Školjka

%V0 There is given a convex quadrilateral $ABCD$. Prove that there exists a point $P$ inside the quadrilateral such that $$\angle PAB + \angle PDC = \angle PBC + \angle PAD = \angle PCD + \angle PBA = \angle PDA + \angle PCB = 90^{\circ}$$ if and only if the diagonals $AC$ and $BD$ are perpendicular. Proposed by Dukan Dukic, Serbia

%V0 Let $ABCD$ be a convex quadrilateral. A circle passing through the points $A$ and $D$ and a circle passing through the points $B$ and $C$ are externally tangent at a point $P$ inside the quadrilateral. Suppose that $\angle{PAB}+\angle{PDC}\leq 90^\circ$ and $\angle{PBA}+\angle{PCD}\leq 90^\circ$. Prove that $AB+CD \geq BC+AD$.

%V0 In triangle $ABC$, let $J$ be the center of the excircle tangent to side $BC$ at $A_{1}$ and to the extensions of the sides $AC$ and $AB$ at $B_{1}$ and $C_{1}$ respectively. Suppose that the lines $A_{1}B_{1}$ and $AB$ are perpendicular and intersect at $D$. Let $E$ be the foot of the perpendicular from $C_{1}$ to line $DJ$. Determine the angles $\angle{BEA_{1}}$ and $\angle{AEB_{1}}$.

%V0 Consider a convex pentagon $ABCDE$ such that $$\angle BAC = \angle CAD = \angle DAE\ \ \ ,\ \ \ \angle ABC = \angle ACD = \angle ADE$$ Let $P$ be the point of intersection of the lines $BD$ and $CE$. Prove that the line $AP$ passes through the midpoint of the side $CD$.

%V0 Let $A_1A_2A_3...A_n$ be a regular n-gon. Let $B_1$ and $B_n$ be the midpoints of its sides $A_1A_2$ and $A_{n-1}A_n$. Also, for every $i\in\left\{2;\;3;\;4;\;...;\;n-1\right\}$, let $S$ be the point of intersection of the lines $A_1A_{i+1}$ and $A_nA_i$, and let $B_i$ be the point of intersection of the angle bisector bisector of the angle $\measuredangle A_iSA_{i+1}$ with the segment $A_iA_{i+1}$. Prove that: $\sum_{i=1}^{n-1} \measuredangle A_1B_iA_n=180^{\circ}$.

%V0 A circle $C$ with center $O.$ and a line $L$ which does not touch circle $C.$ $OQ$ is perpendicular to $L,$ $Q$ is on $L.$ $P$ is on $L,$ draw two tangents $L_1, L_2$ to circle $C.$ $QA, QB$ are perpendicular to $L_1, L_2$ respectively. ($A$ on $L_1,$ $B$ on $L_2$). Prove that, line $AB$ intersect $QO$ at a fixed point. Original formulation: A line $l$ does not meet a circle $\omega$ with center $O.$ $E$ is the point on $l$ such that $OE$ is perpendicular to $l.$ $M$ is any point on $l$ other than $E.$ The tangents from $M$ to $\omega$ touch it at $A$ and $B.$ $C$ is the point on $MA$ such that $EC$ is perpendicular to $MA.$ $D$ is the point on $MB$ such that $ED$ is perpendicular to $MB.$ The line $CD$ cuts $OE$ at $F.$ Prove that the location of $F$ is independent of that of $M.$