Inversion II

E. Full solution. By definition, \(OA\cdot OA^*=6^2=36\). Hence \(OA^*=9\). Point \(A^*\) lies on ray \(OA\). If we invert \(A^*\), its image \(A^{**}\) lies on the same ray and satisfies \(OA^*\cdot OA^{**}=36\). Thus \(9\cdot OA^{**}=36\), so \(OA^{**}=4\). There is only one such point on ray \(OA\), hence \(A^{**}=A\).

E. Full solution. Let the radius of inversion be \(R\). Then \(OA\cdot OA^*=OB\cdot OB^*=R^2\). Hence \(\frac{OA}{OB}=\frac{OB^*}{OA^*}\). Also rays \(OA\) and \(OA^*\) coincide, and rays \(OB\) and \(OB^*\) coincide, so \(\angle AOB=\angle B^*OA^*\). By SAS similarity, \(\triangle OAB\sim\triangle OB^*A^*\). Therefore the corresponding angles are equal: \(\angle OAB=\angle OB^*A^*\).

E. Full solution. Let \(M\in l\). Then \(\angle OCM=90^\circ\). By the similarity lemma for inverse triangles, \(\triangle OCM\sim\triangle OM^*C^*\). Hence \(\angle OM^*C^*=90^\circ\). Therefore all points \(M^*\) lie on the circle with diameter \(OC^*\). Conversely, if \(X\) lies on this circle, line \(OX\) meets \(l\) at a point \(M\), and the same similarity gives \(X=M^*\).

E. Full solution. Since \(OA\) is a diameter of \(\omega\), for every \(M\in\omega\) we have \(\angle OMA=90^\circ\). Under inversion, points \(M\) map to points \(M^*\) on rays \(OM\). By the similarity of inverse triangles, the condition \(\angle OMA=90^\circ\) means that \(M^*\) lies on the line through \(A^*\) perpendicular to \(OA\). The reverse inclusion follows from the invertibility of inversion.

E. Full solution. Points \(A\) and \(B\) are endpoints of a diameter of \(\omega\), so \(\angle AMB=90^\circ\) for every \(M\in\omega\). By the similarity of inverse triangles, the angle between \(A^*M^*\) and \(B^*M^*\) is also right. Hence \(M^*\) lies on the circle with diameter \(A^*B^*\). Since inversion is reversible, the whole circle is the image of \(\omega\).

E. Full solution. Let \(C\) and \(r\) be the center and radius of \(\gamma\). Orthogonality gives \(OC^2=R^2+r^2\). Hence the power of \(O\) with respect to \(\gamma\) equals \(OC^2-r^2=R^2\). If a line through \(O\) meets \(\gamma\) at \(X\) and \(Y\), then \(OX\cdot OY=R^2\). Therefore \(Y\) is the image of \(X\) under the inversion. Thus every point of \(\gamma\) maps to a point of the same circle.

E. Full solution. At the tangency point the circles have a common tangent, so the angle between them is \(0^\circ\). Inversion preserves angles between curves, meaning angles between tangents. Point \(T\) maps to \(T^*\), and the tangents to the images at \(T^*\) form the same angle \(0^\circ\). Therefore the images have a common tangent at \(T^*\), so they are tangent.

E. Full solution. The angle between circles is defined as the angle between their tangents at the intersection point. Let \(l_1\) and \(l_2\) be the tangents to \(\omega_1\) and \(\omega_2\) at \(P\). Inversion sends infinitesimal directions near \(P\) to infinitesimal directions near \(P^*\) and preserves the oriented angle between them; this follows from the similarity of inverse triangles in the limiting position. Hence the angle between the tangents to the images equals the angle between \(l_1\) and \(l_2\).

E. Full solution. Since both circles pass through the inversion center \(A\), each maps to a line. The original circles are tangent at \(A\), so the angle between them is \(0^\circ\). Inversion preserves the angle between curves; hence the angle between the resulting lines is also \(0^\circ\). Two distinct lines with zero angle are parallel.

E. Full solution. Both circles pass through the inversion center \(A\), so each maps to a line. Since \(B\) lies on both circles, its image \(B^*\) must lie on both image lines; hence the lines meet at \(B^*\). Inversion preserves the angle between circles, so it equals the angle between their images, that is, between the two lines.

E. Full solution. Perform an arbitrary inversion centered at \(A\). Line \(l\) maps to a circle \(l^*\) through \(A\). The desired circle passes through the inversion center \(A\), so its image is a line. Since it passes through \(B\), this line must pass through \(B^*\). Tangency to \(l\) is preserved, so the image line must be tangent to circle \(l^*\). Thus draw the tangents from \(B^*\) to \(l^*\), then invert these lines back. Their images are the required circles.

