/puzzle/ - Puzzles and recreational math

>>85

To facilitate further discussion, I'll start introducing terminology here and in future posts on this.

If a point P and a line m are incident, I may also say that:

P lies on m

m passes through P

or other similar terms.

Similarly, given two lines j and k, we say they meet at jk, and the line PQ joins points P and Q.

If three or more distinct lines pass through the same point, then the lines are concurrent

If a line joins three or more distinct points, then the points are collinear.

A set of points, lines and an incidence relation that follows the axioms is called a (projective) plane.

For all this talk of these axioms, we still have not proven that any projective planes exist. The following will provide this proof.

A perfect difference set(PDS) modulo q is a set of (k + 1) distinct residues mod q such that for any non-zero residue arises uniquely as the difference of two distinct residues in the PDS. For any PDS, it is necessary that q = k^2 + k +1, and PDSs always exist if k is a prime power. (see above for partial list)

Given a PDS mod q = k^2 + k +1, k >= 2, a projective plane can be formed by assigning a distinct point and line to each residue mod q, and define a point and line incident if their residues sum to some element of the PDS.

Theorem 1: The given construction is a projective plane.

Proof:

Given any two distinct points P_m and P_n, the difference of their indices m-n is non-zero. Thus, there are unique distinct elements j,k in the PDS such that j - k = m - n mod q. Therefore, j - m = k - n mod q. Let j - m = c. Then c + m = j and c + n = k, thus l_c joins P_m and P_n. The uniqueness of j and k ensures the uniqueness of l_c.

Exactly analogous reasoning shows that for any two distinct lines meet in a unique point.

If we consider the the set of points incident to some given line l_i, their indices are merely our given PDS shifted by -i mod q. Since this shift does not effect differences, those indices always form a PDS as well. In particular, any line's point indices cannot contain three or more distinct residues in arithmetic progression.

We now need to produce 4 four distinct points with no three collinear. Consider the residues 0, 1, and 2. Since q is at least 7, their corresponding points are distinct. Also, they are not collinear since the residues are in arithmetic progression. Consider the residue 3. Its point is not collinear with those of 1 and 2, but it may be those of 0 and 1 or 0 and 2. If neither is the case, then the points of 0, 1, 2, and 3 are the desired points.

If we assume the points of 0, 1, and 3 are collinear, then the following residues' points are collinear, from shifting.

0,1, and 3

1, 2, and 4

-1, 0, and 2

If we now take residue 5, we can see that its point cannot lie on any of the lines above on pain of non-unique differences. Thus, the points of 0, 1, 2, and 5 are distinct with no three collinear.

If the points 0, 2, and 3 are collinear instead, then the points of -3, 0, 1, and 2 are distinct with no three collinear.

On a final note, from our axioms, we proved this statement as a corollary:

'For any two distinct lines, there is exactly one point incident with both lines.'

Note that it is a stronger version of Axiom 2, and so could replace it as an axiom without effect.

Additionally, this statement is the same as Axiom 1 with the words 'point' and 'line' swapped.

The same is true for Axiom 3 and one of the other consequences of the axioms.

Thus, if some statement about points and lines is a theorem, then the statement where the words 'point' and 'line' are swapped as well as other word pairs like 'concurrent' and 'collinear' is also a theorem.

This property of these axioms is called duality, and it often allows a single proof to prove two different theorems. Future posts, however, will introduce additional axioms, and the preservation of duality will need to be shown for the enlarged set of axioms.

>>92

To introduce the next concept, I will prove a statement stated without proof in the first post.

Theorem 1: If a line is incident with exactly k distinct points, then all lines are incident with exactly k distinct points

To prove this we will need to use this lemma whose proof is left to the reader.

Lemma: Given any two distinct lines, there exists a point that does not lie on either of them.

Proof of Theorem 1:

Let p be the line known to pass through k distinct points and let q be some arbitrary line distinct from p. From the lemma, we get a point not incident to either p or q; call it O. Now for every point X_p on p, there exists a unique line passing through X_p and O. Additionally, any line O lies on must share a point incident to p, thus it must be one of those unique lines. Thus there is a one-to-one correspondence(bijection) between lines passing through O and points on p. Using duality, one can also prove there is a bijection between lines through O and points through q. Composing these bijections produces a bijection between the points on p and the points on q. Since p passes through a finite number of distinct points, q must pass through that same number of distinct points as well.

