Minimum Equilateral Triangle Inscribed in Half an Equilateral Triangle

Let the side of equilateral $\triangle DEF$ be $x$ and $\angle EFM = \theta$ . The $MC$ is given by:

$\begin{aligned} MC & = MF + FC \\ 7 & = x \cos \theta + x \cos (120^\circ - \theta) + x\sin (120^\circ - \theta)\cot 60^\circ \\ \implies x & = \frac 7{\cos \theta + \cos (120^\circ - \theta) + \sin (120^\circ - \theta)\cot 60^\circ} \\ & = \frac 7{\cos \theta - \frac 12\cos \theta + \frac {\sqrt 3}2 \sin \theta + \frac 12 \cos \theta + \frac 1{2\sqrt 3} \sin \theta} \\ & = \frac 7{\cos \theta + \frac 2{\sqrt 3} \sin \theta} \\ & = \frac 7{\sqrt{\frac 73}\sin \left(\theta + \tan^{-1} \frac {\sqrt 3}2 \right)} \\ & = \frac {\sqrt{21}}{\sin \left(\theta + \tan^{-1} \frac {\sqrt 3}2 \right)} \end{aligned}$

Therefore, $\min (x) = \sqrt{21}$ , when $\sin \left(\theta + \tan^{-1} \frac {\sqrt 3}2 \right) = 1$ . $\implies n = \boxed{21}$ .

Taking the origin of our coordinate system at point $M$ , and if the center of $\triangle DEF$ is point $C = (C_x, C_y)$ , and its circumradius is $r$ , and the clockwise angle that vector $CD$ makes with the vertical direction is $\theta$ then the coordinates of points $D , E , F$ are given by:

$D = (C_x + r \sin \theta, C_y + r \cos \theta )$

$F = (C_x + r \sin( \theta + 120^{\circ}) , C_y + r \cos( \theta + 120^{\circ} ) )$

$E = (C_x + r \sin( \theta - 120^{\circ} ) , C_y + r \cos ( \theta - 120^{\circ} ))$

Now, we know that point $D$ lies on the line $y = 7 \sqrt{3} - \sqrt{3} x$ , that point $F$ lies on the line $y = 0$ , and the point $E$ lies on the line $x = 0$ . Hence,

$C_y + r \cos \theta = 7 \sqrt{3} - \sqrt{3} ( C_x + r \sin \theta )$

$C_y + r \cos( \theta + 120^{\circ} ) = 0$

$C_x + r \sin ( \theta - 120^{\circ} ) = 0$

Substituting for $C_y$ and $C_x$ from the second and third equations into the first equation, yields,

$- r \cos( \theta + 120^{\circ}) + r \cos \theta = 7 \sqrt{3} - \sqrt{3} ( - r \sin(\theta - 120^{\circ} ) + r \sin \theta )$

Collecting terms containing $r$ , and using the formulas for the sine and the cosine of the sum and difference of angles, yields,

$r ( \cos \theta ( - \cos 120^{\circ} + 1 + \sqrt{3} \sin 120^{\circ} ) + \sin \theta ( \sin 120^{\circ} - \sqrt{3} \cos 120^{\circ} + \sqrt{3} ) )= 7 \sqrt{3}$

Substituting $\cos 120^{\circ}$ and $\sin 120^{\circ}$ , the last equation becomes,

$r = \dfrac{ 7 \sqrt{3} }{ 3 \cos \theta + 2 \sqrt{3} \sin \theta }$

Since we want the minimum $r$ , then we want the maximum of the denominator, which is $\sqrt{ 3^2 + (2 \sqrt{3} )^2 } = \sqrt{ 9 + 12} = \sqrt{21}$

Hence, the minimum circumradius is

$r_{\text{min}} = \dfrac{ 7 \sqrt{3} }{\sqrt{21} } = \sqrt{ 7 }$

Finally the side length of an equilateral triangle corresponding a circumradius $r$ is $s = 2 r \sin 60^{\circ} = \sqrt{3} r$ . Thus, the required side length of the smallest possible inscribed equilateral triangle is $s_{\min} = \sqrt{3} \sqrt{7} = \sqrt{21}$ , so that the answer is $\boxed{21}$ .

We need the pedal triangle of X(15) which is First Isodynamic Point . It is Isogonal Conjugate of Fermat point. Using formulae for Exact Trilinear Coordinates we get $\alpha = 2 \sqrt{3}$ and $\gamma = 3$ . They are perpendicular, so $n= \alpha ^2+ \gamma ^2=21$

By the properties of an equilateral triangle, $CM = 7$ , $AM = 7\sqrt{3}$ , $\cos \angle ACM = \frac{1}{2}$ , and $\cos \angle CAM = \frac{\sqrt{3}}{2}$ .

Let $x = FM$ , $y = EM$ , $z = AD$ , and $s = DE = DF = EF$ . Then $CF = 7 - x$ , $AE = 7\sqrt{3} - y$ , and $CD = 14 - z$ .

By the law of cosines on $\triangle CDF$ , $\triangle ADE$ , and by Pythagorean's Theorem on $\triangle EFM$ :

$s^2 = (7 - x)^2 + (14 - z)^2 - 2 \cdot (7 - x) \cdot (14 - z) \cdot \frac{1}{2}$

$s^2 = (7\sqrt{3} - y)^2 + z^2 - 2 \cdot (7\sqrt{3} - x) \cdot z \cdot \frac{\sqrt{3}}{2}$

$s^2 = x^2 + y^2$

These three equations can be rearranged to $s^2 = \frac{7}{4}(x - 3)^2 + 21$ , which means $s^2$ has a minimum of $21$ and $s$ has a minimum of $\sqrt{21}$ . Therefore, $n = \boxed{21}$ .