圆 O 是过 A、B、C 的圆，如何作一个圆 Q 使之与 AB、AC、圆O 同时相切？

天山草

以下是网上关于 X55 的介绍。不知道说的是啥？

X(55) = INSIMILICENTER(CIRCUMCIRCLE, INCIRCLE)
Trilinears    a(b + c - a) : b(c + a - b) : c(a + b - c)
Trilinears    1 + cos A : 1 + cos B : 1 + cos C
Trilinears    cos2(A/2) : cos2(B/2) : cos2(B/2)
Trilinears    tan(B/2) + tan(C/2) : tan(C/2) + tan(A/2) : tan(A/2) + tan(B/2)
Trilinears    a(a - s) : b(b - s) : c(c - s)
Trilinears    a(cot A/2) : :
Trilinears    a2/(1 - cos A) : :
Trilinears    a(2ar - S) : :
Barycentrics   a2(b + c - a) : b2(c + a - b) : c2(a + b - c) : :
Barycentrics   area(A'BC) : : , where A'B'C' = 1st circumperp triangle
X(55) = R*X(1) + r*X(3)
X(55) = (Ra+Rb+Rc)*X(1) + r*Ja + r*Jb + r*Jc, where Ja, Jb, Jc are excenters, and Ra, Rb, Rc are the exradii
X(55) = center of homothety of three triangles:   tangential, intangents, and extangents. Also, X(55) is the pole-with-respect-to-the-circumcircle of the trilinear polar of X(1). These properties and others are given in

O. Bottema and J. T. Groenman, "De gemene raaklijnen van de vier raakcirkels van een driehoek, twee aan twee," Nieuw Tijdschrift voor Wiskunde 67 (1979-80) 177-182.

Let A', B', C' be the second points of intersection of the angle bisectors of triangle ABC with its incircle. Let Oa be the circle externally tangent to the incircle at A', and internally tangent to the circumcircle; define Ob and Oc cyclically. Then X(55) is the radical center of circles Oa, Ob, Oc. Let A" be the touchpoint of Oa and the circumcircle, and define B" and C" cyclically. The lines A'A", B'B", C'C" concur in X(55). Let Ba, Ca be the intersections of lines CA, AB, respectively, and the antiparallel to BC through a point P. Define Cb, Ab, Ac, Bc cyclically. Triangles ABaCa, AbBCb, AcBcC are congruent only when P = X(55) or one of its 3 extraversions. Let A*B*C* be the incentral triangle. Let La be the reflection of line BC in line AA*, and define Lb and Lc cyclically. Let A''' = Lb∩Lc, and define B''' and C'''. The lines A*A''', B*B''', C*C''' concur in X(55). (Randy Hutson, November 18, 2015)

Let A'B'C' be the extouch triangle. Let A" be the barycentric product of the circumcircle intercepts of line B'C', and define B" and C" cyclically. The lines AA", BB", CC" concur in X(55). (Randy Hutson, July 31 2018)

Let (Oa) be the circumcircle of BCX(1). Let Pa be the perspector of (Oa). Let La be the polar of Pa wrt (Oa). Define Lb and Lc cyclically. The lines La, Lb, Lc concur in X(55). (Randy Hutson, July 31 2018)

X(55) lies on these lines: 1,3   2,11   4,12   5,498   6,31   7,2346   8,21   9,200   10,405   15,203   16,202   19,25   20,388   30,495   34,227   41,220   43,238   45,678   48,154   63,518   64,73   77,1037   78,960   81,1002   92,243   103,109   104,1000   108,196   140,496   181,573   182,613  183,350   184,215   192,385   199,1030   201,774   204,1033   219,284   223,1456   226,516   255,601   256,983   329,1005   376,1056   386,595   392,997   411,962   511,611   515,1012   519,956   574,1015   603,963   631,1058   650,884   654,926   748,899   840,901   846,984   869,893   1026,1083   1070,1076   1072,1074   2195,5452  

X(55) is the {X(1),X(3)}-harmonic conjugate of X(56). For a list of other harmonic conjugates of X(55), click Tables at the top of this page.

