Apologise for typing English, it'll be normal in Thai language by 22 Nov 2007
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Number Theory
5.
1958 putnam
4.
(i) If n is odd greater than 1, that number is even number greater than 2 (not prime)
(ii)If n is even number, pair up like the following :
Since $ 1^{2007} \equiv 1^{2007} (\text{mod}\,(n+1)) $ and $n^{2007} \equiv (-1)^{2007} (\text{mod}\, (n+1)) \Rightarrow 1^{2007}+ n^{2007} \equiv 0 (\,\text{mod}\, (n+1) )$
And consider pairs $(2,n-1) ,(3,n-2), \cdots (\frac{n}{2}, \frac{n}{2}+1) $ , It implies that such sum is divisible by n+1 and thus it's not prime
Geometry
1.
Consider cyclic ABED,we can imply that $ D\hat{A}E =E\hat{A}B $ because EBD is isosceles triangle
Together with other three reasons,
(i) $ D\hat{A}E =D\hat{C}E $ (same arc)
(ii) $ D\hat{C}E =C\hat{F}G$ (parallel line)
(iii) ABCD is cyclic ,
we can easily calculate all angles in triangle CFG and show that it's isosceles triangle
2.
It's easy to show that EFGH is parallelogram, by using property that line segment joining 2 midpoints of triangle is parallel to third side.
Extend DC to RHS of C to ,say,X and join EX , we then have two congruent triangles ABE , ECX
Now we can consider area of triangle AXD instead ,
Since Area of EFGH is half of triangle AED and AE=EX (by such congruence shown above) ,it means that area of EFGH is one-fourth of triangle AXD (=area of ABCD)
3.
Use triangle's inequality for latter inequality. The former one can be deduced by using the fact that,for any convex quadrilateral, sum of its diagonals is more than half of its perimeter (should prove before using this fact)