MathDB

2

Part of 2016 Sharygin Geometry Olympiad

Problems(3)

line from arc midpoint perpendicular to a side

Source: Sharygin Geometry Olympiad 2016 Final Round problem 2 grade 8

7/22/2018
A circumcircle of triangle ABCABC meets the sides ADAD and CDCD of a parallelogram ABCDABCD at points KK and LL respectively. Let MM be the midpoint of arc KLKL not containing BB. Prove that DM⊥ACDM \perp AC.
by E.Bakaev
geometryarc midpointcircumcircleperpendicularparallelogram
Similar Triangles and tangents to the orthocenter

Source: Sharygin Geometry Olympiad, Final Round 2016, Problem 2 grade 9

8/4/2016
Let HH be the orthocenter of an acute-angled triangle ABCABC. Point XAX_A lying on the tangent at HH to the circumcircle of triangle BHCBHC is such that AH=AXAAH=AX_A and XA≠HX_A \not= H. Points XB,XCX_B,X_C are defined similarly. Prove that the triangle XAXBXCX_AX_BX_C and the orthotriangle of ABCABC are similar.
geometrysimilar trianglescircumcircle
Equal Angles

Source: Sharygin geometry olympiad 2016, grade 10, Final Round, Problem 2

8/5/2016
Let II and IaI_a be the incenter and excenter (opposite vertex AA) of a triangle ABCABC, respectively. Let A′A' be the point on its circumcircle opposite to AA, and A1A_1 be the foot of the altitude from AA. Prove that ∠IA1Ia=∠IA′Ia\angle IA_1I_a=\angle IA'I_a.
(Proposed by Pavel Kozhevnikov)
geometryincentercircumcircle