{1,2,...,n}

1 <= i <= j <= m

(a1 + a2 + ... + am) / m = (n +1) / 2

I am sorry I write in English because I HAVE TO enhance my english skills and more my writing skill!
haha
I solved this probelm, here is the answer
ALPHA = 67.5 degrees! --> 90 degree
If I made mistake both in English writing and the solution let me know.

btw I am really interested in geometrical problems, as my proffession is interfering with these kind of problems. so I had to solve this problem and I did, but there may be some general mistakes that I could correct it by resolving the problem, just let me know!
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* seems the answer is not correct, while I resolved again I came to more unbeliveble answer of 90 degrees!
I am going to Scan my solution tomorrow
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it is tomorrow !

again solved the problem with coming to the 90 degree result for alpha and I am sure of it
here goes the solution

look at the picture while I discuss my solution:

consider triangle AEB : A + B + E = 180
consider triangle ADC : A + D + C = 180

ok , now we are going to colculate the angles D and E with the given assumptions

D1 = 2 * a (alpha)
E1 = (180 - 2a) + (180 -(a+O1)
consider the index numbers 1 here O1 is the angle of O which is within triangle ODE

it is part 1 of the solution till here, If u have any problems here just point to this part!

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now part 2
consider the arc which the O1 angle is within it is sum of two arcs of angle B + angle C

so O1 = 2 * (B+C) <eq1>

ok now we will go and colculate B and C angles from the part 1 equations and then come to sobistitute here and calculate the O1

tirangle AEB => we have B = 2 a + O1 - 180
triangle ADC => we have C = 180 - 3a

ok now we sum these and go to <eq1> which will result in:
O1 = 2(2a+O1-3a) = 2(O1 -a)
which will finallt result in O1 = 2a
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part 3

now we will go and compute other part of O = O2 that we can see it either in triangle BOD or COE

triangle OBD : O2 + B + D2 = 180
triangle OCE : O2 + C + E2 = 180

first we must consider and colculate what D2 and E2 are
D2 is easily seen that equals to 180 - 2a
and E2 is equal to 180 - E1
which results in
E2 = 180-E1 = 180 - (360- 5a)

ok now in OBD we will summerize both obd and oce in one equation :
2*O2 + (B+C) + D2 + E2
= 2*O2 + a + 180 - 2a + (180 - (360- 5a)) = 360

=> 2*O2+a+180-2a -180 +5a = 360

2*O2+4a = 360

so we will have 2*(O2 + 2a) = 360
and O2 + 2a = 180
so O2 = 90 -2a
and after correction we'll have 0 = 0, so we must find another equation!
I will do it very soon! _____________-

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finally part for
if we sum o1 and o2 we will have 180
so O1 + 02 = 180
2a + 90 -2a = 180
90 = 180
,,,,,,

again we come to an uqualncy

so it is impossible to have such a circle in a plane!
wether the assumptions are not correct or O is not located at the center!