MathDB

2

Part of 2020 Ukrainian Geometry Olympiad - April

Problems(3)

P interior of ABC, BP > AP and BP > CP => <ABC< 90^o

Source: Ukrainian Geometry Olympiad 2020, VIII p2 , IX p1

4/27/2020
Inside the triangle ABCABC is point PP, such that BP>APBP > AP and BP>CPBP > CP. Prove that ∠ABC\angle ABC is acute.
anglesacutegeometry
equal angles wanted, circles, tangents, symmetric point given

Source: Ukrainian Geometry Olympiad 2020, IX p2, X p1, XI p1

4/27/2020
Let Γ\Gamma be a circle and PP be a point outside, PAPA and PBPB be tangents to Γ\Gamma , A,B∈ΓA, B \in \Gamma . Point KK is an arbitrary point on the segment ABAB. The circumscirbed circle of △PKB\vartriangle PKB intersects Γ\Gamma for the second time at point TT, point P′P' is symmetric to point PP wrt point AA. Prove that ∠PBT=∠P′KA\angle PBT = \angle P'KA.
geometryequal anglescirclestangentsymmetry
tangent wanted, isosceles, 3 circles, tangents given

Source: Ukrainian Geometry Olympiad 2020, XI p2

4/27/2020
Let ABCABC be an isosceles triangle with AB=ACAB=AC. Circle Γ\Gamma lies outside ABCABC and touches line ACAC at point CC. The point DD is chosen on circle Γ\Gamma such that the circumscribed circle of the triangle ABDABD touches externally circle Γ\Gamma. The segment ADAD intersects circle Γ\Gamma at a point EE other than DD. Prove that BEBE is tangent to circle Γ\Gamma .
geometrytangent circlesisoscelestangent