Geometry problem 9 by Dhaval Furia

Let the side lengths of the brick be $a$ , $b$ , and $c$ such that $a<b<c$ . Then $\sqrt{a^2+b^2} : \sqrt{c^2+a^2} : \sqrt{b^2+c^2} = 3: 2\sqrt 3 : \sqrt{15}$ . Since we only need ratio as an answer, we can assume.

$\begin{cases} \sqrt{a^2+b^2} = 3 & \implies a^2+b^2 = 9 & ...(1) \\ \sqrt{c^2+a^2} = 2\sqrt 3 & \implies c^2+a^2 = 12 & ...(2) \\ \sqrt{b^2+c^2} = \sqrt{15} & \implies b^2+c^2 = 15 & ...(3) \end{cases}$

$\begin{aligned} (1)+(2)+(3) : \quad 2(a^2 + b^2 + c^2) & = 36 \\ \implies a^2 + b^2 + c^2 & = 18 & ...(4) \end{aligned}$

$(4)-(3): \ \ a^2 = 3 \implies a = \sqrt 3$ and $(4)-(1): \ \ c^2 = 9 \implies c = 3$ . Then $a:c = \sqrt 3:3 = \boxed{1:\sqrt 3}$ .

Let the edges of the cube be of lengths $a, b, c$ . Then

$\sqrt {a^2+b^2}=3k\implies a^2+b^2=9k^2$ where $k$ is a constant.

$\sqrt {b^2+c^2}=2\sqrt 3k\implies b^2+c^2=12k^2$

$\sqrt {c^2+a^2}=\sqrt 15k\implies c^2+a^2=15k^2$

$\implies a^2+b^2+c^2=18k^2\implies a=\sqrt 6k,b=\sqrt 3k,c=3k$

Hence the shortest edge is of length $\sqrt 3k$ and the longest edge is of length $3k$ , and the ratio of their lengths is $\sqrt 3:3=\boxed {1:\sqrt 3}$ .