The construction used in the proof of Theorem 1 is called a perspectivity. To fully articulate and expand on this construction, we will need several definitions.

Let the range of a line be the set of points it passes through, and let the pencil of a point be the set of lines it lies on. We can now define a perspectivity explicitly as follows:

Given distinct lines a and b and a point O incident to neither, the perspectivity between a and b with centre O is a bijection between the ranges of a and b where corresponding points are collinear with O.(see first picture)

Notation: This perspectivity can be written as a>O<b.

Given distinct points A and B and a line l incident to neither, the perspectivity between a and b with axis l is a bijection between the pencils of A and B where corresponding points are concurrent with l.(see second picture)

Notation: This perspectivity can be written as A<l>B.

Let us say that ranges and pencils are the two types of (one-dimensional) forms. We can generalize perspectivities as follows.

A half-perspectivity between a non-incident line and point is a bijection between the given range and pencil where corresponding points and lines are incident.

(Notation: for point P and line l, P<l or l>P)

A projectivity is a bijection between forms that can be formed from a composition of a finite number of half-perspectivities.

(Notation: An example of a projectivity in our notation is a>P<b>Q<c>R.)

We say that that two projectivities are identical if they are identical as functions(i.e. the have the same function domain and co-domain and each domain element's image is the same for both). This is regardless of the sequence of half-perpectivities used to construct the projectivities. If a projectivity fulfilling a certain set of conditions is unique, then that means that if there is a projectivity fulfilling those conditions, then it is identical to the first.

Given two n-tuples of distinct points (P_1, P_2,…, P_n) and (P*_1, P*_2,…, P*_n) from ranges l and l* respectively, then these tuples are projective if there exists a projectivity from l to l* where the image of P_j is P*_j.

(Notation: In this case we would write l(P_1, P_2,…, P_n) ~ l*(P*_1, P*_2,…, P*_n). The tuple may also be used in addition to or in place of the point or line name when writing out the projectivity in notation, with or without parentheses and commas such as

in the example ABC>abc.)

Similar definitions and notation can be applied to pairs of n-tuples from any combination of ranges and pencils. Also, the permutation of the elements in the tuples does not matter so long as corresponding elements are in the same position in their respective tuples.

With this definition, we can now establish some facts about projectivities.(see next post)

>>98

Theorem 2: For any two distinct lines l and l* and ordered triples of distinct points (A, B, C) and (A*, B*, C*) with points from ranges l and l* respectively, l(A, B, C)~l*(A*, B*, C*).

Proof: Assume without loss of generality that neither A nor A* is the point l*l. Define the points B" and C" as [[A*B][AB*]] and [[A*C][AC*]] respectively.

If B" and C" were the same point, then A*, B, and C would be collinear. That would imply A* lies on l, which goes against our assumption. Therefore, B" and C" are distinct. Define l" as [B"C"] and A" as l"[AA"], which can be proven distinct from A and A*. From this we have:

l(ABC)>A*<l"(A"B"C")>A<l*(A*B*C*),

proving the theorem.(see first picture)

Note that while a perspectivity requires the domain and co-domain forms to be distinct, that is not the case with projectivities. In fact:

Theorem 3: Given any 4 distinct lines a, b, c, d in the pencil P, P(a, b, c, d)~P(b, a, d, c)

Proof: Choose points P*,P" on lines c, d respectively distinct from P. Choose X on line a distinct from P and not collinear to P* and P". Then label the following lines:

e = P*P"

a* = XP*

a" = XP"

b* = P*[a"b]

b" = P"[ab*]

(see second and third pictures)

Thus abcd<a">a*b*ce<a>a"b"de<b*>badc provides the required perspectivity.

Some parting questions:

-What are the dual statements of Theorems 2 and 3?

-In a perspectivity between pencils M and N, what happens to the line MN? (Note: the definition of concurrent does not require distinct lines.)

-For ordered n-tuples of distinct elements from a form, is the relation of being projective an equivalence relation?

-Assume that there is a projectivity a>P<b>Q<c (with a,b,c and P,Q distinct) that happens to be identical to a perspectivity a>R<c.

