Discussion about math, puzzles, games and fun.   Useful symbols: ÷ × ½ √ ∞ ≠ ≤ ≥ ≈ ⇒ ± ∈ Δ θ ∴ ∑ ∫ • π ƒ -¹ ² ³ °

You are not logged in.

## Post a reply

Write your message and submit
|
Options

## Topic review (newest first)

anonimnystefy
2012-12-01 22:31:20

Well, it depends on which line you are rotating around first...

bob bundy
2012-12-01 22:15:09

hi Stefy,

That's what I remembered, but, in this example the situation is confused by clockwise and anticlockwise measuring, so I thought I'd avoid saying anything in case I was wrong.

Bob

anonimnystefy
2012-12-01 21:03:47

Hi Bob

The angle of rotation is twice the angle between the two lines.

bob bundy
2012-12-01 19:40:00

hi Sparky,

Welcome to the forum.

2) Is there a way to get the same rotation for the square using a composition of 2 reflections?  How would this work?

Yes.  See diagram.

How did I get those two lines of reflection?

Well firstly, the centre of rotation doesn't move when rotated.  It is called an invariant point.

So make sure it doesn't move when reflected by making the two lines cross there.

Because the angle of rotation is 90 I started with a line at 45.

(There is a rule for the relationship between the two angles of the reflections and the angle of rotation.  I'd have to experiment a bit to remind myself what it is.  Post again if you are interested.)

So I made a line at 45 though (3,2) and constructed the reflection.  It was then fairly obvious where the second line should go.

I've labelled the half way square B" and so on.

Bob

ps.  Could I have sloped that first 45 line the other way?  I think I could.  I'll leave that as an exercise for you.

scientia
2012-12-01 12:22:46

Suppose the centre of the rotation is (a,b). This rotation takes the point (2,−1) to (0,3). Then the same rotation about the origin would take the point (2−a,−1−b) to (−a,3−b). (And the other points correspondingly, but we only need one point to work out a and b.)

The matrix representing a clockwise rotation of 90° about the origin is

Hence

Solving the simultaneous equations gives
.

Sparky
2012-12-01 11:45:40

A square has coordinates (-3, 2), (1, 3), (2, -1), and (-2, -2).  It is rotated so that:
(-3, 2) goes to (3, 8)
(1, 3) goes to (4, 4)
(2, -1) goes to (0, 3)
(-2, -2) goes to (-1, 7)

I know that the angle of rotation here is 90 degrees because the line going through points (-3, 2) and (1,3) is perpendicular to the line going through points (3, 8) and (4, 4).  However, I have a couple of questions:

1) How do I find the center of rotation?  Is it simply the center of the original square, which is (-1/2, 1/2)?
2) Is there a way to get the same rotation for the square using a composition of 2 reflections?  How would this work?