A T-square allows you to construct a straight line through two points and a line perpendicular to a given line through a given point. Circles C and C' intersect at X and Y. XY is a diameter of C. P is a point on C' inside C. Using only a T-square, find points Q,R on C such that QR is perpendicular to XY and PQ is perpendicular to PR.
Solution
C' seems slightly odd. Why not just say that XY is a diameter of C, P is a point inside C and we want points Q, R on C such that QR is perpendicular etc.? So presumably it is there to help.
Note that PQ perpendicular to PR means that P lies on the circle diameter QR. So take M to be the midpoint of QR. Then MQ = MR = MP. Guided by the hint about C', we extend PM to meet C' at S. Then QM·MR = XM·MY (circle C) = PM·MS (circle C'). So M is also the midpoint of PS. So we want to find S on C' so that XY bisects SP.
Let the line XP meet C at U. Draw the line through U perpendicular to XY to meet C again at V. Draw XV. Draw the line through P perpendicular to XY to meet XV at T. Then XY bisects PT, which is parallel to QR. Now draw the line through T perpendicular to PT (and hence parallel to XY) to meet C' at S. XY must also bisect ST. So we have the required point M. Draw the line through M perpendicular to XY to meet the circle C at Q and R.
© John Scholes
jscholes@kalva.demon.co.uk
26 Aug 2002