MathDB

1

Part of 2007 Turkey MO (2nd round)

Problems(2)

Turkey NMO 2007 Problem 1, m(KNB)=m(BNL)

Source: Turkey NMO 2007 Problem 1

9/27/2011
In an acute triangle ABCABC, the circle with diameter ACAC intersects ABAB and ACAC at KK and LL different from AA and CC respectively. The circumcircle of ABCABC intersects the line CKCK at the point FF different from CC and the line ALAL at the point DD different from AA. A point EE is choosen on the smaller arc of ACAC of the circumcircle of ABCABC . Let NN be the intersection of the lines BEBE and ACAC . If AF2+BD2+CE2=AE2+CD2+BF2AF^{2}+BD^{2}+CE^{2}=AE^{2}+CD^{2}+BF^{2} prove that ∠KNB=∠BNL\angle KNB= \angle BNL .
geometrycircumcirclegeometry unsolved
Turkey NMO 2007 Problem 4, (2^(m-1)-1)/127m is an integer

Source: Turkey NMO 2007 P4

9/27/2011
Let k>1k>1 be an integer, p=6k+1p=6k+1 be a prime number and m=2p−1m=2^{p}-1 .
Prove that 2m−1−1127m\frac{2^{m-1}-1}{127m} is an integer.
modular arithmeticnumber theory unsolvednumber theory