Let be the circumcenter of triangle and let be the circumcenter of triangle Find
Here's my attempt at this tricky problem
Note that angle ABC is inscribed in the first circumcircle
Let the circumcenter = M = O1
Then angle AMC is a central angle intercepting the same arc as angle ABC so its measure = 120°
And triangle AMC is isoceles with AM = CM
And angles MAC and MCA = (180 - 120) / 2 = 30°
Using some trig we can find circumradius AM as
AM / sin 30° = 12 / sin 120°
AM / (1/2) = 12 / ( [sqrt 3 ] / 2 )
AM = 12/sqrt 3 = 4sqrt 3
Now triangle ACD is obtuse and its circumcenter will be exterior to side AD
Angle ADC will be inscribed in the circumcircle of this triangle
Let N be the circumcenter O2
Then angle ANC is a central angle intercepting the same arc as angle ADC so its measure = 60°
And triangle ANC is isosoceles with AN = CN
So angles CAN and ACN = (180 - 60) / 2 = 60°
So triangle ANC is equilateral with circumradius AN = 12
Let A = (0,0)
Then M = O1 has the coordinates ( 4sqrt (3) cos 30° , 4sqrt (3) sin 30°) = (6 , 2sqrt (3) )
And N = O2 has the coordinates ( 12 cos (-60°) , 12 sin (-60°) ) = ( 6 , -6sqrt (3) )
Assuming that O1 O2 is the distance between the circumcenters we have that
O1 O2 = 2sqrt (3) - (-6sqrt (3) ) = 8 sqrt (3)