Math Is Fun Forum

Orion

Guest

Weird problem

Consider rotations of the plane. If you first rotate the whole plane by angle α around some point p, and then rotate it by angle β around the same point, clearly the combined result is a rotation by α + β around p. But what if the two rotations are about different points?
a) For the specific points and angles in the diagram below Part b), what is the combined result of rotating the plane counterclockwise around point p by angle α, followed by rotating it counterclockwise around point q by angle β? (We have deliberately not given any coordinates for the points or measures for the angles; the information is purely geometric. If you decide, for instance, that the result is a rotation, can you find the center of the rotation just from the geometric information given? One way might be with straightedge and compass constructions, but you are not limited to that. You could even measure all the lengths and angles and try to use algebra. If you use only geometry, copy what you need from our diagram onto your own paper; it is ok if it is not absolutely exact.)
b) Same as Part a) but this time rotate first counterclockwise around p (as before) and then rotate clockwise by β around q.

diagram of angles and points

Bob

Administrator

Registered: 2010-06-20

Posts: 10,136

Re: Weird problem

hi Orion

Welcome to the forum.

You've posted this as a 'help Me' but your wording makes it seem more like a challenge to which you know the answer already.  Here's what I know on this topic:

In 2D, if two shapes have all corresponding lengths equal we say they are congruent.  The transformation that will map one onto the other is called, in general, an isometry.  It can be shown that the set of isometries can be divided into four distinct sets as follows:

(1) Rotations.  These are categorised by having a centre of rotation and a single angle of rotation. (In what follows I will take a rotation in an anti clockwise sense as positive and clockwise as negative)

(2)  Translations.  If lines in the second shape are parallel to corresponding lines in the first, then a single vector will move points from the first to the second.

(3)  Reflections.  (1) and (2) preserve the sense of the shape.  That is to say, if you take a clockwise journey around the perimeter of the first, then a journey around the corresponding points in the second is also clockwise.  If the sense is not maintained then one possibility is a reflection of the first in a mirror line to produce the second.

(4) Glide reflection.  An isometry that does not preserve sense and is not a reflection, is a glide reflection.  The first shape is reflected in some line and then translated in the direction of the line.

No other isometries exist.  I could probably prove this if somebody would like to see this.

In the above problem a rotation is followed by another.  Since rotations preserve sense, a single isometry equivalent to this pair must be either another rotation or a translation.  If the first rotation has angle A, and the second B (both measured according to the positive/negative rule above) then the combination will be a translation when A + B is zero or a multiple of 360.  In all other cases, the result is a rotation with angle A + B.  I think I can prove these statements too.

What I cannot yet do is to find a simple rule for locating the centre of the resulting rotation.  But I'll have a go at that too, if someone asks nicely

Rotations and reflections have an invariant point(s).  Translations and glide reflections do not.

Meanwhile, I wish you all a happy Christmas, and hope for peace on Earth.

Bob

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Children are not defined by school ...........The Fonz
You cannot teach a man anything;  you can only help him find it within himself..........Galileo Galilei
Sometimes I deliberately make mistakes, just to test you!  …………….Bob

Bob

Administrator

Registered: 2010-06-20

Posts: 10,136

Re: Weird problem

OK.  proof that same sense isometries are either rotations or translations.

Suppose a line AB is transformed into an equal length line A'B'.

Join AA' and construct the perpendicular bisector DC.  Join BB' and construct the perpendicular bisector EC so that these cross at C.

In triangles ADC and A'DC, DC is common, AD = DA' and angle ADC = A'DC = 90.  So they are congruent and so AC = A'C.

So with C as centre A may be rotated onto A'.

Similarly B may be rotated onto B'

In triangles ABC and A'B'C, AB = A'B', AC = A'C, BC = B'C, so these are congruent.  So angle ACA' = BCB'.  So a single rotation will map A onto A' and B onto B'.

Now suppose the shape has a third point, F.  Rotate F by the same angle to make triangle A'B'F'.  Suppose the isometry that transformed A into A' and B into B' also transforms F into F''.  As ABF is congruent to A'B'F' and to A'B'F'', F' and F'' must coincide.  So this extends the proof to any 2D shape.

The second diagram covers the case where the perpendicular bisectors are parallel and so do not cross.

As AA' and BB' are at right angles to the parallels, they must also be parallel.  So a vector in that direction will map A onto A' and B onto B'.  But is it a vector of the same magnitude?

Mark F on BB' so that BF = AA'.  Then BFA'A is a parallelogram, and AB = A'F.  But AB = A'B', therefore F coincides with B'.

So a single vector maps A onto A' and B onto B'.

Bob

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Children are not defined by school ...........The Fonz
You cannot teach a man anything;  you can only help him find it within himself..........Galileo Galilei
Sometimes I deliberately make mistakes, just to test you!  …………….Bob