the projective completion of the affine plane with four points

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1 Math 3181 Dr. Franz Rothe November 23, 20 1 Test Name: 50 Problem 1.1. Here are once more: a highly symmetric illustration for the Fano plane based on an equilateral triangle the projective completion of the affine plane with four points Denote the seven points in both drawings consistently in a way to show the isomorphism. Figure 1: The symmetric drawing of the Fano-plane is really isomorphic to the projective completion of the affine plane of order 2. Answer. To check that this symmetric illustration is isomorphic to the illustration I have given on the left side, one needs to names the points in both illustrations in a way that the incidence relations hold for the same names. Thus the isomorphism is given by the correspondence of names. To find such an isomorphism, the key observation is that a triangle can be mapped to any triangle, but afterwards the correspondence of the remaining points is uniquely determined. 1

Figure 2: The points and lines of the Fano-plane named in a way to confirm self-duality. Problem 1.2. Here is, once more, a highly symmetric illustration for the Fano plane based on an equilateral triangle.

2 Figure 2: The points and lines of the Fano-plane named in a way to confirm self-duality. Problem 1.2. Here is, once more, a highly symmetric illustration for the Fano plane based on an equilateral triangle. Find the 7 7 matrix which gives the incidence between the lines and points. What does it means geometrically that the matrix is symmetric? Answer. Since the incidence matrix is symmetric, the Fano plane is isomorphic to its dual. A B C D E F G a b c d e f g Problem 1.3. Solve Euler s problem for a 5 5 square: arrange 25 officers to be drawn from 5 different ranks A, B, C, D, E, and at the same time from 5 different regiments α, β, γ, δ, ɛ, so that they are also arranged in a square so that in each line both horizontal and vertical there are 5 officers of different ranks and and regiments. 2

3 Euler s problem is just equivalent to finding two orthogonal Latin squares of side 5 one specifying the rank, the second specifying the regiment for each officer. Answer. Here is the solution I got in two minutes: Clearly, every case of odd order is as easy. Aα Bβ Cγ Dδ Eɛ Bɛ Cα Dβ Eγ Aδ Cδ Dɛ Eα Aβ Bγ Dγ Eδ Aɛ Bα Cβ Eβ Aγ Bδ Cɛ Dα Proposition 1 (Isosceles Triangle Proposition). [Euclid I.5, and Hilbert s theorem 11] An isosceles triangle has congruent base angles. Problem 1.4. Formulate the theorem with specific quantities from a triangle ABC as shown in the drawing. Give a detailed proof. Figure 3: An isosceles triangle Answer. The short statement of the theorem is: If a = b, then α = β. Proof. This is an easy application of SAS-congruence. Assume that the sides AC = BC are congruent in ABC. We need to show that the base angles α = BAC and β = ABC are congruent. Define a second triangle A B C by setting A := B, B := A, C := C (It does not matter that the second triangle is just on top of the first one.) To apply SAS congruence, we match corresponding pieces: (1) ACB = BCA = A C B because the order of the sides of an angle is arbitrary. By axiom III.4, last part, an angle is congruent to itself. Hence ACB = A C B. 3

Finally, we use axiom III.5. Items (1)(2)(3) imply BAC = B A C = ABC. But this is just the claimed congruence of base angles. Problem 1.5. Give the precise definitions of the following terms: angle, vertex of the angle.

4 (2) AC = A C. Question. Explain why this holds. Answer. AC = BC because we have assumed the triangle to be isosceles, and BC = A C by construction. Hence AC = A C. (3) Similarly, we show that (3): BC = B C : Proof. BC = AC because we have assumed the triangle to be isosceles, and congruence is symmetric, and AC = B C by construction. Hence BC = B C. Finally, we use axiom III.5. Items (1)(2)(3) imply BAC = B A C = ABC. But this is just the claimed congruence of base angles. Problem 1.5. Give the precise definitions of the following terms: angle, vertex of the angle. sides of the angle. interior of an angle exterior of an angle Remark. Note the exterior of an angle is defined differently from exterior angle of a triangle. The latter is any supplementary angle to an interior angle of the triangle. Answer. Definition 1 (Angle). An angle BAC is the union of two rays AB and AC with common vertex A not lying on one line. The point A is called the vertex of the angle. The rays AB and AC are called the sides of the angle. Figure 4: Interior and exterior of an angle are intersection and union of two half planes. Definition 2 (Interior and exterior of an angle). The interior of an angle lying in a plane A is the intersection of two corresponding half planes bordered by the sides of the angle, and containing points on the other side of the angle, respectively. The exterior of an angle is the union of two opposite half planes -bordered by the sides of the angle, and not containing the points neither in the interior nor on the legs of the angle. Half planes, interior and exterior of an angle all do not include the lines or rays on their boundary. 4

5 Thus the interior of BAD is the intersection of the half plane of AB in which D lies, and the half plane of AD in which B lies. The exterior of BAD is the union of the half plane of AB opposite to D, and the half plane of AD opposite to B. 5

Definition 1 (Hand-shake model). A hand shake model is an incidence geometry for which every line has exactly two points.

Math 3181 Dr. Franz Rothe Name: All3181\3181_spr13t1.tex 1 Solution of Test I Definition 1 (Hand-shake model). A hand shake model is an incidence geometry for which every line has exactly two points. Definition