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Three circles of radius 1 are externally tangent to each other and internally tangent to a larger circle. What is the radius of the large circle? Express your answer as a common fraction in simplest radical form.

 Sep 3, 2017
 #1
avatar+98005 
+1

 

May be faster ways to do this, but look at the following:

 

 

Let the centers of two of the circles lie at (1,0)  and (-1,0)

 

So....to find the center of the third circle let it be located at  (0,y)   

 

And by the Pythagorean Theorem we have

 

y = sqrt [ GD^2 - HD^2 ]

 

y= sqrt [ 2^2 - 1^2]  =  sqrt (3)

 

So the center of the top circle =  ( 0, sqrt (3) )  call this point  G

 

Let point E be the center of the circle.....and it will be located at (0, c )

 

Now let B be  the point tangent to the top circle and the leftmost circle

 

And its coordinates will be  the midpoint of  CG =   [ -1/2 , sqrt (3)/2 )

 

Let A be the intersection of the apex of the top circle and  the y axis.......and its coordinates  are

 

(0, GA + GH )  =  ( 0, 1 + sqrt (3) ) 

 

And....by similar triangles

 

(BF) / FD  = HE /HD

 

(sqrt [(3)/2] )   / (3/ 2)  =  HE / 1    →   HE  =  sqrt (3)/3

 

So....the radius of the large circle  = [AH - HE]  =  EA =

[1 +  sqrt (3)] -   [sqrt (3)/3]  =

1 + (2/3) sqrt (3) 

 

 

 

cool cool cool

 Sep 3, 2017
edited by CPhill  Sep 3, 2017
edited by CPhill  Sep 3, 2017
 #2
avatar+21816 
+1

Three circles of radius 1 are externally tangent to each other and internally tangent to a larger circle.
What is the radius of the large circle?
Express your answer as a common fraction in simplest radical form.

 

Let Radius of the larger circle = R.
Let Radius of the three circles = r = 1.

 

Let Area of the triangle ABC = \(A_{ABC}\)
Let Area of the triangle AOB = \( A_{AOB}\)

 

Let AC = CB = BA = 2r
Let AO = OB = R-r

 

\(\mathbf{A_{ABC} = \ ?}\)

\(\begin{array}{|lrcll|} \hline (\text{Heron}) & s &=& \frac{2r+2r+2r}{2} \\ & &=& 3r \\ & A_{ABC}^2 &=& s(s-2r)(s-2r)(s-2r) \\ & A_{ABC}^2 &=& s(s-2r)^3 \quad & | \quad s = 3r \\ & A_{ABC}^2 &=& 3r(3r -2r)^3 \\ & A_{ABC}^2 &=& 3rr^3 \\ & A_{ABC}^2 &=& 3r^4 \\ & A_{ABC} &=& r^2 \sqrt{3} \\ \hline \end{array}\)

 

\(\mathbf{A_{AOB} = \ ?} \)

\(\begin{array}{|lrcll|} \hline & A_{AOB} &=& \frac{A_{ABC}}{3} \\ (1) & A_{AOB} &=& \frac{r^2 \sqrt{3}}{3} \\ \hline \end{array}\)

 

\(\mathbf{(\text{Heron})\ A_{AOB} = \ ?}\)

\(\begin{array}{|rcll|} \hline (\text{Heron}) & s &=& \frac{2r+(R-r)+(R-r)}{2} \\ & s &=& \frac{2R}{2} \\ & s &=& R \\ & A_{AOB}^2 &=& s(s-2r)[s-(R-r)][s-(R-r)] \quad & | \quad s = R \\ & A_{AOB}^2 &=& R(R-2r)[R-R+r)][R-R+r] \\ & A_{AOB}^2 &=& R(R-2r)r^2 \\ (2) & A_{AOB} &=&r\sqrt{R(R-2r)} \\ \hline \end{array}\)

 

(1) = (2)

\(\begin{array}{|rcll|} \hline \frac{r^2 \sqrt{3}}{3} &=& r\sqrt{R(R-2r)} \\ \frac{r \sqrt{3}}{3} &=& \sqrt{R(R-2r)} \quad & | \quad \text{square both sides}\\ \frac{r^2}{3} &=& R(R-2r) \\ R^2-2rR -\frac{r^2}{3} &=& 0 \quad & | \quad \cdot 3 \\ 3R^2-6rR - r^2 &=& 0 \\ \\ R &=& \frac{6r \pm \sqrt{36r^2-4\cdot 3\cdot (-r^2) } }{2\cdot 3} \\ R &=& \frac{6r \pm \sqrt{36r^2+12r^2 } }{6} \\ R &=& \frac{6r \pm \sqrt{48r^2 } }{6} \\ R &=& \frac{6r \pm 4r\sqrt{3} }{6} \\ R &=& r \pm \frac23 r \sqrt{3} \quad & | \quad R \gt 0 \\ R &=& r + \frac23 r \sqrt{3} \quad & | \quad r = 1 \\ \mathbf{R} & \mathbf{=} & \mathbf{1 + \frac23 \sqrt{3}} \\ \hline \end{array}\)

 

laugh

 Sep 4, 2017

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