Solutions - Georg Mohr-Konkurrencen

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known formula 4m 2 c = 2a 2 + 2b 2 − c 2 . Hence the inequality of the problem can be rewritten as 4Rmc � a 2 + b 2 , or 8Rmc � 4m 2 c + c 2 . The last inequality is equivalent to |mc − R| � � R 2 − (c/2) 2 , or � � |MC| − |OC| � � � |OM|, which is the triangle inequality for triangle COM . Equality holds if and only if the points C , O , M are collinear. This happens if and only if a = b or � C = 90 ◦ . Remark. Yet another solution can be obtained by setting R = abc (where S 4S denotes the area of the triangle) and expressing S by Heron’s formula. After squaring both sides, cross-multiplying and cancelling a lot, the inequality reduces to (a 2 −b 2 ) 2 (a 2 +b 2 −c 2 ) 2 � 0, with equality if a = b or a 2 +b 2 = c 2 PSfrag replacements A B C . O M E a C b c D O A B Figure 3 12. Let O be the circumcentre of triangle ABC (i.e., the midpoint of BC ) and let AD meet the circumcircle again at E (see Figure 3). Then � BOE = 2� BAE = � CDE , showing that |DE| = |OE|. Triangles ADC and BDE are similar; hence |AD| |CD| = |BD| |DE| , |AD| |BD| = |CD| |DE| |AD| |AD| |CD| |BD| |BC| |BC| + = + = = = 2 |BD| |CD| |DE| |DE| |DE| |OE| 12 and finally
which is equivalent to the equality we have to prove. frag replacements Alternative solution. Let � BAD = α and � CAD = β . By the conditions of the problem, α + β = 90 ◦ A (hence sin β = cos α), � BDA = 2α and � CDA = 2β . By the law of sines, |AD| sin 3α = |BD| sin α = 3 − 4 sin2 α and |AD| sin 3β = |CD| sin β = 3 − 4 sin2 β = 3 − 4 cos 2 B C O M a b c A α . B C Adding O these two equalities we get the claimed one. D E C1 D C2 E C A P F Figure 4 13. Let C1 and C2 be the circumcircles of triangles AED and BCD , respectively. Let DP meet C2 for the second time at F (see Figure 4). Since � ADE = � BDC , the ratio of the lengths of the segments EA and BC is equal to the ratio of the radii of C1 and C2 . Thus the homothety with centre P that takes AE to CB , also transforms C1 onto C2 . The same homothety transforms the arc DE of C1 onto the arc F B of C2 . Therefore � EAD = � BDF = � BDP . The second equality is proved similarly. 14. Since the lines BD and AC are parallel and since AD is the external bisector of � BAC , we have � BAD = � BDA; denote their common size by α (see Figure 5). Also � CAE = � CEA = α, implying |AB| = |BD| and |AC| = |CE|. Let B ′ , C ′ , F ′ be the feet of the perpendiculars from 13 B
Page 1 and 2: Baltic Way 1998 Warsaw, November 8,
Page 3 and 4: 15. Given an acute triangle ABC . P
Page 5 and 6: 1 or −1 modulo 5; so 4c 2 − 3a
Page 7 and 8: 6. The polynomial Q(x) = P (x) −
Page 9 and 10: � Let C be the midpoint of the se
Page 11: 11. Answer: equality holds if a = b
Page 15 and 16: the points B , E , P are collinear.
Page 17 and 18: Indeed: suppose there is no such pl

Solutions - Georg Mohr-Konkurrencen

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