**Office hours:** Tentative: were on MW 15:10–16:00 in 873 Evans.

**Midterm:** Wed Mar 20 (during the class hours)

**Review session:** Fri May 10 (during the class hours, 4pm) Evans 47.

**Office hours (dead weak and afterwards):** next on Thur 3:30.

1.14, 2.19, 5.3, 8.11, 9.4, 10.9, 11.4, 12.5, 14.3, 17.13, 34.3, 34.5, 34.21, 38.6, 39.5, 39.9 (with “limiting parallel P-lines” meaning that the corresponding line/circles intersect at a point of absolute).

**Practice final**: inspect carefully — questions are on page 1, answers on pages 2 and 3.

1,2,11,14 of Section 37. (In problem 14, law of sine [excercise 20.17] follows from a/sinα being the diameter of circumscribed circle — to see this, inscribe in this circle a rectangular triangle with hypotenuse being the side a.)

**12 (due 5/3)** Problems: 17.11, 17.12, 39.2, 39.3, 39.6, 39.8.
In Section 39 problems, consider a P-geometry over reals (the absolute
is a circle in ℝ²); In problem 39.8, "limiting parallel P-lines" means that the corresponding Euclidean lines/circles intersect at a point of the absolute.

**Solutions to selected problems.**
**More solutions (look at the end only — try
to solve the problems suggested above first).**

**11 (due 4/26)** Problems: 13.2, 13.8, 13.10, 13.20, 17.3, 17.5, 17.10.

**Hint for 13.2**: one can do extension of fields in steps so
that every step is a quadratic extension; and any element of a
quadratic extension may be written in the form t+√s.

**Solutions to selected problems.**
**More solutions.**

**10 (due 4/19)** 11.1, 11.3, 11.6, 11.7; 12.1, 12.6.

**Update:** problems involving “Archemedean” are postponed until the next week.

**Solutions to selected problems.**
**More solutions.**

**9 (due 4/12)** 9.1, 9.2, 10.5, 10.8, 10.10, 10.11.

**Solutions to selected problems.**
**More solutions.**

**Correction:** in problem 10.10, CE should have been CD, and
the assumption that B,C are on opposite sides of AD is missing (not
even mentioning that the vertices are not named in a correct order
;-).

**8 (due 4/5)** 8.3, 8.5, 14.5, 14.6, 15.2, 15.3.

**Solutions to selected problems.**

7. Consider a coordinate plane over an ordered associative
algebra A with division (so we know that A has a subset A₊ such that
A₊ × A₊ ⊆ A₊, A₊ + A₊ ⊆ A₊, and A = {0} ⊔ A₊ ⊔ -A₊). So lines are given by {x=const}
or {y=ax+b}. Order the lines {x=const} via y-coordinate, and the
remaining lines via x-coordinate; this gives a betweenness relation on
these lines, so one can define intervals. Consider a triangle with
two vertices on {x=0} and two vertices on {y=0} such that P=(1,1) is
on a side of this triangle. Show that Pasch condition holds for lines
passing through P (remember that multiplication in A may be
non-commutative!). (Reminder: the function y=x+b preserves the order
by definition; the function y=ax preserves the order if a>0, and
reverts it if a<0 - it was done in the class. Hence 1 > 0, -1
< 0.) *( Hint: the only solution I know considers many
different cases...)*

**Hint for 15.3(b):** Consider first the case when F is the
field of rational numbers; then one can order F(√d) since
F(√d) consists of real numbers. Now try to express a+b√d
> 0 in terms of a and b, addition, multiplication and >. After you
do this, see how one can rewrite it in the case of general F.

1. Using ruler and compass for a given point P inside an angle construct points A and B on the sides of the angle such that P is the midpoint of the segment AB. Count the number of steps and justify your construction.

2. Let ABCD be a cyclic quadrilateral such that AB and CD are congruent. Show that the sides BC and AD are parallel. (You may use any proposition from Book 1 - 4 of Euclid.)

3. Construct an incidence plane with 5 points satisfying Playfair's axiom.

4. Consider the real Cartesian (=coordinate) plane with origin removed. Define lines and betweenness on this set in the same way as for usual Cartesian plane. Which of the incidence and betweenness axioms do not hold?

