IMO 1972 SL 10

Origin: NET

Problem

Prove that for each n \geq4 every cyclic quadrilateral can be decomposed into n cyclic quadrilaterals.

Solution

Consider first a triangle. It can be decomposed into k = 3 cyclic quadri- laterals by perpendiculars from some interior point of it to the sides; also, it can be decomposed into a cyclic quadrilateral and a triangle, and it follows by induction that this decomposition is possible for every k. Since every triangle can be cut into two triangles, the required decomposition is possible for each n \geq6. It remains to treat the cases n = 4 and n = 5. n = 4. If the center O of the circumcircle is inside a cyclic quadrilateral ABCD, then the required decomposition is effected by perpendiculars from O to the four sides. Otherwise, let C and D be the vertices of the obtuse angles of the quadrilateral. Draw the perpendiculars at C and D to the lines BC and AD respectively, and choose points P and Q on them such that PQ \parallelAB. Then the required decomposition is effected by CP, PQ, QD and the perpendiculars from P and Q to AB. n = 5. If ABCD is an isosceles trapezoid with AB \parallelCD and AD = BC, then it is trivially decomposed by lines parallel to AB. Otherwise, ABCD can be decomposed into a cyclic quadrilateral and a trape- zoid; this trapezoid can be cut into an isosceles trapezoid and a trian- gle, which can further be cut into three cyclic quadrilaterals and an isosceles trapezoid.

Remark. It can be shown that the assertion is not true for n = 2 and n = 3.