Dynamic Geometry: P57

Let the center of the unit square be $O(0,0)$ , the origin of the $xy$ -plane, Let the vertices of top two variable equilateral triangle be $A$ and $B$ , and the midpoint of $AB$ be $P(x,y)$ . Let the side length of the two triangles be $a$ , then $A = \left(\frac 12 - \frac a2, \frac 12 - \frac {\sqrt 3}2a\right)$ , $B = \left(-\frac 12 + \frac {\sqrt 3}2, \frac 12 - \frac a2 \right)$ , and the coordinates of $P(x,y)$ are:

$\begin{cases} x = \dfrac 12 \left(\dfrac 12 - \dfrac a2 - \dfrac 12 + \dfrac {\sqrt 3}2 a \right) = \dfrac {\sqrt 3 -1}4 a \\ y = \dfrac 12 \left(\dfrac 12 - \dfrac {\sqrt 3}2 a + \dfrac 12 - \dfrac a2 \right) = \dfrac 12 - \dfrac {\sqrt 3 + 1}4 a \end{cases} \\ \begin{aligned} \implies a & = \frac {4x}{\sqrt 3-1} \\ \implies y & = \frac 12 - (2+\sqrt 3)x \end{aligned}$

Therefore the locus of $P$ is a fixed straight line. Let the side length of the orange square be $b$ . Then $b = 2c \cos \theta$ , where $c$ is the $x$ -intercept of the locus and $\theta = \tan^{-1} \dfrac 1{2+\sqrt 3}$ . And $x$ -intercept $c$ is given by:

$\begin{aligned} 0 & = \frac 12 - (2+\sqrt 3)c \\ \implies c & = \frac 1{2(2+\sqrt 3)} \\ b & = \frac {2\cos \theta}{2(2+\sqrt 3)} = \frac 1{2+\sqrt 3} \cdot \frac {2+\sqrt 3}{2\sqrt{2+\sqrt 3}} = \frac 1{2\sqrt{2+\sqrt 3}} \end{aligned}$

Then the area of the orange square $b^2 = \dfrac 1{4(2+\sqrt 3} = \dfrac 12 - \dfrac {\sqrt 3}4$ . Therefore $p+q+m+n = 1+2+3+4 = \boxed{10}$ .