Book 3. Advanced Olympiad Geometry

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 \(B_1,B_2,B_3,B_4\).

E. Full Solution.

Denote the four lines by \(a,b,c,d\). For any transversal not passing through \(O\), the cross-ratio of the intersection points equals the cross-ratio of the pencil:

\[ (A B C D)=\frac{\sin(a,c)}{\sin(b,c)}:\frac{\sin(a,d)}{\sin(b,d)}. \]

The same formula is obtained for \(A_1,B_1,C_1,D_1\), since the angles between \(a,b,c,d\) do not depend on the transversal. Hence the two cross-ratios are equal to the same number.

E. Full Solution.

Let \(X\) be an arbitrary point of the line, and let \(X'\) be its image. Since the transformation is projective, it preserves cross-ratio:

\[ (A B C X)=(A B C X'). \]

For fixed distinct \(A,B,C\), the value \((A B C T)\) uniquely determines the point \(T\) on the line. Hence \(X'=X\). Since \(X\) was arbitrary, every point of the line is fixed.

E. Full Solution.

We have \(f(0)=-\frac{1}{3}\), \(f(1)=\frac{1}{4}\). At \(x=\infty\), the image is the ratio of the leading coefficients, namely \(2\). At \(x=-3\), the denominator is zero, so \(f(-3)=\infty\).

For any \(x_i,x_j\), compute:

\[ f(x_i)-f(x_j)=\frac{(2x_i-1)(x_j+3)-(2x_j-1)(x_i+3)}{(x_i+3)(x_j+3)}=\frac{7(x_i-x_j)}{(x_i+3)(x_j+3)}. \]

When forming the cross-ratio, all factors \(7\) and \((x_i+3)\) cancel. Hence \((f(x_1)f(x_2)f(x_3)f(x_4))=(x_1x_2x_3x_4)\).

E. Full Solution.

Since \(T=a\cap b\), its image \(T'\) is the intersection of the images \(a'\) and \(b'\). But \(T\in s\), and \(s\) maps to the line at infinity, so \(T'\) is a point at infinity. Two ordinary lines meeting at a point at infinity are parallel. Hence \(a'\parallel b'\).

Conversely, if \(a'\parallel b'\), then their intersection is a point at infinity. Its preimage lies on the preimage of the line at infinity, namely on \(s\). Therefore \(a\cap b\in s\).

E. Full Solution.

A central projection is a projective map from the line \(l\) to the line \(m\). Therefore it preserves cross-ratio:

\[ (A_1 B_1 C_1 D_1)=(A B C D). \]

By assumption \((A B C D)=-1\). Hence \((A_1 B_1 C_1 D_1)=-1\), so the image of a harmonic quadruple is harmonic.

E. Full Solution.

The central projection from \(l\) to \(m\) with center \(O\) is a projective map. Similarly, the central projection from \(m\) to \(n\) with center \(P\) is projective.

The resulting map \(l\to n\) is the composition of these two maps. By the definition and basic properties of projective maps, a composition of projective maps is projective.

E. Full Solution.

Choose a projective transformation that sends \(O\) to a point at infinity. Then the images of \(AA_1\), \(BB_1\), \(CC_1\) become parallel.

In this affine model, corresponding vertices of the two triangles lie on parallel lines of one direction. In that model the statement is easily checked by coordinates or by Menelaus: the intersections of corresponding sides are collinear.

Since a projective transformation preserves collinearity, the original points \(P,Q,R\) are also collinear.

E. Full Solution.

The converse of Desargues is dual to the direct form: if we interchange points and lines, concurrence becomes collinearity and collinearity becomes concurrence.

Apply the dual form of the direct theorem to the two triples of sides \(AB,BC,CA\) and \(A_1B_1,B_1C_1,C_1A_1\). Their corresponding intersections \(P,Q,R\) lie on one line, so the joins of the corresponding vertices in the dual configuration are concurrent.

In the original language this means that \(AA_1\), \(BB_1\), \(CC_1\) pass through one point.

E. Full Solution.

By Pascal's theorem for the hexagon \(ABCDEF\), the points

\[ P=AB\cap DE,\quad Q=BC\cap EF,\quad R=CD\cap FA \]

are collinear. But \(AB\parallel DE\), so \(P\) is the point at infinity in the direction of \(AB\). Since the line \(QR\) passes through \(P\), it has the same direction. Hence \(QR\parallel AB\).

E. Full Solution.

Consider the degenerate hexagon \(A,A,B,C,C,D\) inscribed in the given circle. The side between the neighbouring points \(A,A\) is the tangent at \(A\), and the side \(C,C\) is the tangent at \(C\).

The opposite sides of this hexagon meet at \(X\), \(Y\), and \(Z\). By Pascal's theorem these three points are collinear.

E. Full Solution.

Consider the degenerate hexagon \(A,B,B,C,D,D\) on the circle. Its side \(BB\) is the tangent at \(B\), and its side \(DD\) is the tangent at \(D\).

The pairs of opposite sides give:

\[ AB\cap CD=Y,\quad BB\cap DD=X,\quad BC\cap DA=Z. \]

By Pascal's theorem, \(X,Y,Z\) are collinear.

E. Full Solution.

The first step \(X\mapsto Y\) is central projection from the line \(l\) to the circle with center \(M\). The second step \(Y\mapsto X'\) is central projection from the circle to the line \(l\) with center \(N\).

Choose an auxiliary line \(s\) and identify the circle \(\omega\) with \(s\) by a central projection from a point not on \(\omega\). Then both steps become projective maps between lines. Therefore their composition is a projective transformation of \(l\).

Every projective transformation of a line preserves cross-ratio. Hence for any four points \(X_1,X_2,X_3,X_4\) on \(l\), we have \((X_1X_2X_3X_4)=(X'_1X'_2X'_3X'_4)\).

E. Full Solution.

Let \(F_0\) be the second intersection of the line \(e\) with the conic. Apply Pascal to the hexagon \(A,B,C,D,E,F_0\). The points

\[ AB\cap DE=K,\quad BC\cap EF_0=L,\quad CD\cap F_0A \]

are collinear. Since \(E,F_0,L\) lie on \(e\), the point \(BC\cap EF_0\) is indeed \(L\). Hence the third point \(CD\cap F_0A\) lies on the line \(KL\), so it coincides with \(M=KL\cap CD\).

Therefore the line \(AM\) passes through \(F_0\). But the constructed point \(F\) is \(AM\cap e\), and \(F_0\in e\). Thus \(F=F_0\), and \(F\) lies on the conic.

E. Full Solution.

Denote \(D_0=f(D)\). Since \(f\) is projective, it preserves cross-ratio:

\[ (A_1 B_1 C_1 D_0)=(A B C D). \]

By assumption, \((A B C D)=(A_1 B_1 C_1 D_1)\). Therefore,

\[ (A_1 B_1 C_1 D_0)=(A_1 B_1 C_1 D_1). \]

On the line \(m\), the points \(A_1,B_1,C_1\) are distinct. For three fixed distinct points, the value \((A_1 B_1 C_1 X)\) uniquely determines the point \(X\). Hence \(D_0=D_1\). Thus \(f(D)=D_1\).

E. Full Solution.

The union \(l\cup m\) of two lines may be regarded as a degenerate conic. Take the hexagon with vertices \(A,B_1,C,A_1,B,C_1\) on this conic.

The pairs of opposite sides give exactly the points \(P,Q,R\), up to the naming order. By Pascal's theorem for a degenerate conic, these three points are collinear.

E. Full Solution.

The map consists of three steps: projection from \(l\) to the circle with center \(M\), the antipodal map on the circle, and projection from the circle back to \(l\) with center \(M\).

The first and third steps are projective after identifying the circle with a line. The antipodal map is the restriction of a plane motion preserving the circle, and therefore in the same parametrisation it also gives a projectivity.

The composition of three projective maps is a projective transformation of \(l\). Hence \(X\mapsto X'\) is projective.

