Rotate the Square

Date: 09/19/2002 at 13:41:21
From: Gabrielle Lindblad
Subject: Geometry Construction


I have the following construction that I've been trying to figure out.

  
First I've connected A to P and A to Q. Then I've found the midpoint 
of each line segment and constructed the half circles (k and l) 
outside the square, with AP and AQ as their respective diameters.  
I know that B and D must lie on k and l respectively because they are 
guaranteed to have 90-degree angles, by Thales' theorem: if any 
line segment (RS) is the diameter of a circle, any point on the 
circle (T) makes a 90-degree angle when connected to the ends of the 
line segment (angle RTS = 90 degrees).  

The question is: Which points on the half-circles are B and D? 

My teacher mentioned something about transforming the circle l so that 
it intersects with k. Some of my classmates and I cannot figure out 
the transformations. Can you help?  Or do you know of another way to 
construct the square?

Date: 09/19/2002 at 17:31:07
From: Doctor Floor
Subject: Re: Geometry Construction

Hi, Gabrielle,

Thanks for your question.

Let me summarize the question: we have points APQ and must construct 
a square ABCD such that P is on BC and Q is on CD.

I have done this problem before, and I always use a very simple 
construction. It is based on the following:

1. If we rotate square ABCD together with point Q about A through 
   -90 degrees (clockwise) we will get a square attached to AB, 
   and the image of Q, say Q', will lie on the line through B and C. 

2. In the same way if we rotate ABCD together with point P about A 
   through 90 degrees (counterclockwise) we will get a square attached 
   to AD and the image P' of P will lie on the line through D and C.

So the construction is as follows:

   Rotate P about A through 90 degrees, to find P'.
   Rotate Q about A through -90 degrees, to find Q'.
   Line P'Q contains D and C.
   Line Q'P contains B and C.
   So the intersection of P'Q and Q'P gives C.

Having A and C, and the lines through C in the right directions, the 
rest should be easy.

If you have more questions, just write back.

Best regards,
- Doctor Floor, The Math Forum
  http://mathforum.org/dr.math/