## Constant perimeter triangle formed by tangents to circle

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**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* = 2*AB*.

**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*).

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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 *ABC*and *KLM* can be changed by dragging any of the endpoints of segments *AX*, *XB*, *KY* or *YL*).

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Constant perimeter triangles of crossed hexagon circumscribed around circle

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

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Created by Michael de Villiers, 20 April 2013.