1. ## Circles in square...

A circle is inscribed in a square with sides = 40.

A smaller (of course!) circle tangent to the above
circle and 2 sides of the square is inscribed in
one of the corners of the square.

What is the diameter of this circle?

2. ## Re: Circles in square...

The radius of the large circle plus the diameter of the smaller corner circle is equal to the half the length of the diagonal of the square.

$R + d = 20\sqrt{2}$

$R$ of course is just half the length of the side of the square.

$R=\dfrac{40}{2}=20$

$20+d = 20\sqrt{2}$

$d = 20(\sqrt{2}-1)\approx 8.3$

3. ## Re: Circles in square...

..BUT ~8.3 is not the diameter of the inscribed circle: it is the distance
from large circle's circumference to the corner of the square.
In order for circle to be inscribed, its diameter must be shorter than ~8.3; no?

4. ## Re: Circles in square...

Originally Posted by DenisB
..BUT ~8.3 is not the diameter of the inscribed circle: it is the distance
from large circle's circumference to the corner of the square.
In order for circle to be inscribed, its diameter must be shorter than ~8.3; no?
You're right

$d = \dfrac{20(\sqrt{2}-1)}{\sqrt{2}} \approx 5.86$

5. ## Re: Circles in square...

In triangle $OBC$

$\displaystyle (20+r)^2=2(20-r)^2$

so $\displaystyle r=20 \left(3-2 \sqrt{2}\right)$

and the diameter $\displaystyle =40\left(3-2\sqrt{2}\right)\approx 6.86$

6. ## Re: Circles in square...

Thanks Idea; was hoping for that solution!!

7. ## Re: Circles in square...

there is an easier way

$\displaystyle r\sqrt{2}+r+20=20\sqrt{2}$

solve for $r$