E. Full Solution.

Consider polarity with respect to the given circle. Each tangent side of the hexagon corresponds to its point of tangency, and each vertex of the hexagon corresponds to the line joining two neighbouring contact points in the dual configuration.

Pascal's theorem for the six contact points on the circle, in dual form, says that the three lines joining opposite vertices of the tangential hexagon are concurrent.

These three lines are exactly \(AD\), \(BE\), and \(CF\). Hence they pass through one point.

E. Full Solution.

By Pascal's theorem for the hexagon \(ABCDEF\), the points

\[ P=AB\cap DE,\quad Q=BC\cap EF,\quad R_0=CD\cap FA \]

are collinear. The first two are \(P\) and \(Q\), so the third point \(R_0\) lies on the line \(PQ\).

But by definition \(R\) is the intersection of \(PQ\) with \(CD\). The point \(R_0\) also lies on \(PQ\) and on \(CD\). Hence \(R=R_0\), and therefore \(R\in FA\). Thus \(A,F,R\) are collinear.

E. Full Solution.

Applying \(f\) three times, we get:

\[ f^3(A)=A,\quad f^3(B)=B,\quad f^3(C)=C. \]

The map \(f^3\) is a composition of projective transformations, so it is projective. It has three distinct fixed points \(A,B,C\). Therefore, by the basic fact about projectivities of a line, \(f^3\) is the identity.

E. Full Solution.

Cross-ratio is preserved by projective transformations, so we may replace the complete quadrangle by a projectively equivalent convenient model. Choose coordinates so that

\[ A=(-1,0),\quad C=(1,0),\quad B=(0,1),\quad D=(0,-1). \]

Then \(G=AC\cap BD=(0,0)\). The lines \(AB\) and \(CD\) have the same direction, so their intersection \(E\) is the point at infinity of direction with slope \(1\). The lines \(AD\) and \(BC\) are also parallel, so \(F\) is the point at infinity of direction with slope \(-1\).

Therefore the line \(EF\) is the line at infinity, and its intersection with \(AC\) is the point at infinity of the line \(AC\). Thus, in the coordinate on \(AC\), we have \(A=-1\), \(C=1\), \(G=0\), and \(H=\infty\).

Hence

\[ (A C G H)=(-1,1,0,\infty)=\frac{0-(-1)}{0-1}=-1. \]

Since this cross-ratio is preserved by the inverse projective transformation, the original configuration also satisfies \((A C G H)=-1\).

E. Full Solution.

Let \(P=AB\cap DE\), \(Q=BC\cap EF\), and \(R=CD\cap FA\). Choose a projective transformation that sends the conic to a circle and sends \(P\) to a point at infinity. Then the images of the lines \(AB\) and \(DE\) become parallel.

In the new configuration, apply the proved special case of Pascal: the images of \(Q\) and \(R\) lie on a line parallel to the images of \(AB\) and \(DE\), so this line passes through the image of \(P\).

Therefore the images of \(P,Q,R\) are collinear. Since projective transformations preserve collinearity, the original points \(P,Q,R\) are also collinear.

E. Full Solution.

By Pappus' theorem for points \(A,B,C\) on \(l\) and \(A_1,B_1,C_1\) on \(m\), the points

\[ P=AB_1\cap A_1B,\quad Q=BC_1\cap B_1C,\quad R_0=AC_1\cap A_1C \]

are collinear. The first two are the given points \(P\) and \(Q\), so \(R_0\in PQ\).

But by definition \(R\) is the point \(PQ\cap AC_1\). The point \(R_0\) also lies on \(PQ\) and on \(AC_1\). Hence \(R=R_0\). Therefore \(R\in A_1C\), as required.

E. Full Solution.

Assume such a construction exists. Choose a projective transformation of the plane that fixes \(A\) and \(B\), but does not fix the midpoint \(M\) of \(AB\). Such a transformation exists because projective transformations of the line \(AB\) may fix two points and move a third.

Perform the supposed construction once. Then apply the chosen projective transformation to every object appearing at each step. Since a projective transformation sends lines to lines and preserves intersections, the second process is a valid execution of the same instruction starting from the same initial points \(A,B\).

But the result of the second process must be the image of the result of the first process, namely the image of \(M\), which is not \(M\). On the other hand, a universal instruction must again produce the midpoint of \(AB\), namely \(M\). This contradiction proves that a straightedge alone cannot construct the midpoint.

E. Full Solution.

Each step \(X_i\mapsto X_{i+1}\) is a central projection from the line \(l_i\) to the line \(l_{i+1}\) with center \(O_i\). Thus each step is a projective map between lines.

The composition of all \(n\) steps is a projective transformation of the line \(l_1\) to itself. By assumption, this transformation has three distinct fixed points: the three starting points for which the chain closes.

A projective transformation of a line with three fixed points is the identity. Therefore the resulting map sends every point \(X_1\), for which all intermediate projections are defined, back to itself. Hence the chain closes for every such starting point.

E. Full Solution.

The triangles \(OAP\) and \(OBP\) are right triangles, have the common hypotenuse \(OP\), and equal legs \(OA=OB=R\). Hence they are congruent, and \(A\) and \(B\) are symmetric with respect to \(OP\). Therefore \(AB\perp OP\).

In the right triangle \(OAP\), the point \(H\) is the foot of the altitude from \(A\) to the hypotenuse \(OP\). By the leg theorem, \(OA^2=OH\cdot OP\). Thus \(OP\cdot OH=R^2\).

E. Full Solution.

The line \(px+qy=1\) has normal vector \((p,q)\), so it is perpendicular to \(OP\). Let \(H\) be its intersection with \(OP\). Then \(H=t(p,q)\). Substituting into the line equation gives \(t(p^2+q^2)=1\), hence \(t=\frac{1}{p^2+q^2}\).

We have \(OP=\sqrt{p^2+q^2}\), and \(OH=t\sqrt{p^2+q^2}=\frac{1}{\sqrt{p^2+q^2}}\). Therefore \(OP\cdot OH=1\), which defines the polar with respect to the unit circle.

E. Full Solution.

The polar of \(P\) has equation \(px+qy=1\). Since \(Q(u,v)\) lies on it, we get \(pu+qv=1\).

The polar of \(Q\) has equation \(ux+vy=1\). Substituting \(P(p,q)\), we get \(up+vq=1\), the same equality. Therefore \(P\) lies on the polar of \(Q\).

E. Full Solution.

By construction \(l\perp OH\), and since \(P\) lies on \(OH\), we have \(l\perp OP\). Also, \(H=l\cap OP\) and \(OP\cdot OH=R^2\).

These two properties uniquely define the polar of \(P\) with respect to \(\omega(O,R)\): it is perpendicular to \(OP\) and meets \(OP\) at \(H\). Hence \(l\) is the polar of \(P\).

E. Full Solution.

Let \(X\) be the pole of the line \(l\) with respect to \(\omega\). Then \(l\) is the polar of \(X\).

For every point \(P\in l\), the point \(P\) lies on the polar of \(X\). By La Hire's theorem, \(X\) lies on the polar of \(P\). But the polar of \(P\) is the chord of contact \(AB\). Hence every line \(AB\) passes through the same point \(X\).

E. Full Solution.

Since \(TA\) and \(TB\) are tangents, the polar of \(T\) is the line \(AB\). The point \(P\) lies on \(AB\), so \(P\) lies on the polar of \(T\).

By La Hire's theorem, \(T\) lies on the polar of \(P\), as required.

E. Full Solution.

Let the tangents at \(A\) and \(B\) meet at \(X\), and the tangents at \(C\) and \(D\) meet at \(Y\). By the previous fact, \(X\) and \(Y\) lie on the polar of \(P\), so the polar of \(P\) is the line \(XY\).

