Notes
five circles in a square solution

Solution to the Five Circles in a Square Puzzle

Five Circles in a Square

Five circles in a square. The total red area is 2424. What’s the total orange area?

Solution by Trigonometry and Lengths in an Equilateral Triangle

Five circles in a square labelled

In the above picture, let rr be the radius of the orange circles. The points AA and CC are centres of their circles and BB is so that ABA B is horizontal and BCB C is vertical. Let α\alpha be angle BA^CB \hat{A} C.

The length of the line segment ACA C is 4r4 r, so CBC B has length 4rsin(α)4 r \sin(\alpha) and ABA B has length 4rcos(α)4 r \cos(\alpha). Since CDC D has length rr, the horizontal length of the square is 2r+4rcos(α)2 r + 4 r \cos(\alpha).

As CDC D has length rr, EDE D has length rtan(α)r \tan(\alpha) and ECE C has length rsec(α)r \sec(\alpha). So the total height of the square is 2rsec(α)+2rtan(α)+4rsin(α)2 r \sec(\alpha) + 2 r \tan(\alpha) + 4 r \sin(\alpha).

Since it is a square, these two expressions must be equal. Setting them equal and working through goes as follows:

2rsec(α)+2rtan(α)+4rsin(α) =2r+4rcos(α) sec(α)+tan(α)+2sin(α) =1+2cos(α) 1+sin(α)+2sin(α)cos(α) =cos(α)+2cos 2(α) 12cos 2(α)+2sin(α)cos(α) =cos(α)sin(α) cos(2α)+sin(2α) =cos(α)sin(α) sin(2α45 ) =sin(45 α) \begin{aligned} 2 r \sec(\alpha) + 2 r \tan(\alpha) + 4 r \sin(\alpha) &= 2 r + 4 r \cos(\alpha) \\ \sec(\alpha) + \tan(\alpha) + 2 \sin(\alpha) &= 1 + 2 \cos(\alpha) \\ 1 + \sin(\alpha) + 2 \sin(\alpha) \cos(\alpha) &= \cos(\alpha) + 2 \cos^2(\alpha) \\ 1 - 2 \cos^2(\alpha) + 2 \sin(\alpha) \cos(\alpha) &= \cos(\alpha) - \sin(\alpha) \\ -\cos(2\alpha) + \sin(2\alpha) &= \cos(\alpha) - \sin(\alpha) \\ \sin(2\alpha - 45^\circ) &= \sin(45^\circ - \alpha) \end{aligned}

The last step uses the compound angle formula with one of the angles set to 45 45^\circ together with the fact that sin(45 )=cos(45 )\sin(45^\circ) = \cos(45^\circ).

This has solutions 2α45 =45 α+360 k2 \alpha - 45^\circ = 45^\circ - \alpha + 360^\circ k and 2α45 =180 (45 α)+360 k2 \alpha - 45^\circ = 180^\circ - (45^\circ - \alpha) + 360^\circ k, where kk is some integer. The first rearranges to 3α=90 +360 k3 \alpha = 90^\circ + 360^\circ k, so α=30 +120 k\alpha = 30^\circ + 120^\circ k, and the second to α=180 +360 k\alpha = 180^\circ + 360^\circ k. Since α\alpha is acute, the only solution is α=30 \alpha = 30^\circ.

Five circles in a square divided

With α\alpha established as 30 30^\circ, the above diagram shows the relationship between the orange and red circles. Each is the incircle of a 30 30^\circ-60 60^\circ-90 90^\circ triangle, with the side adjacent to α\alpha on the larger circle being the hypotenuse of the smaller. The length scale factor is therefore the ratio of the length of the hypotenuse to the length of that side, which from lengths in an equilateral triangle is 32\frac{\sqrt{3}}{2}. So the area scale factor is 34\frac{3}{4}. That is, the area of each orange circle is 34\frac{3}{4} that of the red. Since the total red area is 2424, each red circle is 1212 and thus each orange circle is 99 meaning that the total orange area is 2727.