Centroid Circle

I tried my best to upload an easy and elegant solution

Now let $BD=3x \therefore BG=2x$ and $GD=x$ {as centroid divides median in $2:1$ ration}

First of all as $AEGD$ is cyclic quadrilateral

$\therefore$ $\angle DAE=\angle EGB$ and $\angle ADG=\angle GEB$

$\therefore \triangle ABD \sim \triangle GBE$

$\therefore \dfrac{AB}{GB}=\dfrac{BD}{BE}=\dfrac{AD}{GE}$

$\therefore \dfrac{11}{2x}=\dfrac{3x}{5.5}~\Rightarrow 12x^2=121$

$x^2=\dfrac{121}{4 \times 3} \Rightarrow x=\dfrac{11 \sqrt 3}{6}$

Means $BD=\dfrac{11 \sqrt 3}{6} \times 3$

$\therefore BD=\dfrac{11 \sqrt 3}{2}$

Now we have median formula for a triangle which states that

Length of Median on side $a=\dfrac { \sqrt { 2b^{ 2 }+2c^{ 2 }-a^{ 2 } } }{ 2 }$

$\therefore BD=\dfrac { \sqrt { 2(BC)^{ 2 }+2(121)-49 } }{ 2 }$

$\Rightarrow \dfrac { 11\sqrt { 3 } }{ 2 } =\dfrac { \sqrt { 2(BC)^{ 2 }+2(121)-49 } }{ 2 } \\ \Rightarrow \dfrac { 121\times 3 }{ 4 } =\dfrac { 2(BC)^{ 2 }+193 }{ 4 } \\ \Rightarrow 170=2(BC^{ 2 })\\ \Rightarrow 85=BC^{ 2 }\\ \Rightarrow \sqrt { 85 } =BC$

$\therefore \color{royalblue}{\huge\boxed{n=85}}$

Let $AF = 3x , CE = 3y , BD = 3z$

Angles - $ADE = AGE = a$

$EDB = GAE = DBC = b$

$DAG = DEG = c$

$DGA = ABC = d$

I hope all the notation is clear from the diagram. This solution does not use median formula.

Using Ptolemy's Theorem to $DGEA$ ,

$(\frac{11}{2})(z) + (\frac{7}{2})(y) = x\sqrt{n}$

$\Rightarrow \sqrt{n} = \frac{1}{2x}(11z + 7y) . . . .(1)$

By angle chasing , in diagram by A-A test ,

$\Delta CDG \sim \Delta CEA \Large\Rightarrow \frac{\frac{7}{2}}{3y} = \frac{z}{\frac{11}{2}} = \frac{2y}{7}$

$\Rightarrow \frac{7}{6y} = \frac{2y}{7} \Rightarrow 49 = 12y^2 \Rightarrow y = \frac{7\sqrt{3}}{6} . . . . (2)$

By substituting value of $y$ from $(2)$ , we also have ,

$\frac{2y}{7} = \frac{2z}{11} \Rightarrow \frac{1}{\sqrt{3}} = \frac{2z}{11} \Rightarrow z = \frac{11\sqrt{3}}{6} . . . . (3)$

We also have $\Delta DGA \sim \Delta EBC$ , Substituting value of $z$ from $(3)$ ,

$\Large\Rightarrow \frac{z}{\frac{11}{2}} = \frac{2x}{\sqrt{n}} = \frac{\frac{7}{2}}{3y}$

$\Rightarrow \frac{11 \sqrt{3}}{6} \times \frac{2}{11} = \frac{2x}{\sqrt{n}}$

$\Rightarrow 2x = \frac{\sqrt{3n}}{3} . . . . (4)$

Substituting $(4)$ in $(1)$ ,

$\frac{3}{\sqrt{3n}} \bigg( \frac{121\sqrt{3}}{6} + \frac{49\sqrt{3}}{6}\bigg) = \sqrt{n}$

$\Rightarrow 3\bigg( \frac{170\sqrt{3}}{6}\bigg) = n\sqrt{3}$

$\Rightarrow \huge\boxed{n = 85}$

By the way, the $F$ in the question should be $G$ . Anyway, I did the long way to come to the answer.

Let $A$ be $(0,0)$ , $B(11,0)$ and $C(x,y)\quad \Rightarrow x^2+y^2 = 7^2\quad \Rightarrow D(\frac {x}{2}, \frac {y}{2})$ $\quad \Rightarrow E(\frac {11}{2}, 0) \quad \Rightarrow G(\frac {11+0+x}{3}, \frac {0+0+y}{3})\Rightarrow G(\frac {x+11}{3}, \frac {y}{3})$ .

Let the center and radius of the concyclic circle be $O(x_o, y_o)$ and $r$ . Then we have the following equations.

