Solutions and scoring for 1st Midterm Exam Geometry 515 Fall Oct 2 Mon 4:00-5:20 pm

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1 Solutions and scoring for 1st Midterm Exam Geometry 515 Fall 2017 Oct 2 Mon 4:00-5:20 pm 1. On the line AB you have the point O between the points A and B (i.e. rays OA and OB form a straight angle AOB). Two other rays OC and OD with the same starting point are given in the same half-plane determined by the line AB such that the angles AOD and COB are congruent and obtuse. Prove that the bisector of the angle COD is perpendicular to the line AB. (8 points) As both rays OC and OD are on the same side of AB and the angles AOD and COB are congruent obtuse, they add up to more than a straight angle: angles AOD and COB overlap by the angle COD. Therefore, AOC = AOD COD = COB COD = DOB. Set α = AOC = DOB and β = COD. Because these angles add up to the straight angle AOB, we have 180 = α + β + α = 2α + β. The bisector of angle COD divides it into two congruent angles of measure β/2; therefore the bisector encloses with both rays OA and OB an angle of measure β/2 + α = 90, so it is perpendicular to AB. 1

2 2. Five given points on the plane are vertices of eight different pentagons. How can this happen? (Draw five points on the plane such that they can be vertices of eight different pentagons, label them, and on eight pictures, draw and label the eight different pentagons with the same fixed configuration of vertices.) (10 points) Vertices of a convex pentagon give rise to a unique pentagon. Vertices of a convex quadrilateral plus a fifth point in its interior give rise to four different pentagons only. These considerations are not necessary to get full credit. 3 points for drawing four pentagons with four points as convex hull - although it has nothing to do with the solution of this problem. Take a triangle ABC and two additional points D and E in its interior. To unify the upcoming discussion, let us assume that line DE intersects sides AC and BC of the triangle ABC. Also, we could assume that all pairwise distances between these five points are different, so all the pentagons on these vertices are pairwise incongruent. This discussion is not needed for full credit. This configuration of points gives rise to eight pentagons ABECD, ADBEC, AEBCD, ADEBC, ABCDE, ABCED, ABDEC, ABEDC. each 2

3 3. Two points A and B are given in the interior of a right angle XOY in a position such that both angles XOA and BOY are strictly less than 45. Find a point P on the side OX and a point Q on the side OY such that the broken line AP QB is the shortest possible. Does your construction always work in the given setting? (10 points) Reflect point A about OX to get A and reflect point B about OY to get B. Set P to be the intersection point of the line A B and the ray OX and set Q to be the intersection point of the line A B and the ray OY. Notice that the two reflections copy angles XOA and BOY outside the right angle XOY, so under the given conditions on the position of A and B, the angle A OB will be strictly less than = 180, that is, less than a straight angle. In this way, the line A B will always intersect the sides of the right angle XOY, so our construction always works under the given conditions. The length of the broken line AP QB is AP + P Q + QB, and notice that by the reflections, AP = A P and BQ = B Q (P and Q by their definition are on the axes of the corresponding reflections so they remain fixed, and reflections preserve distances). So the length of this broken line is A P + P Q + QB = A B (by the definition of the points P and Q), that is, the length of the segment A B. If we take any point P on the ray OX and any point Q on the ray OY, by a similar argument, AP = A P and BQ = B Q, so AP Q B = AP + P Q + Q B = A P + P Q + Q B A B, since every broken line connecting two points is at least as long as the straight segment between them. So the above found broken line is indeed the shortest possible. It is also unique, as equality in this inequality can happen if and only if both P and Q are on the line A B, in which case P = P and Q = Q. 3

4 4. (a) Consider an acute triangle. You can use the fact without proving it that in any acute triangle, the foot of every altitude segment is an internal point of the corresponding side (the foot of every altitude segment is between the vertices on the corresponding side). Arrange the following quantities into non-decreasing order: the sum of altitude segments of the triangle, the sum of internal angle bisector segments of the triangle, the semiperimeter of the triangle, three times the semiperimeter of the triangle. (10 points) (b) Can any of the above proved inequalities be an equality? For what kind of triangles? (2 points) Let the triangle be ABC. Let AA a, BB a and CC a be the altitude segments; let AA b, BB b and CC b be the bisector segments. Write a triangle inequality for triangle CC b A to obtain CC b < AC + AC b and for the triangle CC b B to obtain CC b < BC + BC b. Adding these two together, we get 2CC b < AC + BC + AC b + BC b = AC + BC + AB, for the last equality notice that C b is between A and B (CC b is an internal angle bisector). Similarly obtaining the corresponding inequalities for the other two bisectors and adding the three inequalities together, we get 2(AA b + BB b + CC b ) < 3(AB + BC + CA), that is, the sum of bisectors is less than three times the semiperimeter. Notice that CC a is a perpendicular and CC b is a slant from the point C to the line AB, therefore the former is never greater than the latter: CC a CC b. Writing the corresponding inequalities for the other two altitudes and bisectors and adding them together, we get that AA a + BB a + CC a AA b + BB b + CC b, that is, the sum of altitudes is at most the sum of bisectors. 4

