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MAT 141
Sample Final – solutions
- (4 Points) True or False:
a: Hilbert’s axiom of parallelism is the same as the Eu- clidean parallel postulate. FALSE
b: One of the congruence axioms is the side-angle-side (SAS) criterion for congruence of triangles. TRUE
c: Euclidean geometry is as true today as it was 2300 years ago. TRUE
d: Euclid’s parallel postulate always holds. FALSE
- (4 Points) Use ruler and compass to find the center of a circle.
We did this in class on 5/13/2016. See also: http://www.mathopenref.com/constcirclecenter.html
- (4 Points) Explain, using anything you know, why the construction in Problem 2 yields the desired result.
You construct a chord AB of the circle. You construct the perpendicular bisector b of the chord AB. The circle is symmetric about b, so its center must lie on b. You con- struct another chord CD of the circle and it’s perpendicular bisector b’. Analogously, the center of the circle must lie on b’. The lines b, b’ meet in one point, E. E must be the center of the circle, as it is the only point common to b and b’.
- (4 Points) Invent a model for incidence geometry that has neither the elliptic, hyperbolic nor the Euclidean parallel properties.
Take the Cartesian plane with the origin removed. Points are as usual and lines are restrictions of lines in the Carte- sian plane. Some lines now have two components, for in- stance the line y = 0. Note that the lines that pass through the origin in the Cartesian plane restrict to lines that are parallel. The line y = 1 admits only one parallel line through (0, 2), namely the line y = 2, whereas the line y = 0 admits two parallels lines through (0, 2), namely the line y = 2 and the line x = 0. The elliptic parallel property requires all lines to intersect, it is violated, for instance, by lines y = 1, y = 2. The Euclidean parallel property requires a unique parallel line through a point off the line. It is vi- olated, for instance, by the existence of the two parallels to y = 0 through (0, 2). The hyperbolic parallel property requires at least two parallel lines through a point not on the line. It is violated, for instance, by the existence of the line y = 1 and its unique parallel through (0, 2).
(4 Points)
(4 Points) Consider the Cartesian plane with the origin removed. Interpret points, lines, measurement of segments and angles in the usual ways. Which of the 13 Hilbert plane axioms fail?
B-4 fails, because (0,-1) and (0,1) are on the same side of the line y = 0 as are (0, 1) and (1, 2). But (0, -1) and (1, 2) are on opposite sides of the line.
Solving, we obtain
(1 + m)/(1 − m) =
So
m =
and the midpoint of the segment is
√^3 −^1