E. Full solution. Invert centered at \(A\). The desired circle, passing through \(A\), maps to a line through \(B^*\). Circle \(\omega\), not passing through \(A\), maps to another circle \(\omega^*\). Tangency is preserved, so the image line must be tangent to \(\omega^*\). Therefore draw the tangents from \(B^*\) to \(\omega^*\), then invert them back. The images of these lines are the circles through \(A\) and \(B\) tangent to \(\omega\).

E. Full solution. Invert centered at \(P\). The three given circles pass through the inversion center, so they become three lines. Each such line is the image of a circle passing through three vertices of one triangle of the complete quadrilateral. After inversion, the condition that three points lie on a circle through \(P\) becomes collinearity of their images. For the four original lines these are exactly the corresponding sides of the image complete quadrilateral. If three corresponding triples of images are collinear, the fourth collinearity follows from the usual Miquel theorem for four lines in reverse form. Inverting back, the fourth circle also passes through \(P\).

E. Full solution. Perform an arbitrary inversion centered at \(A\). Circles \(\omega_1\) and \(\omega_2\), passing through the inversion center, become two lines \(l_1\) and \(l_2\) meeting at \(B^*\). The desired circle also passes through \(A\), so its image is a line \(m\). Inversion preserves intersection angles, hence the equal-angle condition becomes: line \(m\) makes equal angles with \(l_1\) and \(l_2\). Therefore \(m\) must be one of the two angle bisectors of \(l_1\) and \(l_2\). Inverting these two bisectors back gives the two desired circles through \(A\).

E. Full solution. Let \(P\) be the point of the arc equidistant from \(A\) and \(B\). Take the inversion centered at \(P\) with power \(PA^2\). It swaps chord \(AB\) and arc \(AB\) of the segment: points \(A\) and \(B\) remain fixed, while the circle through \(P,A,B\) maps to line \(AB\). Circles inscribed in the segment map to circles of the same type. For two such circles, their common chord and its image must pass through the inversion center, because both circles are tangent to the two mutually inverse boundaries. Therefore their common chord \(MN\) passes through \(P\).

E. Full solution. Both circles are inscribed in the same angle, so their centers lie on the angle bisector, and the circles are homothetic with center \(A\). Let the tangency points of the first circle with the sides be \(X_1,Y_1\), and those of the second be \(X_2,Y_2\). The homothety centered at \(A\) sends the first circle to the second and maps \(X_1\) to \(X_2\), \(Y_1\) to \(Y_2\). Therefore segment \(X_1Y_1\) maps to \(X_2Y_2\), and corresponding segments under homothety are parallel. Hence \(X_1Y_1\parallel X_2Y_2\).

E. Full solution. Invert centered at \(A_1\). The circles passing through \(A_1\) become lines. The circle through \(A_1,A_2,A_3,A_4\) also becomes a line. After the transformation, the statement becomes a configuration of two lines and two circles: we must prove that the four images of \(B_i\) lie on one circle or line. This follows from the usual angle criterion: the angle equality expressing cyclicity of \(A_1,A_2,A_3,A_4\) before inversion becomes an equality of angles between lines. Translating this equality back gives cyclicity or collinearity of \(B_1,B_2,B_3,B_4\).

E. Full solution. Choose point \(O\) on the line of centers so that its powers with respect to the two circles are equal. Then the tangent lengths from \(O\) to the two circles are equal. Take inversion centered at \(O\) with power equal to this common power. Under this inversion, both circles transform in a coordinated way into circles whose centers lie on the same line and become concentric. Then common tangents are constructed as inverse images of lines tangent to the two concentric circles, namely lines at a fixed distance from the common center.

E. Full solution. All centers of the circles lie on the angle bisector. For two neighboring circles there is a homothety centered at the vertex of the angle sending the smaller circle to the larger. Their tangency point lies on the line of centers, hence on the angle bisector. Therefore all tangency points of neighboring circles lie on one line: the angle bisector. The third common tangent to two neighboring circles is also perpendicular to the line of centers in the homothetic model, and in the original angle its direction is fixed. Hence it is parallel to the corresponding tangency line.

