Baltic Way 1999 - Georg Mohr-Konkurrencen

eplacementsn = circumcentre 4 O of triangle ABC through point D intersect the circumcirclen = in5D ′ : since ̸ B + ̸ D ′ = 180 ◦ and ̸ B + ̸ D 180 ◦ , then D cannot lien = in6the exterior of the circumcircle.P 1̸ ̸̸ ̸12. Let P 2N be the midpoint of BC and M the midpoint of AC . Let Obe P 3the circumcentre of ABC and I its incentre (see Figure 8). SincePCMO 4= CNO = 90 ◦ , the points C , N , O and M are concyclic (regardlessP 5of whether O lies inside the triangle ABC ). We now have toshow P 6that the points C , N , I and M are also concyclic, i.e I lies onthe A 1same circle as C , N , O and M . It will be sufficient to show thatNCM + NIM = 180 ◦ in the quadrilateral CNIM . SinceA 2A 3|AC| + |BC|O|AB| = = |AM| + |BN| ,2Awe B can choose a point D on the side AB such that |AD| = |AM| and|BD| C = |BN|. Then triangle AIM is congruent to triangle AID , andsimilarly D triangle BIN is congruent to triangle BID . ThereforeO̸ NCM + ̸ NIM = ̸ANCM + (360 ◦ − 2̸ AID − 2̸ BID) =B= ̸ BCA + 360 ◦ − 2̸ AIB =C(= ̸ BCA + 360 ◦ − 2 · 180 ◦ − ̸ BAC− ̸ ABC)D=2 2D ′= ̸ BCA + ̸ ABC + ̸ CAB = 180 ◦ .OCC⃗e 1⃗e 2MIONMHIOGNABAKBFigure 8 Figure 9Alternative solution.Let O be the circumcentre of ABC and I its in-11