Apply degenerate Pascal to the circle with vertices \(A,A,C,D,B,B\). It gives that \(X,Y\) and one diagonal point are collinear; an analogous order gives the other diagonal point. Hence the line \(XY\) coincides with \(QR\). Therefore \(QR\) is the polar of \(P\).

E. Full Solution.

Since \(X\) lies on the polar of \(P\), La Hire's theorem gives that \(P\) lies on the polar of \(X\). Similarly, \(X\) lies on the polar of \(Q\), so \(Q\) lies on the polar of \(X\).

The polar of \(X\) is a line. It passes through \(P\) and \(Q\), hence it is the line \(PQ\).

E. Full Solution.

The line \(AB\) is the polar of \(P\). By assumption \(Q\in AB\), so \(Q\) lies on the polar of \(P\).

By La Hire's theorem, \(P\) lies on the polar of \(Q\). This proves the claim.

E. Full Solution.

Let \(T=AC\cap BD\). For the complete quadrangle \(ABCD\), the diagonal triangle is self-polar, so the polar of \(T\) is the line \(YZ\).

Since \(T,A,C\) are collinear and \(X\) is the intersection of the tangents at \(A\) and \(C\), the point \(X\) lies on the polar of \(T\). Hence \(X\in YZ\). Therefore \(X,Y,Z\) are collinear.

E. Full Solution.

The points \(A,B,C,D\) lie on one circle. In the complete quadrangle \(ABCD\), the diagonal points are

\[ P=AB\cap CD,\quad Q=AC\cap BD,\quad R=AD\cap BC. \]

By self-polarity of the diagonal triangle, the polar of \(P\) is the line \(QR\). Hence both \(Q\) and \(R\) lie on the polar of \(P\).

E. Full Solution.

Let the radius of \(\omega_1\) be \(R\). Since \(A\) and \(B\) lie on \(\omega_1\), the line \(AB\) is perpendicular to the corresponding line of centers of the common chord; from the condition \(O,A,B,C\) in the configuration of \(\omega_2\), angle equalities give similarity of the triangles \(OBP\) and \(OCB\).

The similarity gives \(OP\cdot OC=OB^2=R^2\). The point \(P\) lies on \(OC\). The polar of \(C\) with respect to \(\omega_1\) meets \(OC\) at a point \(P'\) satisfying \(OC\cdot OP'=R^2\). Hence \(P'=P\), so \(P\) lies on the polar of \(C\).

E. Full Solution.

Since \(AB\) and \(AC\) touch the incircle at \(F\) and \(E\), the line \(EF\), joining the contact points of the two tangents from \(A\), is the polar of \(A\).

If \(X\in EF\), then \(X\) lies on the polar of \(A\). By La Hire's theorem, \(A\) lies on the polar of \(X\). Hence the polar of \(X\) passes through \(A\).

E. Full Solution.

Let \(T=BD\cap AC\). For the complete quadrangle \(ABCD\), the polar of \(T\) is the line through the other two diagonal points, namely through \(Y\) and \(Z\).

Since \(T,B,D\) are collinear and \(X\) is the intersection of the tangents at \(B\) and \(D\), the point \(X\) lies on the polar of \(T\). Hence \(X\in YZ\), as required.

E. Full Solution.

For each secant \(AB\), the tangents at \(A\) and \(B\) meet at a point \(T\), whose polar is \(AB\). Since \(P\in AB\), the point \(P\) lies on the polar of \(T\).

By La Hire's theorem, \(T\) lies on the polar of \(P\). Hence all points \(T\) lie on one fixed line - the polar of \(P\) with respect to \(\omega\). Conversely, every admissible point of this line arises from some secant through \(P\), so this is the desired locus in the natural domain.

E. Full Solution.

Consider vertex \(A\). The sides \(AB\) and \(AC\) touch \(\omega\) at two points. The line joining these contact points is, by definition, the polar of \(A\). Hence its pole is \(A\).

The same argument applies to vertices \(B\) and \(C\). Therefore the poles of the three constructed lines are \(A,B,C\), respectively.

E. Full Solution.

Let the sides of the hexagon touch \(\omega\) at \(A_1,B_1,C_1,D_1,E_1,F_1\), respectively. Pascal's theorem for the six contact points on the circle states the collinearity of three intersections of opposite sides of the corresponding inscribed hexagon.

Passing to poles and polars turns the collinearity of three points into the concurrence of the corresponding three lines. In the original tangential hexagon, these lines are exactly \(AD\), \(BE\), and \(CF\). Hence they meet at one point.

E. Full Solution.

The lines \(AB\), \(AC\), \(BC\) and their extensions are tangents to the same excircle. Consider a degenerate tangential hexagon in which some neighbouring sides lie on the same lines \(AB\), \(AC\), \(BC\), and the contact points correspond to \(E,F,D\).

By Brianchon's theorem, the three diagonals of such a tangential hexagon are concurrent. Two of these diagonals give the lines meeting at \(T=BF\cap CE\), and the third is the line \(AD\). Hence \(T\in AD\).

E. Full Solution.

We prove that the polar of \(P\) is \(QR\). The pairs \(A,B\) and \(C,D\) lie on two secants through \(P\). The tangent intersections at \(A,B\) and at \(C,D\) lie on the polar of \(P\). By degenerate Pascal, the same line passes through \(Q\) and \(R\). Hence the polar of \(P\) is \(QR\).

Similarly, using the secants through \(Q\), namely \(AC\) and \(BD\), we get that the polar of \(Q\) is \(PR\). Using the secants through \(R\), namely \(AD\) and \(BC\), we get that the polar of \(R\) is \(PQ\). All three statements are proved.

E. Full Solution.

Let \(R=AD\cap BC\). In the complete quadrangle \(ABCD\), the polar of \(R\) is the line \(PQ\).

On the other hand, \(R\) lies on the chord \(AD\), and the tangents at \(A\) and \(D\) meet at \(S\). Hence, by the secant-and-tangents fact, \(S\) lies on the polar of \(R\). Therefore \(S\in PQ\), so \(P,Q,S\) are collinear.

E. Full Solution.

By the formula \(OP\cdot OH=R^2\) and perpendicularity \(p\perp OP\), the line \(p\) is the polar of \(P\), even though no real tangents can be drawn from \(P\).

Let a chord \(AB\) pass through \(P\), and let the tangents at \(A\) and \(B\) meet at \(T\). The polar of \(T\) is \(AB\). Since \(P\in AB\), La Hire's theorem gives that \(T\) lies on the polar of \(P\), namely on \(p\). This holds for every chord \(AB\) through \(P\).

E. Full Solution.

Let \(Q=AC\cap BD\) and \(R=AD\cap BC\). Since \(A,B,C,D\) lie on the circle and \(P=AB\cap CD\), the points \(P,Q,R\) are the diagonal points of the complete quadrangle \(ABCD\).

By self-polarity of the diagonal triangle, the polar of \(P\) is the line \(QR\). The circle and the point \(P\) are fixed, so the polar of \(P\) is fixed. Therefore, for every choice of the two secants, \(Q\) and \(R\) lie on the same fixed line - the polar of \(P\).

E. Full Solution.

Use a projective transformation that sends \(\omega\) to a circle and the line \(KP\) to the line at infinity. Such transformations preserve tangency and incidence, so it is enough to prove the statement in the new model.

In the new model, the images of \(KA\) and \(KB\) are parallel. The points \(M\) and \(N\), lying on a line through the image of \(P\) at infinity, correspond to parallel cuts between two parallel tangents. The second tangents from \(M\) and \(N\) meet at a point \(X'\) moving along a line parallel to these tangents and passing through the image of \(K\). This follows from symmetry of the circle with respect to the perpendicular to the two parallel tangents.

Returning by the inverse projective transformation, we obtain that the original points \(X\) lie on one line, and this line passes through \(K\).

E. Full Solution.

Let \(P\) be the intersection of the sides \(A_1A_2\) and \(A_4A_5\). The polar of \(P\) passes through the poles of these two sides, namely through \(V_1=a_1\cap a_2\) and \(V_4=a_4\cap a_5\). Therefore the polar of \(P\) is the line \(V_1V_4\).

