2 Solution of Homework II

Transcription

1 Math 3181 Dr. Franz Rothe Name: All3181\3181_spr13h2.tex 2 Solution of Homework II 10 Problem 2.1. Prove the Four-point Theorem directly from the Three-point Theorem and the Plane separation Theorem. Answer. Given are four different points on a line l. By the Three-point Theorem, one of them, which we name C lies between two others. We draw a line m through C different from l and use plane separation. In one half-plane lie two of the remaining points, in the other half-plane only the last remaining point, which we call D. Of the two points in the same half plane, one which we name B lies between C and the other one, finally named A. We have thus obtained the order relations A B C, A C D, B C D To confirm the fourth order relation, we draw a line n through B different from l, and use plane separation once more. Point C and D lie in the same half-plane of n whereas A lies in the opposite half-plane from C. Hence A and D lie in opposite half-planes, confirming the order relation A B D. 1

2 10 Problem 2.2. Given is a triangle and a line through an interior point of the triangle. Show that the line either intersects two sides of the triangle, or it intersects only one side and goes through the opposite vertex. Figure 1: A line through an interior point of a triangle intersects either two sides, or goes through a vertex and intersects the opposite side. Answer. Let the line l go the point P in the interior of triangle ABC. We distinguish two cases: (a) The line goes through a vertex of the triangle. Say the line goes through vertex A. Because point P lies in the interior of angle BAC, the crossbar theorem shows that the ray AP intersects the opposite side BC at some point, say Q. (b) The line does not go through any vertex of the triangle. We draw the ray AP. Because point P lies in the interior of angle BAC, the crossbar theorem shows that the ray AP intersects the opposite side BC at some point, say Q. We now apply Pasch s axiom to the triangle ABQ and the line l. This line intersects side AQ at point P. Hence it intersects a second side, either BQ BC or AB, say at point D. In both cases, the line l intersects one side of the given triangle ABC. By Pasch s axiom it intersects a second side, say at point E. By Bernays lemma, the line l intersects exactly two sides of the given triangle. 2

3 Proposition 1 (Preliminary Converse Isosceles Triangle Proposition). If the two base angles of a triangle are congruent to each other, the triangle is isosceles. With specific quantities: If α = β and β = α hold for the base angles of triangle ABC, then a = b. 10 Problem 2.3. This proposition is proved using ASA-conguence. Define a second triangle A B C by setting A := B, B := A, C := C and argue in a way similar to the proof showing congruence of the base angles of an isosceles triangle. Figure 2: An isosceles triangle, two ways to look at it Answer. Assume that the angles α = CAB and β = ABC are congruent in ABC. We need to show that the two sides a = BC and b = AC are congruent. To apply ASA congruence to the two triangles ABC and A B C, we match corresponding pieces: (1) AB = B A = A B. Hence AB = A B, because the order of the endpoints of a segment is arbitrary, and a segment is congruent to itself. (2) α = α. Indeed, α = β by assumption, and β = ABC = B A C = α by construction. Hence α = α. (3) Similarly, one shows that β = β : Indeed, β = α by assumption, and α = BAC = A B C = β by construction. Hence β = β. Via ASA congruence, items (1)(2)(3) imply that ABC = A B C, and hence especially AC = A C = BC as to be shown. 3

4 Proposition 2 (Angle-Addition and Subtraction). [Theorem 15 in Hilbert] Given are three vertices h, k, l with the common vertex O lying in a plane a, and three vertices h, k, l with the common vertex O lying in a plane a. We assume either the case of angle subtraction or angle addition. angle subtraction: in this case the rays h and k lie in the same half-plane of l, and the rays h and k lie in the same half-plane of l ; angle addition: in this case the rays h and k lie in the different half-planes of l, and the rays h and k lie in different half-planes of l. Given is an angle ABC and a ray BG in its interior, as well as a second angle A B C and a ray B G in its interior. If (h, l) = (h, l ) and (l, k) = (l, k ) then either both h and k are opposite rays, and h and k are opposite rays, too; or (h, k) = (h, k ) 10 Problem 2.4. Explain the case of angle addition. Provide a drawing for the that the two rays k and l lie in different half-planes of ray h, and the two rays k and l lie in different half-planes of ray h. Explain how one proceeds to get back to a simpler case. Answer. We use the opposition rays l of l and l from l. Now the Proposition about the congruence of supplementary angles implies (h, l ) = (h, l ) and (l, k) = (l, k ) Since we are back to case where the two rays k and l lie in the same half-planes of ray h, and the two rays k and l lie in the same half-plane of ray h. From the earlier Proposition, we conclude the claim (h, k) = (h, k ). 4

5 10 Problem 2.5 (Scheduling problem I). Make a 6 day schedule for a school with 25 students. Each day the students are divided in a different way into 5 groups of 5 students. Never are two students in the same group more than one time during the week. Use the coordinate plane Z 5 Z 5, with addition and multiplication modulo 5. Since 5 is a prime, the set Z 5 is a field. Hence the coordinate plane Z 5 Z 5 is an affine plane with order 5. Make a picture of the schedule by drawing the 5 5 pattern of dots separately for every day. The five parallel lines in every pattern can be indicated by different symbols for their points. One needs curved lines to connect all five points of a line, which you may or may not draw. The main point is to show the partition into the five groups clearly, in a separate drawing for every day. Clearly such a picture contains more insight than a bare-bone list. Figure 3: Explain the schedule for five groups of five on six days. Answer. The solution is given in homework one. 5

6 Definition 1 (Subplane). A subplane of an incidence plane is an incidence plane which has as "points" a subset P P from the points P of the given plane, and as "lines" the nonempty intersections of the lines of the original plane with the subset P of the remaining points. The incidence relation is the induced relation. The axioms of incidence (I.1), (I.2),(I.3) have still to hold for the subplane. 10 Problem 2.6. A subplane of the affine coordinate plane Z 5 Z 5 contains the 15 points on three parallel lines l 1, l 2, l 3. Convince yourself that the lines of the subplane are these three parallel lines with 5 points on each, and 25 lines k i with 3 points on each one. How many parallels through a given point has each one of the lines l 1, l 2, l 3. How many parallels through a given point has each one of the lines k 1,... k 25. Answer. The original Z 5 Z 5 plane has = 30 lines. Two lines are deleted. There remain 28 lines, of which 25 have three points each one. In the subplane, there may exist many parallels to a given line through a given point. For a three point line l of the subplane and a point P not lying on l, there exist three parallels to line l through the point P. These parallels are obtained by intersectiong the original parallel m and the lines P A, P B with the subplane. Here A, B be the two points deleted from the extended line l +. 6

7 10 Problem 2.7 (Kirkman s Schoolgirl Problem 1850). Kirkman s schoolgirl problem is a problem in combinatorics proposed by Rev. Thomas Penyngton Kirkman in 1850 as Query VI in The Lady s and Gentleman s Diary (pg.48). The problem states: Fifteen young ladies in a school walk out three abreast for seven days in succession: it is required to arrange them daily so that no two shall walk twice abreast. Draw the schedules for the seven days, for each day in a separate circle of 15, such that one can really see and check the solution easily. You can use a solution obtained from any source. gives the solution Sun ABC DEF GHI JKL MNO Mon ADH BEK CIO FLN GJM Tue AEM BHN CGK DIL FJO Wed AFI BLO CHJ DKM EGN Thu AGL BDJ CFM EHO IKN Fri AJN BIM CEL DOG FHK Sat AKO BFG CDN EIJ HLM Further information is given in