#2 Projective Geometry I

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.