MathDB

7

Part of 1994 All-Russian Olympiad

Problems(3)

trapezoid with points equidistant from the diagonals intersection

Source: All Russian MO 1994 ARO IX P7

7/29/2018
A trapezoid ABCDABCD (AB///CDAB ///CD) has the property that there are points PP and QQ on sides ADAD and BCBC respectively such that ∠APB=∠CPD\angle APB = \angle CPD and ∠AQB=∠CQD\angle AQB = \angle CQD. Show that the points PP and QQ are equidistant from the intersection point of the diagonals of the trapezoid.
(M. Smurov)
geometrytrapezoid
4 circles and non-trivial concurrency through the vertices

Source: All-Russian Olympiad 1993-1994,Final Round,10.7

9/6/2008
Let Γ1,Γ2 \Gamma_1,\Gamma_2 and Γ3 \Gamma_3 be three non-intersecting circles,which are tangent to the circle Γ \Gamma at points A1,B1,C1 A_1,B_1,C_1,respectively.Suppose that common tangent lines to (Γ2,Γ3) (\Gamma_2,\Gamma_3),(Γ1,Γ3) (\Gamma_1,\Gamma_3),(Γ2,Γ1) (\Gamma_2,\Gamma_1) intersect in points A,B,C A,B,C. Prove that lines AA1,BB1,CC1 AA_1,BB_1,CC_1 are concurrent.
geometryincentergeometric transformationhomothetyprojective geometrygeometry proposed
criterion with concurrency of altitudes => circumscribed tetrahedron is regular

Source: All Russian MO 1994 ARO

7/29/2018
The altitudes AA1,BB1,CC1,DD1AA_1,BB_1,CC_1,DD_1 of a tetrahedron ABCDABCD intersect in the center HH of the sphere inscribed in the tetrahedron A1B1C1D1A_1B_1C_1D_1. Prove that the tetrahedron ABCDABCD is regular.
(D. Tereshin)
geometry3D geometrytetrahedronspherealtitudes