Tangential circles inscribed within a square

[dozzer]

In this diagram, a square is inscribed with four circles of equivalent size, all with the radius of 1. Each circle is tangent to two sides of the square, two circles, and the smaller circle. Obviously each side of the square is equal to 4, but what is the radius of the smaller circle?

I think the answer is something like 2^1/2 -1, or .414..., but how would you solve for it?

This is a problem from the SAT before the math section was reformed.

 

[tiny-tim]

Welcome to PF!

Hi dozzer! Welcome to PF!

I can't see your picture yet, but if you divide the square into four squares, doesn't that put the center of the small circle at one corner of each small square?

 

[S_Happens]

Were you provided the answer and don't know how to get there or are you asking something else?

I can't see your pic either, but from the description I came up with the same answer you supplied.

 

[zgozvrm]

Divide the square into 4 even squares and you will have 4 squares, each with a circle of radius 1 circumscribed within.

Now, draw a line from the center of the small circle to one of the corners of the original large square. This line is a diagonal of one of the small squares and is equal to 2 * (R + r), where R is the radius of the large circles and r is the radius of the small circle.

We already know that R = 1, therefore the diagonal is equal to 2 * (1 + r).

The diagonal is the hypotenuse of a right triangle with legs of equal length (in this case, 2), so the length of the diagonal D = (2^2 + 2^2)^1/2, or
D = (4 + 4)^1/2
D = 8^1/2
D = 2 * (2^1/2)

From above, we know that D = 2 * (1 + r), so 1 + r = 2^1/2
and finally, r = 2^1/2 - 1
which is approximately equal to 1.4142 - 1, or 0.4142

 

[dozzer]

Thank you for the explanations, everyone.

 

[zgozvrm]

Just wanted to clean up my earlier post...

Divide the square into 4 even squares and you will have 4 squares, each with a circle of radius 1 circumscribed within.

Now, draw a line from the center of the small circle to one of the corners of the original large square. This line is a diagonal of one of the small squares and is equal to $$2(R + r)$$, where R is the radius of the large circles and r is the radius of the small circle.

We already know that R = 1, therefore the diagonal is equal to $$2(1 + r)$$.

The diagonal is the hypotenuse of a right triangle with legs of equal length (in this case, 2), so the length of the diagonal $$D = \sqrt{(2^2 + 2^2)}$$, or
$$D = \sqrt{(4 + 4)}$$
$$D = \sqrt8$$
$$D = 2\sqrt2$$

From above, we know that $$D = 2(1 + r)$$, so $$1 + r = \sqrt2$$
and finally, $$r = \sqrt2 - 1$$
which is approximately equal to 1.4142 - 1, or 0.4142

 

[tiny-tim]

Welcome to PF!

Hi zgozvrm! Welcome to PF!

Alternatively, instead of LaTeX, have a square-root to copy: √ and try using the X² tag just above the Reply box