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Fifteen unit circles are inscribed in an equilateral triangle in such a way that each circle is externally tangent to its neighbors. What is the area of the triangle?

 

 Jan 8, 2020
 #1
avatar+107098 
+1

Not as  hard as it seems

 

Call the lower left vertex of the  triangle, B

 

Let the center of the bottom left circle  be A

 

Let the radius of this circle intersect the base of the triangle at C

 

So.....triangle  ABC  is a 30 -60 -90   right triangle

 

And   BC  =  radius  of the circle * sqrt (3)  = 1 * √ (3)   =  √ (3)

 

So....using symmetry...the side of the  triangle  =

 

2BC + 8 r  =

 

2sqrt (3)  +  8 r   =

 

2sqrt (3)  +  8  =

 

√12 + 8

 

So....the area of the  triangle =

 

(1/2) ( √12 + 8)^2 * (√3/2) =

 

(√3/4)  ( 12 + 16√12 + 64)  =

 

(√3/4) ( 76 + 16*2√3)   =

 

√3 ( 19 + 8√3)  =

 

19√3 + 24  units^2  ≈  56.9  units^2

 

 

cool cool cool

 Jan 8, 2020
 #2
avatar+331 
+2

All you need is a bottom row:

Circle diameter   >  D = 1

Radius                >  R = 0.5

Triangle side       >  a = 4D + 2[R / tan(30°)]     a = 5.732    a/2 = 2.866

Triangle height    >  H = sqrt[ a² - (a/2)²]      H = 4.964

Triangle area       >  A = H * a/2              A = 14.227 u²   indecision

 Jan 9, 2020

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