be an acute triangle. Points 
,
lie on a line in this order and satisfy 
.
and 
be the midpoints of 
and 
,
is acute, and let 
be its orthocentre. Points 
and 
lie on lines 
and 
,
and 
and 
and 
of positive integers is called central if for every positive integer 
,
terms of the sequence is equal to 
,
,
,
of positive integers such that for every central sequence 
,
,
,
.
are two odd positive integers such that 
.
.
.
be the center of the 
-
be the projection of 
.
and 
intersectlines 
and 
at 
and 
,
and 
intersect at 
is perpendicular to line 
Y by
Problem Statement:
The set S= {1/r : r = 1, 2, 3,...} of reciprocals of the positive integers contains arithmetic progressions of various lengths. For instance, 1/20, 1/8, 1/5 is such a progression, of length 3 (and common difference 3/40). Moreover, this is a maximal progression in S of length 3 since it cannot be extended to the left or right within S (−1/40 and 11/40 not
being members of S).
(i) Find a maximal progression in
S of length 1996.
(ii) Is there a maximal progression in
S of length 1997?
For part (i) I constructed the sequence: 1/1996!, 2/1996!, 3/1996!, ..., 1996/1996!. Which has common difference of 1/1996! and because of the nature of the factorial function, each of them are elements in S, and continuing to the left we get 0, which is not in S, and continuing to the right we get 1997/1996! which is not in S because 1997 is prime. Therefore we have found a maximal progression in S of length 1996 as desired.
For part (ii), let A = (4000!)/(2003!). Now consider the sequence: 2004/A, 2005/A, 2006/A, ... , 4000/A. Each is an element of S, by the nature of the factorial function and the common difference is 1/A. Continuing to the left gives 2003/A, 2003 is prime which is not a factor of A, so 2003/A is not an element of S. Continuing to the right gives 4001/A, 4001 is prime which is not a factor of A, so 4001/A is not an element of S. Therefore we have found a maximal progression in S of length 1997, and so the answer is Yes.
Is this proof sound?
The set S= {1/r : r = 1, 2, 3,...} of reciprocals of the positive integers contains arithmetic progressions of various lengths. For instance, 1/20, 1/8, 1/5 is such a progression, of length 3 (and common difference 3/40). Moreover, this is a maximal progression in S of length 3 since it cannot be extended to the left or right within S (−1/40 and 11/40 not
being members of S).
(i) Find a maximal progression in
S of length 1996.
(ii) Is there a maximal progression in
S of length 1997?
For part (i) I constructed the sequence: 1/1996!, 2/1996!, 3/1996!, ..., 1996/1996!. Which has common difference of 1/1996! and because of the nature of the factorial function, each of them are elements in S, and continuing to the left we get 0, which is not in S, and continuing to the right we get 1997/1996! which is not in S because 1997 is prime. Therefore we have found a maximal progression in S of length 1996 as desired.
For part (ii), let A = (4000!)/(2003!). Now consider the sequence: 2004/A, 2005/A, 2006/A, ... , 4000/A. Each is an element of S, by the nature of the factorial function and the common difference is 1/A. Continuing to the left gives 2003/A, 2003 is prime which is not a factor of A, so 2003/A is not an element of S. Continuing to the right gives 4001/A, 4001 is prime which is not a factor of A, so 4001/A is not an element of S. Therefore we have found a maximal progression in S of length 1997, and so the answer is Yes.
Is this proof sound?
