geometryrectangles
I've been working on a geometrical issue for some time, but I cannot find an answer. I have a rectangle inside another rectangle. The only two values I know is the rotation of the smaller rectangle regarding the bigger rectangle α and also the diameter of the bigger rectangle d.
Is there any way to calculate the dimensions of the inner rectangle or do I need more information?
I don't think there's any "formula" for this and as far as I remember the problem is conjectured to be NP-hard, but not proven to be. For a heuristic that finds very good solutions (conjectured by the authors to be optimal), see this.
The problem is symmetric when the shapes are centered to each other. As such, I can see 4 cases of overlap between a circle of radius $R$ and a rectangle of half lengths $a$ and $b$.
The shaded area corresponding to one quadrant only is calculated as follows: Consider the cases where $b \leq a$ and vary the radius $R$ from small to large.
$$ \text{(Area)} = \begin{cases} \tfrac{\pi}{4} R^2 & R \leq b \\ \tfrac{1}{2} b \sqrt{R^2-b^2} + \tfrac{1}{2} R^2 \sin^{-1}\left( \tfrac{b}{R} \right) & b < R \leq a \\ \tfrac{1}{2} \left( a \sqrt{R^2-a^2} + b \sqrt{R^2-b^2} \right) + \tfrac{1}{4} R^2 \left( 2 \sin^{-1}\left( \tfrac{a}{R} \right) + 2 \sin^{-1} \left( \tfrac{b}{R} \right)- \pi \right) & a < R \leq \sqrt{a^2+b^2} \\ a b & R > \sqrt{a^2+b^2} \end{cases} $$
You can check that at the transition values of $R$ the area results match between adjacent cases. For example area (2) when $R=b$ is $\tfrac{1}{2} b^2 \sin^{-1} \left( \tfrac{b}{b} \right) + \tfrac{1}{2} b \sqrt{b^2-b^2} = \tfrac{\pi}{4} b^2$ which equals the area (1) when $R=b$.
Best Answer
This is not enough information. Imagine sliding the bottom point to the right. The side of the inner rectangle to the right will get smaller and the side to the left will get larger.
However, knowing the diameter of the inner rectangle is enough.
Suppose the sides (width, height) of the big rectangle are $r$ and $s$, and the sides (width, height) of the inner rectangle are $u$ and $v$.
For the triangle on the lower right, the legs are $u \cos a$ and $u \sin a$.
For the triangle on the lower left, the legs are $v \cos a$ and $v \sin a$.
On the bottom, we get $u \cos a+v \sin a =r$.
Similarly, on the upper right, the remaining part of the height is $v \cos a$, so $u \sin a+v \cos a = s$.
Squaring these, $r^2 =(u \cos a+v \sin a)^2 =u^2\cos^2a + 2uv\cos a\sin a + v^2 \sin^2 a $ and $s^2 (u \sin a+v \cos a )^2 =u^2\sin^2a+2uv\sin a \cos a + v^2\cos^2 a $.
Adding these, $d^2 = r^2+s^2 =u^2+v^2+4uv\sin a \cos a =u^2+v^2+2uv\sin 2a $.
From the top, we have $r =v \sin a + u \cos a $ and from the left we have $s =u \sin a + v \cos a $.
However, these are the same as before, so we can not deduce anything further.
However, if we know the diameter of the inner rectangle, say $c^2 = u^2+v^2$, then we have $d^2 =c^2+2uv\sin 2a $ so we can get $uv$.
Knowing $uv$ and $u^2+v^2$, we can get $u+v$ and $u-v$ ($(u+v)^2 = u^2+v^2+2uv$ and $(u-v)^2 = u^2+v^2-2uv$) and thus get $u$ and $v$.