Math 311 Test III, Spring 2013 (with solutions)


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1 Math 311 Test III, Spring 2013 (with solutions) Dr Holmes April 25, 2013 It is extremely likely that there are mistakes in the solutions given! Please call them to my attention if you find them. This exam will begin at 1030 am and end officially at 1145 am (at which time you will get a five minute warning to hand your papers in). There is an ID number on the first inside page of your paper by which your grade will be posted. You are allowed your test paper, drawing instruments, and your writing instrument. Cell phones must be turned off and inaccessible to you. One of the nine questions (the one on which you do worst or which you choose to omit) will be dropped. There may be other adjustments for length as seems appropriate. For the last two proof questions, there are alternative proof questions in brackets that you may do instead if you prefer. 1
2 1. Triangle congruence and notation (a) Write down the six congruence facts which make up the statement ABC = DEF. AB = DE BC = EF AC = DF BAC = EDF ABC = DEF BCA = EF D (b) List three of these which imply the full congruence by the SAS axiom. for example. AB = DE BC = EF ABC = DEF three possible answers (c) Explain why ABC = DEF is a different statement from ABC = DF E. For example the first statement requires AB to be congruent to DE and the second requires AB to be congruent to DF. Similar statements about angles work. I did not give credit for statements about points corresponding to each other: these do not express any information about the geometry. 2
3 (d) Describe a situation in which both of the two congruence statements in the previous part would be true. Both would hold if ABC = DEF and in addition DF were congruent to DE (and also AB congruent to BC) so both triangles are isosceles. Saying the triangles could be equilateral works. Saying that the triangles have to be equilateral for this to be true earns a deduction, as they do not have to be. 3
4 2. Describe a counterexample to the ASA Theorem in taxicab geometry, with angle measure defined as in the usual geometry of the plane (hint: use triangles with different orientations. Give all relevant details and calculations). The typically given counterexample involves two triangles differently oriented. For example A = (0, 0), B = (1, 1), C = (2, 0); D = (0, 0), E = (0, 1), F = (1, 1). The information you should specify is that both BAC and EDF have measure 45, and so do BCA and DF E, and both AC and DF have measure 2. This gives the conditions for ASA, but the triangles are not congruent (as for example AB has length 2 and DE has length 1. You lose a lot of points for giving an SAS counterexample, and a lot of points for using the square metric instead of the taxicab metric. 4
5 3. We give an interpretation of part of our geometrical language on a circle. A point is a point on the circumference of the circle. For two points A and B, d(a, B) is the length of the shorter of the two arcs between A and B (or of either of them if they are the same length). Define A B C as d(a, B) + d(b, C) = d(a, C). There is only one line in this interpretation, containing all the points on the circle. For each of the points A, B, A (at the opposite end of a diameter of the circle from A), X, Y, and Z, determine using the usual definition of ray with the new definition of betweenness whether it is on the ray AB. Explain your answer briefly for each one. You may judge betweenness facts visually. Sketch the ray AB on the diagram given. Lots of you know the facts (the ray is the semicircle from A to A inclusive including B and including its endpoints). Few of you gave the reasons, which were a huge part of the value. A and B are of course on ray AB by definition. X is on the ray because A X B. Y is on the ray because A B Y (nothing to do with A, which has no role in the definition of ray AB). A is on the ray because A B A. Z is not on the ray because neither A Z B nor A B Z is true. (nothing to do with halfplanes, which have no role in the definition of rays). 5 points for the bare facts about what the ray is; one more for many of you who did write A X B or say X was between A and B. More points if you gave more of the above justifications. 5
