The centers of four circles are at the vertices of a square of sidelength 100m. Each circle has the radius of 100m. Which is the area of their intersection?
[Math] Area of intersection between 4 circles centered at the vertices of a square
areacirclesgeometry
Related Solutions
A picture would help. I hope it will be clear how to draw it.
Let the two circles have equations $(x+1)^2+(y-1)^2=1$ and $(x-1)^2+(y-1)^2=1$. So they are tangent to the $x$-axis, and symmetrical about the $y$-axis. The centre of the bottom side of the square is then at $(0,0)$.
Let the square have side $2t$. Then the point $(t,2t)$ is on the right-hand circle.
It follows that $(t-1)^2+(2t-1)^2=1$. This simplifies to $5t^2-6t+1=0$. The relevant root is $t=\frac{1}{5}$. So the square has area $\frac{4}{25}$.
Remark: We could strip away the coordinatization, and write the solution using only the Pythagorean Theorem. But coordinatization is a (provably) powerful tool in geometry, so one might as well use it.
If we really want to use Pythagorean Theorem only, let $O$ be the centre of the right-hand circle, and let $P$ be the point where the top right corner of the square meets the circle. Drop a perpendicular from $O$, towards the $X$-axis, and let $Q$ be the point where this perpendicular meets the top edge of the square, extended.
Then $PQ=1-t$ and $OQ=1-2t$. By the Pythagorean Theorem, $(PQ)^2+(OQ)^2=(OP)^2$, and therefore $(1-t)^2+(2-t)^2=1^2$. Expand, and continue as in the main solution.
Firstly, to remove any doubts, you are correct so far in that:
- The radius of the circle is $10$ (though you kind of went about this in a roundabout method)
- The diagonal is $10\sqrt{2}$
- The square has area $100$
- The area of each quarter-circle is $25\pi$ for a total of $50\pi$
However, you forget that they overlap. This means that while the quarter circles sum to $50\pi$ in area, some of it just doesn't matter, since the other circle is already covering up that bit of space. (If you've ever heard of the inclusion-exclusion principle for stuff like probabilities and counting problems, it has a very similar feel to it.) Consider the below picture, which only has one of the circles in it. Imagine rotating a copy of this square $180^\circ$ and superimposing it on itself: you obviously end up with the original picture you have, but with overlap:
In fact, it is the area of the overlap you are tasked to find. That overlap is the blue area in the picture below:
With the picture color-coded like this, it should not be difficult to convince you that
$$\text{(The blue area)} = \text{(The square's area)} - \text{(The two bits of green area)}$$
In a similar vein, the first picture should show you that the area of one portion of the green area can be obtained by taking the square, and subtracting the area of the quarter circle. Double that for the green area, and subtract that from the area of the square. Then you have your result, the area of the overlapping region.
Best Answer
I think this question becomes much easier if we can know the four intersection points. And we can do this through observing the equilateral triangles if we join lines between centers of the circles.
If my calculation right, C $(\frac{\sqrt3}{2}r,\frac{1}{2}r)$, B $(\frac{1}{2}r,\frac{\sqrt3}{2}r)$, and you can work out the leftmost and downmost ones because of symmetry about the center $(\frac{1}{2}r,\frac{1}{2}r)$ but not even necessary.
One fourth of the area (suppose the upperright quarter) can be represented as:
$$\frac{S}{4} = \pi r^2\times\frac{30}{360}-\frac{(\frac{\sqrt3}{2}r-\frac{1}{2}r)\frac{1}{2}r}{2}\times 2$$
Simplified:
$$S=\frac{\pi}{3}r^2-(\sqrt3-1)r^2$$
Graph and the hilarious font :)
Look! This is integration-free!