Here we suggest a formal discussion of our problems in order to show to some readers that Hilbert’s formalism is more clear than a didactic approach, and to other readers that a formal approach can be difficult at the beginning and the didactic one is necessary.

Following Hartshorne (2000), we recall the Incidence and Betweenness Axioms formulated by Hilbert.

*Incidence axioms*:

- (I1)
For any two distinct points *A*, *B*, there exists a unique line \(\ell\) containing *A*, *B*.

- (I2)
Every line contains at least two points.

- (I3)
There exist three noncollinear points, that is, three points not all contained in a single line.

A set whose elements are called

*points*, together with a set of subsets called

*lines*, satisfying axioms (I1), (I2), (I3), will be called an

*incidence geometry*. If a point

*P* belongs to a line

\(\ell\), we say that

*P* *lies* on

\(\ell\), or that

\(\ell\) *passes through* *P*.

An easy consequence of the incidence axioms is the following:

(\(\star\)) Two distinct lines can have at most one point in common.

Assuming the Incidence Axioms (I1), (I2), (I3), we postulate a

*relation* between sets of three points

*A*,

*B*,

*C*, called

*B is between A and C*, subject to the following

*Betweenness axioms*:

- (B1)
If *B* is between *A* and *C*, written \(A*B*C\), then *A*, *B*, *C* are three distinct points on a line, and also \(C*B*A\).

- (B2)
For any two distinct points *A*, *B*, there exists a point *C* such that \(A*B*C\).

- (B3)
Given three distinct points on a line, one and only one of them is between the other two.

- (B4)
Let *A*, *B*, *C* be three noncollinear points, and let \(\ell\) be a line not containing any of *A*, *B*, *C*. If \(\ell\) contains a point *D* lying between *A* and *B*, then it must also contain either a point lying between *A* and *C* or a point lying between *B* and *C*, but not both.

If

*A* and

*B* are distinct points, we define the

*line segment* (or

*segment* for short)

*AB* to be the set consisting of the points

*A*,

*B* and all points lying between

*A* and

*B*. We define a

*triangle* to be the union of the three line segments

*AB*,

*BC*,

*AC* whenever

*A*,

*B*,

*C* are three noncollinear points. The points

*A*,

*B*,

*C* are the

*vertices* of the triangle and the segments

*AB*,

*BC*,

*AC* are the

*sides* of the triangle.

We say that a point *P* is *inside* the triangle \(A_1 A_2 A_3\) if there exists a vertex \(A_i\) and a point *Q* on the side \(A_j A_k\) (\(j,k \ne i\)) such that \(A_i*P*Q\). If *P* is not inside nor on the sides, we say that *P* is *outside* the triangle.

An important consequence of the Incidence and Betweenness Axioms is the following:

(\(\star \star\)) Every line has infinitely many distinct points.

Using the above axioms one can rewrite “Arranging Stones and Masters”, starting from the following:

This definition is well-posed due to axiom (B3).

We state and prove Problem 1 with this formalism.

The Solution of Problem 2 can be rewritten by using the axiom (I1) and the properties (\(\star\)) and (\(\star \star\)). In Problem 3 it is crucial to involve the axiom (I3) and, as before, (I1) and (\(\star \star\)). In order to show the relevance of (B4) we propose the detailed solution of Problem 4. We mention to the reader that (B4), known as *Pasch’s axiom*, was introduced only in 1882. More precisely, Pasch discovered that this property was implicitly used by Euclid, tough it cannot be derived from the postulates stated by Euclid.

### **Solution 4**

The existence of the triangle \(M_1 M_2 M_3\) follows from the axiom (I2) and, by (\(\star \star\)), there exists \(S_4\) inside the triangle \(M_1 M_2 M_3\) (see Fig. 3c). For every \(i=1,2,3\), we choose \(S_i\) such that \(M_i*S_i*S_4\). In this way \(S_4\) cannot be seen by any of the \(M_i\). To complete the proof, up to permuting the indices, it is enough to show that \(M_1\) can see \(S_2,S_3\). Assume by contradiction that \(M_1\) does not see \(S_2\). Then either \(M_1*S_1*S_2\) or \(M_1*S_3*S_2\).

In the first case, \(M_1,S_1,S_2\) lie on the same line, which contains also \(S_4\) by (I1) and the assumption \(M_1*S_1*S_4\). Since \(M_2\) belongs to the line through \(S_2\) and \(S_4\), it follows that \(S_1,S_2,S_4\) lie on the line segment \(M_1 M_2\), against the fact that \(S_4\) is inside the triangle \(M_1 M_2 M_3\).

In the second case, \(M_1*S_3*S_2\) and, moreover, \(M_2*S_2*S_4\). Let us consider the triangle \(M_1 M_2 S_2\) and the point \(S_4\). By the axiom (B4), the line \(\ell\) through \(S_4\) and \(S_3\) intersects the side \(M_1 M_2\) in a point *P* such that \(S_4*S_3*P\). Notice that the line cannot intersect the segment \(M_2S_2\), indeed (\(\star\)) is satisfied by the point \(S_4\) and the lines through \(S_2,S_4\) and \(S_3,S_4\), where \(S_3 \ne S_2\).

Considering *P* in the triangle \(M_1 M_2 M_3\) and the line \(\ell\) containing \(S_3,S_4\), we conclude that \(S_4\) and \(S_3\) are outside the triangle \(M_1 M_2 M_3\), against the construction. Figure 4a shows a possible configuration.

Hence \(M_1\) can see \(S_2\) and, for the same reason, \(S_3\).

We do not give a formal proof of Problem 5 since it is a generalization of Problem 4. In Problem 6 and 7 it is important to choose the position of the Masters and of the stones, hence property (\(\star \star\)) plays a crucial role. Thus all incidence and betweenness axioms are needed. A formal discussion of “Polygonal arrangements” involves other Euclid and Hilbert axioms.