MathDB
tg^2 a + tg^2 \b + tg^2 c=12.$

Source: Polish MO Finals 1960 p2

August 30, 2024
geometry3D geometrytetrahedrontrigonometry

Problem Statement

A plane is drawn through the height of a regular tetrahedron, which intersects the planes of the lateral faces along 3 3 lines that form angles α \alpha , β \beta , γ \gamma with the plane of the tetrahedron's base. Prove that tg2α+tg2β+tg2γ=12. tg^2 \alpha + tg^2 \beta + tg^2 \gamma =12.
Back to ProblemsView on AoPS