Solutions to Practice Problems

Higher Geometry Final Exam Tues Dec 11, 5-7:30 pm Practice Problems (1) Know the following definitions, statements of theorems, properties from the notes: congruent, triangle, quadrilateral, isosceles triangle, equilateral triangle, trapezoid, parallelogram, rhombus, rectangle, square, the triangle congruence theorems (SSS,SAS,AAS,ASA), the triangle similarity theorems (same ones), parallel line/transversal theorems, the fifth axiom of Euclid, be able to give 2 examples of the first 4 axioms of Euclid, Playfair s Postulate, incenter, circumcenter, orthocenter, median, centroid, altitude, perpendicular bisector, cyclic polygon, supplementary angles, complementary angles, excircle, Pythagorean theorem, tangent to a circle, orthogonal, power of a point with respect to a circle, Ceva s theorem, Menelaus Theorem, nine-point circle, polar coordinates, slopes of parallel and perpendicular lines, definitions of the trig functions (unit circle), right triangle formulas for trig functions, the important trig identities, Law of Sines, Law of Cosines, definition of vector dot product, matrix addition, matrix scalar multiplication, matrix multiplication, rotation and reflection matrices, translation, isometry, isometric, Euclidean motion, rigid motion, geodesic, Poincaré disk, sphere, lune. (2) Be able to do problems similar to those on the three tests. (3) Derive the formula v w = v w cos θ, where v and w are two vectors in R n. (4) Derive the trig identity cos (2A) = 2 cos 2 (A) 1. (You may assume the angle sum identity.) (5) Derive the Law of Sines. (6) Perform the following matrix multiplication, and explain how to use this to prove the formulas for cos (A + B) and sin (A + B): cos (A) sin (A) cos (B) sin (B) sin (A) cos (A) sin (B) cos (B) (7) Prove Thales Theorem: If A, B, and C are points on a circle where the line AC is a diameter of the circle, then the angle ABC is a right angle. (8) What are the types of isometries of the plane? Give an interesting example of each type. (9) State the theorem concerning collapsing and rigid compasses. (10) Prove that the angle bisectors of a triangle are concurrent. (11) Prove that the perpendicular bisectors of the sides of a triangle are concurrent... (a) with a proof that uses analytic geometry. (b) with a proof that does not use analytic geometry. (12) State Menelaus Theorem. (13) Suppose that three circles of different sizes in the plane are tangent to each other (at three points). Prove that the tangent lines to the circles at these three points are concurrent. (14) Identify all possible geodesics in the Euclidean plane, on a sphere of radius 1, and in the Poincaré Disk model of hyperbolic space. (15) What is the maximum length of a geodesic on a sphere of radius 1? 1

2 (16) Discuss differences in the geometry of parallel lines in Euclidean, hyperbolic, and spherical geometry. (17) Give formulas for the areas of triangles in Euclidean, hyperbolic, and spherical geometry. (18) Suppose that a triangle on the sphere of radius 1 has three right angles. (a) Draw a picture of such a triangle. (b) Find the area of the triangle. (c) Is this triangle equilateral? If so, what is the side length? (19) Suppose that a triangle in the Poincaré disk has three sides that are infinitely long. (a) Draw a picture of such a triangle. (b) Find the angles of this triangle. (c) Find the area of the triangle. (20) Derive the formula for the area of a lune in spherical geometry. (21) Suppose that a quadrilateral in the Poincaré Disk is bounded by four infinitely long geodesics. Find the area of this hyperbolic quadrilateral. (22) Prove that the area of a geodesic quadrilateral on a sphere of radius 1 with angles whose radian measures are A, B, C, and D is A + B + C + D 2π.

