MathDB

2008.11.4

Part of Kyiv City MO Seniors 2003+ geometry

Problems(1)

SA=SB=SC wanted in a tetrahedron (Kyiv City Olympiad 2008 11.4)

Source:

6/30/2020
In the tetrahedron SABCSABC at the height SHSH the following point OO is chosen, such that: ∠AOS+α=∠BOS+β=∠COS+γ=180o,\angle AOS + \alpha = \angle BOS + \beta = \angle COS + \gamma = 180^o, where α,β,γ\alpha, \beta, \gamma are dihedral angles at the edges BC,AC,ABBC, AC, AB , respectively, at this point HH lies inside the base ABCABC. Let A1, B1, C1{{A} _ {1}}, \, {{B} _ {1}}, \, {{C} _ {1}} be the points of intersection of lines and planes: A1=AO∩SBC{{A} _ {1}} = AO \cap SBC , B1=BO∩SAC{{B} _ {1}} = BO \cap SAC , C1=CO∩SBA{{C} _ {1}} = CO \cap SBA . Prove that if the planes ABCABC and A1B1C1{{A} _ {1}} {{B} _ {1}} {{C} _ {1}} are parallel, then SA=SB=SCSA = SB = SC .
(Alexey Klurman)
geometry3D geometrytetrahedronanglesequal segments