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Discovery of Non-Euclidean Geometry

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Discovery of Non-Euclidean Geometry Discovery of Non-Euclidean Geometry April 24, 2013 1 Hyperbolic geometry J¶anosBolyai (1802-1860), Carl Friedrich Gauss (1777-1855), and Nikolai Ivanovich Lobachevsky (1792-1856) are three founders of non-Euclidean geometry. Hyperbolic geometry is, by de¯nition, the geometry that assume all the axioms for neutral geometry and replace Hilbert's parallel postulate by its negation, which is called the hyperbolic axiom. Hyperbolic axiom (Negation of Hilbert axiom). There exists a line l and a point P not on l such that at least two distinct lines parallel to l pass through P . Theorem 1.1. In hyperbolic geometry, all triangles have angle sum less than 180±, and all convex quadrilaterals have angle sum less than 360±. In particular, there is no rectangle. Proof. Trivial. Theorem 1.2 (Universal Hyperbolic Theorem). In hyperbolic geometry, for every line l and every point P not on l there pass through P at least two distinct lines parallel to l. In fact there are in¯nitely many lines parallel to l through P . t t' P m S n S' n' l Q R R' Figure 1: Existence of in¯nite parallels Proof. Drop segment PQ perpendicular to l with foot Q on l. Erect line m at P perpendicular to PQ. Then l; m are parallel. Pick a point R on l other than Q, and erect line t at R perpendicular to l. Drop line n through P perpendicular t, intersecting t at S. If S is on m, then S is the intersection m and t, and subsequently ¤P QRS is a rectangle, which is impossible in hyperbolic geometry. So point S is not on m. Hence m; n are distinct lines through P , both are parallel to l. See Figure 1. Let R0 be point on l other than Q; R, and t0 be line through R0 perpendicular to l. There exists line n0 through P perpendicular to t0, intersecting t0 at S0. If PS = PS0, then ¤RR0S0S is a rectangle, which is impossible. So P S; PS0 are distinct lines. Thus for all points R on l other than Q, the lines PS through P perpendicular to l are all distinct. 1 De¯nition 1. Two triangles are ¢ABC and ¢A0B0C0 said to be similar if their vertices can be put in on-to-one correspondence so that corresponding angles are congruent, i.e., \A; \B; \C are congruent to angles \A0; \B0; \C0 respectively. Theorem 1.3 (AAA criterion for congruence of hyperbolic triangles). In hyperbolic geom- etry, if two triangles are similar then they are congruent. A A' B C B'' C'' B' C' Figure 2: Similar triangles are congruent in hyperbolic geometry Proof. Given similar triangles ¢ABC and ¢A0B0C0. Suppose the statement is not true, i.e., ¢ABC is not congruent to ¢A0B0C0. Then AB 6»= A0B0, AC 6»= A0C0, BC 6»= B0C0; otherwise ¢ABC »= ¢A0B0C0 by ASA. We may assume AB < A0B0 and AC < A0C0. Lay o® segment AB on ray r(A0;B0) to have point B00 such that AB »= A0B00 and A0 ¤ B00 ¤ B0, and lay o® segment AC on ray r(A0;C0) to have point C00 such that AC »= A0C00 and A0 ¤ C00 ¤ C0. See Figure 2. Then ¢ABC »= ¢A0B00C00 by SAS. Hence \A0B00C00 »= \B »= \B0; \A0C00B00 »= \C »= \C0: It follows that lines B00C00; B0C0 are parallel because of Congruent Corresponding Angles. So ¤B0C0C00B00 is a convex quadrilateral. Since angles \A0B00C00; \B0B00C00 are supplementary, angles \A0C00B00; \C0C00B00 are supplementary, and \B0 »= \A0B00C00; \C0 »= \A0C00B00, then the angle sum of ¤B0C0C00B00 is 360±. This is a contradiction. Theorem 1.3 says that in hyperbolic geometry it is impossible to magnify or shrink a triangle without distortion. So in hyperbolic world photography would be inherently surrealistic. Another consequence of Theorem 1.3 is that the length of a segment may be determined by angles in hyperbolic geometry. For example, an angle of an equilateral triangle determines the length of a side uniquely. This fact is sometimes referred to that hyperbolic geometry has an absolute unit length. 