COURSE OF PROJECTIVE GEOMETRY

EP: hints:

1.

a) ψ is the product of the elementary perspectivities from *l* via *A* and *K* and *B* onto *l*, then the restriction to *l* of the polarity, and finally the
elementary perspectivity from *L* onto *l*.

b) Make a picture, and show that each of *C*, *D* and *O*: λ(0,0,1) is mapped to itself. Since ψ has three fixed points, ψ is identity.

c) (Simple. For the construction of the polar line, see §22.)

d) Check that *C* and *D* are always fixed points.

e) If *X* = *AE* **.** *l*, then *S _{X}* =

f) (The same as the answer to

g)

2. π is the product of the restriction to *l* of the polarity related to *K* and the elementary perspectivity from *L* onto *m*.

If *p* is the Pappus line, then π(*l* **.** *m*) = *p* **.** *m* and π(*l* **.** *p*) = *l* **.** *m*.

So *LM* **.** *m* = *p* **.** *m* and *LP* **.** *m* = *l* **.** *m*.

So *LM* goes through *p* **.** *m* and (since *l* **.** *m* lies on *LP*) through *l* **.** *p*.

So *p* = *LM*.

3. If *A*, *B*, *C*, *D*, *E* are the given points, we can take for φ the projectivity between pencils of lines from *A* to *B* that maps
*AC* to *BC*, *AD* to *BD*, *AE* to *BE*. The intersection points we have to construct are the fixed points of the projectivity induced on *l*.
We can construct these using an auxiliary circle as in the fragment Kaper-Westerman in the answer to *O102*.

4. Let *A*, *B*, *C*, *D* be the given proper intersection points, and *O* the given point at infinity.

Construct the perpendicular to *AO* from *A*. Let *S* be the other intersection point of this perpendicular and the conic. Construct *S* by applying Pascal to hexagon ABC-DOS.

Construct the perpendicular on [*AS*] that goes through the middle of [*AS*]. This line goes through *O* and the top *T* we have to construct. Construct *T* by applying Pascal to
hexagon AOD-CBT.

5. We can do this using the methode of the false positions. (See the answer to *O102*.) So:

Start with an arbitrary point *X* on *l*. Connect this point with *A* and intersect the connecting line with the line at infinity. Connect this intersection point at infinity with
*C* and intersect the connecting line with *m*. Connect this intersection point with *D* and intersect the connecting line with the line at infinity. Connect this intersection point at
infinity with *B* and intersect the connecting line with *l*. Call this last intersection point *X* '.

The mapping *X* → *X* ' is a projectivity from *l* onto itself. Construct for two more points *X* the image point *X* '.

Now construct the fixed points of this projectivity, as in the first fragment of the answer to *O102*. (Use for *J* a circle, then it's still a construction with compasses and ruler.)
There are zero or one or two fixed points.

If we begin with a fixed point *X*, then the construction produces a solution to the problem (and also the point *Y* on *m*).

6.

*L* ∈ *s* ⇒ *S* ∈ *l*
*L* ∈ *c* ⇒ *C* ∈ *l*

So *l* = *SC* (and *LD* is tangent to *K* in *D*).

So if *T* := *AB* **.** *SC*, then {*L*, *T*} and {*A*, *B*} separate harmonically.

By projection from *S* we then see that {*L*, *C*} and {*E*, *F*} separate harmonically.

By projection from *D* we then see that {*L*, *C*} and {*G*, *H*} separate harmonically.

Let *J* be the conic through *A*, *B*, *D*, *E* and *F*.

Because of the first and second of the forementioned harmonic situations, *SC* is also polar line of *L* with respect to *J*. Since *D* lies on this polar line, *L*
must lie on the tangent to *J* in *D*.

So *LD* is tangent to both *K* and *J* in *D*.

7. A conic, because φ is a projectivity and induces a projectivity from A onto B. In concrete cases, we can simply construct the conic point after point.

8. The polar line of the top is the tangent x_{2} = 0 with projective coordinates λ(0,1,0). The polar line of the point at infinity on the axis is the tangent x_{3} = 0 with
projective coordinates λ(0,0,1).
The intersection point of these two tangents is the point at infinity on x_{2} = 0, so λ(1,0,0). So that's the pole of the axis of the parabola.

9. *K* is generated by φ : *A* → *B* with φ(*AS*) = *BS*, φ(*AT*) = *BT*, φ(*AC*) = *BC*.

*K* ' is generated by ψ : *A* → *B* with ψ(*AU*) = *BU*, φ(*AV*) = *BV*, ψ(*AC*) = *BC*.

Construct ψ^{-1}(*BS*) and ψ^{-1}(*BT*) using the theorem of Steiner.

Thus we find three images of θ := ψ^{-1}φ : *A* → *A*, namely the images of *AS*, *AT* and *AC* (the last one is invariant). If we find yet another
invariant line *x*, then φ(*x*) = ψ(*x*), so *x* **.** φ(*x*) = *x* **.** ψ(*x*), so this then yields a fourth
intersection point of *K* and *K* '.

We can construct such a second invariant line as follows:

θ induces on an arbitrarily chosen line *l* that doesn't go through *A* a projectivity ω := ε^{-1}θε where ε is the elementary perspectivity
of connection from *l* onto *A*. So we also know a fixed point and two other pairs of point and image point of ω. In general there is one more fixed point (otherwise, *K* and *K* '
are tangent or they coincide). Construct this second fixed point as in *O52*.

10. Apply Brianchon to *d* *b* α - *a* δ *c*. Then we see that α **.** δ lies on the fixed line ((*D*-(*a* **.** *b*)) **.**
(*A*-(*c* **.** *d*))) - (*b* **.** *c*).

11. If we apply Pascal to *A* *C* *D*_{1} - *D* *B* *A*_{1} and to *B*_{1} *C* *D*_{1} -
*C*_{1} *B* *A*_{1}, we see that *C**A*_{1} **.** *B**D*_{1} lies on *OL* and *OM*.

12.

a) If we apply Brianchon to *a* *b* *c* - *c* *a* *b* .

b) If we apply Brianchon to *a* *b* *c* - *b* *a* *l*, then we see that the lines *A*_{1}*B*_{1} all go through
((*b* **.** *l*) - (*a* **.** *c*)) **.** ((*a* **.** *l*) - (*b* **.** *c*)) .

c) *AA*_{1} → *BB*_{1} is a projectivity from *A* onto *B* (it's the product of the elementary perspectivity from *A* onto *BC*, the perspectivity
of part b) from *BC* onto *AC*, and the elementary perspectivity from *AC* onto *B*).

This projectivity φ induces the conic mentioned in the problem.

This conic goes through *A* and *B* and, since φ(*AC*) = *BC*, also through *C*.