+0

# test

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test

Jul 12, 2019
edited by Guest  Jul 12, 2019
edited by Guest  Jul 14, 2019

#1
+5 If we connect the centers of the three smaller circles, we get an equilateral triangle because the length of each side is twice the radius of a smaller circle. Then if we connect the center of the equilateral triangle to each of its three vertices, we get three new congruent triangles. Since these new triangles are congruent to each other,

m∠CBD  =  (1/2)(60°)   =   30°

And let's say

radius of smaller circle  =  AB  =  BD  =  r

And so  △CBD  is a  30-60-90  triangle where

the side across from the  60°  angle  =  r

the side across from the  30°  angle  =  r / √3

the side across from the  90°  angle  =  2r / √3

BC  =  2r / √3

Because the radius of the bigger circle  =  1

AB + BC  =  1

r  +  2r / √3   =  1

r( 1 + 2/√3 )  =  1

r   =   1 / ( 1 + 2/√3 )

r   =   2√3 - 3

Jul 13, 2019
edited by hectictar  Jul 13, 2019

#1
+5 If we connect the centers of the three smaller circles, we get an equilateral triangle because the length of each side is twice the radius of a smaller circle. Then if we connect the center of the equilateral triangle to each of its three vertices, we get three new congruent triangles. Since these new triangles are congruent to each other,

m∠CBD  =  (1/2)(60°)   =   30°

And let's say

radius of smaller circle  =  AB  =  BD  =  r

And so  △CBD  is a  30-60-90  triangle where

the side across from the  60°  angle  =  r

the side across from the  30°  angle  =  r / √3

the side across from the  90°  angle  =  2r / √3

BC  =  2r / √3

Because the radius of the bigger circle  =  1

AB + BC  =  1

r  +  2r / √3   =  1

r( 1 + 2/√3 )  =  1

r   =   1 / ( 1 + 2/√3 )

r   =   2√3 - 3

hectictar Jul 13, 2019
edited by hectictar  Jul 13, 2019
#2
+1

Very nice, hectictar  !!!   CPhill  Jul 13, 2019