2 Parallels that admit a common perpendicular Given lines l; l0 and points A; B; C; : : : on l. Drop perpendiculars AA0; BB0;CC0;::: from A; B; C; : : : to l0 with feet A0;B0;C0;::: on l0 respectively. We say that A; B; C; : : : are equidistant from l0 if all these perpendicular segments are congruent to one another. See Figure 3. Theorem 2.1 (At most two points equidistant). Given two distinct parallels l; l0 in hyperbolic geometry. Then any set of points on l equidistant from l0 contains at most two points. Proof. Suppose it is not true, i.e., there is a set of three points A; B; C on l equidistant from l0. Then quadrilaterals ¤ABB0A0; ¤ACC0A0; ¤BCC0B0 are Saccheri quadrilaterals (the base 2 ABC l l' A' B' C' Figure 3: No more than two equidistant points between two parallels angles are right angles and the sides are congruent). Then the summit angles of the Saccheri quadrilaterals are congruent, i.e., \BAA0 »= \ABB0; \CAA0 »= \ACC0; \CBB0 »= \BCC0: Thus \ABB0 »= \CBB0. Since \ABB0; \CBB0 are supplementary, they must be right angles. Hence all ¤ABB0A0, ¤ACC0A0, ¤BCC0B0 are rectangles, which is impossible. Lemma 2. Given a Saccheri quadrilateral ¤ABB0A0 with base right angles \A0; \B0 and equal opposite sides AA0; BB0. Let M; M 0 be the middle points of AB; A0B0 respectively. Then segment MM 0 is perpendicular to both lines AB and A0B0. Proof. Draw segments A0M and B0M. Note that AA0 »= BB0, \A »= \B, and AM »= BM. Then ¢A0AM »= ¢B0BM by SAS. So A0M »= B0M. Hence ¢A0MM 0 »= ¢B0MM 0 by SSS. We then have \A0M 0M »= \B0M 0M. Subsequently, \A0M 0M and \B0M 0M are right angles. So MM 0 is perpendicular to the base A0B0. Note that \A0MM 0 »= \B0MM 0 and AM B A' M' B' Figure 4: Perpendicular middle point segment \AMA0 »= \BMB0. Then \AMM 0 »= \BMM 0 by angle addition. Subsequently, \AMM 0 and \BMM 0 are right angles. So MM 0 is perpendicular to the summit AB. Theorem 2.2 (Divergent and symmetric parallels). Let l; l0 be two lines perpendicular to a segment MM 0 with M 2 l; M 0 2 l0. (a) Then jMM 0j < jXY 0j for all X 2 l; Y 0 2 l0 with XY 0 6= MM 0. (b) If M is the middle point of a segment AB on l, then A; B are equidistant from l0. (c) If M ¤ B ¤ D on l and BB0; DD0 are segments perpendicular to l0 with feet B0;D0 2 l0, then BB0 < DD0. Proof. (a) It is clear that MM 0 < MY 0 for all Y 0 on l0 with Y 0 6= M 0. Let X be a point on l with X 6= M. Let XX0 be segment perpendicular to l0 with foot X0 on l0. Then ¤MM 0XX0 is a Lambert quadrilateral. Thus MM 0 < XX0 by properties of Lambert quadrilaterals. Since XX0 < XY 0 for Y 0 on l0 with Y 0 6= X0. We see that MM 0 < XY 0. (b) Let AA0; BB0 be segments perpendicular to l0 with A0;B0 2 l0. Draw segments AM 0 and BM 0. Then ¢AMM 0 »= ¢BMM 0 by SAS. So AM 0 »= BM 0 and \AM 0M »= \BM 0M. 3 l C D A B M l' C'A'M' B' D' Figure 5: Divergent parallels are symmetric Subsequently, \AM 0A0 »= \BM 0B0 by angle subtraction. Thus ¢AA0M 0 »= ¢BB0M 0 by SAA. Hence AA0 »= BB0 and A0M 0 »= B0M 0. (c) Note that ¤M 0B0BM and ¤M 0D0DM are Lambert quadrilaterals. Then \B0BM and \D0DM are acute angles. So \B0BD is obtuse, for it is supplementary to \B0BM. Hence \D0DB = \D0DM < \B0BD. Therefore BB0 < DD0 by the property of ¤B0D0DB. Proposition 2.3 (Asymptotic and monotonic parallels). Given parallels l; l0 in hyperbolic geometry, no two points of l are equidistant from l0. Let AA0; BB0;CC0 be perpendicular segments to l0 with A ¤ B ¤ C on l and A0;B0;C0 2 l0. See Figure 6. (a) If AA0 < BB0, then BB0 < CC0. (b) If BB0 < CC0, then AA0 < BB0. C l B A l' A' B' C' Figure 6: Monotone distance between asymptotic parallels Proof. Consider quadrilaterals ¤A0B0BA and ¤B0C0CB. (a) Since AA0 < BB0, then \ABB0 < \BAA0. Since the angle sum of ¤A0B0BA is less than 360±, it follows that \ABB0 is acute. So \CBB0 is obtuse. Hence \BCC0 must be acute, since the angle sum of ¤B0C0CB is less than 360±. Of course \BCC0 < \CBB0, subsequently, BB0 < CC0 by the property of ¤B0C0CB. (b) Fix a point D on l with A ¤ B ¤ C ¤ D. For each X on the open ray ºr(D; A) we write jDXj = x and de¯ne f(x) = jXX0j, where XX0 is perpendicular to l0 with foot X0 on l0. We claim that f(x) is a continuous function for x > 0. In fact, ¯x an x0 with point X0 on l 0 0 0 0 such that jBX0j = x0. Let X0X0 be segment perpendicular to l with X0 2 l . Note that 0 0 0 0 0 0 jXX j · jXX0j · jX0X0j + jXX0j; jX0X0j · jX0X j · jXX j + jXX0j: Then ½ 0 0 0 0 jXX j ¡ jX0X0j if jXX j ¸ jXX0j jf(x) ¡ f(x0)j = 0 0 0 0 jX0X0j ¡ jXX j if jXX j < jXX0j · jXX0j = jx ¡ x0j: Clearly, f(x) is continuous at x0. So f(x) is a continuous function for x > 0. Suppose AA0 > BB0. Note that AA0 6»= CC0.
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