Reach for the Summit - M-S5-A2

Setup

We want to find configurations that maximize or minimize the difference between the area of a square and the area of an equilateral triangle inscribed in it. To find these configurations, we will take the following approach.

We will take an equilateral triangle of side length $d$ . Then, we will try to fit squares around it. Our goal will be to find the largest and the smallest fitting squares.

To that end, we will hinge one vertex (namely, $A$ ) of the equilateral triangle at the origin. Other vertices ( $B, C$ ) of the triangle will me movable. The x-axis will contain the bottom side of the fitting square, and this bottom side will have a vertex, which is at origin, on it. The top side will be parallel to the x-axis. The left and the right side of the square will be parallel to the y-axis, and will contain at least one vertex each.

Case 1

The enclosing square here is named $APQR$ . Let $a$ denote the side length of the square. Also let $\angle BAP = k$ . So, $\angle CAP = \frac{\pi}{3} + k$ .

Fig-1: 0 <= k < pi/6 for case 1

In this case, $k$ will vary from $0$ to $\frac{\pi}{6}$ .

From the triangle $\Delta ABP$ , $\cos k = \frac{AP}{AB}$ $\implies a = AP = AB \cos k = d \cos k$

But this square will not always circumscribe $\Delta ABC$ because the vertex $C$ needs to be on $QR$ . This condition can be expressed as $a = d \cos k = AC \sin (\frac{\pi}{3} + k) = d \sin (\frac{\pi}{3} + k)$ $\implies \sin (\frac{\pi}{2} - k) = \sin (\frac{\pi}{3} + k)$ $\implies (\frac{\pi}{2} - k) = (\frac{\pi}{3} + k)$ $\therefore k = \frac{\pi}{12}$

Therefore, there is only one valid value of $k$ for $0 \leq k < \frac{\pi}{6}$ . And the corresponding side length is $a_1 = d \cos (\frac{\pi}{12})$

Case 2

The enclosing square here is $JKLM$ . The side length is denoted by $a$ as before. Let $\angle BAK = k$ and $\angle CAK = \frac{\pi}{3} + k$ .

Fig-2: pi/6 <= k <= pi/3 for case 2

In this case, $k$ will vary from $\frac{\pi}{6}$ to $\frac{\pi}{3}$ . Now we will get an expression for the side length. $a = JK = AK + JA = d \cos k - d \cos (\frac{\pi}{3} + k)$ $a = 2 d \sin (\frac{\pi}{6}) \sin (\frac{\pi}{6} + k)$ $\therefore a = d \sin (\frac{\pi}{6} + k)$

We notice that the $a$ needs to clear the height of $AC$ . Otherwise, $LM$ will enter inside $\Delta ABC$ . So, $a \geq AC \sin (\frac{\pi}{3} + k)$ $d \sin (\frac{\pi}{6} + k) \geq d \sin (\frac{\pi}{3} + k)$ $\frac{\pi}{6} + k \geq \pi - \frac{\pi}{3} - k$ $\therefore k \geq \frac{\pi}{4}$

So, the valid values of $k$ lies on the range $[ \frac{\pi}{4} , \frac{\pi}{3} ]$ . Also, $a = \sin (\frac{\pi}{6} + k)$ is an increasing function for $k \in [ \frac{\pi}{4} , \frac{\pi}{3} ]$ . Hence, the extrema are $a_2 = d \sin (\frac{\pi}{6} + \frac{\pi}{4}) = d \sin (\frac{5 \pi}{12})$ $a_3 = d \sin (\frac{\pi}{6} + \frac{\pi}{3}) = d \sin (\frac{\pi}{2}) = d$

Other cases

All the configurations of other values of $k$ are just mirror images of the two cases mentioned above. So, we have effectively found all the optimum configurations.

Results

We notice that $a_{min} = a_1 = d \cos (\frac{\pi}{12}) = d \sin (\frac{5 \pi}{12}) = a_2$ $\implies a_{min} = d \sin (\frac{5 \pi}{12})$ $a_{max} = d$

Fig-3: The difference between the area of the square and the area of the inscribed triangle is minimized.

Fig-4: The difference between the area of the square and the area of the inscribed triangle is maximized.

In our case, the reverse is happening: the square is of fixed length, and the triangle length is variable. Hence, $d_{min} = a = 1$ $d_{max} = \frac{a}{ \sin (\frac{5 \pi}{12}) } = \csc (\frac{5 \pi}{12})$

Therefore, $\lfloor 10^6 (S_{max} - S_{min}) \rfloor = \lfloor 10^6 \frac{\sqrt{3}}{4} (\csc^2 \frac{5 \pi}{12}- 1) \rfloor = \boxed{31088}$

P.S. If you want to play with the setup, feel free to use my Desmos project .

The largest equilateral triangle one that is symmetrical about the diagonal of the square as indicated by the blue triangle. The smallest equilateral triangle is one that is symmetrical about the line joining the midpoints of two opposite sides with a vertex at the midpoint of a side as indicated by the red triangle.

The largest triangle has a side of $a_{max} = \dfrac 1{\cos 15^\circ}$ and hence an area of $S_{max} = \dfrac {a_{max}^2 \sin 60^\circ}2 = \dfrac {\sin 60^\circ}{2\cos^2 15^\circ} = \dfrac {\sin 60^\circ}{1+\cos 30^\circ} = \dfrac {\sqrt 3}{2+\sqrt 3}$ .

The smallest triangle has a side of $a_{min} = \dfrac {\frac 12}{\sin 30^\circ}= 1$ and hence an area of $S_{min} = \dfrac {a_{min}^2 \sin 60^\circ}2 = \dfrac {\sqrt 3}4$ .

Therefore, $\left \lfloor 10^6 \left(S_{max} - S_{min} \right) \right \rfloor = \left \lfloor 10^6 \left(\dfrac {\sqrt 3}{2+\sqrt 3} - \dfrac {\sqrt 3}4 \right) \right \rfloor = \boxed{31088}$ .