PROJECTIVE GEOMETRY COURSE

§ 27: *Projective pencils on a conic*

If *X*, *Y*, *A*, *B*, *C*, *D* are points on a conic *J*, then *J* is the conic generated by the projectivity φ from *X* onto
*Y* with φ(*XA*) = *YA*, φ(*XB*) = *YB*, φ(*XC*) = *YC*. We also have φ(*XD*) = *YD*.

Hence (*XA*, *XB*; *XC*, *XD*) = (*YA*, *YB*; *YC*, *YD*), so we can define:

*Definition:* The cross ratio of four points *A*, *B*, *C*, *D* on a conic *J* is the number (*XA*, *XB*; *XC*, *XD*), with *X* on
*J* arbitrary.

*Definition:* A projectivity from a conic *J* onto itself is a bijection from *J* onto *J* that preserves cross ratio.

*Fundamental Theorem:* Given three points *A*, *B*, *C* on a conic *J* and three points *A* ', *B* ', *C* ' on *J*.

Then there exists exactly one projectivity φ: *J* → *J* with φ(*A*) = *A* ', φ(*B*) = *B* ', φ(*C*) = *C* '.

*Proof:*

i) Existence:

Choose *D* on *J*. Let ψ be the projectivity from the pencil of lines *D* onto *D* with ψ(*DA*) = *DA '*, ψ(*DB*) = *DB '*,
ψ(*DC*) = *DC '*. Define φ as follows: for *X* on *J*, let φ(*X*) be the intersection point unequal to *D* of ψ(*DX*) and *J*. Check that this φ
meets the requirements.

ii) Unicity:

Suppose φ_{1} and φ_{2} both meet the requirements.

Then we have for *X* on *J*: (*A*, *B*; *C*, *X*) = (φ_{1}(*A*), φ_{1}(*B*); φ_{1}(*C*), φ_{1}(*X*)) =
(*A* ', *B* ' ; *C* ', φ_{1}(*X*)), and likewise: (*A*, *B*; *C*, *X*) = (*A* ', *B* ' ; *C* ', φ_{2}(*X*)).

So (*DA*, *DB*; *DC*, *DX*) = (*DA* ', *DB* ' ; *DC* ', *D*φ_{1}(*X*)) = (*DA* ', *DB* ' ; *DC* ',
*D*φ_{2}(*X*)).

So *D*φ_{1}(*X*)) = *D*φ_{2}(*X*)), and since φ_{1}(*X*)) and φ_{2}(*X*)) both lie on *J*, it follows that
φ_{1}(*X*) = φ_{2}(*X*).

*Note:* If *P* is a point, then the mapping that, for *X* on *J*, maps *X* to the other intersection point of *PX* and *J*, is in general *not* a projectivity.

*Theorem of Steiner:* Let *J* be a conic and φ a projectivity from *J* onto *J*.

Then all the points *X*φ(*Y*)**.** *Y*φ(*X*), with *X*, *Y* on *J*, lie on one and the same line (the Pascal line of φ).

*Proof:* Choose three points *A*, *B*, *C* on *J*. We are going to prove that, for *X* and *Y* on *J*, *X*φ(*Y*)**.** *Y*φ(*X*) lies
on the Pascal line of hexagon *ABC*φ(*A*)φ(*B*)φ(*C*).

Let ψ be the projectivity from the pencil of lines *A* onto the pencil of lines φ(*A*) with ψ(*A*φ(*B*)) = φ(*A*)*B*, ψ(*A*φ(*C*)) = φ(*A*)*C*,
ψ(*A*φ(*A*)) = φ(*A*)*A*.

Since *A*φ(*A*) is invariant, this is a perspectivity whose axis is the line through *A*φ(*B*))**.**φ(*A*)*B* and
*A*φ(*C*))**.**φ(*A*)*C*, this is the Pascal line of hexagon *ABC*φ(*A*)φ(*B*)φ(*C*).

Since (*B*, *C*; *A*, *X*) = (φ(*B*), φ(*C*); φ(*A*), φ(*X*)), we have
(φ(*A*)*B*, φ(*A*)*C*; φ(*A*)*A*, φ(*A*)*X*)) = (*A*φ(*B*), *A*φ(*C*); *A*φ(*A*), *A*φ(*X*)).

So, since ψ preserves cross ratio, and because of *O38*, we find ψ(*A*φ(*X*)) = φ(*A*)*X*.

This implies that *A*φ(*X*)**.** *X*φ(*A*) lies on the Pascal line of hexagon *ABC*φ(*A*)φ(*B*)φ(*C*).

Likewise we deduce that *A*φ(*Y*)**.** *Y*φ(*A*) lies on that line.

Now if we apply the theorem of Pascal to hexagon *AXY*φ(*A*)φ(*X*)φ(*Y*), then we see that *X*φ(*Y*)**.** *Y*φ(*X*) also lies on the same
line.

*O102* Study the two texts in in the answers of this problem. Together, they form an *application* of the theory above.

*O103* Given two triangles *ABC* and *UWV*. Construct triangle *PQR* whose sides go through *U*, *V* and *W* and whose vertices lie on the sides of triangle
*ABC*.