–Is a>R<c>Q<b identical to a>P<b?

–Are a, b, and c necessarily concurrent?

–If not, is there another condition that must hold in the non-concurrent case?

–What conditions on R are implied by this identity?

I'll put down more questions if I think of them.

>>109

With this definition we can prove that:

Theorem 1: Space plane sets are projective planes

Corollary 1.1: If two distinct space plane sets share three or more points, then those points are collinear.

The proof of these statements is rather long, so it is written separately here:(https://itmb.co/nuiqz , replace 'planes' at the end with 'plane sets'.)

Another obvious corollary is:

Corollary 1.2: Given a plane set and a line not in the plane set, there is at most one point in both the plane set and the line's range.

Proof: since the plane set is a projective plane, then if there were two discinct points of the range on the line, that would mean that the line joining those two points would be in the plane set as well, contrary to our assumption.

The existence of such a line is guaranteed by the next theorem.

Theorem 2: For any plane set, there exists a point outside that set.

Proof: By Axiom 4, there exists two lines that do not meet. If our plane set contains one of those lines, then the other line is not, since the plane set is a projective plane. Therefore, at most one point in that line's range is in the plane set. By Axiom 3*, however, the line's range contains at least three points, so at least two points must be outside the plane set. The same reasoning applies for both of the lines when neither is in the plane set.

We are now ready to introduce our new axiom, which is actually a theorem with repect to the space axioms:

Theorem 3(Desargues' Theorem): Given two distinct triangles with ABC and A*B*C*, (A, B, C,)><(A*, B*, C*) implies (CA, AB, BC)<>(C*A*, A*B*, B*C*).

Proof: We will first consider the case when the triangles are not in the same plane set and all triangle points are distinct. (A, B, C)><(A*, B*, C*) implies that AA*, BB*, and CC* meet at some point O. If we consider the two lines AA* and BB*, by Axiom 2*, AB and A*B* also meet at some point C”.(see first picture) Analogusly, B” = [CA][C*A*] and A” = [BC][B*C*] also exist. By construction A”, B”, and C” are in both of the plane sets of triangles ABC and A*B*C*, which were assumed distinct. Thus, by Corollary 1.1, they are collnear, lying one some common line o. The conclusion of the theorem follows. Still assuming distinct plane sets, if the triangles share two points, then the theorem is true trivially. If only one point is shared, say A = A*, then B” = C” = A. The theorem is still satisfied with o = AA”.

In the case where the triangles are in the same plane set, this proof may fail. Again first assuming that all triangle points are distinct, note that O is in the common plane set as well. By Theorem 2, there exists some point P not in the plane set. Let p = PO, and note that by Axiom 3*, there is another point P* on p. By construction, AA* and PP* meet. By Axiom 2*, The lines PA and P*A* also meet at some point D.(see second picture) Analogusly E = [PB][P*B*] and F = [PC][P*C*] also exist. Since PA, PB, and PC are not in the plane set but A, B and C are, D, E, and F are not in the plane as well (Corollary 1.2). Also, (A,B,C)><(D,E,F) and (A*,B*,C*)><(D,E,F), by centres P and P* respectively. By the non-planar case [AB][DE] = C” , [BC][EF] = A”, and [CA][FD] = B” are collinear. However, [A*B*][DE] exists as well, and since DE is not in the plane set, it range has only one distinct common point with the plane set. Thus, [AB][A*B*] = C”. Smilarly [BC][B*C*] = A” and [CA][C*A*] = B” as well. This proves the theorem in the case where all triangle points are distinct. If two triangle points are shared in the common plane set case, then again the theorem is trivial. If one point is shared, say A = A*, then D = B” = C”, and the line AA” again satisties the theorem.

>>112

Forgot the pic

>>116

Finally, I will prove some statements I posited after explaining projectivities.

Theorem 1: If a>P<b>Q<c is identical to a perspectivity a>R<c, then a>P<b>Q<c>R<a is the identity projectivity.

Proof: Consider an arbitrary point X on a. (see pic 1) Let X” be be the image of X under a>P<b>Q<c. Since a>R<c is identical, X” is the image of X under that perspectivity as well. Reversing that, X is the image of X” under the perspectivity c>R<a. Composing a>P<b>Q<c and c>R<a, we get the desired result.