Similarly, the polars of the other two Pascal points are \(V_2V_5\) and \(V_3V_6\). By Pascal, the three Pascal points are collinear. The pole of this line lies on the polars of all three points. Hence the lines \(V_1V_4\), \(V_2V_5\), and \(V_3V_6\) pass through one point.

E. Full Solution.

Since \(A_1\in BC\) and \(PA_1\perp BC\), we have \(\angle PA_1C=90^\circ\). Similarly, \(B_1\in CA\) and \(PB_1\perp CA\), so \(\angle PB_1C=90^\circ\).

Thus \(A_1\) and \(B_1\) lie on the circle with diameter \(PC\). Therefore \(P,A_1,C,B_1\) are cyclic.

E. Full Solution.

By the previous fact, \(A_1\) and \(B_1\) lie on the circle with diameter \(PC\), so \(A_1B_1\) is a chord of this circle. Similarly, \(B_1\) and \(C_1\) lie on the circle with diameter \(PA\), and \(C_1\) and \(A_1\) lie on the circle with diameter \(PB\).

Thus all three sides of the pedal triangle are naturally connected with the circles on \(PA,PB,PC\) as diameters.

E. Full Solution.

Since \(H\) lies on the altitude from \(A\), the line \(AH\) is perpendicular to \(BC\). Therefore the foot of the perpendicular from \(H\) to \(BC\) coincides with the foot of the altitude from \(A\). Similarly, the projections of \(H\) onto \(CA\) and \(AB\) coincide with the feet of the altitudes from \(B\) and \(C\).

E. Full Solution.

The side \(BC\) is a chord of the circumcircle. The perpendicular from the center \(O\) to the chord \(BC\) bisects it. Hence the projection of \(O\) onto \(BC\) is the midpoint of \(BC\). The same applies to \(CA\) and \(AB\).

E. Full Solution.

The right angles give cyclic quadrilaterals \(P,A_1,C,B_1\) and \(P,B_1,A,C_1\). Hence \(\angle A_1B_1C=\angle A_1PC\), and \(\angle C_1B_1A=\angle C_1PA\) in the directed sense.

Since \(A,B,C,P\) lie on one circle, angles subtending the corresponding arcs give equality of these two directed angles. Thus the rays \(B_1A_1\) and \(B_1C_1\) lie on one line. Therefore \(A_1,B_1,C_1\) are collinear.

E. Full Solution.

As in the direct theorem, the right angles give cyclic quadrilaterals \(P,A_1,C,B_1\) and \(P,B_1,A,C_1\). The collinearity of \(A_1,B_1,C_1\) gives equality of the corresponding directed angles.

Transferring these angles to the circles with diameters, we get \(\angle PCA=\angle PBA\). This is the cyclicity criterion for \(P,C,A,B\). Hence \(P\) lies on the circumcircle of triangle \(ABC\).

E. Full Solution.

The pedal triangle has zero area if and only if its vertices \(A_1,B_1,C_1\) are collinear.

If \(P\) lies on the circumcircle, then by the Simson theorem \(A_1,B_1,C_1\) are collinear. Conversely, if they are collinear, then by the converse Simson theorem \(P\) lies on the circumcircle. The claim follows.

E. Full Solution.

Let \(A_1,B_1,C_1\) be the ordinary perpendicular feet from \(P\). By the Simson theorem, they are collinear. The new points \(A_2,B_2,C_2\) on \(BC,CA,AB\) are obtained from \(A_1,B_1,C_1\) by the same change of angle at \(P\): from the perpendicular direction to the direction making angle \(\alpha\) with the side.

For each side, the correspondence \(A_1\mapsto A_2\), \(B_1\mapsto B_2\), \(C_1\mapsto C_2\) is given by one spiral homothety centered at \(P\) with angle \(90^\circ-\alpha\). Such a transformation sends a line to a line. Therefore \(A_2,B_2,C_2\) are collinear.

E. Full Solution.

Let \(A_1,B_1,C_1\) be the projections of \(P\) onto \(BC,CA,AB\). The direction of the Simson line can be expressed, for instance, through the angle between \(B_1C_1\) and \(AB\). From the cyclicity of \(P,B_1,A,C_1\), this angle equals the corresponding angle between \(PA\) and \(PC\).

Since \(PQ\perp BC\), the inscribed angles defining the direction of \(AQ\) coincide with the angles defining the direction of \(B_1C_1\). Therefore \(B_1C_1\parallel AQ\). Hence the whole Simson line of \(P\) is parallel to \(AQ\).

E. Full Solution.

Let \(M\) be the midpoint of \(PH\), and let \(B_1,C_1\) be the projections of \(P\) onto \(CA\) and \(AB\). Since \(BH\perp AC\), and \(PB_1\perp AC\), we have \(PB_1\parallel BH\). Similarly, \(PC_1\parallel CH\).

Under the homothety with center \(P\) and ratio \(\frac12\), the point \(H\) maps to \(M\), while the directions \(BH\) and \(CH\) correspond to the directions through \(B_1\) and \(C_1\). From the parallelism of corresponding segments, we get \(M\in B_1C_1\). But \(B_1C_1\) is the Simson line, so it passes through \(M\).

E. Full Solution.

The direction of the Simson line changes by half of the angular displacement of the point on the circumcircle. Points \(P\) and \(Q\) are antipodal, so the radii \(OP\) and \(OQ\) differ by a rotation of \(180^\circ\).

Therefore the directions of their Simson lines differ by \(90^\circ\). Hence the two lines are perpendicular.

E. Full Solution.

Let \(H\) be the orthocenter, and let \(M,N\) be the midpoints of \(PH\) and \(QH\). By the Simson line property, the Simson line of \(P\) passes through \(M\), and the Simson line of \(Q\) passes through \(N\). The points \(M,N\) lie on the nine-point circle, because this circle is the image of the circumcircle under the homothety with center \(H\) and ratio \(\frac12\).

By the previous problem, the Simson lines are perpendicular. In the standard nine-point circle configuration, the angle \(MXN\) subtends a diameter \(MN\) of this circle, so \(X\) lies on it. Equivalently, check by angles: \(\angle MXN=90^\circ\), and \(MN\) is the corresponding diameter.

E. Full Solution.

When a point moves along the circumcircle by an arc of angular measure \(2\varphi\), the direction of the Simson line changes by \(\varphi\). Thus if the Simson lines of \(P\) and \(Q\) are parallel, their directions coincide modulo \(180^\circ\). Hence the arcs between \(P\) and \(Q\) have angular measure \(0^\circ\) or \(360^\circ\), so \(P=Q\).

The converse is immediate: if \(P=Q\), the Simson lines coincide, hence are parallel.

E. Full Solution.

By earlier facts, the projections of the circumcenter \(O\) onto the sides of \(ABC\) are the side midpoints, while the projections of the orthocenter \(H\) are the altitude feet.

The nine-point circle of triangle \(ABC\) passes through the side midpoints and the altitude feet. Therefore both pedal triangles have all their vertices on the same nine-point circle.

E. Full Solution.

Let \(A_1\) and \(B_1\) be the projections of \(P\) onto \(CB\) and \(CA\). The points \(A_1\) and \(B_1\) lie on the circle with diameter \(PC\). The angle \(\angle A_1CB_1\) equals \(\angle ACB\), which is constant by assumption.

Therefore the chord \(A_1B_1\) of the circle with diameter \(PC\) has constant length. The Simson line is the line \(A_1B_1\). All chords of a fixed circle with the same length are tangent to a concentric circle of fixed radius. Hence all these Simson lines are tangent to one fixed circle.

E. Full Solution.

By the Simson theorem, \(A_1,B_1,C_1\) are collinear. Consider the angle between the Simson line and \(AB\). Through the cyclic quadrilateral \(P,B_1,A,C_1\), it equals the angle between \(PB_1\) and \(PA\), that is, the angle between a perpendicular to \(AC\) and a line parallel to \(BC\).

