from ETC. Note that, whatever the status of the assertions in ETC, it is easy to verify that centres P and Q are inverse with respect to a known circle E. From the equation of E, we can obtain the equation of the polar l of P. Then we need only check that Q is the intersection of l and the line joining P to the centre of E.
In the course of preparing this note, we found that our argument has been anticipated by Conway [C]. He indicated that this proves the existence of the Lester circle, and another circle through X[3],X[5],X[6],X[115].
Name | Centres | Line | Name | Centres | Line |
O[1] | X[3],X[5] | e | O[8] | X[379],X[857] | e |
O[2] | X[6],X[115] | f | O[9] | X[382],X[546] | e |
O[3] | X[13],X[14] | f | O[10] | X[383],X[1080] | e |
O[4] | X[25],X[427] | e | O[11] | X[405],X[442] | e |
O[5] | X[27],X[469] | e | O[12] | X[406],X[475] | e |
O[6] | X[297],X[458] | e | O[13] | X[470],X[471] | e |
O[7] | X[378],X[403] | e | O[14] | X[472],X[473] | e |
O[15] | X[868],X[1316] | e |
Theorem 2
Suppose that ΔABC is not isosceles.
For 1 ≤ m < n ≤ 15, the pairs O[m],O[n] define a line or circle orthogonal to O.
This family of objects consists of two lines and 26 distinct circles.
Proof The first part follows from the remarks in §3.
Of the 105 pairs m,n, 78 (m,n ≠ 2,3) define the line e. The pair m = 2, n = 3 defines f.
The remaining 26 pairs must define circles since each contains two points on each of e,f.
Since three pairs must contain four collinear points, these circles are distinct.
The family contains the Lester circle, from the pair O[1],O[3], and Conway's example O[1],O[2].
Name | Centres | Line | Name | Centres | Line |
C[1] | X[1],X[36] | l[1,3] | <C[11] | X[35],X[484] | l[1,3] |
C[2] | X[2],X[23] | e | C[12] | X[54],X[1157] | l[3,54] |
C[3] | X[4],X[186] | e | C[13] | X[55],X[1155] | l[1,3] |
C[4] | X[6],X[187] | b | C[14] | X[56],X[1319] | l(1,3) |
C[5] | X[10],X[1324] | l(3,10) | C(15) | X(58),X(1326) | b |
C[6] | X[15],X[16] | b | C(16) | X(237),X(1316) | e |
C[7] | X[21],X[1325] | e | C[17] | X[352],X[353] | l[3,352] |
C[8] | X[22],X[858] | e | C[18] | X[667],X[1083] | l[3,667] |
C[9] | X[24],X[403] | e | C(19) | X(1054),X(1283) | l(3,1054) |
C[10] | X[25],X[468] | e | C[20] | P(1),P(2) | l(1,3) |
Theorem 3
Suppose that ΔABC is not isosceles.
For 1 ≤ m < n ≤ 20, the pairs C[m],C[n] define a line or circle orthogonal to C.
This family of objects consists of 3 lines and at most 154 circles.
Proof The first part of the proof of Theorem 2 can be adapted to show that the family includes the only the three lines e,b and l(1,3). These account for 34 of the 190 pairs. Thus the family includes 156 circles. But any two of {2,6,17} give the Parry circle, so we have at most 154 distinct circles.
Name | Centres | Line | Name | Centres | Line |
N[1] | X[2],X[858] | e | N[4] | X[427],X[468] | e | N[2] | X[4],X[403] | e | N[5] | X[623],X[624] | l[5,141] |
N[3] | X[141],X[625] | l[5,141] |
Theorem 4
Suppose that ΔABC is not isosceles.
For 1 ≤ m < n ≤ 5, the pairs N[m],N[n] define a line or circle orthogonal to N.
This family of objects consists of two lines and 6 distinct circles.
This may be proved exactly as for Theorem 2.
