Baltic Way 1999 - Georg Mohr-Konkurrencen

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DEFKHwhichGimplies the required equality.IEGCEGCF ′ADADMFBFBG ′Figure 12 Figure 13Alternative solution. As in the first solution, we need to show that|BD| + |CG||BF | + |CE| = |AD||AE| .Let M be the midpoint of BC , and let F ′ and G ′ be the points symmetricto F and G, respectively, relative to M (see Figure 13). Since CG isparallel to AB , then point G ′ lies on the line AB , and |BG ′ | = |CG|.Similarly point F ′ lies on the line AC , and |CF ′ | = |BF |. It remains toshow that|DG ′ ||EF ′ | = |AD||AE| ,which follows from DE and F ′ G ′ being parallel.Another solution. Express the areas of the quadrilaterals asand[DBCG] = [ABC] − [ADE] + [ECG][F BCE] = [ABC] − [ADE] + [DBF ] .The required equality can now be proved by direct computation.15. Consider a point F on BC such that |CF | = |BD| (see Figure 14). Since̸ ACF = 60 ◦ , triangle ACF is equilateral. Therefore |AF | = |AC| = |CE|14
andIABCDAFEB = ECD = 120 ◦ . Moreover, |BF | = |CD|. This implies thattrianglesFAF B and ECD are congruent, and |AB| = |DE|.GF ′ EG ′MC60 ◦ FFigure 14DABAlternative solution. The cosine law in triangle ABC implies|AB| 2 = |AC| 2 + |BC| 2 − 2 · |AC| · |BC| · cos ̸ ACB == |AC| 2 + |BC| 2 − |AC| · |BC| == |AC| 2 + (|BD| + |DC|) 2 − |AC| · (|BD| + |DC|) == |AC| 2 + (|AC| + |DC|) 2 − |AC| · (|AC| + |DC|) == |AC| 2 + |DC| 2 + |AC| · |DC|On the other hand, the cosine law in triangle CDE gives|DE| 2 = |DC| 2 + |CE| 2 − 2 · |DC| · |CE| · cos ̸ DCE == |DC| 2 + |CE| 2 + |DC| · |EC| == |DC| 2 + |AC| 2 + |DC| · |AC| .Hence |AB| = |DE|.16. Answer: 14.Assume that there are integers n, m such that k = 19 n − 5 m is a positiveinteger smaller than 19 1 − 5 1 = 14. For obvious reasons, n and m must bepositive.Case 1: Assume that n is even. Then the last digit of k is 6. Consequently,we have 19 n − 5 m = 6. Considering this equation modulo 3 implies that m15
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Baltic Way 1999 - Georg Mohr-Konkurrencen

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