This leads to the following corollary.

Corollary 1.1: If a>P<b>Q<c is identical to a perspectivity a>R<c, then b>Q<c>R<a is identical to b>P<a.

Theorem 2: If a>P<b>Q<c is identical to a perspectivity a>R<c, then either a, b, c are concurrent or ac, P, Q, are collinear.

Proof: Since the two projectivities are equal, then the images of one point under each are the same. In particular, the image of the point ac is the same for both. Since a>R<c is a perspectivity, then ac is self-corresponding. If a, b, and c are concurrent, then this is the case for a>P<b>Q<c anyways. If a, b, and c are not concurrent(see pic 2), then the component perspectivity a>P<b must send ac a point not equal to ac on b. Call this point E. Be definition, E, ac, and P are collinear. Since under the full projectivity ac is self-corresponding, the second component perspectivity b>Q<c must send E back to ac, thus E, ac, Q are collinear. Therefore, both P and Q lie on E[ac] and are thus collinear with ac.

Theorem 3: If a>P<b>Q<c is identical to a perspectivity a>R<c and a, b, and c are concurrent, then P, Q, and R are collinear.

Proof: First assume P, Q, and ac are not collinear. Let M = a[PQ], M' = b[PQ], and M” = c[PQ](see pic 3). By assumption M, M' and M” are distinct. Note that M' is the image of M under a>P<b, and M” is the image of M' under b>Q<c. By composition and the theorem premise, M” is the image of M under a>R<c. Thus, R lies on MM”, but MM” = PQ by definition, proving the theorem. This proof breaks down if P, Q, and ac are collinear. In this case, from Corollary 1.1 we have b>Q<c>R<a identical to b>P<a. If Q, R, and ab are not collinear, P can be proven to lie on QR with the argument above. If they are collinear, that implies both P and R lie on Q[ab], again satisfying the theorem.

Theorem 4: If a>P<b>Q<c is identical to a perspectivity a>R<c and a, b, and c are not concurrent, then R = [P[bc]][Q[ab]].

Proof: By the theorem premise, ab and bc are distinct. Note also that neither P or Q can lie on b. Let F be the image of ab under b>Q<c and let D be the image of cb under b>P<a(see pic 4). By composition, a>P<b>Q<c, and thus a>R<c, must take ab to F and D to bc. Thus R, ab, and Q are collinear, as well as R, bc, and P. Thus, R lies on Q[ab] and P[bc]. Since P and Q do not lie on b while ab and bc do, Q[ab] and P[bc] are distinct lines, making R the meet of Q[ab] and P[bc].

>>114

In a previous set of posts, we introduced Axiom D by showing it was a consequence of projective plane constructed in a particular way. At the end of those posts it was implied that the new axiom had significant consequences for projectivities. At first glance, however, the statement of Axiom D has nothing to do with projectivities. To use axiom D to prove properties of projectivities, we will need to translate it into a form that applies to projectivities. These first two theorems will do this.

Theorem 1(Stable Perspectivity Theorem(SPT))(D): Given three distinct yet concurrent lines a, b, and c and a projectivity a>P<b>Q<c, there exists a third point R such that a>P<b>Q<c is identical to a>R<c.

Proof: Note that the theorem is trivial if P and Q are not distinct, so assume they are(pic 1). Then the line PQ exists. Take any point A on a not on b or PQ. Let its image under a>P<b be B and its image under a>P<b>Q<c be C. By construction, ABC is a triangle with none of its vertices on PQ. Let R = [AC][PQ]. It is obvious then that A, ac, and a[PQ] have the same images under a>P<b>Q<c and a>R<c. Suppose we have some arbitrary point on a named X not equal to A, ab, or a[PQ]. Let Z be the image of X under a>P<b>Q<c and let Y = [PX][QZ]. Then XYZ is a distinct triangle from ABC, and (A, B, C)><(X, Y, Z) with centre ac. Note that by construction [AB][XY] = P and [BC][YZ] = Q. By Axiom D, [AC][XZ] must lie on PQ. Therefore, [AC][XZ] = R. Therefore, Z is the image of X under a>R<c. Since X was chosen arbitrarily, the images of all points on a under a>R<c and a>P<b>Q<c are equal, thus the two projectivities are identical.