This angle equals the angle between the altitude from \(A\) and \(AB\). Therefore the Simson line has the same direction as the altitude from \(A\), so it is parallel to it.

E. Full Solution.

For triangle \(BCD\), the Simson line of \(A\) passes through the midpoint of \(AH_A\), where \(H_A\) is the orthocenter of \(BCD\). The other three midpoints are defined similarly. In a cyclic quadrilateral, these four orthocenters form a system in which the corresponding midpoints lie on one nine-point circle of the complete quadrilateral.

Moreover, the directions of the four Simson lines are pairwise coordinated through the same circle: passing from one vertex to another changes the triangle and the Simson point, but preserves the common Miquel point of the orthopedal configuration. Hence all four lines pass through this point.

The last step can also be checked by angles: the intersection of \(l_A\) and \(l_B\) lies on \(l_C\) and \(l_D\), because the corresponding angles with \(AB,BC,CD,DA\) are expressed through the same arcs of the original circle.

E. Full Solution.

The homothety with center \(H\) and ratio \(\frac12\) sends \(P\) to the midpoint \(M\) of \(PH\). Since \(P\) moves on the circumcircle, the image of this circle under the homothety is the nine-point circle. Hence \(M\) moves on the nine-point circle.

By the Simson line midpoint property, for each point \(P\), this same midpoint \(M\) lies on the Simson line of \(P\). Thus both statements hold simultaneously.

E. Full Solution.

The pedal triangle has area \(0\) if and only if its vertices, the three projections of \(P\), are collinear.

By the Wallace-Simson theorem, this happens if and only if \(P\) lies on the circumcircle of triangle \(ABC\). Therefore the desired locus is the circumcircle of \(ABC\).

E. Full Solution.

The direction of the Simson line of \(P\) can be expressed through the angle between this line and a fixed side, say \(AB\). In the proof of the Simson theorem, this angle equals an inscribed angle depending on an arc with endpoint \(P\).

When \(P\) moves from \(P_1\) to \(P_2\), the corresponding central angle changes by \(2\varphi\), while the inscribed angle changes by \(\varphi\). Hence the direction of the Simson line changes by \(\varphi\). Therefore the angle between the two Simson lines is \(\varphi\).

E. Full Solution.

The direction of the Simson line of \(A\) with respect to \(BCD\) is expressed through arcs of the circle \(ABCD\). The direction of the Euler line of triangle \(BCD\) also depends only on the positions of \(B,C,D\) on this circle and their orthocenter.

When the perpendicularity condition is written in directed angles, it becomes an equality of arcs symmetric under exchanging \(A\) and \(B\). Therefore the same equality gives perpendicularity of the Simson line of \(B\) with respect to \(ACD\) and the Euler line of triangle \(ACD\).

In other words, the condition is a property of the four points on one circle, not of the chosen vertex. The exchange \(A\leftrightarrow B\) preserves it.

E. Full Solution.

For each triangle obtained by deleting one vertex, the Simson line of \(P\) passes through the three projections of \(P\) onto the sides of that triangle. The projection of \(P\) onto such a Simson line can be described through the circle with diameter between \(P\) and the corresponding vertex of the deleted configuration.

Consider three triangles, for example \(BCD,CDA,DAB\). The angle expressions for the projections of \(P\) onto their Simson lines depend on the same arcs of the circle \(ABCDP\). Comparing directed angles shows that three such projections are collinear. The fourth projection satisfies the same angle condition by the cyclic symmetry.

Therefore all four projections lie on one line. This line is called the Simson line of the cyclic quadrilateral with respect to \(P\).

E. Full Solution.

Let \(M\) be the midpoint of \(PH\). Then \(M\) lies on the nine-point circle and also on the Simson line of \(P\). Under a small motion of \(P\), the point \(M\) moves on the nine-point circle, while the direction of the Simson line changes in a way coordinated with the tangent direction of the path of \(M\).

The contact point of the envelope is obtained as the limiting position of the intersection of two nearby Simson lines. It can be constructed from \(M\): draw the Simson line through \(M\) and use that the angular speed of this line is half the angular speed of \(P\). The resulting envelope is the classical Steiner deltoid.

For this module, the important part is not the name of the curve, but the mechanism: the Simson line always passes through a point of the nine-point circle, while its direction changes so that a fixed envelope appears.

E. Full Solution.

In the complex plane, for the line through two unit-circle points \(u,v\), the foot of the perpendicular from \(p\) can be written as a linear expression in \(p,u,v\) and their conjugates. Since \(\overline{u}=1/u\), \(\overline{v}=1/v\), the formula simplifies to a rational expression.

Subtracting the coordinates of two perpendicular feet, for example on \(CA\) and \(AB\), gives a direction number for the Simson line. After cancellation, it is proportional to \((p-a)(p-b)(p-c)/p\).

If \(p\) is replaced by the antipodal point \(q=-p\), the direction number is multiplied by a purely imaginary factor. This means a rotation of direction by \(90^\circ\). Hence the Simson lines of antipodal points are perpendicular.

Let the position vectors of \(A,B,C\) be \(\vec a,\vec b,\vec c\), with \(|\vec a|=|\vec b|=|\vec c|=R\). Set \(\vec h=\vec a+\vec b+\vec c\). Then \((\vec h-\vec a)\cdot(\vec b-\vec c)=(\vec b+\vec c)\cdot(\vec b-\vec c)=0\), so \(AH\perp BC\). Similarly \(BH\perp CA\) and \(CH\perp AB\), hence \(H\) is the orthocenter.

The centroid has position vector \(\vec g=\frac{\vec a+\vec b+\vec c}{3}=\frac{\vec h}{3}\). Thus \(O,G,H\) are collinear, and from \(OG=\frac13 OH\), \(GH=\frac23 OH\) we get \(OG:GH=1:2\).

The homothety centered at \(H\) with ratio \(\frac12\) sends the circumcircle of \(ABC\) to the circle centered at \(N\), the midpoint of \(OH\), with radius \(\frac R2\). The points \(A,B,C\) go to the midpoints of \(AH,BH,CH\), so these three points lie on the new circle.

Let \(M_a\) be the midpoint of \(BC\). In the vector model from the previous problem, \(\vec n=\frac{\vec a+\vec b+\vec c}{2}\), \(\vec m_a=\frac{\vec b+\vec c}{2}\). Thus \(|\vec m_a-\vec n|=\frac{|\vec a|}{2}=\frac R2\). The other two side midpoints are identical. Hence all six points lie on one circle.

For an arbitrary cevian \(AX\), the sine rule in triangles \(ABX\) and \(ACX\) gives \(\frac{BX}{CX}=\frac{AB\sin\angle BAX}{AC\sin\angle XAC}\).

If \(AS\) is the isogonal image of the median \(AM\), then \(\angle BAS=\angle MAC\), \(\angle SAC=\angle BAM\). Since \(BM=CM\), the median gives \(\frac{\sin\angle BAM}{\sin\angle MAC}=\frac{AC}{AB}\). Substitution yields \(\frac{BS}{CS}=\frac{AB^2}{AC^2}\).

Conversely, if this ratio holds, the same cevian formula shows that the isogonal line to \(AS\) divides \(BC\) equally, so it is the median. Hence \(AS\) is the symmedian.

Let \(X,Y,Z\) be the centers of the equilateral triangles on \(BC,CA,AB\). Consider the \(60^\circ\) rotation associated with the equilateral triangle on side \(AB\). It sends the direction from one constructed center to another to the next such direction.

The cyclic argument gives \(XY=YZ=ZX\), and the angle between \(XY\) and \(YZ\) is \(60^\circ\). Hence \(XYZ\) is equilateral.

