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The area of equilateral triangle inscribed in the circle \[x^2+y^2-4x+2y+1=0\] is of the form \[\frac{a√b}{c}\]
Find \[a+b+c\]

 Feb 9, 2021
 #1
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This is a  circle  centered  at  (2, -1)  with a  radius  of 2

 

The  side^2, S^2,  of the  triangle   can be found as

 

S^2 =  2^2  + 2^2  - 2 ( 2 * 2) cos (120 degrees)

 

S^2  = 8  -  8 (-1/2)

 

S^2  =  8  + 4  =  12

 

The area of this triangle    =

 

(1/2) S^2  sin (60 degrees)   =

 

(1/2) 12 * sqrt (3)  / 2   =

 

6 sqrt (3)   / 2  =     3 sqrt (3)  / 1

 

a +  b +  c  =     3  +  3  +  1   =    7

 

cool cool cool

 Feb 9, 2021

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