Looking for the smallest circle which will hold three 1×1 squares. What is its radius?
By visualisation, I think that this is the optimal arrangement, but I'm unable to find the radius. Any hints would be appreciated.
(source: stetson.edu)
[Math] 3 squares in a circle
algebra-precalculusgeometry
Related Solutions
There are two solutions.
- The circle covering four squares (arranged in a diamond) with $$R=\sqrt{0.5^2+0.5^2}=\sqrt{2}/2$$
- The larger circle covering eight squares (also in a diamond, 3 per side) with $$R=\sqrt{1.5^2+0.5^2}=\sqrt{10}/2$$
It helps to draw a diagram like this:
or this:
I don't think any bigger would fit. But how do you prove this statement???
So what you are asking is: find the smallest square (rotated 45 degrees with respect to axes) containing $2$ or $3$ points. Automatically the smallest square will have at least two points on the bounday.
It is simpler if you make a 45 degree rotation of your problem so that square have sides parallel to the axes.
The solution is not unique, even if you request the smallest square. For example if you take points $(0,0),(1,0),(2,0)$ (in the rotated frame) any square of size $2$ which contains $(0,0)$ and $(2,0)$ will be a solution.
Clearly the side of the minimal square will be given by this formula: $$ s = \max\{|x_1-x_2|,|x_1-x_3|,|x_2-x_3|,|y_1-y_2|,|y_1-y_3|,|y_2-y_3|\} $$ where $(x_i,y_i)$ are the coordinates (in the rotated frame) of your points.
To find a particular solution you can always take the lower-left vertex of your square to be the point with coordinates: $$ \begin{cases} x = \min\{x_1,x_2,x_3\}\\ y = \min\{y_1,y_2,y_3\} \end{cases} $$
The 45-degree rotation is given by $$ \begin{cases} x' = \frac{\sqrt 2}{2}(x-y)\\ y' = \frac{\sqrt 2}{2}(x+y)\\ \end{cases} $$
Best Answer
The side lengths of the highlighted triangle are $1$, $\sqrt{2^2+\left(\frac{3}{2}\right)^2}$ and $\sqrt{2^2+\left(\frac{1}{2}\right)^2}$ by the Pythagorean theorem. Since its area is $1$, we have: $$ \color{red}{R} = \frac{abc}{4\Delta} = \frac{1}{4}\sqrt{1\cdot\frac{25}{4}\cdot\frac{17}{4} }=\color{red}{\frac{5}{16}\sqrt{17}}\approx 1.28847. $$