Baltic Way 1999 - Georg Mohr-Konkurrencen

"!n = 4n = 5n = 6P 1P 2P 3the centre along that perpendicular bisector (and changing its radius at thesame P 4 time, so that points A and B remain on the circle) we arrive at asituation P 5 where at least one of the remaining two points will also be on thecircle P 6 (see Figure 5).A 1A 2¡ §¡§¡¨¡¨ ¡#¡# $¡$Areplacements 3D CD CODC¡¡n ¡= ¡4 ¡OOn¡= 5O¡ ¡n = 6P©¡© ¡¡ ¡1¡ ¡A BABAP 2BP 3P 4Figure 5P 5Alternative P 6 solution. The quadrangle with its vertices in the four points canbe convex or non-convex.A 1A 2If the quadrangle is non-convex, then one of the points lies in the interiorof A 3 the triangle defined by the remaining three points (see Figure 6) – thecircumcircle O of that triangle has the required property.ABCCDODADAD ′ O)()C%&'('BBFigure 6 Figure 7Assume now that the quadrangle ABCD (where A, B, C, D are the fourpoints) is convex. Then it has a pair of opposite angles, the sum of whichis at least 180 ◦ — assume these are at vertices B and D (see Figure 7).We shall prove that point D lies either in the interior of the circumcircleof triangle ABC or on that circle. Indeed, let the ray drawn from the10