Institute of Mathematics University of the Philippines-Diliman. 1. In, the sum of the measures of the three angles of a triangle is π.

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1 Institute of Mathematics University of the Philippines-iliman Math 140: Introduction to Modern Geometries Second Exam MHX 14 ebruary 2008 Name: Student I..: I. ill-in the blanks. ill-up the blank in each question below with,euclidean geometry, Hyperbolic geometry, or Elliptic geometry so that the resulting statements will be true. 1. In, the sum of the measures of the three angles of a triangle is π. 2. In, all Saccheri quadrilaterals have an acute angle. 3. In, there exists two triangles which are similar but not congruent 4. In, a Lambert quadrilateral has an obtuse angle 5. In, it is possible for a third line to intersect two parallel lines and yet not form a pair of congruent alternate interior angles. II. True or alse Write if the statement is true in the specified geometry. Otherwise, write LSE. If S B is a Saccheri quadrilateral in a hyperbolic plane then B <. In neutral geometry, if and are points on the opposite sides of l(b, ) and B = B In neutral geometry, if and are points on the same side of l(b, ) and m B+m B < π then the lines l(, B) and l(, ) intersect at some point. In neutral geometry, if two lines are parallel then they have a common perpendicular. In neutral geometry, if and are points on the same side of l(b, ) and m B+m B = π III. Problem Solving nswer the following question in details. (5 points each). The following questions are to be answered in the setting of Euclidean Geometry. 1. onsider a triangle B which is not isosceles. Let be a point on l(b, ) such that is between B and and let the line t bisect angle and intersect l(, B) at. The line t is called the external angle bisector of B at. Prove that B = B. Hint: raw a line through parallel to l(, ). 1

2 B = 2. Use the preceeding problem(question 1) and eva s theorem to prove that the external angle bisector at and and the internal angle bisector at B intersect at a point. Note the conclusion in the preceeding problem(question 1) holds tivially true if B is isosceles. 3. Prove that for any triangle, the radius of the nine point circles of is half of that of the circumcircle. 4. Suppose that l and m are parallel lines in hyperbolic geometry. Prove that there exists a unique line perpendicular to l and m. Total points: 30 ; passing: 18 2

3 nswer Key for Exam I. ill-in the blanks. ill-up the blank in each question below with,euclidean geometry, Hyperbolic geometry, or Elliptic geometry so that the resulting statements will be true. 1. In Euclidean geometry, the sum of the measures of the three angles of a triangle is π. 2. In Hyperbolic Geometry, all Saccheri quadrilaterals have an acute angle. 3. In Euclidean geometry, there exists two triangles which are similar but not congruent 4. In Elliptic geometry, a Lambert quadrilateral has an obtuse angle 5. In Hyperbolic Geometry, it is possible for a third line to intersect two parallel lines and yet not form a pair of congruent alternate interior angles. II. True or alse Write if the statement is true in the specified geometry. Otherwise, write LSE. LSE If S B is a Saccheri quadrilateral in a hyperbolic plane then B <. In neutral geometry, if and are points on the opposite sides of l(b, ) and B = B LSE In neutral geometry, if and are points on the same side of l(b, ) and m B+m B < π then the lines l(, B) and l(, ) intersect at some point. In neutral geometry, if two lines are parallel then they have a common perpendicular. In neutral geometry, if and are points on the same side of l(b, ) and m B+m B = π III. Problem Solving nswer the following question in details. (5 points each). The following questions are to be answered in the setting of Euclidean Geometry. 1. onsider a triangle B which is not isosceles. Let be a point on l(b, ) such that is between B and and let the line t bisect angle and intersect l(, B) at. The line t is called the external angle bisector of B at. Prove that B = B. Hint: raw a line through parallel to l(, ). 1

4 B = nswer: Let l be the line through parallel to t and let G be the point of intersection of l and l(b,, ). Then B = B G. Since l and t are parallel and l(, ) is a transveral then G = = G. Hence, G is isosceles with = G. The conclusion follows. B G l G = = G 2. Use the preceeding problem(question 1) and eva s theorem to prove that the external angle bisector at and and the internal angle bisector at B intersect at a point. Note the conclusion in the preceeding problem(question 1) holds tivially true if B is isosceles. nswer: Let E be the intersection of the internal angle bisector at B and l(, ). Let be the point of intersection of the external angle bisector at and l(b, ) and be the intersection of the external angle bisector at. pplying a theorem discussed in class we have E E = E E = B B. rom the previous problem we have B = B = B and B = B = B. Hence, B E B E B B = ( )( = E E. 3. Prove that for any triangle, the radius of the nine point circles of is half of that of the circumcircle. nswer: Let K be the center of the circumcircle of B and let, E and be the respective midpoints of the sides B, and B. Then l(k, ), l(k, E) and l(k, ) are the respective perpendicular bisectors of the sides B, and B. 2

5 Let O be the orthocenter of B. Then l(o, ) is an altitude, so l(o, ) l(b, ). Since, l(k, ) is also perpendicular to l(b, ), then l(o, ) is parallel to l(k, ). Let P be the midpoint of O P. and let H be the center of the nine-point circle. Thus, P is on the nine point circle and HP is a radius of the nine-point circle. It was discussed in class that OP = K and since l(, O) l(k, ) then OPK is a parallelogram and the diagonal P is a diameter of the nine-point circle. Since P is the midpoint of O the P = P O = K. Hence, P K is also a parallelogram and therefore K = P = 2 PH. Since, K is a radius of the circumcircle, we obtain the desired result. 4. Suppose that l and m are parallel lines in hyperbolic geometry. Prove that there exists a unique line perpendicular to l and m. nswer: The existence of the common perpendicular was already shown in class. Suppose there are two common perpendiculars. Let and B be the feet of these common perpendicular on line l and let and be the feet of these common perpendiculars at line m. Since the line perpendicular to a given point on a line is unique, = B or c = implies l = m. If B and then L B is a Lambert quadrilateral with four right angles which cannot occur in hyperbolic geometry. Therefore, we must have = B and = and l = m. 3