Baltic Way 1999 - Georg Mohr-Konkurrencen

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centre, and let G, H and K be the points where the incircle touchesthe sides BC , AC and AB of the triangle, respectively. Also, let N bethe midpoint of BC and M the midpoint of AC (see Figure 9). Since̸ CMO = ̸ CNO = 90 ◦ , points M and N lie on the circle with diameterOC . We will show that point I also lies on that circle. Indeed, wehave|AH| + |BG| = |AK| + |BK| = |AB| =|AC| + |BC|2= |AM| + |BN| ,implying |MH| = |NG|. Since MH and NG are the perpendicular projectionsof OI to the lines AC and BC , respectively, then IO must beeither parallel or perpendicular to the bisector CI of angle ACB . (Toformally prove this, consider unit vectors ⃗e 1 and ⃗e 2 defined by the raysCA and CB , and show that the condition |MH| = |NG| is equivalent to( ⃗e 1 ± ⃗e 2 ) · −→ IO = 0.)If IO is perpendicular to CI , then ̸ CIO = 90 ◦ and we are done. IfIO is parallel to CI , the the circumcentre O of triangle ABC lies onthe bisector CI of angle ACB , whence |AC| = |BC| and the condition2|AB| = |AC| + |BC| implies that ABC is an equilateral triangle. Hencein this case points O and I coincide and the claim of the problem holdstrivially.13. Answer: 60 ◦ .Let F be the point of the side AB such that |AF | = |AE| and |BF | = |BD|(see Figure 10). The line AD is the angle bisector of ̸ A in the isoscelestriangle AEF . This implies that AD is the perpendicular bisectorof EF , whence |DE| = |DF |. Similarly we show that |DE| = |EF |.This proves that the triangle DEF is equilateral, i.e. ̸ EF D = 60 ◦ .Hence ̸ AF E + ̸ BF D = 120 ◦ , and also ̸ AEF + ̸ BDF = 120 ◦ . Thus̸ CAB + ̸ CBA = 120 ◦ and finally ̸ C = 60 ◦ .Alternative solution. Let I be the incenter of triangle ABC , and let G,H , K be the points where its incircle touches the sides BC , AC , ABrespectively (see Figure 11). Then|AE| + |BD| = |AB| = |AK| + |BK| = |AH| + |BG| ,implying |DG| = |EH|. Hence the triangles DIG ja EIH are congruent,12
MNKHGIand⃗e 1⃗e 2 ̸ DIE = ̸ GIH = 180 ◦ − ̸ C .OCCEIDEHIGDAFBAKBFigure 10 Figure 11On the other hand,̸ DIE = ̸ AIB = 180 ◦ − ̸ A + ̸ B2Hence̸ C = ̸ A + ̸ B2which gives ̸ C = 60 ◦ .= 90 ◦ − ̸ C2 ,.14. The quadrilaterals DBCG and F BCE are trapeziums. The area of atrapezium is equal to half the sum of the lengths of the parallel sides multipliedby the distance between them. But the distance between the parallelsides is the same for both of these trapeziums, since the distance from Bto AC is equal to the distance from C to AB . It therefore suffices to showthat|BD| + |CG||CE| + |BF | = |AD||AE|(see Figure 12).similar, we haveNow, since the triangles BDF , ADE and CGE are|BD||BF | = |CG||CE| = |AD||AE| , 13
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Baltic Way 1999 - Georg Mohr-Konkurrencen

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