#3 Poles and Polars

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.