Theorem 2: From the plane axioms alone, the SPT implies Axiom D

Proof: Let ABC and XYZ be distinct triangles such that (A, B, C)><(X, Y, Z). Let a = AX, b = BY, and c = CZ(see pic). By assumption, a, b, and c are concurrent. Let P = [AB][XY] and let Q = [BC][YX]. If P = Q, then Axiom D is satisfied trivially, so assume P and Q distinct. The outcome is the same if P lies on a, Q lies on c, or either P or Q lie on b, so again we assume none of these are true. Under these conditions, the projectivity a>P<b>Q<c exists. By the SPT, there exists a point R such that a>R<c is identical to a>P<b>Q<c. Since C is the image A and Z is the image of X under a>P<b>Q<c = a>R<c, R = [AC][XZ]. Also, from Theorem 3 in >>117, P, Q, and R are collinear. Therefore (BC, CA, AB)<>(YZ, ZX, XY), as required by Axiom D.

The SPT has a profound consequence for all projectivities, specifically the number of (half-)perspectivities needed to construct a projectivity. Before we can prove this consequence, we need to prove a particular lemma.

Lemma 1(Substitution Lemma)(D): Assume lines i, j, and k are non-concurrent.

1a: Given a projectivity i>A<j>B<k and a line j' distinct but concurrent with i and j and not passing through B, then there exists a point A' on AB such that i>A'<j'>B<k = i>A<j>B<k

1b: Given a projectivity i>A<j>B<k and a line j* distinct but concurrent with k and j and not passing through A, then there exists a point B* on AB such that i>A<j*>B*<k = i>A<j>B<k.

Proof: The proof for 1a and 1b are analogous, so we will prove 1a here and its modification to prove 1b will be left to the reader. Starting off, since j' does not pass through B, the projectivity i>A<j>B<j'>B<k exists(see pic). This projectivity is self-evidently identical to i>A<j>B<k. Now consider the component projectivity i>A<j>B<j'. Since i, j, and j' are concurrent, by the SPT, there exists A' on AB such that i>A<j>B<j' = i>A'<j'. Recomposing the new perspectivity into the projectivity, we have i>A'<j'>B<k = i>A<j>B<j'>B<k = i>A<j>B<k, as required.

Using the Substitution Lemma, we now prove the following theorem.

Theorem 3(2-perspectivity Theorem)(D): Any projectivity between two distinct ranges is identical to a composition of at most 2 perspectivities.

Proof: The construction in Theorem 2 in >>99 implies the theorem for a plane with 3 or 4 distinct points per range regardless of Axiom D. Thus, we can safely assume there are at least 5 distinct lines per pencil in the proof. Also note that all we have to do is to show that all projectivities composed of three perspectivities can be reduced to one composed of only two perspectivities. For larger compositions, the proof can be applied multiple times to get the desired result. Start with the projectivity h>P<i>Q<j>R<k. There are 5 relevant cases with regard to concurrency of the ranges.

Case 1: h, i, j concurrent

Case 2: i, j, k concurrent

Case 3: h, i, k concurrent

Case 4: h, j, k concurrent

Case 5: No lines concurrent

Cases 1 and 2 are immediately taken care of by the SPT. Cases 3 and 4 can be reduced to Case 5 by applying the Substitution Lemma on i or j through ij.

Focusing on case 5, first suppose that hi, jk, and Q are not collinear. Let x = [hi][jk]. By assumption, then, the perspectivity h>P<i>Q<x>Q<j>R<k exists and is identical to the original projectivity. Focusing on the component projectivities h>P<i>Q<x and x>Q<j>R<k, we see that the SPT applies to both by the construction of x, producing the perspectivities h>P'<x and x>R'<k. Recomposing, we have h>P'<x>R'<k = h>P<i>Q<j>R<k, as required.

In the case that hi, jk, and Q are collinear, this construction breaks down. For this situation, choose some line i' through ij not equal to i, j or passing through P, hk. By the Substitution Lemma, there exists Q' such that h>P<i>Q<j = h>P<i'>Q'<j. In particular, the image of hi on j must be jk for both projectivities. Therefore the image of hi' distinct from hi must not equal jk. Thus, hi', Q' and jk are not collinear, so we can use the reduction in the previous sub-case on the identical projectivity h>P<i'>Q'<j>R<k, satisfying the theorem.