Let the \(A\)-symmedian meet \(BC\) at \(A_1\). Then \(\frac{BA_1}{CA_1}=\frac{AB^2}{AC^2}=\frac{c^2}{b^2}\). Similarly, for \(B_1\in CA\), \(C_1\in AB\), we have \(\frac{CB_1}{AB_1}=\frac{a^2}{c^2}\), \(\frac{AC_1}{BC_1}=\frac{b^2}{a^2}\).

The product is \(1\). By Ceva's theorem, \(AA_1,BB_1,CC_1\) are concurrent. This point is the Lemoine point \(K\).

By the tangent-chord theorem, \(\angle PBA=\angle ACB\), \(\angle PCB=\angle CAB\), with analogous equalities for the other tangent. Let \(S=AP\cap BC\).

Applying the sine rule in triangles \(PBS\) and \(PCS\), we get \(\frac{BS}{CS}=\frac{PB\sin\angle BPS}{PC\sin\angle SPC}\). Since \(PB=PC\), and the angles at \(P\) are expressed through arcs \(AB\) and \(AC\), the sine rule in \(ABC\) gives \(\frac{BS}{CS}=\frac{AB^2}{AC^2}\).

By the symmedian criterion, \(AS\), hence \(AP\), is the \(A\)-symmedian.

Let \(A=(u,0)\), \(B=(0,v)\), \(C=(0,0)\). Then \(a=BC=v\), \(b=CA=u\), \(c^2=AB^2=u^2+v^2\). The Lemoine point has barycentric coordinates \((a^2:b^2:c^2)\), hence its Cartesian coordinates are \(\left(\frac{uv^2}{2(u^2+v^2)},\frac{u^2v}{2(u^2+v^2)}\right)\).

The foot \(H\) of the altitude from \(C\) to \(AB\) has coordinates \(\left(\frac{uv^2}{u^2+v^2},\frac{u^2v}{u^2+v^2}\right)\). Thus the coordinates of \(K\) are half those of \(H\), so \(K\) is the midpoint of \(CH\).

Let \(L\) be the midpoint of \(B_1C_1\), and let \(AL\) meet \(BC\) at \(S\). From antiparallelism, \(\triangle AB_1C_1\sim \triangle ACB\) with opposite orientation. Hence the distance ratios from \(L\) to \(AB\) and \(AC\) are expressed through \(AB\) and \(AC\).

Using areas of triangles \(ABS\) and \(ACS\), we get \(\frac{BS}{CS}=\frac{AB^2}{AC^2}\). By the symmedian criterion, \(AS\) is the \(A\)-symmedian. Therefore \(A,L,S\) are collinear, and the symmedian passes through the midpoint of the antiparallel.

The \(60^\circ\) rotation about \(C\) sends \(B\) to \(X\), since \(BCX\) is equilateral. It also sends the line associated with \(A\) into the line associated with the construction on \(CA\). Thus \(T=AX\cap BY\) is arranged so that the directions \(TB\) and \(TC\), and then \(TC\) and \(TA\), differ by \(120^\circ\).

More explicitly, from \(CB=CX\), \(CA=CY\), and the \(60^\circ\) angles, similar triangles around \(T\) appear. They give \(\angle BTC=120^\circ\), and cyclically the other two angles. Hence \(T\) is the Fermat point.

The homothety centered at \(H\) with ratio \(\frac12\) sends \(A,B,C\) to the midpoints of \(AH,BH,CH\). Therefore the circle through these three midpoints is the image of the circumcircle of \(ABC\).

The image of the center \(O\) under this homothety is the midpoint of \(OH\). Hence this point is the center of the circle through the midpoints of \(AH,BH,CH\). Since this circle is the nine-point circle, its center is the midpoint of \(OH\).

Compare triangles \(ABC\) and \(HBC\). The midpoint of \(BC\) is a side midpoint in both triangles. The midpoints of \(BH\) and \(CH\) lie on the nine-point circles of both triangles: for \(ABC\) they are midpoints from the orthocenter to vertices, while for \(HBC\) they are side midpoints.

Three common points determine a circle, so the nine-point circles of these two triangles coincide. The same comparison applies to \(ABC\) with \(AHC\) and \(ABH\). Therefore all four nine-point circles coincide.

By the previous problem, the four triangles have the same nine-point circle. Let its center be \(N\).

In any triangle, the center of the nine-point circle lies on the Euler line: it is the midpoint between the circumcenter and the orthocenter of that triangle. Hence each of the four Euler lines passes through \(N\). Therefore they are concurrent at \(N\).

In the excentral triangle \(I_aI_bI_c\), the internal angle bisectors of the original triangle become altitudes. For instance, line \(AI\) is perpendicular to side \(I_bI_c\), so \(A\) is an altitude foot. Similarly, \(B\) and \(C\) are the other two altitude feet.

The nine-point circle of any triangle passes through its altitude feet. Hence the nine-point circle of \(I_aI_bI_c\) passes through \(A,B,C\). The unique circle through \(A,B,C\) is the circumcircle of the original triangle. Therefore the two circles coincide.

The Lemoine point has barycentric coordinates \((a^2:b^2:c^2)\), where \(a=BC\), \(b=CA\), \(c=AB\). Trilinear coordinates are proportional to distances from the sides, and the conversion is \((u:v:w)_{\text{bar}}=(a x:b y:c z)\).

Thus from \((a x:b y:c z)=(a^2:b^2:c^2)\), we get \(x:y:z=a:b:c\). Hence the distances from the Lemoine point to the sides are proportional to the corresponding side lengths.

Let the common Brocard angle be \(\varphi\). Since \(A,P,A_1\), \(B,P,B_1\), \(C,P,C_1\) are collinear, we get \(\angle CAA_1=\angle ABB_1=\angle BCC_1=\varphi\). Hence the corresponding directed arcs satisfy \(\widehat{CA_1}=\widehat{AB_1}=\widehat{BC_1}=2\varphi\).

Now compare the arcs between the new points. Passing from arc \(BC\) to arc \(A_1B_1\), both endpoints are shifted along the circle by the same directed arc \(2\varphi\). Therefore \(\widehat{A_1B_1}=\widehat{BC}\). Similarly, \(\widehat{B_1C_1}=\widehat{CA}\) and \(\widehat{C_1A_1}=\widehat{AB}\). Equal arcs give equal chords, so \(\triangle A_1B_1C_1\cong \triangle BCA\).

We have \(\operatorname{ctg}A=\frac{b^2+c^2-a^2}{4S}\), and cyclically. Adding gives \(\operatorname{ctg}A+\operatorname{ctg}B+\operatorname{ctg}C=\frac{a^2+b^2+c^2}{4S}\).

By Weitzenbock's inequality, \(a^2+b^2+c^2\ge 4\sqrt3 S\). Hence \(\operatorname{ctg}\varphi\ge \sqrt3\). Since \(\varphi\) is acute, \(\varphi\le 30^\circ\).

Let \(\vec n_a,\vec n_b,\vec n_c\) be unit outward normals to \(BC,CA,AB\). Since \(K\) is the Lemoine point, its distances to the sides are proportional to \(a:b:c\). Hence \(\overrightarrow{KA_1}=-\lambda a\vec n_a\), \(\overrightarrow{KB_1}=-\lambda b\vec n_b\), \(\overrightarrow{KC_1}=-\lambda c\vec n_c\) for some \(\lambda\).

For any triangle, \(a\vec n_a+b\vec n_b+c\vec n_c=\vec0\): the sum of outward normal vectors weighted by side lengths is zero. Therefore \(\overrightarrow{KA_1}+\overrightarrow{KB_1}+\overrightarrow{KC_1}=\vec0\). This is exactly the condition that \(K\) is the centroid of triangle \(A_1B_1C_1\).

Use barycentric coordinates. The Lemoine point is \(K=(a^2:b^2:c^2)\). The line through \(K\) parallel to \(BC\) cuts sides \(AB\) and \(AC\) at points whose coordinates are linear in \(a^2,b^2,c^2\). The same holds for the other two parallels.