E. Full solution. Let the centers of the given circles be \(O_1,O_2\), their radii \(R_1,R_2\), and let a circle centered at \(X\) with radius \(r\) be orthogonal to both. Then \(XO_1^2=R_1^2+r^2\) and \(XO_2^2=R_2^2+r^2\). Subtracting gives \(XO_1^2-R_1^2=XO_2^2-R_2^2\). This means that the powers of \(X\) with respect to \(\omega_1\) and \(\omega_2\) are equal. Therefore \(X\) lies on their radical axis.

E. Full solution. A circle passing through the inversion center \(P\) maps to a line. Consider the three circles through \(P\) and pairs of neighboring vertices of the complete quadrilateral. Their images are three lines. These lines meet pairwise at the images of the corresponding vertices, because inversion preserves incidence. Therefore the three image lines are the sides of a triangle whose vertices are the images of the three remaining vertices of the complete quadrilateral.

E. Full solution. Take inversion centered at \(A\) with radius \(R\), where \(R^2=k\cdot AB^2\). Let \(X\) map to \(X^*\). From the similarity of inverse triangles we get \(\frac{XB}{XA}=\frac{X^*B^*}{AB^*}\), where \(B^*\) is the image of \(B\). The condition \(\frac{XA}{XB}=k\) is equivalent to \(\frac{XB}{XA}=\frac1k\), hence \(X^*B^*=\frac{AB^*}{k}\). Thus \(X^*\) lies on the circle centered at \(B^*\) with fixed radius \(\frac{AB^*}{k}\). This circle does not pass through center \(A\), so its inverse image is a circle. Therefore the original locus of points \(X\) is a circle.

E. Full solution. After inversion centered at \(A\), circles \(R_1\) and \(R_2\) become parallel lines. Each circle \(S_i\), tangent to both, maps to a circle tangent to these two parallel lines. Hence all images \(S_i^*\) have the same radius, and their centers lie on the midline between the parallel lines. If neighboring equal circles are tangent, their tangency points lie on the same midline. Thus points \(T_i^*\) are collinear. The inverse image of a line not passing through the inversion center is a circle through \(A\); if the line passes through \(A\), it remains a line. Therefore the original points \(T_i\) lie on one circle through \(A\), or on one line.

E. Full solution. There is an inversion followed by a homothety sending \(R_1\) and \(R_2\) to concentric circles. Tangencies are preserved. In the annular region between two concentric circles, every circle tangent to both has the same radius, so all chain circles become equal. A closed chain of \(n\) equal circles between concentric circles means that moving from one circle to the next is a rotation by a fixed angle \(\varphi\), with \(n\varphi=360^\circ\). If we start with any other such circle, the same rotation constructs the next circle; after \(n\) steps the full rotation returns to the starting circle. Applying the inverse transformation proves the statement for the original circles.

E. Full solution. Invert centered at \(A\). Since \(R_1\) and \(R_2\) are tangent at the inversion center, their images are two parallel lines. Each circle \(S_i\), tangent to both given circles, maps to a circle tangent to the two parallel lines. Hence all \(S_i^*\) have the same radius, and their centers lie on the midline between the parallel lines. If two equal circles are tangent, their tangency point lies on the line of centers; therefore all points \(T_i^*\) lie on that midline. The inverse image of this line is a circle passing through the inversion center \(A\). Hence the original points \(T_i\) lie on one circle through \(A\).

E. Full solution. There is an inversion, possibly followed by a homothety, sending \(R_1\) and \(R_2\) to concentric circles. Tangencies are preserved. Between two concentric circles, all circles tangent to both are equal; their centers lie on a circle concentric with the given ones. If one closed chain of \(n\) circles exists, then passing from the center of one chain circle to the next is a rotation by a fixed angle \(\varphi\), with \(n\varphi=360^\circ\). Hence, starting from any other position of the first circle, we repeat the same rotation; after \(n\) steps we make a full turn and return to the beginning. The inverse transformation gives the required chain for the original circles.

E. Full solution. Invert centered at \(A_1\). Circles \(S_1\), \(S_2\), and the circle \((A_1A_2A_3A_4)\) become lines. Hence the images of \(A_2,A_3,A_4\) are collinear. Circles \(S_3\) and \(S_4\) become circles meeting these lines at the corresponding image points. The statement now reduces to the standard angle criterion in a configuration of two lines and two circles: from the collinearity of the images of \(A_2,A_3,A_4\), the angles under which the segments between the images of \(B_i\) are seen are equal or supplementary. Therefore \(B_1^*,B_2^*,B_3^*,B_4^*\) lie on one circle or line. Inverting back gives the same conclusion for \(B_1,B_2,B_3,B_4\).