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IV.11

Part of Ukrainian TYM Qualifying - geometry

Problems(1)

S^2_{ABM}=CM^2/CD^2 S^2_{ABD}+(1- CM^2/CD^2)S_{ABC}, tetrahedron

Source: IV All-Ukrainian Tournament of Young Mathematicians, Qualifying p11

5/19/2021
In the tetrahedron ABCDABCD, the point EE is the projection of the point DD on the plane (ABC)(ABC). Prove that the following statements are equivalent: a) C=EC = E or CE∥ABCE \parallel AB b) For each point M belonging to the segment CDCD, the following equation is satisfied S△ABM2=CM2CD2⋅S△ABD2+(1−CM2CD2)S△ABC2S^2_{\vartriangle ABM}= \frac{CM^2}{CD^2}\cdot S^2_{\vartriangle ABD}+\left(1- \frac{CM^2}{CD^2} \right)S^2_{\vartriangle ABC} where S△XYZS_{\vartriangle XYZ} means the area of ​​triangle XYZXYZ.
geometry3D geometrygeometric inequalitytetrahedronUkrainian TYM