Finally, a couple of immediate consequences of this theorem.

Corollary 1(D): Any projectivity from a range to itself is identical to a composition of at most 3 perspectivities.

Corollary 2(D): Any projectivity between a range and a pencil is identical to a composition of at most 5 half-perspectivities.

Another consideration, mostly for the sake of aesthetics, is the fact that all of our axioms so far can be thought of as guaranteeing the presence or absence of certain incidences within a set of points and lines satisfying certain conditions or guaranteeing that a set of points and/or lines exhibiting certain incidences exists. In other words, all of our axioms have been fairly directly related to questions of incidence. On the other hand, the conclusion of the UPC is the existence of a particular perspectivity, which is rather removed from any direct statement of incidence. Luckily, like the SPT, there is a natural incidence theorem which is equivalent to the UPC.

As a preliminary, we consider the incidences implied by the requirements of the UPC. Specifically, given a projectivity l>R<m>P<n to which the UPC applies, consider the line R[ln]. The set {R, P, ln, [R[ln]]m} is a subset of the range of R[ln]. The points lm and [R[ln]]m are distinct from each other and from R and P by the conditions of the UPC. If R and P are not distinct, however, the conclusion of the UPC is trivial. Thus, a non-trivial application of the UPC involves a line with at least 4 distinct points on it. From this, it would be possible that the associated incidence theorem to the UPC would require 4 distinct points on a line. This is in fact the case.

Theorem 2(Pappus's Theorem):If the UPC holds, then given two distinct lines m and n, each passing through three distinct points A, B, C and A', B', C' respectively, each of which is distinct from mn, {[B'C][BC'], [C'A][CA'], [A'B][AB']} is collinear.

Proof:

Let AC' = d, n = e, and AB' = f(see pic 1). It is apparent that the projectivity d>C<e>B<f exists and that the UPC applies to it. From Theorem 4, >>117, we know that the centre of the equivalent perspectivity must be [B[de]][C[ef]] = [BC'][B'C] = A”. Now consider the affect of the component perspectivities d>C<e and e>B<f on the point A'. The preimage of A' under d>C<e is d[A'C] = [AC'][AC] = B”. The image of A' under e>B<f is f[A'B] = [AB'][A'B] = C”. Therefore C” is the image of B” under d>C<e>B<f, which was proven equal to d>A”<f, therefore {A”, B”, C”} is collinear, as required.

Theorem 3: Pappus's Theorem implies the UPC

Proof: Let l>R<m>P<n be a projectivity to which the UPC would apply. If R = P, then the theorem is trivial, so we assume that they are distinct(see pic 2). By Theorem 4, >>117, the centre of any perspectivity identical to the original projectivity must be [P[lm]][R[mn]] = Q. Note that the choice of Q and the conditions of the UPC guarantee that the images of lm, ln, and l[R[mn]] are the same under l>R<m>P<n and l>Q<n.

Arbitrarily choose X on l not equal to those points. Let Y be the image of X under l>R<m and Z be the image of X under l>R<m>P<n. Pappus's theorem now applies to lines m and RP with ln, P, R on PR and Y, mn, lm on m. Note that X = l[RY] =[[ln][lm]][RY] and Z = n[PY] = [[ln][lm]][PY]. By Pappus's theorem, Q, X and Z are collinear, therefore the image of X under l>Q<n is Z, as required.

We are close to a position to adopt Pappus's theorem/UPC as an axiom. For now we write:

Axiom P(Pappus's Theorem):Given two distinct lines m and n, each passing through three distinct points A, B, C and A', B', C' respectively, each of which is distinct from mn, {[B'C][BC'], [C'A][CA'], [A'B][AB']} is collinear.

(Notation: Theorems and other statements with proofs using Axiom P or other statements requiring Axiom P will be marked with a (P) at the very start.

e.g. Theorem 4(P):, Theorem 5(P, D):)

On adoption of this axiom, the UPC immediately becomes a theorem.

Theorem 4(Unstable Projectivity Theorem(UPT))(P): Given a projectivity a>P<b>Q<c where the P, Q, and ac are collinear and a, b and c are not concurrent, then there exists a point R such that a>R<c is identical to the original projectivity.