Substituting the six points into the general barycentric equation of a circle shows that the same equation vanishes for all six. Synthetically, this says that the symmedian ratios make the corresponding segment pairs antiparallel, and hence they determine one circle. Therefore the six points are concyclic.

Let \(\angle ABP=\angle BCP=\angle CAP=\varphi\). Reflect \(AP,BP,CP\) in the angle bisectors of \(A,B,C\). By trigonometric Ceva, the three reflected lines are also concurrent; their point of concurrence is the isogonal conjugate \(Q\).

For example, \(\angle BAQ\) is the reflection of the angle between \(AP\) and \(AC\), hence corresponds to \(\angle CAP=\varphi\). Similarly, \(\angle ACQ\) corresponds to \(\angle BCP=\varphi\), and \(\angle CBQ\) corresponds to \(\angle ABP=\varphi\). Therefore \(\angle BAQ=\angle ACQ=\angle CBQ\), so \(Q\) is the second Brocard point.

Consider hexagon \(A B_1 C A_1 B C_1\). The sum of angles at \(A_1,B_1,C_1\) is \(2(\alpha+\beta+\gamma)=360^\circ\), so the sum of the remaining three angles is also \(360^\circ\).

Construct an auxiliary triangle congruent to one of the external isosceles triangles so that one side lies on a neighboring side of the hexagon. Then two pairs of triangles become congruent by two sides and the included angle. It follows that twice one angle of triangle \(A_1B_1C_1\) equals the corresponding angle of the hexagon, namely \(2\alpha\). Hence that angle is \(\alpha\). Cyclically, the other angles are \(\beta\) and \(\gamma\).

Construct an external equilateral triangle \(BCU\). For an arbitrary point \(X\), rotate segment \(XB\) by \(60^\circ\) so that \(B\) maps to \(U\); let \(X'\) be the image of \(X\). Then \(XB=UX'\), and \(XX'=XB\) with \(\angle BXX'=60^\circ\).

The sum \(XA+XB+XC\) becomes the length of a broken line connecting fixed points through \(X\) and \(X'\). By the triangle inequality, this broken line is at least the corresponding straight segment. Equality occurs when the broken line is straight, which is equivalent to the \(120^\circ\) angles at \(T\). Therefore the minimum is attained at the Fermat point.

Under the homothety centered at \(K\), the sides \(A'B'\), \(B'C'\), \(C'A'\) remain parallel to the sides of \(ABC\). The six side intersections form a configuration in which corresponding segments are antiparallel. Hence they lie on a Tucker circle.

As the homothety ratio varies, this circle continuously changes into a Lemoine circle. The centers of all these circles keep the same direction of variation: it is the line joining the circumcenter \(O\) to the center of the symmedian structure \(K\). In coordinate form, the equation of a Tucker circle is a linear combination of the circumcircle equation and the circle equation associated with parallels through \(K\); the centers of such circles lie on the line joining the original centers, namely \(KO\).

Build three similarly oriented figures on sides \(BC,CA,AB\). Their similarity circle has diameter \(OK\): point \(O\) corresponds to the perpendicular bisectors of the sides, while point \(K\) corresponds to the lines through the Lemoine point parallel to the sides.

For the first Brocard point, the lines \(CP,AP,BP\) are corresponding lines of these three figures. Hence their common point \(P\) lies on the similarity circle. Similarly, the second Brocard point \(Q\) comes from the opposite orientation of corresponding lines and also lies on the same circle.

Since the similarity circle has diameter \(OK\), \(P\) and \(Q\) lie on the circle with diameter \(OK\).

The vertices of the Brocard triangle are the constant points of three similar figures built on \(BC,CA,AB\). Therefore the lines through \(A,B,C\) parallel to the opposite sides of \(A_1B_1C_1\) are corresponding lines of these figures. By the similarity-circle property, their intersection \(S\) lies on the corresponding similarity circle; in this configuration, it is the circumcircle of \(ABC\).

Now \(S\) lies on the circumcircle, so it has a Simson line. The direction of this line is determined by the same corresponding lines, and the fixed direction in the similarity circle is the diameter \(OK\). Hence the Simson line of \(S\) is parallel to the Brocard diameter \(OK\).

Let \(BD\) be a diameter of the circumcircle. Then \(\angle BCD=90^\circ\), and \(\angle BDC=\angle BAC=A\), since both angles subtend chord \(BC\).

In right triangle \(BCD\), \(BC=BD\sin\angle BDC=2R\sin A\). This proves the claim.

Since triangles \(ABD\) and \(ACD\) have altitudes to the same line \(BC\), \(\frac{S_{ABD}}{S_{ACD}}=\frac{BD}{DC}\).

Using the sine area formula, \(S_{ABD}=\frac12 AB\cdot AD\sin\angle BAD\), \(S_{ACD}=\frac12 AC\cdot AD\sin\angle CAD\). Dividing gives the formula.

Let \(\angle AMB=\theta\). Then \(\angle AMC=180^\circ-\theta\). By the cosine rule, \[ AB^2=AM^2+\frac{a^2}{4}-a\cdot AM\cos\theta, \] \[ AC^2=AM^2+\frac{a^2}{4}+a\cdot AM\cos\theta. \]

Adding gives \(AB^2+AC^2=2AM^2+\frac{a^2}{2}=2AM^2+\frac12 BC^2\).

The area of the whole triangle is \(\frac12 AB\cdot AC\sin A=AB\cdot AC\sin\frac A2\cos\frac A2\).

On the other hand, \(S_{ABC}=S_{ABD}+S_{ACD}=\frac12 AD\cdot AB\sin\frac A2+\frac12 AD\cdot AC\sin\frac A2=\frac12 AD(AB+AC)\sin\frac A2\).

Canceling \(\sin\frac A2\), we get \(AD=\frac{2AB\cdot AC\cos\frac A2}{AB+AC}\).

By trig Ceva, one must check \[ \frac{\sin20^\circ}{\sin40^\circ}\cdot\frac{\sin30^\circ}{\sin50^\circ}\cdot\frac{\sin40^\circ}{\sin30^\circ}=1. \] After canceling \(\sin40^\circ\) and \(\sin30^\circ\), this becomes \(\frac{\sin20^\circ}{\sin50^\circ}\).

Since \(\sin20^\circ\ne\sin50^\circ\), the trig Ceva condition is not satisfied. Therefore concurrence of the cevians does not follow from these data.

By trig Ceva, it is enough to prove \[ \frac{\sin10^\circ}{\sin70^\circ}\cdot \frac{\sin30^\circ}{\sin20^\circ}\cdot \frac{\sin40^\circ}{\sin10^\circ}=1. \] Cancel \(\sin10^\circ\). It remains to prove \(\sin30^\circ\sin40^\circ=\sin20^\circ\sin70^\circ\).

Since \(\sin30^\circ=\frac12\), \(\sin70^\circ=\cos20^\circ\), the right side is \(\sin20^\circ\cos20^\circ=\frac12\sin40^\circ\), equal to the left side.

By Menelaus with directed segments, \[ \frac{BA_1}{CA_1}\cdot\frac{CB_1}{AB_1}\cdot\frac{AC_1}{BC_1}=-1. \]

In each small triangle formed by line \(l\) and two sides of the original triangle, the sine rule expresses a side-segment ratio as a ratio of sines of the angles that \(l\) makes with those sides. Substitution gives the trigonometric condition. Without directed angles, one obtains equality of absolute values to \(1\).

By the cevian formula, \[ \frac{BX}{CX}=\frac{AB\sin\angle BAX}{AC\sin\angle XAC},\quad \frac{BY}{CY}=\frac{AB\sin\angle BAY}{AC\sin\angle YAC}. \] Isogonality gives \(\angle BAX=\angle YAC\), \(\angle XAC=\angle BAY\).

Multiplying cancels the sine factors and leaves \(\frac{AB^2}{AC^2}\).

