Rotate the SquareDate: 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/ |
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