5 Write a triangle inequality for CC a A to obtain AC < AC a + CC a and for the triangle CC a B to obtain BC < BC a + CC a. Adding these two together, we get AC + BC < 2CC a + AC a + BC a = 2CC a + AB, for the last equality notice that C a is between A and B (see the assumption). Similarly obtaining the corresponding inequalities for the other two altitudes and adding the three inequalities together, we get 2(AB + BC + AC) < 2(AA a + BB a + CC a ) + (AB + BC + CA), that is, the sum of altitudes is greater than the semiperimeter: (AB + BC + AC) < 2(AA a + BB a + CC a ). So the answer is: semiperimeter, sum of altitudes, sum of bisectors, three times the semiperimeter. Notice that we used triangle inequalities for actual triangles, so the first and the last inequality is always strict. The middle one can be an equality if and only if every slant considered happens to be a perpendicular, so every bisector is an altitude. But then the triangle has to be isosceles from three different viewpoints: the middle inequality is an equality if and only if the triangle is equilateral. 5

6 5. In a triangle, an angle bisector is a median. Prove that the triangle is isosceles. (10 points) Solution 1: Let the triangle be ABC, with CM being the bisector being the median (that is, assume CM bisects the angle BCA and M is the midpoint of AB: MA = MB). Reflect the figure with respect to the line CM. C and M remain fixed; denote the image of A by A and the image of B by B. Because CM is a bisector, A is on CB and B is on CA, and because reflection preserves distances, CA = CA and CB = CB. So if we assume indirectly that CA CB, then A B and B A. To simplify the discussion, we can assume without loss of generality that AC < BC. Notice that MA = MB = MA = MB (M is the midpoint of AB and reflection keeps distances). Therefore triangles MAB and MA B are both isosceles with bases AB and A B, respectively. Isosceles triangles have congruent angles at their bases, so MAB = AB M and BA M = MBA. But triangles MAB and MA B are not just both isosceles, they are in fact congruent, because they are symmetric with respect to line CM (or, you can use side-angle-side with the vertical angles at M). So in fact MAB = AB M = BA M = MBA. But MAB is an exterior angle of the original triangle ABC at A, while MBA is an internal angle of the same triangle ABC at B, so the former should be strictly larger than the latter; a contradiction. Our indirect assumption that CA CB has to be false: the triangle ABC is isosceles with base AB (therefore A = A, B = B, the triangles MAB and MA B are in fact degenerate, our previous argument fails and so there is no contradiction in this case). 6

7 Solution 2: Let the triangle be ABC, with CM being the bisector being the median (that is, assume CM bisects the angle BCA and M is the midpoint of AB: MA = MB). Consider the ray CM and mark a point C past M such that CM = C M. Notice that triangle AMC is congruent to triangle BMC by side-angle-side: AM = MB because M is the midpoint, the two triangles have congruent (vertical) angles at M and CM = C M by the choice of C. So MCA = MC B and AC = BC. Similarly, triangle MBC is congruent to triangle MAC by side-angle-side. So BCM = AC M and BC = AC. Because CM is also an angle bisector, MCA = BCM. So eventually we obtain s s MCA = MC B = BCM = AC M. Take triangle C CA, by the previous identities, it has congruent angles on side CC, therefore the opposite sides are congruent: AC = AC. By the congruence of the above mentioned triangles, this means eventually s AC = AC = BC, that is, triangle ABC is isosceles. s 7

Mathematics 2260H Geometry I: Euclidean geometry Trent University, Winter 2012 Quiz Solutions

Mathematics 2260H Geometry I: Euclidean geometry Trent University, Winter 2012 Quiz Solutions Quiz #1. Tuesday, 17 January, 2012. [10 minutes] 1. Given a line segment AB, use (some of) Postulates I V,