Math 444 Quiz 2

  • Part A: Answer all 4 questions in Part A.
  • Part B: Answer ONE of 3 questions 5A, 5B, or 5C.  Notice 5A and 5B count more points than 5C, so one of them is a better choice if the answer is correct.

Part A

Problem 1 (20 points) Composition of functions

In each case, tell what kind of isometry the product could be as precisely as you can from the given information.  Write ALL possible kinds but do not include kinds that you can exclude with the given information.  The answers should be taken from the list of names of the types of isometries of the plane.

 

Answers below are in blue.  (Additional explanatory comments, not expected in student answers, are in parentheses in red.)

 

a)      If S is a glide reflection and T is a rotation, what kind of isometry is ST?

 

ST is either a glide reflection or a line reflection.

 

b)      If U is a point reflection and V is a translation, what kind of isometry is UV?

 

UV is a point reflection. 

(The answer "rotation" will be counted partly right, since it is correct but not as precise as possible.)

 

c)      If E and F are glide reflections, what kind of isometry is EF?

 

EF is a rotation or translation.

(This includes the identity, but you can list identity separately also.  EF is a rotation if the invariant lines of the glide reflections are not parallel and is a translation if they are parallel.)

 

d)      If A is a rotation and B is a line reflection, what kind of isometry is AB?

 

Glide reflection or line reflection. 

(Glide reflection if A is not on the mirror line of B; line reflection if A is on the mirror line of B.)

 

Problem 2 (10 points) Write the equations of two lines u and v so that RvRu maps the shape on the left to the shape on the right. (Notation: Ru means reflection in line u.)

 

The isometry that that moves the shapes is translation by 8 units horizontally to the right.

One answer: Equation for u is x = 0. Equation for v is x = 4.

Other Answers: For any choice of constant c there is a correct answer with u equation being x = c and v equation being x = c+4.

 

 

Problem 3 (25 points) Image of an angle

 

a)      Write the definition of isometry.

 

An isometry is a transformation that preserves distance.

 

b)      Let T be an isometry. Suppose A, B, C be non-collinear points, and A' = T(A), B' = T(B), C' = T(C).  Prove that the angles ABC and A'B'C' are congruent.

 

Note:  You can use the definition of isometry and facts about figures in the plane, but no other theorems about isometries without proving them.

 

Since T is an isometry, by definition the distances AB = A'B', BC = B'C', CA = C'A'.  Therefore, by SSS, triangle ABC is congruent to triangle A'B'C'.  By corresponding parts, then angle ABC is congruent to angle A'B'C'.

 

Problem 4 (20 points) Constructing defining data of an isometry

The two triangles in the figure are congruent.

 

Label the vertices of the second triangle A'B'C' corresponding to the vertices of the congruent triangle ABC (you can figure out the congruent angles in this example by inspection).

 

a)      Tell what kind of isometry T will map triangle ABC to the other triangle.

 

T is a rotation with center D and angle BDB'.

 

b)      Construct with straightedge and compass the geometrical defining data for this isometry T. 

 

D is the intersection of any two of the perpendicular bisectors of AA', BB' and CC'.  These perpendicular bisectors are concurrent, so this confirms that the isometry is a rotation (in addition to the visualization that one can do informally).

 

Then any of these angles is the angle of rotation: angles ADA', BDB', CDC'.

 

c)      Make clear from labeling or comments what your method is and what the data are, but you do not have to justify your method.

 

Defining Data: D and angle BDB' as described above. (Angle BDB' is shown in yellow.  It is actually 130 degrees, but there was no way of finding this numerical measure on the quiz.)

 

 

Part B.

Answer ONE of Problems 5A (25 points) or 5B (25 points) or 5C (15 points).  You can only count one.  Cross out the others if you work on more than one

 

Problem 5A (25 points) Composition of rotations

Given the points A and B below; let S be rotation with center A by 60 degrees and let T be rotation with center B by 180 degrees. 

 

a)      Tell what is the isometry U = ST.  Be as explicit as you can.

 

U is a rotation by 240 degrees with center C constructed below.

 

b)      Construct the geometric data defining U.

 

Since 180/2 = 90 and 60/2 = 30, one must construct lines making angles of 90 degrees at B and 30 degrees at A.  Then C is the intersection of these lines and U is EF, where F is reflection in line BC and E is reflection in line AC (see Brown, p. 69).  The angle between the lines is 120 degrees, so the rotation is by angle 240 degrees.  (This is illustrated in the figure by the sector from D to D''.  It was not something that was necessary to answer the problem on the quiz.)

