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$\overline{BC}$ is a chord of a circle with center $O$ and area $48\pi$. Point $A$ is inside $\triangle BCO$ such that $\triangle ABC$ is equilateral and $A$ is the circumcenter of $\triangle BCO$. What is the area of triangle $ABC$?

 
michaelcai  Dec 5, 2017
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+2

Look at the following diagram to get a feel for this :

 

 

The  larger circle will have the equation

 

x^2 + y^2  = 48

 

Since  A will be the circumcenter for BOC.....then the distance from A to O will be the same distance as from A to B  and from A to C

 

And since ABC is equilateral, then BAC  = 60°

Then major angle  BAC  = 300°

And using symmetry, angle OAC  = 150°

 

But AO = AC....so...triangle  OAC is isosceles....   and angle AOC  = 15°

And since OC is a radius of the larger circle it equals √48

 

So......we can find a side of the equilateral triangle - AC - thusly

 

OC / sin (OAC) = AC/ sin (AOC)

√48 / sin (150)  =  AC / sin (15)

AC  =  √48sin (15) / sin (150)  = 

AC = √48   ( sin (45 -30) / (1/2) =

2√48 (sin45sin30 - sin30cos45) =

2√48 (√2√3 / 4   - √2 / 4)  =

√48/2 (√6 - √2 )  =

2√3 ( √6  - √2 )

√12 ( √6 - √2)  =

√72  - √24   =

6√2 - 2√6  (exact value)

 

Then  the equation for the circle with A as a center  passing through the vertices of BOC  is

 

x^2  +  (y^2 +6√2 - 2√6)^2  = ( 6√2 - 2√6)^2

 

And the area of equilateral triangle ABC  =  (√3 /4)s^2  =

 

(√3/4) (6√2 - 2√6)^2  =

 

(√3 / 4) ( √72 - √24)  =

 

(√3/ 4) ( 72  - 2 √(72*24) + 24) =

 

(√3/ 4) ( 96 - 2√1728)  =

 

(√3/ 4) (96 - 48√3)  =

 

12√3 ( 2 - √3)  =

 

24√3 - 36   units ^2   ≈  5.57  units^2

 

 

cool cool cool

 
CPhill  Dec 5, 2017
edited by CPhill  Dec 5, 2017
edited by CPhill  Dec 5, 2017

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