5. Show that an incidence plane with betweenness cannot be finite.

6. In △ABC, mark points B', C' on extensions of sides AB and AC beyond B and C. Show that bisectors of the angles ∠BCC', ∠CBB' and ∠BAC intersect at one point.

**7 (due 3/18)** 7.2, 7.5, 7.9, 7.10, 7.11, 7.14.

**NOTE: Postponed until Mon Mar 18**. Solutions to selected problems.

**6 (due 3/8)** Definitions:

On a projective plane, given two lines l,l' and a point P not on these lines, the projection l --> l' with center at P is the mapping which sends a point L on l to the unique point L' on intersection of l' with line PL.

In what follows, a triangle is 3 points not on the same line. A configuration C on an incidence plane I is a collection of lines and points of I. An automorphism of C is a permutation f of points of C such that if points P,Q from C are on a line belonging to C, then f(P),f(Q) are also on a line belonging to C.

**1.** On a projective plane, consider two triangles: ABC and A'B'C' such
that lines a=AA', b=BB', and c=CC' intersect at one point P, and 7
points P,A,B,C,A',B',C' are distinct. Then there is a unique point C''
such that projection p_{c} with center at C' from a to b sends A,A' to
B,B'; likewise, there is a unique point A'' such that projection p_{a}
with center at A'' from b to c sends B,B' to C,C'; and a unique point
B'' such that projection p_{b} with center at B'' from a to c sends A,A'
to C,C'. Show that the following statements are equivalent:

1) for any choice of A,B,C,A',B',C' as above, the composition of p_{c} and p_{a}
coincides with p_{b};

2) for any choice of A,B,C,A',B',C' as above, A'', B'', C'' are on the same line.

**Simplification:** when you show (1) ==> (2), you may assume that
P, A'', C'' are not on the same line.

**2.** If A'',B'',C'' are on the same line, then the configuration of 10 points
A,B,C,A',B',C',P,A'',B'',C'' and 10 lines a,b,c, ABC'', ACB'', BCA'', A'B'C'', A'C'B'',
B'C'A'', A''B''C'' is called the Desargues configuration D. (Actually, "other
coincidences" may happen; for example, P may be on the line A''B''C''; we
assume they do not happen.) Show that for
any two points X,Y of D, there is an automorphism of D which
sends X to Y.

**3.** Calculate the number of automorphisms of the Desargues configuration D from
the preceding problem. (Hint: given a point X of D, what is the number of
automorphisms which send X to X?)

**4.** A "flat" on a Desargues configuration D is a set of 6 distinct points
R,S,T,U,V,W of D such that V,R,S are on the same line of D, and
same holds for subsets {V,T,U}, {W,S,T}, {W,R,U}. Show that the number of
flats in D is 5. Show that for any permutation f of the set of flats, there
is a unique automorphism of D which gives this permutation of flats.

**Solutions. 1.** Sketch (2) ==> (1): Look what changes when one replaces a point A' by
a point A''', and chooses points B''', C''' so that A'''B''' passes through
C'' (same as A'B' does), and B'''C''' passes through A'' (same as B'C').
(Essentially, we replaced A'B'C' by A'''B'''C''' without changing points
A'' and C''). The point B'', would it change? (We must replace it
by the intersection of AC and A'''C'''.) By (2), it must remain on the line
A''C''; and it remains on the line AC. Unless the latter two lines coincide,
the point B'' does not move (which proves (1)). If they coincide,
the point A'' of intersection of BC and B'C' is on AC; so A''=C, so B'C'
passes through C - contradiction.

Sketch (1) ==> (2): Consider the points of intersection of A''C'' with
lines a,b,c; composition of p_{a} and p_{c} sends one of them to another (call
them X and Y). By (1) X,Y,B'' are on the same line; unless X=Y, this
finishes the proof. If X=Y, then they coincide with P; so P,A'',C'' are
on the same line. Obviously, they are distinct, hence unless B'' is
on this line, the line A''B'' does not pass through P. Now replace A''C''
in the argument above by A''B''.

