MathDB

2010.16

Part of Ukrainian TYM Qualifying - geometry

Problems(1)

regular tetrahedron if ABxACxADxBCxBDxCD= 512/27, A,B,C,D on sphere radius=1

Source: 2010 XIII All-Ukrainian Tournament of Young Mathematicians, Qualifying p16

5/19/2021
Points A,B,C,DA, B, C, D lie on the sphere of radius 11. It is known that AB⋅AC⋅AD⋅BC⋅BD⋅CD=51227AB\cdot AC\cdot AD\cdot BC\cdot BD\cdot CD=\frac{512}{27}. Prove that ABCDABCD is a regular tetrahedron.
geometry3D geometrytetrahedronsphereRegularUkrainian TYM