### Metropolises 2016-18 (IOM) (Russia) 6p

geometry problems from International Olympiad of Metropolises (IOM)
with aops links in the names

2016 - 2018

Let A1A2 . . .An be a cyclic convex polygon whose circumcenter is strictly in its interior. Let B1, B2, …, Bn be arbitrary points on the sides A1A2, A2A3, …, AnA1, respectively, other than the vertices. Prove that $$\frac{{{B}_{1}}{{B}_{2}}}{{{A}_{1}}{{A}_{3}}}+\frac{{{B}_{2}}{{B}_{3}}}{{{A}_{2}}{{A}_{4}}}+...+\frac{{{B}_{n}}{{B}_{1}}}{{{A}_{n}}{{A}_{1}}}>1$$

A convex quadrilateral ABCD has right angles at A and  C. A point E lies on the extension of the side AD beyond D so that < ABE = < ADC. The point K is symmetric to the point C with respect to point A. Prove that < ADB = < AKE.

Let ABCD be a parallelogram in which the angle at B is obtuse and AD > AB. Points K and L are chosen on the diagonal AC such that < ABK = < ADL (the points A, K, L, C are all different, with K between A and L). The line BK intersects the circumcircle ω of triangle ABC at points B and E, and the line EL intersects ω at points E and F. Prove that BF // AC.

Let ABCDEF be a convex hexagon which has an inscribed circle and a circumscribed circle. Denote by ωA, ωB, ωC, ωD, ωE, and ωF the inscribed circles of the triangles FAB, ABC, BCD, CDE, DEF, and EFA, respectively. Let lAB be the external common  tangent of ωA and ωB other than the line AB; lines lBC, lCD, lDE, lEF , and lFA are analogously defined. Let A1 be the intersection point of the lines lFA and lAB , B1 be the intersection point of the lines lAB and lBC , points C1, D1, E1, and F1 are analogously defined. Suppose that A1B1C1D1E1F1 is a convex hexagon. Show that its diagonals A1D1, B1E1, and C1F1 meet at a single point.

A convex quadrilateral $ABCD$ is circumscribed about a circle $\omega$. Let $PQ$ be the diameter of $\omega$ perpendicular to $AC$. Suppose lines $BP$ and $DQ$ intersect at point $X$, and lines $BQ$ and $DP$ intersect at point $Y$. Show that the points $X$ and $Y$ lie on the line $AC$.

Géza Kós
The incircle of a triangle $ABC$ touches the sides $BC$ and $AC$ at points $D$ and $E$, respectively. Suppose $P$ is the point on the shorter arc $DE$ of the incircle such that $\angle APE = \angle DPB$. The segments $AP$ and $BP$ meet the segment $DE$ at points $K$ and $L$, respectively. Prove that $2KL = DE$.

Dušan Djukić

source: megapolis.educom.ru/en