Six circles are tangent to each other and an equilateral triangle is inscribed around them as shown. What percent of the area of triangle is NOT shaded?

- geerky42

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- schrodinger

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- geerky42

Bad drawing, sorry. Hopefully you know what I'm trying to draw...|dw:1352593402510:dw|

- anonymous

Do you know any other information?

- geerky42

No, this is all given information I have.

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## More answers

- anonymous

Well it's equilateral so all of it is the same...

- geerky42

I guess so.

- anonymous

The area of an equilateral triangle is :
\[\frac{ s^2\sqrt{3} }{ 4 }\]

- geerky42

And circles are tangent to each other, so they all should be the same.

- anonymous

Mhmm... Let me see. You post very intrsting questions.

- geerky42

And how can I applying to it?

- anonymous

I am thinking...

- geerky42

Sorry for interruption, but I need your help... @Hero @tcarroll010 @AccessDenied @AriPotta

- geerky42

Any ideas, hints, tips?

- anonymous

Are all those circles of the same radius?

- anonymous

if you know the area of the triangle and the area of the circle(s) what can you say about the area not covered by the circles?

- geerky42

Well, we only know that they are tangent to each other and in the shown image on my paper, they appear that they also tangent to the sides of triangle too, so I guess yeah.
This is literally the only given information I have...

- geerky42

Go on... @jon.stromer.galley

- anonymous

Well if they are tengential to each other I assume they have the same radius.

- geerky42

I suppose.

- geerky42

So where should we start?

- anonymous

I am exploring a relationship between the triangle and a second triangle formed by connecting the centers of the circles.

- geerky42

Why not the relationship between the area of circle to the area of triangle? I think this is good start, but I'm not sure.

- geerky42

I don't know, lol.

- anonymous

I am thinking about saying that:
|dw:1352594259221:dw|

- anonymous

I don't think so though...

- geerky42

I doubt it. |dw:1352594316081:dw|

- anonymous

Yeah I know. Was just wondering...

- AccessDenied

A better picture perhaps: http://puu.sh/1oLWD

- geerky42

I think @tcarroll010 has a good point, perhaps we should determine the relationship between the triangle and a second triangle formed by connecting the centers of the circles.

- geerky42

Much better. @AccessDenied

- anonymous

LOT better.

- AccessDenied

Oh wow, this is sort of interesting:
I draw in all of the tangents of the circles to the exterior triangle and an interior triangle:
http://puu.sh/1oM08

- anonymous

Where are you getting these questions from? O_o .

- anonymous

Hmm!

- anonymous

Similar triangles perhaps?

- anonymous

Ok, so all you know for sure is that the triangle is equilateral and that the circles have a radius r. That is enough to express the combined circle area, but to answer the question you need to know how big the triangle is (how long the sides are) in terms of R.
Let's look at the lower left corner.

- anonymous

|dw:1352594743279:dw|

- anonymous

We don't know that...

- AccessDenied

Hint: you have an equilateral triangle for the exterior, so you know some angles. ;)

- anonymous

I've got what might be a good idea. You use trig. And 30-60-90 right triangles in the corners. I was independently working a similar diagram to Accessd.

- anonymous

let's draw it again...|dw:1352594832325:dw|

- anonymous

Ohh yeah!!!

- anonymous

Isn't that also one of those special triangles?

- anonymous

At this point if you have been shown you have enough data to figure out the base of the triangle.
The length of a side of the over all triangle will be then 3r + 2( length of the base of the triangle we just drew)

- anonymous

It looks like a whole bunch of us about simultaneously hit on the right idea! Fun trig at this point.

- geerky42

I have sense I'm getting closer to solution, yet I have no idea where to start. I have to find the relationship between the radius and length of triangle side, right?

- anonymous

|dw:1352594992142:dw|

- anonymous

Now we know the radius of all those circles are 1!

- AccessDenied

not necessarily. The 30-60-90 rule only expresses a ratio. :P

- geerky42

And from this, we can find the ratio of the total area of six circle to the area of triangle, right?

- anonymous

Can we not assume? O_o .

- anonymous

the radius is still "1" r
but now the base can be expressed in terms of r as well

- anonymous

Not sure is the side length would just be 2r...

- anonymous

To be pure, you would use "r" for radius, but you are free to call radius "1" for your purposes, so, yes you can make the assumption, with that proviso. And then it's simple to get your triangle side lengths. Nice little problem!

- anonymous

Well looking at @AccessDenied Drawing... :
Since the radius is 1 we can say the side length's of the smaller, inside triangle are 3, 3 and 2 going clockwise.

- anonymous

4 4 and 4 sorry.

- anonymous

|dw:1352595331990:dw|

- anonymous

I am assuming the height would be 3...

- anonymous

|dw:1352595405601:dw|

- anonymous

Now I can't get a ratio....

- anonymous

With a circle radius of "r", being the "short leg", then the "long leg" will be r times sqrt(3). So a triangle side will be 2r(sqrt(3)) + 2r.

- anonymous

So it would just be 2root(3)+2 .

- anonymous

If you are going with your assumption or requirement that r is "1", then, yes.

- anonymous

So, you should be able to get the altitude using Pythagorean and then area of triangle, and areas of circles. You're just about done!

- anonymous

I calculate that the length of a side of the master triangle is 6.4641r
Anyone else?
tanks to:
adjacent = r / tan(30)

- anonymous

or alternatively by expressing the length of the line as:
2r * root(3) + 3r
that plugs in nicely to the formula for the area of an el triangle:
(length of side * root(3)) / 4
so that tells us the area is...

- AccessDenied

I missed the answer placed by the original poster, but my work in ms paint is here:
http://puu.sh/1oNgt

- geerky42

I got the same answer.
Thank you, everyone, for your times. I appreciate it.

- AccessDenied

You're welcome! :)

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