A plane meets a set of three mutually perpendicular planes in the sides of a triangle whose angles are A, B, C. Show that the first plane makes with the other three planes angles, the squares of whoso cosines are cot B cot C, cot C cot A, cot A cot B
I am looking for a more simpler approach where analytical geometry concepts can be used. I have no knowledge of the properties of triangles, except the definitions of sine, consine etc
Assume that $A=(a,0,0), B=(0,b,0), C=(0,0,c)$ with $a,b,c>0$. The equation of the plane $\pi$ through $A,B,C$ is clearly $\frac{x}{a}+\frac{y}{b}+\frac{z}{c}=1$, hence $\pi$ is orthogonal to the vector $\left(\frac{1}{a},\frac{1}{b},\frac{1}{c}\right)$. The side lengths of $ABC$ are $AB=\sqrt{a^2+b^2},AC=\sqrt{a^2+c^2}$ and $BC=\sqrt{b^2+c^2}$, corresponding to the following planar configuration (it is useful to recall that the height through $A$ is the locus of points $P$ such that $PB^2-PC^2=AB^2-AC^2$):
Unfolded trirectangular tetrahedron. Red segment $=a$, Blue segment $=b$, Green segment $=c$.
Let $H$ be the orthocenter of $ABC$, $H_A H_B H_C$ the orthic triangle and $V_A,V_B,V_C$ the points on the heights "seeing" the sides of $ABC$ under a right angle. The cosine of the angle between $\pi$ and $OAB$ equals $\frac{HH_C}{H_C V_C}$. Your 3d-problem is now a 2d-problem. Can you finish it? Consider that:
$$ \cot B\cot C =\left(\frac{b c}{2\Delta}\right)^2. $$