@Allan methods it's nice! here is my answer: since $$\left(\dfrac{1-a^2}{2a}\right)\cdot\left(\dfrac{1-b^2}{2b}\right)\left(\dfrac{1-c^2}{2c}\right)=1$$$$\left(\dfrac{1-a^2}{2a}\right)\left(\dfrac{1-b^2}{2b}\right)\left(\dfrac{1-c^2}{2c}\right)=1$$ so let $$\dfrac{x}{y}=\dfrac{1-a^2}{2a},\dfrac{y}{z}=\dfrac{1-b^2}{2b},\dfrac{z}{x}=\dfrac{1-c^2}{2c}$$$$\dfrac{x}{y}=\dfrac{1-a^2}{2a},\;\dfrac{y}{z}=\dfrac{1-b^2}{2b},\;\dfrac{z}{x}=\dfrac{1-c^2}{2c}$$ and solving for $a,b,c$, $$a = \frac{-x+\sqrt{x^2+y^2}}{y},\;\;b = \frac{-y+\sqrt{y^2+z^2}}{z},\;\; c = \frac{-z+\sqrt{x^2+z^2}}{x}$$ it is easy to see $x^2+y^2,y^2+z^2,z^2+x^2$ aremust be square.then so we use Euler bricks solution $$x=u|4v^2-w^2|,y=v|4u^2-w^2|,z=4uvw$$ then it is not hard to find to give solution
$$(a,b,c)=\left(\dfrac{p^2-4p+1}{p^2+4p+1},\dfrac{p^2+1}{2p^2-2},\dfrac{3p^2-1}{p^3-3p}\right).$$
