Solutions - Georg Mohr-Konkurrencen

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PABCD̸ CDEM= ̸ ADB = 45 ◦ it follows that ̸ AEO = 45 ◦ . Since |AO| = |EO|,we haveN̸ AOE = 90 ◦ and AO ‖ DM .KFrom theXequality |BD| · |CD| = |AD| 2 we obtain |AD| = |DE|, which impliesthatA|OM| = |ME|. Therefore |BO| = |BE| and also |BO| = |EO|.Hence the triangle BOE is equilateral. This gives ̸BBAE = 30 ◦ , so̸ BACC= 60 ◦ . Summing up the angles of the triangle ABD we obtain̸ ABC = 105 ◦ and from this ̸DACB = 15 ◦ .PQAOBDMCEFigure 511. Answer: 0 and 1 2 .Obviously the constant functions f(n) = 0 and f(n) = 1 provide solutions.2We show that there are no other solutions. Assume f(2001) ≠ 0. Since2001 = 3 · 667 and gcd(3, 667) = 1, thenf(2001) = f(1) · (f(3) + f(667)) ,and f(1) ≠ 0. Since gcd(2001, 2001) = 2001 thenf(2001 2 ) = f(2001)(2 · f(1)) ≠ 0 .Also gcd(2001, 2001 3 ) = 2001, sof(2001 4 )=f(2001) · (f(1) + f(2001 2 ))=f(1)f(2001)(1 + 2f(2001)) .On the other hand, gcd(2001 2 , 2001 2 ) = 2001 2 andf(2001 4 )=f(2001 2 ) · (f(1)+f(1))=2f(1)f(2001 2 )=4f(1) 2 f(2001) .So 4f(1) = 1 + 2f(2001) and f(2001) = 2f(1) − 1 2 .Exactly the same8
argument starting from f(2001 2 ) ≠ 0 instead of f(2001) shows thatf(2001 2 ) = 2f(1) − 1 2 . So2f(1) − 1 (2 = 2f(1) 2f(1) − 1 ).2Since 2f(1) − 1 2 = f(2001) ≠ 0, we have f(1) = 1 , which implies2f(2001) = 2f(1) − 1 2 = 1 2 .12. By Hölder’s inequality,n∑ n∑(∑ n ) 3/5 (∑ n ) 2/5a 3 = (a i · a 2 i ) a 5/3i · (a 2 i ) 5/2 .i=1 i=1i=1i=1We will show thatn∑(∑ n ) 5/3 a i . (4)i=1a 5/3iLet S =n∑i=1( aiSi=1n∑a i , then (4) is equivalent toi=1) 5/3 1 =n∑i=1a iS ,which holds since 0 < a iS 1 and 5 (3 > 1 yield ai) 5/3 a i S S . So,n∑(∑ n ) (∑ n ) 2/5a 3 i a i · a 5 i ,i=1which givesi=1n∑i=113. Consider the equation( 7) x ( 1 +9 9i=1a i 35 2/5 > 3 2 , since 25 > 5 2 and hence 2 > 5 2/5 .) x= 1 .9
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Solutions - Georg Mohr-Konkurrencen

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