geometry problems from Grand Duchy of Lithuania Mathematical Contest (Baltic Way TST) with aops links in the names
started in 2009
aops forum collection here
collected inside aops here
2009-22
2009 Grand Duchy of Lithuania p4 (Irish MO 2008 p1)
A triangle $ABC$ has an obtuse angle at $B$. The perpendicular at $B$ to $AB$ meets $AC$ at $D$, and $CD = AB$ . Prove that $ AD^2 = AB \cdot BC$ if and only if $\angle CBD = 30^o$.
In the triangle $ABC$ angle $C$ is a right angle. On the side $AC$ point $D$ has been found, and on the segment $BD$ point K has been found such that $\angle ABC = \angle KAD = \angle AKD$. Prove that $BK = 2DC$.
In the cyclic quadrilateral $ABCD$ with $AB = AD$, points $M$ and $N$ lie on the sides $CD$ and $BC$ respectively so that $MN = BN + DM$. Lines $AM$ and $AN$ meet the circumcircle of $ABCD$ again at points $P$ and $Q$ respectively. Prove that the orthocenter of the triangle $APQ$ lies on the segment $MN$.
The base $AB$ of a trapezium $ABCD$ is longer than the base $CD$, and $\angle ADC$ is a right angle. The diagonals $AC$ and $BD$ are perpendicular. Let $E$ be the foot of the altitude from $D$ to the line $BC$. Prove that $$\frac{AE}{BE} =\frac{ AC \cdot CD}{AC^2 - CD^2}$$
Let $ABC$ be an isosceles triangle with $AB = AC$. The points $D, E$ and $F$ are taken on the sides $BC, CA$ and $AB$, respectively, so that $\angle F DE = \angle ABC$ and $FE$ is not parallel to $BC$. Prove that the line $BC$ is tangent to the circumcircle of $\vartriangle DEF$ if and only if $D$ is the midpoint of the side $BC$.
An isosceles triangle $ABC$ with $AC = BC$ is given. Let $M$ be the midpoint of the side $AB$ and let $P$ be a point inside the triangle such that $\angle PAB = \angle PBC$. Prove that $\angle APM + \angle BPC = 180 ^o$
Let $\omega_1$ and $\omega_2$ be two circles , with respective centres $O_1$ and $O_2$ , that intersect each other in $A$ and $B$. The line $O_1A$ intersects $\omega_2$ in $A$ and $C$ and the line $O_2A$ inetersects $\omega_1$ in $A$ and $D$. The line through $B$ parallel to $AD$ intersects $\omega_1$ in $B$ and $E$. Suppose that $O_1A$ is parallel to $DE$. Show that $CD$ is perpendicular to $O_2C$.
Let $ABC$ be an isosceles triangle with $AB = AC$. Let $D, E$ and $F$ be points on line segments $BC, CA$ and $AB$, respectively, such that $BF = BE$ and such that $ED$ is the angle bisector of $\angle BEC$. Prove that $BD = EF$ if and only if $AF = EC$.
Let ABC be a triangle with $\angle A = 90^o$ and let $D$ be an orthogonal projection of $A$ onto $BC$. The midpoints of $AD$ and $AC$ are called $E$ and $F$, respectively. Let $M$ be the circumcentre of $\vartriangle BEF$. Prove that $AC\parallel BM$.
The altitudes $AD$ and $BE$ of an acute triangle $ABC$ intersect at point $H$. Let $F$ be the intersection of the line $AB$ and the line that is parallel to the side BC and goes through the circumcenter of $ABC$. Let $M$ be the midpoint of the segment $AH$. Prove that $\angle CMF = 90^o$
Let $ABC$ be an acute triangle with orthocenter $H$ and circumcenter $O$. The perpendicular bisector of segment $CH$ intersects the sides $AC$ and $BC$ in points $X$ and $Y$ , respectively. The lines $XO$ and $YO$ intersect the side $AB$ in points $P$ and $Q$, respectively. Prove that if $XP + Y Q = AB + XY$ then $\angle OHC = 90^o$
The tangents of the circumcircle $\Omega$ of the triangle $ABC$ at points $B$ and $C$ intersect at point $P$. The perpendiculars drawn from point $P$ to lines $AB$ and $AC$ intersect at points$ D$ and $E$ respectively. Prove that the altitudes of the triangle $ADE$ intersect at the midpoint of the segment $BC$.
Let $ABCD$ be a convex quadrilateral satisfying $\angle ADB + \angle ACB = \angle CAB + \angle DBA = 30^o$, $AD = BC$. Prove that there exists a right-angled triangle with side lengths $AC$, $BD$, $CD$.
The center $O$ of the circle $\omega$ passing through the vertex $C$ of the isosceles triangle $ABC$ ($AB = AC$) is the interior point of the triangle $ABC$. This circle intersects segments $BC$ and $AC$ at points $D \ne C$ and $E \ne C$, respectively, and the circumscribed circle $\Omega$ of the triangle $AEO$ at the point $F \ne E$. Prove that the center of the circumcircle of the triangle $BDF$ lies on the circle $\Omega$.
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