Topics in geometry Exam 1 Solutions 7/8/4

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1 Topics in geometry Exam 1 Solutions 7/8/4 Question 1 Consider the following axioms for a geometry: There are exactly five points. There are exactly five lines. Each point lies on exactly three lines. Each line goes through exactly three points. Prove that this geometry has at least two inequivalent models. Does the geometry have more than two inequivalent models? Explain. What further axiom would distingush between these models? Is each axiom dependent or independent of the other three axioms? Prove your results. Are the following theorems true (explain): No parallel lines exist. No parallel points exist. We begin by constructing two models explicitly: p q r s t A B C D E p q r s t A B C D E The row and column sums are always 3, proving that each of these is a model.

2 The models are inequivalent because the following axiom is true for the first model and false for the second: No two lines have three points in common. We prove that there are no other models. First assume that a model has two lines, say s and t, with three points in common. Then the other two points (by axiom one), say A and B, cannot lie on either s or t, by the fourth axiom, so by the second and third axioms must go through every other line, say lines p, q and r. Then each of p, q and r must go through one additional point, so through one of the points of the lines s and t. Say p goes through point A on s and t. Then A has three lines through it, so can have no more, so q does not go through A. Say q goes through point B on s and t, distinct from A. Then each of A and B has three lines through it, so can have no more, so r goes through neither A nor B. Say r goes through point C on s and t, distinct from A and B. Then all points of s and t are accounted for and all five lines and points have been obtained obeying the axioms. We have arrived at the second model. Now assume that no two lines have three points in common. Let p be a line, with points three points on it. Let B and C be the other two points. So p goes through neither B nor C. The four other lines must each go through at least one of the points B or C, since otherwise our hypothesis that no two lines have three points in common will be violated. But three lines go through B and three go through C, so exactly two lines must go through both B and C, say liens r and s These two lines must each go through one point of the line p, and these points must be distinct, since otherwise our hypothesis is violated. Let r go through point A on p and s go through point D on p. Let E be the other point on p.

3 Question In Euclidean geometry, let C be a circle and P a point outside the circle. Give, with proof, a construction for the tangents from P to the circle. Show that each tangent is perpendicular to the radius vector at its intersection with the circle. Also prove that each of the tangents in question cannot intersect the circle in more than one point. Also explain why there are no more than two tangents from P to the circle. We define a tangent from a given point outside a given circle to be a straight line through the point, intersecting the circle at exactly one point. To construct the tangents through P : Pick any three points R, S and T on the circle. Construct the perpendicular bisectors L and M of the segments RS and ST. The intersection point O of the lines L and M is the center of the circle C. Now construct the perpendicular bisector N of the line OP. Then N meets OP at a point Q, the midpoint of the segment OP. Now draw the circle D, centered at Q with radius OQ. This circle intersects the circle C at two points, say J and K. The lines JP and KP are the desired tangents. Since OP is a diameter of D, and since J and K are points of D, we know that the lines OJ and JP are perpendicular, as as the lines OK and KP. But OJ and OK are radii of the circle C, so each tangent is perpendicular to the radius vector at its intersection with the circle, as required. Let X be any point of the line P J that is not J. Then by Pythagoras Theorem, applied to the right-angled triangle OJX, we have OX = OJ + JX > OJ, since JX > 0. If X were on the circle C, we would have OX = OJ and OX = OJ contradicting the fact that OX > OJ. So X is not on the circle C. So the line P J meets the circle C at exactly one point, the point J. Similarly the line P K meets the circle C at exactly one point, the point K. We have proved that each tangent intersects the circle at exactly one point, as required. 3

4 An alternative proof, not using the Theorem of Pythagoras is by contradiction: Suppose the line JP meets the circle at an additional point H, apart from J. Then the triangle HJO has the angles OJH and OHJ the same since OH = OJ, so the triangle is isosceles with base HJ. But we know that angle OJH is a right-angle, since OJ is perpendicular to JH. So the triangle OJH has two of its angles right-angles, so the sum of its angles exceeds two-right-angles, contradicting the fact that the sum of the angles of any triangle is exactly two right-angles. 4

5 Question 3 In the Z 11 affine geometry, find the equation of the line L through the points (1, 3) and (4, 5). Also find the other points of the line L and the intersection of the line L with the line M whose equation is 5x + 4y + 1 = 0. The line P is given by the determinant formula: 0 = det x y z = x(3 5) y(1 4) + z(5 1) = x + 3y 7z, 0 = 8x + 1y 8z = 8x + y 6, y = 8x + 6. Alternatively the slope is 5 3 = = equation of a line, we get: = 8, so by the point slope form of the y 3 = 8(x 1), y = 8x = 8x 5 = 8x + 6. The various points of the line are obtained by varying x from 0 to 10: {(0, 6), (1, 3), (, 0), (3, 8), (4, 5), (5, ), (6, 10), (7, 7), (8, 4), (9, 1), (10, 9)}. The two lines meet where both y = 8x + 6 and 5x + 4y + 1 = 0, which gives: 0 = 5x + 4(8x + 6) + 1 = 37x + 5, x = 5 37 = 3 4 = 8 4 =, y = 8() + 6 = = 0. So the required meeting point is (, 0) which is easily checked to lie on both lines. 5