3 Solutions to Practice Problems (1) (2) n/a (3) n/a (4) Derive the formula v w = v w cos θ, where v and w are two vectors in R n. From class notes: Construct a triangle with two adjacent sides corresponding to the vectors v and w, with tails at one vertex. Let θ be the angle between the vectors. Then the third side corresponds to the vector v w. Then the Law of Cosines states that Then v w 2 = v 2 + w 2 2 v w cos θ. (v w) (v w) = v v + w w 2 v w cos θ, and by using the linearity and symmetry of the dot product we get (v w) (v w) = v v w v v w + w w = v v 2v w + w w. Substituting back into the previous equation, we get v v 2v w + w w = v v + w w 2 v w cos θ. Subtracting v v + w w from both sides, we get 2v w = 2 v w cos θ, and thus we have v w = v w cos θ. (5) Derive the trig identity cos (2A) = 2 cos 2 (A) 1. (You may assume the angle sum identity.) From class notes: Given the identity cos (A + B) = cos A cos B sin A sin B, we have cos (2A) = cos (A + A) = cos 2 (A) sin 2 (A). Since sin 2 (A) + cos 2 (A) = 1, we can substitute sin 2 (A) = 1 cos 2 (A) to get cos (2A) = cos 2 (A) ( 1 cos 2 (A) ) = 2 cos 2 (A) 1. (6) Derive the Law of Sines. From class notes: Start with a triangle with one side (length c) horizontal and adjacent angles A and B, corresponding to opposite side lengths a and b, respectively. Drop an altitude from the vertex opposite the horizontal side, which intersects the line containing that side. From this you get two right triangles, and you can solve for the height of the vertex in two different ways (in terms of the sine formula).

4 (7) Perform the following matrix multiplication, and explain how to use this to prove the formulas for cos (A + B) and sin (A + B): cos (A) sin (A) cos (B) sin (B) sin (A) cos (A) sin (B) cos (B) cos (A) cos (B) sin (A) sin (B) cos (A) sin (B) sin (A) cos (B) =. sin (A) cos (B) + cos (A) sin (B) sin (A) sin (B) + cos (A) cos (B) x This product of matrices, if applied to a vector, first rotates around the y origin by the angle B, then it rotates around the origin by the angle A. But this is equivalent to rotating around the origin by the angle (A + B). Therefore, the matrix we computed must be the same as ( cos (A + B) sin (A + B) sin (A + B) cos (A + B) From the computation above, this means that ). cos (A + B) = cos (A) cos (B) sin (A) sin (B), sin (A + B) = sin (A) cos (B) + cos (A) sin (B). (8) Prove Thales Theorem: If A, B, and C are points on a circle where the line AC is a diameter of the circle, then the angle ABC is a right angle. Proof: Let D be the center of the circle, which is the midpoint of the diameter AC. Construct the segment BD, and notice that AD = BD = CD since they are all radii of the circle. Thus, triangles ADB and CDB are isosceles, so that the angles satisfy m DAB = m DBA, m DCB = m DBC. Since the angles in the triangles add to 180 degrees, 2m DBA + m ADB = 180 = 2m DBC + m CDB. Also m ADB + m CDB = 180, because they add to a straight angle. Substituting 180 2m DBA = m ADB and 180 2m DBC = m CDB into the previous equation, we get so that Dividing by two, we get QED. 180 2m DBA + 180 2m DBC = 180, 2m DBC + 2m DBA = 180. m CBA = m DBC + m DBA = 90. (9) What are the types of isometries of the plane? Give an interesting example of each type. They are translations, rotations, reflections, glide reflections. Examples: (a) (b) Translation of all points by the vector ( 7, 2): F : R 2 R 2, defined by F (x, y) = (x 7, y + 2).

( x (c) Rotation by π around the origin: G : 3 R2 R 2, defined by G y ( ) ( 1 1 x 1y 3 2 2 3 2 2 3 1 2 2 ) ( x y = 1 y + 1x 3 2 2 (d) Reflection across the y-axis: R : R 2 R 2, defined by R (x, y) = ( x, y). (e) Glide reflection, reflecting across the y-axis, then sliding in the direction of the axis 11 units: Q : R 2 R 2, defined by Q (x, y) = ( x, y + 11). (10) State the theorem concerning collapsing and rigid compasses. Any construction done with a rigid compass and unmarked straightedge can also be done with a collapsing compass and unmarked straightedge alone. (11) Prove that the angle bisectors of a triangle are concurrent. Let ABC be any triangle, and let the point Z be the intersection of the angle bisectors from the vertices A and B. Let D be the point on AB that is the foot of the perpendicular segment from Z, and similarly let E be the foot of the perpendicular segment from Z to BC and F the foot of the perpendicular segment from Z to AC. Observe that the angles of AZD and AZF are the same, since AZ bisects DAF and one other angle is a right angle. Also, the triangles AZD and AZF share side AZ, so that by the ASA Congruence Theorem, AZD = AZF. Therefore, ZD = ZF. Using a similar argument at vertex B, ZD = ZE. Thus, ZE = ZF. Since CZE and CZF are right triangles and since the corresponding sides satisfy CZ = CZ, ZE = ZF, we also have that CE = (CZ) 2 (ZE) 2 = (CZ) 2 (ZF ) 2 = CF. Thus, by the SSS congruence theorem, CZE = CZF. Therefore, since ZCF = ZCE, CZ bisects BCA, and thus the angle bisectors of ABC are concurrent. (12) Prove that the perpendicular bisectors of the sides of a triangle are concurrent... (a) (b) with a proof that uses analytic geometry. Proof: We place an arbitrary triangle ABC so that it has vertices A = ( a, 0), B = (a, 0), C = (b, c), and note that a, c 0. Then the y-axis is the perpendicular bisector of AB. We now calculate the equation for the perpendicular bisector of AC. The slope of AC is (including the possibility of c b+a b + a = 0, where the slope is undefined), and so the slope of the perpendicular bisector is b+a (which is always defined). The midpoint of AC is ( b a, c c 2 2), so that the equation of the perpendicular bisector of AC is y c 2 = b + a c y c 2 = b + a c ). ( x b a ), or 2 x + b2 a 2, or 2c y = b + a x + b2 a 2 + c c 2c 2. ) 5 =

