Vigilant Triangle(s)

Let $BC = 4w$ , $AC = 4h$ , and $C(0,0)$ be the origin of the $xy$ -plane. Then the coordinates of the other points are $A(0,4h)$ , $B(4w,0)$ , $D(w,3h)$ , $E(2w,2h)$ , and $F(3w, h)$ . Then we have:

$\begin{aligned} CD^2+CE^2+CF^2 & = 350 \\ (w^2 + 9h^2) + (4w^2 + 4h^2) + (9w^2 + h^2) & = 350 \\ 14(w^2+h^2) & = 350 \\ \implies w^2 + h^2 & = 25 \end{aligned}$

And $AB = \sqrt{BC^2+AC^2} = \sqrt{16w^2 + 16h^2} = \sqrt{16 \times 25} = \boxed{20}$ .

Denote the lengths of $AB$ , $BC$ , $CA$ by $c$ , $a$ and $b$ . Then, $AD=DE=EF=FB=\dfrac{c}{4}$ .

By Pythagorean theorem on $\triangle ABC$ , ${{c}^{2}}={{a}^{2}}+{{b}^{2}} \ \ \ \ \ (1)$ Moreover, for the mediean $CE$ of the right $\triangle ABC$ it holds $CE=\frac{AB}{2}=\dfrac{c}{2}$ Using Apollonius’s theorem on $\triangle CAE$ and $\triangle CEB$ we have:

$\begin{matrix} C{{A}^{2}}+C{{E}^{2}}=2C{{D}^{2}}+{{\dfrac{AE}{2}}^{2}}\Rightarrow {{b}^{2}}+\dfrac{{{c}^{2}}}{4}=2C{{D}^{2}}+{{\dfrac{c}{8}}^{2}} \\ C{{E}^{2}}+C{{B}^{2}}=2C{{F}^{2}}+{{\dfrac{EB}{2}}^{2}}\Rightarrow \dfrac{{{c}^{2}}}{4}+{{a}^{2}}=2C{{F}^{2}}+{{\dfrac{c}{8}}^{2}} \\ \end{matrix}$ Adding and rearranging,

$2\left( C{{D}^{2}}+C{{E}^{2}}+C{{F}^{2}} \right)={{a}^{2}}+{{b}^{2}}+\dfrac{3{{c}^{2}}}{2}\overset{\left( 1 \right)}{\mathop{\Rightarrow }}\,2\times 350=\dfrac{7{{c}^{2}}}{4}\Rightarrow {{c}^{2}}=400\Rightarrow c=\boxed{20}$

Let $AD=DE=EF=FB=x$ and draw lines through $D,E$ and $F$ parallel to base $BC$ . Let them intersect $AC$ at $G, H$ and $I$ respectively. Since the triangles formed are similar, $AG=GH=HI=IC=y$ $DG=z,\; EH=2z,\; FI=3z,\; BC=4z$

Using the Pythagorean Theorem in $\triangle CGD,\; \triangle CHE$ and $\triangle CIF$ , $\begin{aligned} 9y^2+z^2&=CD^2 \\ 4y^2+4z^2&=CE^2 \\ y^2+9z^2&=CF^2 \end{aligned}$ Adding the above equations, $14y^2+14z^2=CD^2+CE^2+CF^2=350$ $\implies y^2+z^2=25$ In right-triangle $ABC$ , $AC^2+BC^2=AB^2$ $\implies AB=\sqrt{16y^2+16z^2}=\sqrt{16\cdot 25}=\boxed{20}$

By the law of cosines on $\triangle DAC$ , $CD^2 = d^2 + b^2 - 2bd \cos A$ .

By the law of cosines on $\triangle EAC$ , $CE^2 = 4d^2 + b^2 - 4bd \cos A$ .

By the law of cosines on $\triangle FAC$ , $CF^2 = 9d^2 + b^2 - 6bd \cos A$ .

By right $\triangle BAC$ , $b = 4d \cos A$ .

Adding the first three equations and substituting $b = d \cos A$ , $CD^2 + CE^2 + CF^2 = 14d^2 + 3b^2 - 12bd \cos A = 14d^2 + 3b^2 - 3b^2 = 14d^2 = 350$ .

Therefore, $d = 5$ and $AB = 4d = 4 \cdot 5 = \boxed{20}$ .

Put $C$ at the origin, $A$ on the $y$ -axis and $B$ on the $x$ -axis. Letting the sides of the triangle be $a=BC$ , $b=CA$ , $c=AB$ as usual, the coordinates of the points are

so that $CD^2+CE^2+CF^2=\frac{1}{16} \left[a^2+9b^2+4a^2+4b^2+9a^2+b^2\right]=\frac78 \left[a^2+b^2 \right]$

Since we're told in the question that this is $350$ , and since $a^2+b^2=c^2$ , we find $c^2=400$

so the length of the hypotenuse is $\boxed{20}$ .