The secret to this challenge is to realize that what we're given are points on a circle,with the offset on the x axis being the key and the offset on the y axis is the iv.The solution is found by Googling "find circle with 3 points" and copy-pasing the equations from a random website.
```pyimport randomimport secretsimport mathfrom decimal import Decimal, getcontextfrom Cryptodome.Cipher import AES
getcontext().prec = 50
ct = "838371cd89ad72662eea41f79cb481c9bb5d6fa33a6808ce954441a2990261decadf3c62221d4df514841e18c0b47a76"
x1, y1 = (Decimal("45702021340126875800050711292004769456.2582161398"), Decimal("310206344424042763368205389299416142157.00357571144"))x2, y2 = (Decimal("55221733168602409780894163074078708423.359152279"), Decimal("347884965613808962474866448418347671739.70270575362"))x3, y3 = (Decimal("14782966793385517905459300160069667177.5906950984"), Decimal("340240003941651543345074540559426291101.69490484699"))
# Source: http://www.ambrsoft.com/TrigoCalc/Circle3D.htmA = x1*(y2 - y3) - y1*(x2 - x3) + x2*y3 - x3*y2B = (x1**2 + y1**2)*(y3 - y2) + (x2**2 + y2**2)*(y1 - y3) + (x3**2 + y3**2)*(y2 - y1)C = (x1**2 + y1**2)*(x2 - x3) + (x2**2 + y2**2)*(x3 - x1) + (x3**2 + y3**2)*(x1 - x2)D = (x1**2 + y1**2)*(x3*y2 - x2*y3) + (x2**2 + y2**2)*(x1*y3 - x3*y1) + (x3**2 + y3**2)*(x2*y1 - x1*y2)
# The center of the circle is the key and ivkeynum = int(round(-B/(2*A)))ivnum = int(round(-C/(2*A)))
key = int.to_bytes(keynum, 16, "big")iv = int.to_bytes(ivnum, 16, "big")
cipher = AES.new(key, AES.MODE_CBC, iv=iv)enc = cipher.decrypt(bytes.fromhex(ct))print(enc)```