1. ## Circles..

Sorry unsure where to put this.

Not sure if I should draw it out... I'll try to explain...

There is a circle. It has a radius of 10pi

There are four equal circles within the circles, taking up the exact same space and such, just at the four quarters of the circle.

If you imagine it, in the middle between these 4 circles is a diamond shaped thingy.

What is the area of the largest circle than can fit in there?

The answer is triple square route of 25pi but how does that work? And why?

2. Originally Posted by Mukilab
Sorry unsure where to put this.

Not sure if I should draw it out... I'll try to explain...

There is a circle. It has a radius of 10pi
radius or area? It would be peculiar for the radius to involve "pi".

There are four equal circles within the circles, taking up the exact same space and such, just at the four quarters of the circle.

If you imagine it, in the middle between these 4 circles is a diamond shaped thingy.

What is the area of the largest circle than can fit in there?

The answer is triple square route of 25pi but how does that work? And why?
Draw a two perpendicular diameters, dividing the circle into 4 equal areas. In order to fit into one of those quarters, the small circle must have center where the distance from that center to the large circle is equal to the distance from the center perpendicular to one of those diameters. If we let the distance from the center of the large circle to the feet of those perpendiculars be "x", then the distance from the center of the large circle to the center of the small circle is $\displaystyle \sqrt{2}x$ and the distance from the center of the small circle out to the large circle is $\displaystyle R- \sqrt{2}x$ (R is the radius of the large circle). Then we must have $\displaystyle x= R- \sqrt{2}x$ so $\displaystyle (1+\sqrt{2})x= R$ and $\displaystyle x= \frac{R}{1+ \sqrt{2}}$. That small circle, then, has radius $\displaystyle \frac{R}{1+ \sqrt{2}}$ and so area $\displaystyle \frac{\pi R^2}{(1+ \sqrt{2})^2}= \frac{\pi R^2}{3+ 2\sqrt{2}}$.

Now look at the square formed by those four centers of the small circles. It has sides of length $\displaystyle \frac{2R}{1+\sqrt{2}}$ and so area $\displaystyle \frac{4R^2}{3+2\sqrt{2}}$.

The region you want is that square with four quarter small circles taken out. But those four quarter circles together have area equal to the a single small circle. The area of the center region is the area of the square minus the area of the small circle:
$\displaystyle \frac{4R^2}{3+2\sqrt{2}}-\frac{\pi R^2}{3+2\sqrt{2}}= \frac{(4-\pi)R^2}{3+ 2\sqrt{2}}$

I'm not clear on what you mean by "triple square root" but I don't that could be "triple square root of 25 pi".

3. You need to find the area of the largest circle that will fit into the shaded area (imagine it is drawn properly...)

Sorry, does this explain anything? Oh and I meant 10cm

Using that I worked out the area of one of the four circles as 25pi.... it shouldn't work but it does.

The actual answer is roughly 1.72. You get the by doing the square route of 25pi, square routing that and square routing that.

4. Originally Posted by Mukilab
...
There is a circle. It has a radius of 10pi

There are four equal circles within the circles, taking up the exact same space and such, just at the four quarters of the circle.

If you imagine it, in the middle between these 4 circles is a diamond shaped thingy.

What is the area of the largest circle than can fit in there?
The answer is triple square route of 25pi but how does that work? And why?
Sorry, ... I meant 10cm
You have 3 size circles:

Middle size. 4 identical circles: Radius = $\displaystyle 10(\sqrt{2}-1) = 4.142135$

Small inner circle: Radius = $\displaystyle 10 - 2 \times 4.142135 = 1.715728753$

Area of small circle: $\displaystyle 1.715729^2 \pi = 9.248 \text{ sq.cm}$

$\displaystyle \sqrt{ \sqrt{ \sqrt{ 25\pi} } } = 1.725385$

Are you looking for the radius or the area?
.

5. area, but please explain how you came about to your first answer, the square route of 2-1.... (how do I type in maths equations?)

6. Originally Posted by Mukilab
area, but please explain how you came about to your first answer, the square route of 2-1.... (how do I type in maths equations?)
Move your mouse pointer to a formula in a post & right click it.
You should see a pop-up that has that formula in text form.
When you "quote" to reply to a post, you can see the formulas displayed in text form.

::
How did I get $\displaystyle \sqrt{2}-1$

From the attached image:

R = radius of the larger circle
r = radius of the smaller circle

Diameter of the square is R-r (that should be obvious)

$\displaystyle r^2 + r^2 = (R-r)^2 = R^2 - 2Rr + r^2$ which results in a simple quadratic equation.

Suppose r=1
Then the diameter of the square is $\displaystyle \sqrt{2}$ ( which is R-r )
Thus expressing R in terms of r
R = $\displaystyle r(\sqrt{2}+1)$

& similar
Suppose R=1
Expressing r in terms of R
r = $\displaystyle R(\sqrt{2}-1)$

Hope that helps.
.

7. Thanks, that's pretty darn awesome. Nobody in my class (Year 9, according to England) could get that, I got the answer but I had no idea why it worked. Thanks for explaining this! I'll be sure to show it to my classmates.

8. I hope you feel elated.

Somehow, a person with a phd in Mathematics couldn't work this out. My friend couldn't understand my explanation so he gave it to his mother (the Dr) who couldn't solve it....

:P

9. Does anyone have any similar puzzles? I found this quite enjoyable ^^