Originally Posted by woollybull
I can see that by looking at a sketch that the co-ordinates of A will increase as C decreases (rotating clockwise).
It depends how the sketch is drawn, you can't assume that.
A could be (0,q)
C could be (p,0)
The origin O and a pt B (p,q) are the opposite vertices of the square OABC. Find the co-ordinates of the pts A and C.
I can see that by looking at a sketch that the co-ordinates of A will increase as C decreases (rotating clockwise). So then (p+q)/2, (p-q)/2 will be the co-ordinates of A and reversing x and y will give the co-ordinates of C?
I tried other ways to do this but the algebra got very messy.
Thanks for any ideas.
Originally Posted by woollybull
I can see that by looking at a sketch that the co-ordinates of A will increase as C decreases (rotating clockwise).
It depends how the sketch is drawn, you can't assume that.
A could be (0,q)
C could be (p,0)
This assertion is unfounded..
Plus, p and q are supposed to be fixed, so it is nonsense to say that the coordinates of A "will increase" ... ?
- let M be the midpoint of OB. Find its coordinates.
- *property of a square* MA=MC=MO=MB ---> MA˛=MC˛=MO˛=MB˛. You only need MO˛ or MB˛, both will be unuseful. This gives a first set of equations with A and C's coordinates.
- *property of a square* + Pythagoras theorem : OA˛=OM˛+MA˛=2OM˛ and similarly OC˛=2OM˛. These two equations are actually identical, but they contain 2 solutions.
You have enough equations involving the coordinates of A and C to be able to finish. If not (because I didn't count them), it is easy to find some more, using the properties of a square :
Attributes
Vertex The vertex (plural: vertices) is a corner of the square. Every square has four vertices.
Perimeter The distance around the square. All four sides are by definition the same length, so the perimeter is four times the length of one side, or:
perimeter = 4s
where s is the length of one side. See also Perimeter of a square.
Area Like most quadrilaterals, the area is the length of one side times the perpendicular height. So in a square this is simply:
area = s2
where s is the length of one side. See also Area of a square.
Diagonals Each diagonal of a square is the perpendicular bisector of the other. That is, each cuts the other into two equal parts, and they cross and right angles (90°).
The length of each diagonal is
s√2
where s is the length of any one side.
Thanks for responding.
I drew several sketches so I could see what was happening. As I rotated the square clockwise from a diamond position, the co-ordinates of A moved in the positive x and y, while C moved in the negative x and y.
I found the midpt, the equation of the line joining A and C as well as the length of the diagonal and sides of the square. None of this seemed to help me as the algebra became far too messy for what I assume is to be an easy problem. I worked out that the side of the square had length sqr-root((p^2+q^2)/2). The co-ordinates of C would have to satisfy this. (1/2(p+q, 1/2(p-q)) does just that for all positions that I sketched for the square (rotated from the starting position). However, I still can't get to this result without this deduction.