6 0, a2 +c 2 +b 2 2c 0, b2 a 2 + c = 2c 2 Thus, this perpendicular bisector of AC intersects the y-axis at the point 0, a2 +c 2 +b 2. By a similar calculation (interchanging a with a in all places), 2c ) the perpendicular bisector of BC intersects the y-axis at the point (0, ( a)2 +c 2 +b 2 = 2c. Thus, the perpendicular bisectors of ABC are concurrent. (c) with a proof that does not use analytic geometry. Proof: Given any triangle ABC, let D, E, F be the midpoints of AB, BC, and AC, respectively. Let Z be the intersection of the perpendicular bisectors of segments AB and BC. Construct the segments AZ, BZ, and CZ. Observe that since AD = DB, ZD = ZD, and since ADZ and BDZ are both right angles, by the SAS Congruence Theorem ADZ = BDZ. Then AZ = BZ. Similarly, by considering the perpendicular bisector EZ, we deduce that BZ = CZ. Thus, AZ = CZ. Now, since AZ = CZ, ZF = ZF, and AF = CF, by the SSS Congruence Theorem, AF Z = CF Z. Then AF Z = CF Z, and these two angles are supplementary, and thus they are two right angles. Thus, F Z is on the perpendicular bisector of AC, and hence the perpendicular bisectors of ABC are concurrent. (13) State Menelaus Theorem. Given a triangle ABC, fix a directional orientation for each of the segments AB, BC, and CA. Consider the transversal line XZ that crosses AB, BC and CA at points X, Y, and Z respectively. Let AX, XB, etc. denote the signed lengths of the line segments, where for example AX is considered positive if A is before X with respect to the fixed directional orientation of AB. If the points X, Y, Z are not vertices of the triangle, then (AX) (BY ) (CZ) (ZA) (XB) (Y C) = 1. (14) Suppose that three circles of different sizes in the plane are tangent to each other (at three points). Prove that the tangent lines to the circles at these three points are concurrent. (Note: the last part of this problem is a bit harder than anything I would ask you on the final.) Proof: Let A, B, and C be the centers of the three circles S A, S B, S C that are tangent, and construct the segments AB, BC, and AC. At the point D of tangency between the circles S A and S B, the tangent line is perpendicular to both AD and DB, and thus A, D, B are collinear. Thus, the point of tangency is on the segment AB. Similarly, define the points E and F on BC and AC be the corresponding points of tangency between the circles. Now, let Z be the point of intersection of the tangent lines through D and F, and construct the line AZ. Observe that AD = AF since they equal the radius of S A, and thus DZ = (AZ) 2 (AD) 2 = (AZ) 2 (AF ) 2 = F Z by the Pythagorean theorem. Thus the triangles ADZ and AF Z are congruent by the SSS Congruence Theorem. Then DAZ = F AZ, and so AZ actually bisects BAC. By a similar argument, the intersection of any two of these tangent lines must lie on an angle bisector of ABC.

