Find the area of this larger circle

[grandpa2390]

Homework Statement

I'm helping to run a math competition at my school and we pulled a problem from this site that we are having trouble solving.

Eight circles of radius 1 have centers on a larger common circle and adjacent circles are tangent. Find the area of the common circle. See the illustration below.

Relevant Equations

Not sure. It has been a long time since I've done this kind of math.

if it helps, the answer is supposed to be

my colleagues and I can't figure out how to come to that answer. It's probably something simple.

edit: I tried to solve it by inscribing an octagon, and then finding the distance from the center of the octagon to the side of the octagon. but I got 1 + sqrt(2) as the radius... rather than the 2*sqrt(2+sqrt(2))

 

[BvU]

if it helps, the answer is supposed to be

The $\pi$ shouldn't be in there -- as you later seem to know...

$\theta = \pi/8\quad\& \quad R = 1/\sin\theta$ so: Area = $\pi \over \sin^2\theta$

$\cos 2\theta = \cos^2 {\theta} - \sin^2 {\theta}=1- 2\sin^2 {\theta} \quad \Rightarrow \sin^2 {\theta} = {1 - \cos2\theta\over 2}$ and with $\cos {\pi\over 4}= {1\over 2}\sqrt 2$ we get ${1/ \sin^2 {\pi\over 8}} = {\displaystyle {4\over 2-\sqrt 2}} = 2(2+\sqrt 2)$

 

[grandpa2390]

The $\pi$ shouldn't be in there -- as you later seem to know...

View attachment 250517
$\theta = \pi/8\quad\& \quad R = 1/\sin\theta$

$cos 2\theta = \cos^2 {\theta} - \sin^2 {\theta}=1- 2\sin^2 {\theta} \quad \Rightarrow \sin {\theta} = {1 - \cos2\theta\over 2}$ and with $\cos {\pi\over 8}= {1\over 2}\sqrt 2$ we get ${1/ \sin {\pi\over 8}} = {\displaystyle {4\over 2-\sqrt 2}} = 2(2+\sqrt 2)$

why shouldn't the pi be there? Area = pi*r^2

 

[BvU]

Oh sorry, fixed on determining R

Have to edit my $\TeX$ some more, patience ...

 

[grandpa2390]

Oh sorry, fixed on determining R

Have to edit my $\TeX$ some more, patience ...

that's ok. I solved it with sin(22.5) = 1/h. for some reason I didn't get it earlier. but now it's working out.

no way I could have gotten it in the form you got it though. the form the solution wants.

edit: I see my mistake. I thought the radius would be 2*sqrt(2+sqrt(2)), but it is sqrt(2*(2+sqrt(2))

 

[BvU]

Post #2 fixed now, I hope...

Half-angle formula clear ?

 

[grandpa2390]

Post #2 fixed now, I hope...

Half-angle formula clear ?

you helped me see my error. I miscalculated what the radius was supposed to be.
I also inscribed my octagon improperly. I should have been looking for the distance to an angle rather than the distance to a side.
thanks!

 

[archaic]

The $\pi$ shouldn't be in there -- as you later seem to know...

View attachment 250517
$\theta = \pi/8\quad\& \quad R = 1/\sin\theta$ so: Area = $\pi \over \sin^2\theta$

$\cos 2\theta = \cos^2 {\theta} - \sin^2 {\theta}=1- 2\sin^2 {\theta} \quad \Rightarrow \sin^2 {\theta} = {1 - \cos2\theta\over 2}$ and with $\cos {\pi\over 4}= {1\over 2}\sqrt 2$ we get ${1/ \sin^2 {\pi\over 8}} = {\displaystyle {4\over 2-\sqrt 2}} = 2(2+\sqrt 2)$

I can't see how $R$ is the radius of the big circle, it looks like just a portion of it, could you explain more please?

 

[BvU]

$R$ is what the sketch starts out with. Its relation to the radius of the small circles is the object of the analysis. Is it clear to you where the $\pi/8$ comes from ?

 

[archaic]

$R$ is what the sketch starts out with. Its relation to the radius of the small circles is the object of the analysis. Is it clear to you where the $\pi/8$ comes from ?

My bad, for some reason, it wasn't clear to me in your picture that the intersection point was the center of the common circle.