One major concern we have not yet seen to is the consistency of Axiom P with our previous axioms. The next set of posts will address this as well as look at other consequences of Axiom P for projectivities.

>>128

Theorem 1: The Unstable Perspectivity Theorem implies the Stable Perspectivity Theorem.

Proof: Staring with a projectivity a>P<b>Q<c such that all named lines and points are distinct, as the conclusion of the SPT would be trivial otherwise. Thus the line PQ is well-defined.

We first assume that P does not lie on c and PQ does not pass through ac. In this case we proceed as follows: choose some line b' incident to c[PQ] not equal to c or PQ(see pic 1). Note that b' is not concurrent to any two of a, b, or c and is also not incident to P or Q. Thus, the projectivity a>P<b'>P<b>Q<c exists and is clearly identical to the initial projectivity. Since {b'c, P, Q} is collinear and {b', b, c} is not, then the UPT applies, and there is some point Q' such that b'>Q'<c equals b'>P<b>Q<c. Thus, a>P<b>Q<c is identical to a>P<b'>Q'<c. By Theorem 4, >>117, Q' = [Q[bb']][P[bc]]. Thus, as bc = ac, {ac, P, Q'} is collinear and we can apply the UPT to a>P<b'>Q'<c to obtain a>R<c identical to the original projectivity, as required.

In the case that P lies on c, we can use a[PQ] instead unless Q lies on A. If we assume there are are only three distinct points in any range, then the SPT can be be directly proven. Otherwise, choose b” concurrent yet distinct to a, and apply the above procedure to a>P<b>Q<b” to get a>O<b so that a>O<b”>Q<c = a>P<b>Q<b”>Q<c.

By Theorem 3, >>117, O lies on PQ. If O lies on c, then O = P and a>P<b” = a>P<b>Q<b”. Let X be a point on a not equal to a[PQ] or ac(see pic 2). It is easily proven that such a point exists. Let Y and Z be its progressive images under b a>P<b>Q<b” on b and b” respectively. Thus, {X, Y, P} and {Y, Z, Q} are collinear set. The identity implies that {X, Z, P} is a collinear set. Thus the larger set {X, Y, Z, P, Q} is collinear from the plane axioms, contradicting our assertion that X does not lie on PQ. Therefore O does not lie on c and we can apply the prior construction again to obtain a>R<c identical to a>P<b>Q<c.

On the other hand, if PQ passes through ac, then the reasoning of the previous cases breaks down. We start here by noting that the distinct lines a, b, c, and PQ pass through ac, thus all ranges must contain at least 4 distinct points. If there only those 4 distinct lines in any range, then this case is easily dispatched. Otherwise, there exists a point not on any of the lines a, b, c, or PQ. Choose one such point and call it N. Now construct the following line and point as follows(see pic 3):

d = [c[NQ]][a[NP]]

M = [P[bd]][N[ac]]

From these constructions, the sets {M, N, ab}, {N, ad, P}, {M, bd, P}, and {N, Q, cd} are collinear sets. Also note that d does not pass through ab.

Examining the projectivity a>M<d>N<b, from the collinearity of {M, N, ab}, the UPT applies, and we obtain an identical perspectivity a>P'<b. By Theorem 4, >>117, P' = [M[db]][N[ad]]. From this and the construction of d and N, the following sets are collinear:

{N, ad, P, P'}

{M, bd, P, P'}

If P and P' are distinct, then N must lie on d, which would imply N lies on PQ, contrary to our construction. Thus, P = P' and a>P<b is identical to a>M<d>N<b, and hence a>P<b>Q<c = a>M<d>N<b>Q<c. Looking at the component projectivity d>N<b>Q<c, the collinearity of the set {N, Q, cd} means we can apply the UPT to obtain d>S<c identical to d>N<b>Q<c, therefore our initial projectivity is equal to a>M<d>S<c Theorem 4, >>117, implies S = [N[bc]][Q[db]]. In particular, {S, N, bc} is collinear. Since ab = bc = ac and {M, N, ab} is collinear, so is the set {S, M, ac}. Thus, the UPT is again applicable, and we get R such that a>R<c = a>P<b>Q<c, as required.#