The question remains whether the tangent lines are concurrent. Suppose not, and call Z 2 the point of intersection of the tangent lines through D and E, and call Z 3 the point of intersection of the tangents through E and F. Suppose that DZ < DZ 2, which then implies that F Z 3 < F Z. (If instead DZ 2 < DZ, the same argument works but with a relabeling of points.) We label a = DZ, b = EZ 2, c = F Z 3, α = ZZ 2, β = Z 2 Z 3,γ = Z 3 Z. Using DZ = F Z and the corresponding facts for the other intersection points, we get the equations from which we have that a = c + γ a + α = b b + β = c, a = b + β + γ = a + α + β + γ, so that α + β + γ = 0, which is impossible since by assumption α, β, γ > 0. Since the original assumption must be wrong, Z = Z 2 = Z 3, and the three tangent lines are concurrent. (15) Identify all possible geodesics in the Euclidean plane, on a sphere of radius 1, and in the Poincaré Disk model of hyperbolic space. Euclidean plane: the geodesics are lines, rays, and line segments. Sphere of radius 1: the geodesics are great circles (of radius 1) or parts of great circles. Poincaré Disk: the geodesics are either diameters of the unit circle or arcs of circles within the disk that are perpendicular to the boundary circle, or parts of these diameters and perpendicular circles. (16) What is the maximum length of a geodesic on a sphere of radius 1? The maximum length is the length of a great circle, which is 2π (circumference of circle of radius 1). (17) Discuss differences in the geometry of parallel lines in Euclidean, hyperbolic, and spherical geometry. Euclidean geometry: Given a line l in R 2 and a point p not on that line, there exists a unique line through the point p that is parallel to l. Hyperbolic geometry: Given a geodesic α in the Poincaré Disk and a point P not on α, there exist an infinite number of geodesics through P that do not meet α. Spherical geometry: Given a great circle β on the sphere and a point Q not on β, then every great circle through Q intersects β. (18) Give formulas for the areas of triangles in Euclidean, hyperbolic, and spherical geometry. Euclidean geometry: many formulas; for example A = 1 ab sin θ, where a and b are 2 the lengths of two of the sides of the triangle and θ is the angle between the sides. Hyperbolic geometry: Area = π A B C, where A, B, C are the three angle measures in radians. Spherical geometry: Area = A + B + C π, where A, B, C are the three angle measures in radians. 7

8 (19) Suppose that a triangle on the sphere of radius 1 has three right angles. (a) (b) Draw a picture of such a triangle. Should look like one fourth of the upper hemisphere of the sphere. (c) Find the area of the triangle. π 2 + π 2 + π 2 π = π 2. (Alternately, it is 1 8 of the whole sphere, which is 1 8 (4π) = π 2.) (d) Is this triangle equilateral? If so, what is the side length? Yes. Using the picture of the one fourth of the upper hemisphere, the side length is 1 4 of a great circle, or π 2. Note that in spherical geometry, the angles determine the triangle, so any such triangle must be equilateral and have that side length. (20) Suppose that a triangle in the Poincaré disk has three sides that are infinitely long. (a) (b) Draw a picture of such a triangle. Should look like http://en.wikipedia.org/wiki/file:ideal circles.gif (There are actually three different triangles like this in the picture, the red one, blue one, and green one.) (c) Find the angles of this triangle. All zero. (d) Find the area of the triangle. Area = π 0 0 0 = π. (21) Derive the formula for the area of a lune in spherical geometry. A lune with angle A is the portion of the sphere between two halves of great circles intersecting with an angle A. The area is A times the area of the whole sphere, 2π which is A (4π) = 2A. 2π (22) Suppose that a quadrilateral in the Poincaré Disk is bounded by four infinitely long geodesics. Find the area of this hyperbolic quadrilateral. The four geodesics must touch at vertices on the boundary circle. If the vertices are (in order) A, B, C, D, connect A and C with a geodesic, so that the quadrilateral is split into two triangles, both of which have all zero angles. Thus the area of the quadrilateral is Area = (π 0 0 0) + (π 0 0 0) = 2π. (23) Prove that the area of a geodesic quadrilateral on a sphere of radius 1 with angles whose radian measures are A, B, C, and D is A + B + C + D 2π. Given any quadrilateral on the sphere, connect opposite vertices with a geodesic so that the quadrilateral is the union of two triangles with no overlap that intersect on that geodesic. Then, if the radian measures of the angles of the first triangle are θ 1, θ 2, θ 3 and those of the second triangle are β 1, β 2, β 3, then the area of the

quadrilateral is Area = (θ 1 + θ 2 + θ 3 π) + (β 1 + β 2 + β 3 π) = θ 1 + θ 2 + θ 3 + β 1 + β 2 + β 3 2π. The sum of the radian measures of the angles of the two triangles is the same as the sum of the measures of the four angles of the quadrilateral, so that Area = A + B + C + D 2π. 9