### Almaty Olympiad 2008-18 (Kazakhstan) 14p

geometry problems from Almaty Olympiads (final stage, from Kazakhstan)
with aops links in the names
2008 - 2018

The diagonals $AC$ and $BD$ of the convex quadrilateral $ABCD$ intersect at the point $E$, $M$ the midpoint of the segment $AE$ and $N$ the midpoint of the segment $CD$. It is known that the diagonal $BD$ is the bisector of the angle $ABC$. Prove that  quadrilateral $ABCD$ is cyclic  if and only if  quadrilateral $MBCN$ is cyclic.

The extensions of the sides $AB$ and $CD$ of the inscribed quadrilateral $ABCD$ intersect at the point $P$, and the extensions of the sides $BC$ and $AD$ at the point $Q$. Prove that the intersection points of the bisectors of the angles $AQB$ and $BPC$ with the sides of the quadrilateral are vertices of a rhombus.

Given a triangle $ABC$, in which $AB \ne AC$. Denote on its sides $AB$ and $AC$, the points $M$ and $N$ , respectively, such that $BM = CN$ The circumcircle of the triangle $AMN$ intersects the circumcircle of the triangle $ABC$ at $D$, other than $A$. Prove that $DM = DN$.

The circle $\omega$ is circumscribed around the quadrilateral $ABCD$. The lines $AB$ and $CD$ intersect at the point $K$, and the lines $AD$ and $BC$ intersect at the point $L$. The line passing through the center of the circle $\omega$ and perpendicular on $KL$ intersects the lines $KL$, $CD$ and $AD$ at the points $P$, $Q$ and $R$, respectively. Prove that the lines $QL$, $BP$ and $KR$ intersect at one point.

In $ABC$  bisector  $BK$ is drawn. The tangent at $K$ to the circumcircle $\omega$ of the triangle $ABK$ intersects the side $BC$ at the point $L$. The line $AL$ intersects $\omega$ at $M$. Prove that the line $BM$ passes through the middle of the segment $KL$.

The tangents $SA$ and $SB$ are drawn to the circle $\omega$ with center $O$, from the point $S$. Points $C$ and $C '$ on the circle $\omega$ such that $AC \parallel OB$ and $CC'$ is the diameter of $\omega$. Let lines $BC$ and $SA$ intersect at $K$, and lines $KC '$ and $AC$ at $M$. Prove that in the triangle $MKC$ the altitude from the vertex $M$ divides the altitude from the vertex $C$ in half if the angle $BMK$ is straight.

On the coordinate plane $xOy$, a parabola $y = {{x} ^ {2}}$ is drawn. Let $A$, $B$ and $C$ be different points of this parabola. We define the point ${{A} _ {1}}$ as the intersection point of the line $BC$ and the $Oy$ axis. Similarly, we define the points ${{B} _ {1}}$ and ${{C} _ {1}}$. Prove that the sum of the distance from $A$, $B$ and $C$ to the $Ox$ axis is greater than the sum of the distance from ${{A} _ {1}}$, ${{B} _ {1}}$ and ${{C} _ {1}}$ to the $Ox$ axis.

In the isosceles triangle $ABC$ $(BC = AC)$ on the bisector of $BN$ there was a point $K$ such that $BK = KC$ and $KN = NA$. Find the angles of the triangle $ABC$.

On a line containing the altitude $A {{A} _ {1}}$ of the triangle $ABC$ $(\angle B \ne 90^\circ)$, a point $F$ is taken that is different from $A$ and ${{A} _ {1}}$ such that the lines $BF$ and $CF$ intersect the lines $AC$ and $AB$ at the points ${{B} _ {1}}$ and ${{C} _ {1}}$ respectively. Perpendiculars $BP$, $BQ$, $FS$, $FR$ on lines ${{A} _ {1}} {{B} _ {1}}$ and  ${{A}_{1}}{{C}_{1}}$ are pass through points $B$ and $F$. Prove that the lines $PQ$, $SR$ and $B {{B} _ {1}}$ intersect at the same point.
a) Solve the problem when $F$ is the intersection point of the altitudes of the triangle $ABC$.
b) Solve the problem for an arbitrary point $F$.

The line $l$ is the tangent to the circle circumscribed around the acute-angled triangle $ABC$, drawn at the point $B$. The point $K$ is the projection of the orthocenter of the triangle onto the line $l$, and the point $L$ is the midpoint of the side $AC$. Prove that the triangle $BKL$ is isosceles.

The altitudes $AA_1$ and $CC_1$ of the acute-angled triangle $ABC$ intersect at the point $H$. On the altitude $AA_1$, lies point $P$ such that $A_1P = AH$. On the altitude $CC_1$, lies point $Q$ such that $C_1Q = CH$. Prove that the perpendiculars on the lines $AA_1$ and $CC_1$ passing through the points $P$ and $Q$, respectively, intersect on the circumcircle of the triangle $ABC$.
The altitudes of $A {{A} _ {1}}$, $B {{B} _ {1}}$ and $C {{C} _ {1}}$ are drawn in the acute triangle $ABC$. To the circumcircle of the triangle $ABC$, the tangents at the points $A$ and $C$ intersecting at the point $Q$ are drawn. A straight line passing through the midpoint of the side $AC$ and the orthocenter of the triangle $ABC$ intersects the line ${{A} _ {1}} {{C} _ {1}}$ at the point $F$. Prove that the points $Q$, ${{B} _ {1}}$, and $F$ lie on the same line.
The circle $\omega$ passing through the vertices $A$ and $B$ of the triangle $ABC$ intersects the sides $AC$ and $BC$ at the points $E$ and $F$, respectively. The circle $\Gamma$ tangent to the segment $EF$ at the point $P$ and the arc $AB$ of the circumcircle of the triangle $ABC$ at the point $Q$. Prove that $C$, $P$, $Q$ lie on one line.
In the triangle $ABC$: $BC\ge CA \ge AB$.  From the vertices $B$ and  $C$ let  $BK$ and $CL$ be angle bisectors respectively. Inside the triangle $AKL$ a point $X$ is selected from which perpendiculars  $XY$ and  $XZ$ are dropped to $AB$ and  $AC$ respectively. Prove that $XY+YZ+ZX < AC$ .