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is a cyclic quadrilateral inscribed in a circle of radius one unit. If 
, prove that is a square.
+1 w
, 
, 
, 
satisfy 
and 
. Prove that there exists a permutation 
of 
such that 
.
be a cyclic quadrilateral. A circle centered at 
passes through 
and 
and meets lines 
and 
again at points 
and 
(distinct from 
). Let 
denote the orthocenter of triangle 
Prove that if lines 
are concurrent, then triangle 
and 
are similar.
such that 
, where 
be the given sequence and let 
.
and 
for all 
.
terms.
is the answer, just take 16 real numbers satisfying
.
and 
for 
. Thus we found all sequences with the given properties.
terms exists. Consider the matrix![\[\begin{bmatrix} x_1 & x_2 & x_3 & x_4 & x_5 & x_6 & x_7 & x_8 & x_9 & x_{10} \ \ \ x_{11} \\ x_2 & x_3 & x_4 & x_5 & x_6 & x_7 & x_8 & x_9 & x_{10} & x_{11} \ \ \ x_{12} \\ x_3 & x_4 & x_5 & x_6 & x_7 & x_8 & x_9 & x_{10} & x_{11} & x_{12} \ \ \ x_{13} \\ x_4 & x_5 & x_6 & x_7 & x_8 & x_9 & x_{10} & x_{11} & x_{12} & x_{13} \ \ \ x_{14} \\ x_5 & x_6 & x_7 & x_8 & x_9 & x_{10} & x_{11} & x_{12} & x_{13} & x_{14} \ \ \ x_{15} \\ x_6 & x_7 & x_8 & x_9 & x_{10} & x_{11} & x_{12} & x_{13} & x_{14} & x_{15} \ \ \ x_{16} \\ x_7 & x_8 & x_9 & x_{10} & x_{11} & x_{12} & x_{13} & x_{14} & x_{15} & x_{16} \ \ \ x_{17} \end{bmatrix}.\]](http://latex.artofproblemsolving.com/6/c/5/6c529101af135c3f0589e2c954664d9967cf001f.png)
works 
are any two distinct positive integers such that none of them divides the other, and if a sequence has the property that the sum of any 
consecutive terms is always positive and the sum of any 
consecutive terms is negative then the maximal length of this sequence is 
where 
. One can find a construction without too much difficulty.
-sum the sum of consecutive terms. If we suppose, for contradiction, that the sequence has terms, then every 
-sum can be written as an 
-sum with a 
-sum subtracted, and must hence be positive. Similarly, every 
-sum can be written as a -sum with a -sum subtracted, and must therefore be negative. And, finally, every term can be written as a -sum with a -sum subtracted, and must therefore be positive again. However, if every term is positive, then it's impossible for every -sum to be negative, and we've reached a contradiction.
![\[ 5,5,-13,5,5,5,-13,5,5,-13,5,5,5,-13,5,5. \]](http://latex.artofproblemsolving.com/5/c/5/5c549f6c5c9138ff7540fab045cef2fb23cf40f7.png)
But how do you find such an example?
, meaning that if we can make it so that we get exactly two negative numbers of pretty large magnitude in each block of length and in each block of length we would be done by making the other numbers all equal to some positive number. It's not so obvious to pick the numbers 
and 
, but maybe this is easier?![\[20, 20, -51, 20, 20, 20, -51, 20, 20, -51, 20, 20, 20, -51, 20, 20.\]](http://latex.artofproblemsolving.com/f/8/1/f815c033f368c8adb22c2f14e3c3930db334866f.png)
reaches from the first 
to right before the last 
.
. For 
, we can consider any subset of 
.
, then suppose 
. Then, 
and 
. Therefore,
which is clearly not possible. So, any sequence of real numbers has seven consecutive terms which sum to a negative value and has eleven consecutive terms which sum to a positive value. Thus, the maximum number of terms in the sequence is 
.
then take the first 17 numbers 
and write them as follows.
, and 
can be constructed (posted in other posts above).
