Begin with a square. Using the side length as a radius, construct four circles, centered at each vertex. These circles divide the square into nine regions: four of one shape, four of another shape, and one unique one in the center. Have we got a name for any of these shapes? Is is possible to calculate their areas without using calculus?

(I have already calculated the areas, using calculus. The large central region has area $1+\frac{\pi}{3}-\sqrt3$, the four regions sharing its boundary each have area $\frac{\pi}{12}+\frac{\sqrt3}{2}-1$, and the four remaining regions, along the edges of the square, each have area $1-\frac{\pi}{6}-\frac{\sqrt3}{4}$, if I didn't make any mistakes.)

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    $\begingroup$ This was the Missouri State Problem Corner #8 The second solution uses just geometry. A discussion of the 3D problem is here $\endgroup$ Dec 2, 2017 at 15:33
  • $\begingroup$ That geometric solution is lovely! Thank you. :) $\endgroup$ Dec 2, 2017 at 15:40
  • $\begingroup$ The 3D problem took me a double integral and was a mess. The 4D problem has a cute simple solution. The link from the Missouri solution page is broken, though. $\endgroup$ Dec 2, 2017 at 15:45
  • $\begingroup$ @RossMillikan As the dimension increases, doesn't that central region eventually not exist anymore? Even for 4 dimensions, the main diagonal has length $\sqrt4=2$, so I don't see how there's anything to talk about. Is that the "cute, simple solution"? $\endgroup$ Dec 2, 2017 at 15:48
  • $\begingroup$ How do we even visualise the 4D problem? $\endgroup$
    – jonsno
    Dec 2, 2017 at 15:55

2 Answers 2


I’m not sure about the names of these regions, but we can find the areas of each using nothing more than trigonometry.

Let the square have sides length $1$. Let the region that shares a side with the square have area $a$, the other region which appears four times have area $b$ and the unique central region have area $c$. We would like to find three linearly independent equations that relate these variables.

Two are easy to find. The area of the square yields $$4a+4b+c=1.$$ Furthermore, exactly $1/4$ of the area of one of the circles is contained in the square, which gives us that $$2a+3b+c=\pi/4.$$

To find a third equation, consider two of the circles that are centered at neighboring vertices of the square. The area of the intersection of these two circles is $\frac{2\pi}3-\frac{\sqrt{3}}2$ and this can be found with trigonometry. (A nice little explanation of how to do this in general is found here: http://jwilson.coe.uga.edu/EMAT6680Su12/Carreras/EMAT6690/Essay2/essay2.html ) Half of this area lies inside our square, which gives us our third equation, $$a+2b+c= \frac{2\pi}6-\frac{\sqrt{3}}4.$$

At this point, we just need to solve the system. I got solutions $$a=1-\sqrt{3}/4-\pi/6,$$ $$b=\sqrt{3}/2+\pi/12-1,$$ $$c=1+\pi/3-\sqrt{3}.$$

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    $\begingroup$ Nice approach. We don't even really need trigonometry for the third equation; in this special case of the intersection of circles, you can use the Pythagorean Theorem to get the area of an equilateral triangle, and use the dissection of a regular hexagon into six equal equilateral triangles to deduce the area of each circular sector. $\endgroup$
    – David K
    Dec 2, 2017 at 16:45
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    $\begingroup$ This is just what I was looking for. I had two equations, and you've shown me how to find a third one. Thank you! $\endgroup$ Dec 2, 2017 at 17:02
  • $\begingroup$ @GTonyJacobs No problem. If this satisfactorily answers your question, feel free to accept it. $\endgroup$ Dec 2, 2017 at 17:23

Here's a relatively direct approach. Construct the square and inscribe the circular arcs about the four vertices as shown below.

enter image description here

Since $\triangle ADE$ is equilateral and since $\angle BAD$ is a right angle, $\angle BAE = \frac\pi6$ and the area of the circular sector with edges $AB$ and $AE$ is $\frac\pi{12}.$ Likewise, $\angle ADF = \frac\pi6,$ so $\angle AEF = \frac\pi6,$ and therefore $\triangle ADF \cong \triangle AEF \cong \triangle ABE$ and $DF = EF = BE$; nine other pairs of points symmetric to these pairs also span the same distance.

Since $E$ is on the horizontal line of symmetry of the square, $\triangle ABE$ has base $1$ and height $\frac12,$ so its area is $\frac14.$ The area of the circular segment between the line segment $\overline{BE}$ and the arc $\overset{\frown}{BE}$ (the shaded region in the figure) is therefore $S = \frac\pi{12}-\frac14.$

To get the areas of the various regions bounded by the circular arcs and/or the edges of the square, we can take the areas of the eight triangles and the square that we would obtain by "straightening the sides" of these regions, and add or subtract an appropriate multiple of the area $S$ to represent the addition or deletion of circular segments on the edges of those figures.

For $\triangle ABH,$ the base is $1$ and the height is $1 - \frac{\sqrt3}2,$ so the area of the triangle is $\frac12 - \frac{\sqrt3}4$ and the corresponding region bounded by the line $AB$ and the arcs $\overset{\frown}{AH}$ and $\overset{\frown}{BH}$ has area $$\frac12 - \frac{\sqrt3}4 - 2S = 1 - \frac{\sqrt3}4 - \frac\pi6. \tag1$$

The diagonal of square $EFGH$ is $1 - 2\left(1 - \frac{\sqrt3}2\right) = \sqrt3 - 1,$ so the side of the square is $GH = \frac{\sqrt2}2(\sqrt3 - 1).$ The triangle $\triangle AGH$ is therefore equilateral with side $\frac{\sqrt2}2(\sqrt3 - 1)$, so its area is $$ \frac{\sqrt3}4\left(\frac{\sqrt2}2(\sqrt3 - 1)\right)^2 = \frac{\sqrt3}2 - \frac34 $$ and the area of the region bounded by three arcs is $$ \frac{\sqrt3}2 - \frac34 + S = \frac{\sqrt3}2 - 1 + \frac\pi{12}.\tag2 $$

The area of the central region bounded by four arcs is the area of the square $EFGH$ plus four circular segments, that is, $$ \left(\frac{\sqrt2}2(\sqrt3 - 1)\right)^2 + 4S = (2 - \sqrt3) + \left(\frac\pi3 - 1\right) = 1 - \sqrt3 + \frac\pi3. \tag3 $$

Equations $(1),$ $(2),$ and $(3)$ confirm the results previously found by other methods.


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