6 4. Prove that if line L is perpendicular to line M, then all four angles formed by a ray from each line are right angles (hint: this is all about the Linear Pair Theorem). If L is perpendicular to M, then (A being their point of intersection) there are points B on L and C on M such that BAC is right. Choose B such that B A B. Angles BAC and B AC form a linear pair, so their measures add to 180 degrees, so the measure of B AC is also 90 degrees. Choose C so that C A C. Angles BAC and BAC form a linear pair, so their measures add to 180 degrees, so the measure of BAC is also 90 degrees. Similarly angles B AC and B AC form a linear pair, so their measures add to 180 degrees, so the measure of B AC is also 90 degrees. You lose two points for starting by saying something like Let L and M be perpendicular lines intersecting at A; let B be on L and C be on M; by definition of perpendicular lines the measure of BAC is 90 degrees : in fact, this is the result we are trying to prove. All you know is that one of the four angles is 90 degrees, and you have to start by specifying one of them as the 90 degree angle provided by the definition (without implying that any such angle is 90 degrees by definition, which is not in the definition). 6
7 5. Exterior angle theorem (a) Draw a triangle. Draw one of its exterior angles (and name it, which requires labelling of appropriate points). State what the remote interior angles for this exterior angle are. Not set up to draw here. (b) Prove that if one of the angles of a triangle is obtuse, the other two angles are acute. Suppose that ABC is a triangle. Suppose without loss of generality that ABC is obtuse (has measure greater than 90). Choose D so that B C D. Angles DBC and ABC form a linear pair, so their measures add to 180, so the measure of DBC is less than 90 degrees. Angles BAC and BCA have measure less than the measure of DBC by the exterior angle theorem, so they have measure less than 90, so they are acute. 7
8 6. Prove that if P is a point not on line L, and M is a line incident on P and perpendicular to L, then M is the only line incident on P and perpendicular to L. (uniqueness of dropped perpendiculars). You are not being asked to construct a perpendicular. Start by supposing that there is another one. If you use the diagram that I provide you need to label it appropriately and possibly justify some of the facts implied by the way it is drawn. You needed to say everything in words that was indicated on the diagram. Let Q be the point at which M intersects L. Let N be another line through P perpendicular to L (for the sake of a contradiction); let R be the intersection of N and L. Choose X so that Q R X. Then angle P RX is external to triangle P QR and so greater in measure than angle P QX by the exterior angle theorem. But both P RX and P QX have measure 90 by facts about perpendicular lines proved in another problem on this test, which is a contradiction. 8
9 7. Prove the existence of angle bisectors: for each angle BAC there is a ray AD which is an angle bisector of BAC. You are not being asked to prove uniqueness. You do need to prove that it has all the defining properties of an angle bisector: just constructing the right ray and saying that it is an angle bisector will not carry that much credit. You may use either method of proof. Since there are two methods that you have seen, I give both proofs. There is an error in the statement of the problem, caught by just one student: it is necessary to provide that the angle is not a zero angle. First proof: Let ABC be an angle which is not a zero angle. ABC has measure r > 0. There is a unique ray BD such that D is on the same side of line BC as A and DBC has measure r. The ray 2 BD is between BA and BC by the betweenness theorem for rays (not by the Protractor Postulate). Thus the sum of the measures of DBC and DBA is r, the measure of ABC, so the measure of DBA is also r. We have shown that DBA and DBC are 2 congruent and BD is between BA and BC, that is, that BD is an angle bisector for ABC. Second proof: Let ABC be an angle which is not a zero angle. Choose a point A on BA with BA = BC (point construction postulate). Let D be the midpoint of A C. Note that A BC is isosceles, so BCA = BA C (Isosceles Triangle Theorem); further A D = DC (since D is chosen as the midpoint of A C), and A B = CB by choice of A : this is the evidence for A BD = CBD (SAS). Now by this congruence A BD = ABD is congruent to CBD. Further, since A D C we have BD between BA = BA and BC (3.3.10), so BD is an angle bisector for ABC. It is really important to show that the bisector is between the two rays that make up the original angle as well as showing that it makes angles of the same measure with the two sides of the original angle. 9
