Constant perimeter triangle formed by tangents to circle

Theorem: If two tangents are drawn to a circle from a point A outside the circle and these tangents touch the circle at points B and C respectively, and a third tangent intersects AB in P and AC in R, and touches the circle at Q, then the perimeter of triangle APR = 2AB.

Investigate: Drag point Q to move tangent PR, and the centre O to change the size of the circle. You can also change the length of AB by dragging either of X or Y (since XY was used to determine the length of AB).

Please install Java (version 1.4 or later) to use JavaSketchpad applets.Constant perimeter triangle formed by tangents to circle

This theorem was used as Problem no. 13 in the 2nd Round of the 2013 Senior South African Mathematics Olympiad (SAMO)1. Can you explain why (prove that) the theorem is true?

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A straight forward, direct application of the theorem is demonstrated in the figure below, which shows two triangles ABC and KLM overlapping, and circumscribed to the same circle. If either one, or both of these two triangles are rotated around the circle, then the perimeters of the coloured triangles APU, KQP, etc. remain constant (provided none of the points P, Q, R, etc. move onto the extensions (outside) of any of the sides of ABC and KLM).

Investigate: Drag point X or Y to rotate triangles ABC and KLM. You can also drag the point V to change the size of the circle. The lengths of the sides of triangles ABCand KLM can be changed by dragging any of the endpoints of segments AX, XB, KY or YL).

Please install Java (version 1.4 or later) to use JavaSketchpad applets.Constant perimeter triangles of crossed hexagon circumscribed around circle

1Note: Past papers with solutions from 1997 of the Junior and Senior South African Mathematics Olympiad are available at: SAMO