$\begin{cases} x^2+y^2 = 49 &...(1) \\ x_o^2+y_o^2 = r^2 &...(2) \\ (x_o-\frac {11}{2})^2+y_o^2 = r^2 &...(3) \\ (x_o-\frac {x}{2})^2+(y_o-\frac {y}{2})^2 = r^2 &...(4) \\ (x_o-\frac {x+11}{3})^2+(y_o-\frac {y}{3})^2 = r^2 &...(5) \end{cases}$

Eqn 3 - Eqn 2:

$\quad \Rightarrow x_o^2 - 11x_o + (\frac {11}{2})^2 + y_o^2 - x_o^2 - y_o^2 = 0\quad \quad x_o = \frac {11}{4}$

Eqn 4 - Eqn 2:

$\quad \Rightarrow x_o^2 - x_ox + \frac {1}{4}x^2 + y_o^2 - y_oy + \frac {1}{4}y^2 - x_o^2 - y_o^2 = 0$

$\quad \quad x_ox + y_oy = \frac {x^2+y^2}{4} = \frac {49}{4} \quad ... (6)$

Eqn 5 - Eqn 2:

$\quad \Rightarrow (\frac {11}{4}-\frac {x+11}{3})^2+(y_o-\frac {y}{3})^2 - (\frac{11}{4})^2 - y_o^2 = 0$

$\quad \quad (-\frac {11}{12}-\frac {x}{3})^2+(y_o-\frac {y}{3})^2 - \frac {121}{16} - y_o^2 = 0$

$\quad \quad \frac {121}{144}+\frac{11}{18}x+\frac{1}{9}x + y_o^2 - \frac {2}{3} y_oy + \frac {1}{9}y^2 - \frac {121}{16} - y_o^2 = 0$

$\quad \quad -\frac {121}{18}+\frac{11}{18}x- \frac {2}{3} y_oy = 0\quad \Rightarrow \frac {11}{6}x - 2y_oy = \frac {26}{6}\quad ... (7)$

Eqn 6 $\times$ 2 + Eqn 7:

$\quad \Rightarrow \frac {11}{2} x + 2y_oy + \frac {11}{6}x - 2y_oy = \frac {49}{2} + \frac {26}{6} \quad \Rightarrow x = \frac {85}{22}$

Now, $BC = \sqrt{n}$

$\quad \Rightarrow n = (x-11)^2 + y^2 = x^2 - 22x + 121 + y^2$

$\quad \quad \quad = 49 - 22(\frac {85}{22} + 121 = \boxed {85}$

To solve this problem, I'll use barycentric coordinates with respect to $\triangle ABC$ , assuming $BC=a$ , $CA=b$ and $AB=c$ . Note that the points $A,D,E,G$ have a very simple representation in this system of homogeneous coordinates: $A=(1,0,0)\ \ \ \ D=(1,0,1)\ \ \ \ G=(1,1,1)\ \ \ \ E=(1,1,0)$ Recalling that the equation of a circle is $-a^2yz-b^2zx-c^2xy+(ux+vy+wz)(x+y+z)=0$ for some values of the parameters $u,v,w$ , we plug the coordinates of three points to find the equation of the circle. $A) \ \ \ \ -a^2\cdot 0\cdot 0-b^2\cdot0\cdot1-c^2\cdot 1\cdot 0+(u\cdot 1+v\cdot 0+w\cdot0)(1+0+0)=0 \Rightarrow u=0$ $D)\ \ \ \ -a^2\cdot 0\cdot 1-b^2\cdot1\cdot 1-c^2\cdot1\cdot0+(u\cdot 1+v\cdot 0+w\cdot 1)(1+0+1)=0\Rightarrow -b^2+2(u+w)=0$ But we already found that $u=0$ , so this condition becomes $w=\frac{b^2}{2}$ . Using symmetry, we can also find that imposing the passage through $E$ leads to $v=\frac{c^2}{2}$ . So we have the equation of the circle! $-a^2yz-b^2zx-c^2xy+\left(\frac{c^2}{2}y+\frac{b^2}{2}z\right)(x+y+z)=0$

Now we plug $G=(1,1,1)$ in this equation to obtain $-a^2-b^2-c^2+3\left(\frac{b^2}{2}+\frac{c^2}{2}\right)=0\Rightarrow a^2=\frac{b^2+c^2}{2}\Rightarrow a=\sqrt{\frac{b^2+c^2}{2}}$ Recalling that $b=7$ and $c=11$ $BC=a=\sqrt{\frac{121+49}{2}}=\sqrt{\frac{170}{2}}=\sqrt{85}$ So the answer is $\boxed{85}$ .

Sorry for my terrible English.