X(55) = reflection of X(i) in X(j) for these (i,j): (1478,495), (2099,1)
X(55) = isogonal conjugate of X(7)
X(55) = isotomic conjugate of X(6063)
X(55) = complement of X(3434)
X(55) = anticomplement of X(2886)
X(55) = centroid of curvatures of circumcircle and excircles
X(55) = circumcircle-inverse of X(1155)
X(55) = antigonal conjugate of polar conjugate of X(37767)
X(55) = X(i)-Ceva conjugate of X(j) for these (i,j): (1,6), (3,198), (7,218), (8,219), (9,220), (21,9), (59,101), (104,44), (260,259), (284,41)
X(55) = cevapoint of X(42) and X(228) for these (i,j)
X(55) = X(i)-cross conjugate of X(j) for these (i,j): (41,6), (42,33), (228,212)
X(55) = crosspoint of X(i) and X(j) for these (i,j): (1,9), (3,268), (7,277), (8,281), (21,284), (59,101)
X(55) = crosssum of X(i) and X(j) for these (i,j): (1,57), (2,145), (4,196), (11,514), (55,218), (56,222), (63,224), (65,226), (81,229), (177,234), (241,1362), (513,1086), (905,1364), (1361,1465)
X(55) = crossdifference of every pair of points on line X(241)X(514)
X(55) = X(i)-Hirst inverse of X(j) for these (i,j): (6,672), (43,241)
X(55) = X(1)-line conjugate of X(241)
X(55) = X(i)-beth conjugate of X(j) for these (i,j): (21,999), (55,31), (100,55), (200,200), (643,2), (1021,1024)
X(55) = insimilicenter of the intangents and extangents circles
X(55) = insimilicenter of the intangents and tangential circles
X(55) = exsimilicenter of then extangents and tangential circles
X(55) = X(22)-of-intouch-triangle
X(55) = trilinear pole of line X(657)X(663) (polar of X(331) wrt polar circle)
X(55) = pole wrt polar circle of trilinear polar of X(331)
X(55) = X(48)-isoconjugate (polar conjugate) of X(331)
X(55) = homothetic center of ABC and Mandart-incircle triangle
X(55) = inverse-in-Feuerbach-hyperbola of X(1001)
X(55) = inverse-in-circumconic-centered-at-X(1) of X(1936)
X(55) = {X(1),X(40)}-harmonic conjugate of X(65)
X(55) = trilinear square of X(259)
X(55) = Danneels point of X(100)
X(55) = vertex conjugate of PU(48)
X(55) = vertex conjugate of foci of Mandart inellipse
X(55) = excentral isotomic conjugate of X(2942)
X(55) = homothetic center of the reflections of the intangents and extangents triangles in their respective Euler lines
X(55) = perspector of ABC and extraversion triangle of X(56)
X(55) = trilinear product of PU(104)
X(55) = barycentric product of PU(112)
X(55) = bicentric sum of PU(112)
X(55) = PU(112)-harmonic conjugate of X(650)
X(55) = perspector of ABC and unary cofactor triangle of 7th mixtilinear triangle
X(55) = perspector of 4th mixtilinear triangle and unary cofactor triangle of 7th mixtilinear triangle
X(55) = perspector of unary cofactor triangles of 3rd, 4th and 5th extouch triangles
X(55) = {X(3513),X(3514)}-harmonic conjugate of X(56)
X(55) = perspector of ABC and cross-triangle of ABC and extangents triangle
X(55) = perspector of ABC and cross-triangle of ABC and Hutson extouch triangle
X(55) = homothetic center of ABC and cross-triangle of ABC and 2nd Johnson-Yff triangle
X(55) = Thomson-isogonal conjugate of X(5657)
X(55) = homothetic center of midarc triangle and 2nd-circumperp-of-1st-circumperp triangle (which is also 1st-circumperp-of-2nd-circumperp triangle)
X(55) = homothetic center of 2nd midarc triangle and 1st-circumperp-of-1st-circumperp triangle (which is also 2nd-circumperp-of-2nd-circumperp triangle)
X(55) = Cundy-Parry Phi transform of X(942)
X(55) = Cundy-Parry Psi transform of X(943)
X(55) = X(4)-of-1st-Johnson-Yff-triangle
X(55) = homothetic center of anti-Hutson intouch triangle and anti-tangential midarc triangle
X(55) = barycentric product of circumcircle intercepts of excircles radical circle

kanyikan

kanyikan

luyuanhong

楼上 kanyikan 的解答很好！已收藏。

数学小白新

各位大神的解法真精彩啊，我继续来凑下热闹。

luyuanhong

楼上 数学小白新 的解答很好！已收藏。

数学小白新

风花飘飘 发表于 2022-6-8 08:50
直线是无限放大的圆弧，对于尺规作图来讲的话就是这样。
CCC,CCL,CLL,CLC,LCC,LCL,LLL,...都是一样的。
...

老师谦虚了！我虽然来论坛不久，但也知道老师是数学中国的泰山北斗。以后还望老师多多指教。