**2 and 3**. Accumulate some examples. First, one can exchange A with B, A'
with B', and A'' with B''. (Likewise for B and C; or for A and C). Second,
one can exchange A,B,C with A',B',C' correspondingly. This still leaves
P fixed, and A'', B'', C'' not mixing with A,B,C,A',B',C'; but it shows that:
(i) all 3 lines passing through P may be freely exchanged, and (ii) on any of
these 3 lines two points (those which are not P) may be exchanged simultaneously
with exchanging corresponding pairs points on two other lines passing through P
(and remaining 3 points being fixed). We must get a way to mix these 3
groups together.

Make a leap of faith in the fact that all the points are created equal: try to apply the same construction to A instead of P: there are 3 lines passing through A (a, AC'' and AB''), and there is a pair of points not equal to A on all of them; try exchanging these pairs: P with A', B with C'', C with B'', and leaving the remaining 3 points B',C',A'' fixed. This gives a permutation of 10 points; we MUST check that any of 10 "lines" (i.e., a triple of points of configuration) is permuted to another line. (Here is the proof: the "other" 3 points are on the same line; it is fixed; 3 lines through A are OK - fixed - by the construction; BCA'' <--> C''B''A''; A'B'C'' <--> PB'B; A'C'B'' <--> PC'C.

That's it: THIS example shows that P may be mixed into the subset A,B,C,A',B',C', and the same for C''. Hence any point may be sent to P. We saw that there are 6*2 automorphisms which send P |--> P (6 from permutation of a,b,c, and 2 from permutation of not-equal-to-P points on these lines). Since P can be send to any of 10 points, the total is 120 automorphisms.

**4**. Assume this checked: any point X on a flat has 2 lines passing through it;
each of these lines has 3 points of the flat (denote other points on these
lines Y,Y' and Z,Z'). For a flat which contains P, these lines are either
a,b, or b,c, or a,c. If they are a,b, then {Y,Y'}={A,A'} and {Z,Z'}={B,B'};
hence the 6th point of the flat must be on AB and A'B', so it must be C''.

Hence there is a unique flat which contains P and has a,b as lines of the flat; hence there are 3 flats containing P. P can be replaced by any of 10 points, so there are 30 pairs a flat + a point on the flat; therefore there are 5 flats. Since flats are defined completely in terms of "a line contains a points", automorphisms permute flats; we obtain a homomorphism from the group of automorphisms into the group of permutations of flats.

Comparing 120 with 5!, it is enough to check that this homomorphism has no kernel. An element of kernel would send a flat to itself; in particular, it preserve 3 flats which contain P; but P is the only point of intersection (check!) hence it would preserve P. Likewise, it would preserve any point.

**5 (due 3/1)** Problems: 6.3, 6.5, 6.6, 6.7, 6.8, 6.9, 6.10.

**Solutions to selected problems. More solutions.**

**4 (due 2/22)** Problems: 4.2, 4.3, 5.4, 5.6, 5.10, 5.11, 5.12. (Some of these may be quite tricky too.)

**3 (due 2/15)** Problems: 4.6, 5.2, 5.3,
5.13, 5.15, 5.18. (Be warned, some of these may take a lot of work!)

**CLARIFICATION/CORRECTION:** For this homework, you can still use
high-school level of being rigorous (since we did not introduce yet a proper rigid foundation).

**2 (due 2/8)** Chapter 1. [Euclid : Books I–IV.] Problems: 3.4,
3.5, 3.10, 4.7.

**Solutions to selected problems in HW 2,3.**

**1 (due 1/26)**
Read sections 1.1–1.3. [and: Euclid: Book I, Propositions 1–34, Book III,
Propositions 1–34.] Problems: 1.5, 1.9, 1.13, 2.5, 2.7, 2.12, 2.17, 2.22.

For homeworks 1,2,3,4 one can use intuitive arguments for discussion
why points have relative positions the way you assume (why points are
on the same/opposite sides of some line; or why some ray is
inside/outside a certain angle etc.) For ruler-and-compass
constructions, you do not need to use *actual* ruler and compass
(but using them has a chance to improve your drawings; but maybe not; use common sense).

Starting from homework 5, all the proofs must be complete: based on existing axioms, definitions, theorems etc, and not on "intuition as in high school".