 

The angle of 30 degrees can be constructed by bisecting an angle of 60 degrees, which is constructed by constructing an equilateral triangle (BAE in the figure below).

 

 

 

 

 

 

 

Problem 5B (25 points) Let M1, M2, M3 be line reflections in the lines m1, m2, m3 below.  Let N = M1 M2 M3.

 

a)      Tell what isometry is N.  Be as explicit as you can.

 

N is a glide reflection.  The invariant (special) line of N is the line through the feet of the altitudes through A1 and A3.

 

b)      Construct the geometric data defining N.

 

Here are 3 solutions.  The first two solutions are based on the idea of mirror adjustment as in Problem 19, page 54.  This is a proof that reflection in 3 lines, not concurrent or parallel, is always a glide reflection.

 

The third solution is short because it assumes this theorem and so begins with the knowledge that N is a glide reflection.

Solution 1

Construct the altitude through A1 and call this line m2'.  Let C be the foot of this altitude.

 

Then construct m3' so that M2'M3' = M2M3 (denoting reflection in m2' and m3' by M2' and M3').  In effect this means that one applies mirror adjustment and rotates line m3 by angle A2A1C so that the angle from m2' to m3' is the same as from m2 to m3.

 

Then N = M1(M2 M3) = M1(M2'M3') = (M1M2')M3' = C180 M3'.

 

This uses the observation that double reflection in the perpendicular lines (M1M2') is a rotation by 180 degrees with center at C, i.e., a point symmetry with center C.

 

But in an assignment it was shown that a line reflection followed by a point symmetry is a glide reflection with special line g = line through C perpendicular to m3'.

 

Let C' be the reflection of C in m3'.  Then N(C') = C180 M3'(C') = C180 (C) = C.  So the glide vector of N is C'C. .

 

So N is a glide reflection with invariant line CC' and glide vector C'C. Thus N = GC'C in our usual glide reflection notation.

 

 

Solution 2

This is very much like Solution 1 except that one constructs the altitude m2" through A3 and then m1" through A3 so that M1"M2" = M1M2 (M1" and M2" being the reflections in lines m1" and m2").

 

Let D be the foot of the altitude so that the perpendicular lines m3 and m2" intersect at D.

 

Then as before N = (M1M2) M3 = (M1"M2") M3 = M1"(M2"M3) = M1"D180.

 

This is a glide reflection with invariant (special) line g through D perpendicular to m1".

 

If D' is defined as M1"(D), then N(D) = M1"D180 (D) = M1" (D) = D'. 

 

So N is a glide reflection with invariant line DD' and glide vector DD'. Thus N = GDD' in our usual glide reflection notation.

 

Solution 3

This solution is based on already knowing that N is a glide reflection and then solving for the invariant line by finding images of two points and then constructing the midpoints.

 

Let A1' be the reflection of A1 in line m1 and let A3' be the reflection of A3 in line m3.  Then since A1 is on both lines m3 and m2,

N(A1) = M1(M2(M3(A1))) = M1(M2(A1)) = M1(A1) = A1'.

 

Similarly N(A3') = M1(M2(M3(A3'))) = M1(M2(A3)) = M1(A3) = A3.

 

Let C be the midpoint of A1A1' and let D be the midpoint of A3A3'. 

 

Then the invariant (special) line g of N passes through these midpoints.  Notice that as before, C and D are the feel of two altitudes, so this is the same line as before (as it must be).

 

A glide vector of N can be found by constructing perpendicular lines to g through A1 and its N image A1'.  If the feet of these perpendiculars are E and F, then EF is a glide vector for N.

 

So N is a glide reflection with invariant line DC and a glide vector of N is EF. Thus N = GEF in our usual glide reflection notation.

 

 

Problem 5C (15 points) Let M1, M2, M3 be line reflections in the lines m1, m2, m3 below.  Let N = M1 M2 M3.

 

a)      Tell what isometry is N.  Be as explicit as you can.

 

Since N = M1 M2 M3 for concurrent lines, then N is a line reflection.

 

Since N = M1 M2 M3, then N M3 = M1 M2.  (It is also true that M1 N = M2 M3. There is also a solution based on this equation.)

 

 

b)      Construct the geometric data defining N.

 

Label as O the point of concurrence of the lines.  Drawing points A1, A2, A3 on the lines in the figure, construct a point A4 on N by copying angles so that angle A2OA1 = A3OA4.