MathDB

4

Part of 2017 Ukrainian Geometry Olympiad

Problems(3)

right angle wanted starting with a parallelogram and one circumcircle

Source: Ukrainian Geometry Olympiad 2017, IX p4 , XI p3

12/12/2018
Let ABCDABCD be a parallelogram and PP be an arbitrary point of the circumcircle of ΔABD\Delta ABD, different from the vertices. Line PAPA intersects the line CDCD at point QQ. Let OO be the center of the circumcircle ΔPCQ\Delta PCQ. Prove that ∠ADO=90o\angle ADO = 90^o.
geometryparallelogramcircumcircleCircumcenterright angle
concyclic wanted, incircle, circumcircle, midline related in a right triangle

Source: Ukrainian Geometry Olympiad 2017 X p4

12/12/2018
In the right triangle ABCABC with hypotenuse ABAB, the incircle touches BCBC and ACAC at points A1{{A}_{1}} and B1{{B}_{1}} respectively. The straight line containing the midline of ΔABC\Delta ABC , parallel to ABAB, intersects its circumcircle at points PP and TT. Prove that points P,T,A1P,T,{{A}_{1}} and B1{{B}_{1}} lie on one circle.
geometrycircumcircleincirclemidlineConcyclic
tangent circumcircles wanted, symmedian related

Source: Ukrainian Geometry Olympiad 2017 XI p4

12/12/2018
Let ADAD be the inner angle bisector of the triangle ABCABC. The perpendicular on the side BCBC at the point DD intersects the outer bisector of ∠CAB\angle CAB at point II. The circle with center II and radius IDID intersects the sides ABAB and ACAC at points FF and EE respectively. AA-symmedian of ΔAFE\Delta AFE intersects the circumcircle of ΔAFE\Delta AFE again at point XX. Prove that the circumcircles of ΔAFE\Delta AFE and ΔBXC\Delta BXC are tangent.
geometrysymmedianangle bisectortangent circlescircumcircle