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Let $O_1$ be the circumcenter of triangle $ABC,$ and let $O_2$ be the circumcenter of triangle $ACD.$ Find $O_1 O_2.$

[asy] size(5cm);pair A,B,C,D;A=(0,0);C=(4,0);B=(3,3.5);D=(6,-2.5);draw(A--B--C--D--A--C);label(

 Apr 14, 2022
 #1
avatar+83 
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Can you explain more clearly?

What does "O1O2" MEAN?

Does it mean distance?

 Apr 15, 2022
 #2
avatar+128631 
+1

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)

 

cool cool cool

 Apr 15, 2022

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