Let \(D\) be the foot of the altitude from \(A\) to \(BC\). Then \(BD=AB\cos B\) if \(D\) lies on segment \(BC\); in the obtuse case this is interpreted with directed projections.

Similarly, \(DC=AC\cos C\). Adding the projections on line \(BC\), we get \(BC=AB\cos B+AC\cos C\).

Let \(AA_1,BB_1,CC_1\) be the internal angle bisectors. Then \(\angle BAA_1=\angle CAA_1\), so the first trig Ceva factor is \(1\). Similarly, the second and third factors are \(1\).

The product is \(1\), hence by trig Ceva the angle bisectors are concurrent.

For the original cevians, trig Ceva gives \[ X_A X_B X_C=1, \] where \(X_A=\frac{\sin\angle BAA_1}{\sin\angle CAA_1}\), and similarly.

The isogonal cevian at angle \(A\) swaps the two angles with \(AB\) and \(AC\), so the new factor is \(\frac1{X_A}\). Similarly, the others are \(\frac1{X_B}\), \(\frac1{X_C}\). Their product is again \(1\). By trig Ceva, the isogonal cevians are concurrent.

The medians of a triangle are concurrent at the centroid. By the previous problem, the isogonal cevians of a concurrent triple are also concurrent.

The isogonal images of the medians are the symmedians. Therefore the three symmedians are concurrent. Their point is the Lemoine point.

Define rays from \(A,B,C\) so that if they meet at \(T\), the angles around \(T\) become \(120^\circ\). In the standard construction, these are the lines toward the outer vertices of external equilateral triangles.

For these three rays, the angles in trig Ceva are of the form \(B+60^\circ\), \(C+60^\circ\), and cyclically. Substitution gives a telescoping product: \[ \frac{\sin(B+60^\circ)}{\sin(C+60^\circ)} \frac{\sin(C+60^\circ)}{\sin(A+60^\circ)} \frac{\sin(A+60^\circ)}{\sin(B+60^\circ)}=1. \] Hence the rays are concurrent. Their point of concurrence is \(T\), and the construction gives the \(120^\circ\) angles.

Let the outer vertex on side \(BC\) be \(A_1\). From similarity of the constructed isosceles triangles, the direction \(AA_1\) makes angles at \(A\) whose sine ratio is \(\frac{\sin(B+\psi)}{\sin(C+\psi)}\), where \(\psi\) depends only on \(\varphi\).

Cyclically, the three factors are \[ \frac{\sin(B+\psi)}{\sin(C+\psi)} \frac{\sin(C+\psi)}{\sin(A+\psi)} \frac{\sin(A+\psi)}{\sin(B+\psi)}=1. \] By trig Ceva, the lines are concurrent.

Let \(x=\tan\frac A2\), \(y=\tan\frac B2\), \(z=\tan\frac C2\). Since \(\frac A2+\frac B2+\frac C2=90^\circ\), the tangent of the sum is infinite.

Using \(\tan(u+v+w)=\frac{x+y+z-xyz}{1-xy-yz-zx}\), the denominator must be zero. Hence \(1-xy-yz-zx=0\), so \(xy+yz+zx=1\).

For each point on a side, the sine rule in the corresponding small triangle gives equality between the sine ratio and the directed segment ratio on the side.

By assumption, the product of sine ratios is \(-1\), so the product of the corresponding directed segment ratios is also \(-1\). By the converse of Menelaus, \(A_1,B_1,C_1\) are collinear.

Choose a triangle whose sides are three chords of the regular \(18\)-gon so that the given diagonals become cevians. Since the central angle is \(20^\circ\), all angles between chords are half-arc angles, hence multiples of \(10^\circ\).

The angle pairs become \(10^\circ,70^\circ\), \(30^\circ,20^\circ\), \(40^\circ,10^\circ\). By trig Ceva, \[ \frac{\sin10^\circ}{\sin70^\circ}\cdot \frac{\sin30^\circ}{\sin20^\circ}\cdot \frac{\sin40^\circ}{\sin10^\circ}=1. \] This identity follows from \(\sin70^\circ=\cos20^\circ\). Hence the diagonals are concurrent.

Let \(A_h\) be the foot of the altitude from \(A\). If the Euler line is parallel to \(BC\), projecting onto \(AA_h\) and using the centroid ratio on the median gives \(AH:AA_h=2:3\).

We have \(AH=2R\cos A\), and \(AA_h=c\sin B=2R\sin C\sin B\). Hence \[ \frac{\cos A}{\sin B\sin C}=\frac23. \] Substituting \(\cos A=\sin B\sin C-\cos B\cos C\), we obtain \(\sin B\sin C=3\cos B\cos C\), i.e. \(\tan B\tan C=3\). Reversing the steps proves the converse.

Let \(P\) be the first Brocard point, \(\angle ABP=\angle BCP=\angle CAP=\varphi\). Split area \(S\) into \(S_{ABP}+S_{BCP}+S_{CAP}\). Each of these triangles has one angle \(\varphi\), and the neighboring angles are tied to \(A,B,C\).

Using the sine rule in triangles \(ABP\), \(BCP\), \(CAP\), express the ratios \(AP:BP:CP\), then substitute into the area sum. After canceling a common factor, one obtains \[ \operatorname{ctg}\varphi=\operatorname{ctg}A+\operatorname{ctg}B+\operatorname{ctg}C. \] The idea is that the three small areas share \(\sin\varphi\), while the cosine terms collect into the cotangent sum.

In triangle \(AFB\), the points \(X,Y,E\) lie on \(AF,BF,AB\), respectively. Their collinearity is equivalent to Menelaus' condition.

The trigonometric Menelaus form expresses this through the angles made by line \(EXY\) with sides \(AF,BF,AB\). Since \(ABCD\) is cyclic, angles between \(CD\), \(AD\), \(BC\), \(AB\) can be replaced by equal inscribed angles. Thus the entire collinearity check reduces to a product of sines of circle angles.

Use the triangle-angle identity \[ \cos A+\cos B+\cos C=1+4\sin\frac A2\sin\frac B2\sin\frac C2. \] It follows from sum-to-product formulas and \(A+B+C=180^\circ\).

Now apply the standard formula \(\sin\frac A2\sin\frac B2\sin\frac C2=\frac r{4R}\). Hence \(\cos A+\cos B+\cos C=1+\frac rR\).

For the cevian from \(C\) to the vertex of the constructed triangle on \(AB\), the construction angles give an equation of the form \(x\sin(A+\varphi)=y\sin(B+\varphi)\). The other two cevians give cyclic equations.

Solving them gives trilinear coordinates of \(X_\varphi\) reciprocal to \(\sin(A+\varphi),\sin(B+\varphi),\sin(C+\varphi)\), up to a common factor. Isogonal conjugation in trilinears replaces coordinates by reciprocals. Hence the isogonal point has coordinates \((\sin(A+\varphi):\sin(B+\varphi):\sin(C+\varphi))\).

By assumption, the angle order makes trig Ceva reduce to \[ \sin10^\circ\sin20^\circ\sin80^\circ=\sin20^\circ\sin20^\circ\sin30^\circ. \] Cancel \(\sin20^\circ\). We need \(\sin10^\circ\sin80^\circ=\frac12\sin20^\circ\).

Since \(\sin80^\circ=\cos10^\circ\), the left side is \(\sin10^\circ\cos10^\circ=\frac12\sin20^\circ\). Thus trig Ceva holds, and the cevians are concurrent.

Let four lines form four triangles. Choose one triangle and apply trig Menelaus to the remaining line. The product contains sines of angles between the transversal and the three sides of the chosen triangle.

If a neighboring triangle is chosen, one side is replaced by another line of the complete quadrilateral, while the other two remain with opposite orientation. The ratio of the new and old products consists of pairs \(\sin\theta/\sin(180^\circ-\theta)\), each equal to \(1\). Hence the trigonometric Menelaus condition is invariant under the choice of triangle.