Geometry Solutions EXAM I

Rutgers University Geometry Spring 2007 Solutions EXAM I

The exam is open book, which means you can use your notes, past home- work and of course the book. Do not forget to write your name on the blue book.

1. Euclid assumed the compass to be collapsible. That is, given two points P and Q, the compass can draw a circle with center P passing through Q; however, once the spike is moved, the compass collapses, and one cannot draw a circle with the same radius. Given three distinct points P, Q and R. Construct with a straightedge and collapsible compass a rectangle P QST with P Q as a side and such that P T ∼= P R (see ﬁgure below).

Solution:

(a) Draw a circle with radius P R and center at P . Mark the points −→ of intersection of the circle with the line P Q, and call them A and B. (b) Draw the perpendicular bisector of AB through P . (This can be done by drawing the circles with center at A and radius AB and with center at B and radius AB and joining the points of intersection). (c) Mark the intersection of the line of part (b) with the circle of part (a) and call it T . (d) Draw the perpendicular to P T through T . (e) Draw the perpendicular to P Q through Q. (f) Call the intersection of the last 2 lines S.

2. (a) Negate Incidence Axiom 2. That is what is ¬( For every line l there exist at least two distinct points incident with l)?

1 (b) Construct a model or interpretation in which incidence axiom 1 and 3 are satisﬁed, but not incidence axiom 2. Solution:

(a) There exist a line such that no two points are incident with it. (b) Consider the following model: Points: A, B, C Lines: {A, B}, {A, C}, {C, B}, {A}. Incidence Axiom 1 is satisﬁed as well as axiom 3, but axiom 2 is not since the line {A} is incident only with one point.

3. Let A be a finite affine plane (i.e. it has only a finite number of points). Assume that all the lines in A have the same number of points lying on them; call this number n. Prove the following

(a) For each point P in A there exist n + 1 lines passing through it. (b) The total number of points in A is n2. (c) The total number of lines in A is n(n + 1).

Solution:

(a) Let P be a point in A. There exist a line ℓ not incident with P . By hypothesis, ℓ has n points incident with it, say A1,...,An. Any line through P either intersects ℓ (in one of the Ai’s) or does not. By the parallel property, there is a unique line m parallel to ℓ through P . Therefore, the total number of lines through P is: n + 1.

2 (b) For any point Q =6 P in A, there exist a unique line joining P and Q. Therefore the number of points in A is: (n + 1)(n − 1)+1= n2, which is (number of lines through P )(number of points on each line minus P ) + P . (c) Let ℓ be a line as above. total number of lines in A = # of lines that intersect ℓ + # lines parallel to ℓ. # lines that intersect ℓ = (# points on ℓ)( # lines through each point -1) +1 = (n)(n +1 − 1)+1= n2 + 1. # lines parallel to ℓ = # of points on a line intersecting ℓ −1 = n − 1. Therefore total number of lines in A = n2 +1+ n − 1= n2 + n.

4. In the usual Euclidean model, deﬁne the betweenness relation in the following way: A∗B ∗C if B is the midpoint of the segment AC. Prove that the ﬁrst two betweenness axioms hold but Betweenness axiom 3 fails. Solution:

BA1: If B is the midpoint of AC, then A,B,C are three distinct colinear points. BA2: Given points B and D, we can ﬁnd C the midpoint of BD. We can also ﬁnd A, such that d(A, B) = d(B,D), and a point E such that d(B,D)= d(D, E).

3 BA3: This axiom is not satisﬁed by the following points: A = (0, 0), B = (1, 0) and C = (10, 0).

5. Let △ABC be a triangle. Assume that ∢ABC =∼ ∢BCA, prove that AB =∼ AC. Solution:

−→ (a) By CA1, there exist a unique point A′ on the ray BA such that BA′ ∼= CA.

(b) Consider the following triangles: △A′B′C and △ACB. Then A′B ∼= AC (by step (a)), BC ∼= CB, and ∢A′BC ∼= ∢ABC ∼= ACB (by hypothesis). (c) Therefore, by CA6, △A′BC ∼= △ACB. (d) Which implies that: ∢BCA′ ∼= ∢CBA =∼ ∢BCA. (e) By CA4, the point A is unique, so A′ = A. Therefore BA ∼= CA.