Name | Centres | Line | Name | Centres | Line |
B[1] | X[2],X[110] | l[2,98] | B[12] | X[371],X[372] | b | B[2] | X[15],X[16] | b | B[13] | X[389],X[578] | b |
B[3] | X[32],X[39] | b | B[14] | X[379],X[857] | b |
B[4] | X[50],X[566] | b | B[15] | X[500],X[582] | b |
B[5] | X[52],X[569] | b | B[16] | X[567],X[568] | b |
B[6] | X[58],X[386] | b | B[17] | X[570],X[571] | b |
B[7] | X[61],X[62] | b | B[18] | X[572],X[573] | b |
B[8] | X[111],X[353] | l[111,182] | B[19] | X[575],X[576] | b |
B[9] | X[125],X[184] | l[2,98] | B[20] | X[580],X[581] | b |
B[10] | X[187],X[574] | b | B[21] | X[583],X[584] | b |
B[11] | X[284],X[579] | b | B[22] | X[1151],X[1152] | b |
Theorem 5
Suppose that ΔABC is not isosceles.
For 1 ≤ m < n ≤ 22, the pairs B[m],B[n] define a line or circle orthogonal to B.
This family of objects consists of three lines and 57 distinct circles.
Proof Of the 231 pairs, 171 define b and m = 1, n = 9 gives l(2,98). Thus, we have 59 circles. Any two of {1,2,8} give the Parry circle, so we have at most 57 distinct circles. If k ≠ 1,8,9, and k' = 1,8 or 9, then B[k],B[k'] give a circle, and two such pairs will give distinct circles unless they have the same value of k. For a fixed k, The pairs B[k],B[1], B[k],B[9] are distinct since they contain different points on l[2,98]. If B[k],B[1] and B[k],B[8] coincide, then they define the Parry circle - given by B[1],B[8]. Then the common circle contains B[2], so we must have k = 2. If B[k],B[8], B[k],B[9] coincide, then they give the circle E given byB[8],B[9]. This can happen for only one value of k, since otherwise E contains four points of b. Thus the 57 pairs B[k],B[k'] define at least 55 distinct circles, and at most 56 as the Parry circle occurs twice. If this collection includes the circle E then we have 55 in total. Otherwise, the collection has 56 distinct circles, andE is another. Tedious calculation could be used to show that none of the points on the Brocard axis lies on E. A Cabri figure shows that, for some triangles, E and b are disjoint, so we have exactly 57 circles.
Name | Centres | Line | Name | Centres | Line |
I[1] | X[7],X[1323] | l(1,7) | I[4] | X[80],X[1387] | l(1,5) | I[2] | X[36],X[942] | l(1,3) | I[5] | X[354],X[1155] | l(1,3) |
I[3] | X[65],X[1319] | l(1,3) | I(6) | B[1],B[2] | l(1,3) |
Theorem 6
Suppose that ΔABC is not isosceles.
For 1 ≤ m < n ≤ 6, the pairs I[m],I[n] define a line or circle orthogonal to I.
This family of objects consists of one line and 8 or 9 distinct circles.
This may be proved as before for Theorem 2. Clearly, the cases including one of I[1],I[4} give two sets of 4 distinct circles. The pair I[1],I[4] may give rise to a circle identical to one of these. Cabri sketches suggest that all 9 circles are distinct.
As remarked earlier, the first named centre on the incircle is the Feuerbach point, X[11]. This is the point of tangency of I and N, and has the function f(a,b,c) = bc(b+c-a)/(b-c)2. We shall show that the centres on I can be obtained from those on C in a simple geometrical way. The corresponding algebra is also straight-forward.
In the course of his work, Brisse proved the following beautiful result. For internal consistency in this paper, we shall restate it in terms of trilinears.
Theorem
If P = p(a,b,c) lies on the circumcircle, then S = a/(b+c-a)p2(a,b,c) lies on the incircle.
The geometry of the proof is neat. Let the tangents from P to I meet C again in Q,R. Then, by the porism, QR touches I. The contact point is S. Of course, the hard part is the ingenious algebra. The geometry makes it clear that each S on I arises from a unique point on C. The tangent to I at S meets C in two points Q',R'. The other tangents to I from Q' and R' meet at P' on C, again by the porism. Here the algebra is less clear - we have to cope with the choice of roots. From the geometry, we have the
Corollary
Each point on the incircle arises from a unique point on the circumcircle.
For example, the point X[109] = a/(b-c)(b+c-a) on C corresponds to the point (b-c)2(b+c-a)/a on I. This is X[11]. The point X[100] = 1/(b-c) corresponds to S = a(b-c)2/(b+c-a) on I. Although the trilinears are fairly simple, S = X(1357).