MathDB

g5

Part of Indonesia MO Shortlist - geometry

Problems(4)

<AMD = <CMD if tangents at B,D of cyclic ABCD concurrent with AC

Source: Indonesia INAMO Shortlist 2008 G5

8/25/2021
Let ABCDABCD be quadrilateral inscribed in a circle. Let MM be the midpoint of the segment BDBD. If the tangents of the circle at B B, and at DD are also concurrent with the extension of ACAC, prove that ∠AMD=∠CMD\angle AMD = \angle CMD.
geometryequal anglescyclic quadrilateral
right angle wanted, starting with 2 intersecting circles

Source: Indonesia INAMO Shortlist 2009 G5 https://artofproblemsolving.com/community/c1101409_

12/11/2021
Two circles intersect at points AA and BB. The line ℓ\ell through A intersects the circles at CC and DD, respectively. Let M,NM, N be the midpoints of arc BCBC and arc BDBD. which does not contain AA, and suppose that KK is the midpoint of the segment CDCD . Prove that ∠MKN=90o\angle MKN=90^o.
geometrycirclesright angle
collinear wanted, <A=60^o, semicircle of diameter BC. another circle related

Source: Indonesia INAMO Shortlist 2010 G5

8/27/2021
Given an arbitrary triangle ABCABC, with ∠A=60o\angle A = 60^o and AC<ABAC < AB. A circle with diameter BCBC, intersects ABAB and ACAC at FF and EE, respectively. Lines BEBE and CFCF intersect at DD. Let Γ\Gamma be the circumcircle of BCDBCD, where the center of Γ\Gamma is OO. Circle Γ\Gamma intersects the line ABAB and the extension of ACAC at MM and NN, respectively. MNMN intersects BCBC at PP. Prove that points AA, PP, OO lie on the same line.
geometrycollinear
BE = BF wanted, tangents to a circle (O)

Source: Indonesia INAMO Shortlist 2017 G5 https://artofproblemsolving.com/community/c1101409_indonesia_shortlist__geometry

11/15/2021
Given a circle (O)(O) with center OO and PP a point outside (O)(O). AA and BB are points on (O)(O) such that PAPA and PBPB are tangents to (O)(O). The line ℓ\ell through PP intersects (O)(O) at points CC and DD, respectively (CC lies between PP and DD). Line BFBF is parallel to line PAPA and intersects line ACAC and line ADAD at EE and FF, respectively. Prove that BE=BFBE = BF.
geometryequal segments