# Area of Bulging Square

Here's the problem:

Let $ABCD$ be a square. Let $A$ be the center of the circle with radii $AB$ and $AD;$ $B$ be the center of the circle with radii $BA$ and $BC;$ $C$ be the center of the circle with radii $CB$ and $CD;$ and $D$ be the center of the circle with radii $DA$ and $DC.$ When the two arcs $AC$ and the two arcs $BD$ are drawn, a bulging square region is formed, with center the center of $ABCD.$ Find the area of this region.

I've tried to split the region into multiple pieces, but I'm not able to create a division that leads to an easy area calculation. For example, drawing $BD$ doesn't remove the curved lines on the other side.

Any hints would be appreciated.

• Try subtracting area from a larger region. – Justin Benfield Mar 14 '16 at 0:16
• What areas should I subtract? I tried subtracting the region by the corners but that didn't work. Also, please post your hint an answer so I can check it if correct @Justin Benfield – K. Jiang Mar 14 '16 at 0:20
• Finally figured it out, it's very clever. It would help to have a drawing of the figure. – Justin Benfield Mar 14 '16 at 0:43
• I'm not exactly sure how to make a good image of this - I would have to draw it freehand. So how would you do it? Could you please make it an answer? – K. Jiang Mar 14 '16 at 0:44

You asked for hints, so I'll give you only hints. Your "bulging square" is made of the regular square $EFGH$ and four circle segments like the one in red between points $E$ and $H$.

By the construction, it is clear that $\angle EAH=30°$ and $AE=AH=1$. (I'm assuming the side of the original square is $1$.) You can then easily calculate the length of $EH$, the area of triangle $AEH$, and the area of the circular sector $AEH$. The area of the segment is the difference between the areas of the sector and the triangle.

Since you already found $EH$, find the area of the square and add that to four times the area of the segment, and you are done.

Let me know if you need more beyond these hints.

• You might want to explain how that angle is really $30^{\circ}$ (it's not obvious imo). – Justin Benfield Mar 14 '16 at 0:54
• Nice answer btw. – Justin Benfield Mar 14 '16 at 0:54
• And also that angle is $30^\circ$ by considering two equilateral triangles $\Delta AEB$ and $\Delta AHD$, as it seems. – Amir Naseri Mar 14 '16 at 1:20
• @AmirNaseri: You are correct; that is how I got the $30°$. I considered adding those equilateral triangles to the diagram but thought it would look too cluttered. I also thought it was clear: it looks like I was wrong. – Rory Daulton Mar 14 '16 at 10:10