6 Question 4 In the Z 7 projective geometry, find the equation of the line P through the points (, 3, ) and (4, 4, 1). Find the other points of the line P and the point where P intersects the line at infinity. Find the intersection of the line P with the line Q, whose line co-ordinates are (1, 4, ). Sketch the affine parts of the lines P and Q in the Z 7 affine plane. The line P is given by the determinant formula: 0 = det x y z = x(3 8) y( 8) + z(8 1) = 5x + 6y 4z. So P has the line co-ordinates ( 5, 6, 4) = (, 1, 3). Solving for y, we get: y = 5x 4z = x + 3z. Putting z = 0, we get the required point at infinity on the line (x, x, 0) or just (1,, 0). Putting z = 1 and varying x from 0 to 6, we get the other seven points of the line: {(0, 3, 1), (1, 5, 1), (, 0, 1), (3,, 1), (4, 4, 1), (5, 6, 1), (6, 1, 1)} Note that the given point (, 3, ) = (8, 1, 8) = (1, 5, 1) is in this list. The intersection point of P and Q is given by the vanishing of the determinant formula: a b c 0 = det = a( 1) b(4 3) + c(8 + 1) = b + c. So the intersection point is (0, 1, ) = (0, 4, 8) = (0, 3, 1). This point has vanishing dot product with each of (, 1, 3) and (1, 4, ) so does indeed lie on both lines. 6

7 Question 5 Find the precise image under inversion in the circle of radius one, centered at the origin, of the following geometrical figures: The circle center P = (3, 4), radius 5 units. The point P is a distance of = 5 from the origin, so the circle goes through the origin, so its inversion is a straight line not through the origin. Note that this circle has points both inside and outside the inversion circle, so meets it at exactly two points. To get the equation of the line, we can either directly compute the circle s inverse, or see where the circle meets the inversion circle, or pick some points and invert them: The given circle has equation (x 3) +(y 5) = 5, which expanded gives: 1(x + y ) 6x 8y + 0 = 0, so its inverse has the equation: 0(x + y ) 6x 8y + 1 = 0, or the line 6x + 8y = 1. The circle x + y 6x 8y = 0 meets the circle x + y = 1, where both equations hold at once, so where 6x + 8y = 1. This is the equation of a line which passes through the two intersection points, so is the line we seek. The point Q = (6, 0) lies on the given circle as does the point R = (0, 8). Q has inverse S = ( 1, 0) and R has inverse T = (0, 1 ), so the required 6 8 line is the line ST, which has slope 3 and intercept (0, 1 ), so has the 4 8 equation: y = 3x + 1 or 6x + 8y = 1, as before

8 The triangle ABC, with vertices the points A = (, 0), B = ( 1, 3) and C = ( 1, 3). For this case, what are the angles of intersection of the image curves of the sides of the triangle? Explain. Sketching the triangle ABC, we see that it is equilateral, with all angles sixty degrees and circumscribes the inversion circle, with each side tangential to the circle at the mid-point. These tangency points, which are their own inverse are: D = ( 1, 0), the midpoint of BC, E = ( 1, 3), the midpoint of CA, F = ( 1, 3), the midpoint of AB. The inverses of the points A, B and C, which are each at distance from the origin are the following points: α = ( 1, 0), the inverse of A, β = ( 1, 3), the inverse of B, 4 4 γ = ( 1, 3), the inverse of C. 4 4 So the required inverse is three circular arcs, joined at their end-points: The arc βdγ. This lies on the inverse of the line BC: x = 1, which is a circle through the origin O, with diameter OD, so its center is at ( 1, 0) and its radius is 1. The arc γeα. This lies on the inverse of the line CA, which is a circle through the origin O, with diameter OE, so its center is at ( 1, 3) and its radius 4 4 is 1. The arc αf β. This lies on the inverse of the line AB, which is a circle through the origin O, with diameter OF, so its center is at ( 1, 3) and its radius is 4 1. The original lines meet at inner angles of 60 degrees, or π 3 radians. The inverted arcs meet at outer angles of 60 degees, or π 3 radians. Each arc subtends an angle of 4π 3 at the center of the circle that it lies on. 4 8

Inversive Plane Geometry

Inversive Plane Geometry An inversive plane is a geometry with three undefined notions: points, circles, and an incidence relation between points and circles, satisfying the following three axioms: (I.1)