10 8. Carefully state and prove the AngleSideAngle Theorem. [You may state and prove AngleAngleSide if you prefer.] A number of students proved AAS. There is a proof of AAS on the class announcements page. You cannot prove AAS without invoking the exterior angle theorem (no, the uniqueness part of the Angle Construction Postulate does not work), so proofs of AAS with the same gap that I call attention to in the proof of ASA are even more deficient than proofs of ASA with the same gap. Theorem (ASA): Let ABC and DEF be triangles with BC = EF, ABC = DEF and ACB = DF E: it follows that ABC = DEF. Proof of Theorem: Let A be chosen on BA such that BA = ED. It follows that A BC = DEF by SAS. Thus A CB = DF E. But also ACB is congruent to DF E by our initial assumptions, so ACB = A CB. A and A are on the same side of line BC by the ray theorem. By the uniqueness portion of the Angle Construction Postulate, it follows that µ( ACB) = µ( A CB) forces CA and CA to be the same ray. This is the main move of the proof. This does not give you A = A in one step trying to make this leap loses you a lot of points. Giving me an equation between segments instead of rays also loses you a lot of points, for the same reason. Now this shows that A and A are both on line CA as well as both being on line BA by construction, so they are the same point by the incidence axiom. Since A = A and we have already shown A BC = DEF, we get ABC = DEF by substitution of equals for equals. 10
11 9. Prove the Scalene Inequality in one direction: if ABC is a triangle and d(a, B) > d(b, C) then µ( ACB) > µ( (BAC). [You may prove existence of a perpendicular to a line L through a point P not on L if you prefer.] Let ABC be a triangle. Suppose d(a, B) > d(b, C). Choose D on BA such that BD = BC (point construction postulate). B D A by , so CD is between CA and CB by , so µ( BCD) + µ( DCA) = µ( ACB) by the Angle Addition Postulate, so BCD is smaller in measure than ACB. BCD is isosceles, so BCD has the same measure as BDC by the isosceles triangle theorem. Angle BDC is exterior to ADC so greater in measure than DAC = BAC by the exterior angle theorem. So we have shown µ( ACB) > µ( BCD) = µ( BDC) > µ( DAC) = µ( BAC) which completes the proof. No one came close to proving the bracketed theorem; there is a proof in the book. 11
12 1 Selected axioms, theorems and definitions: There is no intention here that every axiom theorem or definition will be given, or even every one that will be useful to you will be given. If you think I have omitted something that you think ought to be here, you may ask me during the exam, but I am most likely to tell you that I left it out on purpose. The form in which I state an axiom theorem or definition will not necessarily be literally identical to what is in the book: the meaning should be the same! taxicab distance: d((a, b), (c, d)) = a c + b d definition of a ray: Omitted on purpose. You should know it or be able to figure it out. It is defined in terms of betweenness : If B lies on AC then A B C iff d(a, B) < d(a, C). Point Construction Postulate: For any segment AB and ray CD, there is a unique point E on the ray CD such that AB = CE. definition of a linear pair: Angles ABD and DBC make up a linear pair iff A B C. Linear Pair Theorem: The sums of the measures of the two angles in a linear pair is exactly 180 degrees. interior of an angle, betweenness of rays: A point D is in the interior of an angle BAC iff D is on the same side of line AB as C and D is on the same side of line AC as B. A ray AD is between AB and AC iff D is in the interior of BAC : If BAC is an angle and D is on line BC, AD is between AB and AC iff B D C. Betweenness Theorem for Rays: Let A, B, C, D be distinct, with C, D on the same side of line AB. Then µ( BAD) < µ( BAC) iff AD is between AB and AC. Crossbar Theorem: If BAC is an angle and AD is between AB and AC then AD intersects BC at some point G. Protractor Postulate: There is a function µ from the set of all angles to the set of all real numbers with the following properties: 12
13 1. 0 µ( BAC) < 180 for any angle BAC. 2. µ( BAC) = 0 iff AB = AC. 3. Angle Construction Postulate: for each pair of distinct points A, B and each half plane H determined by line AB and each real number r such that 0 r < 180, there is a unique point C such that C is in H and µ( BAC) = r. 4. Angle Addition Postulate: If AD is between AB and AC then µ( BAD) + µ( DAC) = µ( BAC). 13
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