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Ukrainian Geometry 2017,2020 22p

geometry problems with aops links in the names
from Ukrainian Geometry Olympiad (thematic olympiads)

(19 November 2017, 21 April 2020 the dates)


2017 problems in pdf in English here
2017 collected inside aops here
2020 collected inside aops here


                                                   2017 (8p)

grade IX

In the triangle $ABC$, ${{A}_{1}}$ and ${{C}_{1}} $  are the midpoints of  sides $BC $ and $AC$ respectively. Point $P$ lies inside the triangle. Let $\angle BP {{C}_{1}} = \angle PCA$. Prove that $\angle BP {{A}_{1}} = \angle PAC $.

Point $M$ is the midpoint of the base $BC$ of trapezoid $ABCD$. On base $AD$, point $P$ is selected. Line $PM$ intersects line $DC$ at point $Q$, and the perpendicular from $P$ on the bases intersects line $BQ$ at point $K$. Prove that $\angle QBC = \angle KDA$.

Circles ${w}_{1},{w}_{2}$ intersect at points ${{A}_{1}} $ and ${{A}_{2}} $. Let $B$ be an arbitrary point on the circle ${{w}_{1}}$, and line $B{{A}_{2}}$ intersects circle ${{w}_{2}}$ at point $C$. Let $H$ be the orthocenter of $\Delta B{{A}_{1}}C$. Prove that for arbitrary choice of point $B$, the point $H$ lies on a certain fixed circle.

Let $ABCD$ be a parallelogram and $P$ be an arbitrary point of the circumcircle of $\Delta ABD$, different from the vertices. Line $PA$ intersects the line $CD$ at point $Q$. Let $O$ be the center of the circumcircle $\Delta PCQ$. Prove that $\angle ADO = 90^o$.

grade X

In the triangle $ABC$, ${{A}_{1}}$ and ${{C}_{1}} $  are the midpoints of  sides $BC $ and $AC$ respectively. Point $P$ lies inside the triangle. Let $\angle BP {{C}_{1}} = \angle PCA$. Prove that $\angle BP {{A}_{1}} = \angle PAC $.

Circles ${w}_{1},{w}_{2}$ intersect at points ${{A}_{1}} $ and ${{A}_{2}} $. Let $B$ be an arbitrary point on the circle ${{w}_{1}}$, and line $B{{A}_{2}}$ intersects circle ${{w}_{2}}$ at point $C$. Let $H$ be the orthocenter of $\Delta B{{A}_{1}}C$. Prove that for arbitrary choice of point $B$, the point $H$ lies on a certain fixed circle.

On the hypotenuse $AB$ of a right triangle $ABC$, we denote a point $K$ such that $BK = BC$. Let $P$ be a point on the perpendicular from the point $K$ to line $CK$, equidistant from the points $K$ and $B$. Let $L$ be the midpoint of $CK$. Prove that line $AP$ is tangent to the circumcircle of $\Delta BLP$.

2017 Ukrainian Geometry Olympiad X p4
In the right triangle $ABC$ with hypotenuse $AB$, the incircle touches $BC$ and $AC$ at points ${{A}_{1}}$ and ${{B}_{1}}$ respectively. The straight line containing the midline of $\Delta ABC$ , parallel to $AB$, intersects its circumcircle at points $P$ and $T$. Prove that points $P,T,{{A}_{1}}$ and ${{B}_{1}}$ lie on one circle.

grade XI

In the triangle $ABC$, ${{A}_{1}}$ and ${{C}_{1}} $  are the midpoints of  sides $BC $ and $AC$ respectively. Point $P$ lies inside the triangle. Let $\angle BP {{C}_{1}} = \angle PCA$. Prove that $\angle BP {{A}_{1}} = \angle PAC $.

On the side $AC$ of a triangle $ABC$, let a $K$ be a point such that $AK = 2KC$ and $\angle ABK = 2 \angle KBC$. Let $F$ be the midpoint of $AC$, $L$ be the projection of $A$ on $BK$. Prove that $FL \bot BC$.

Let $ABCD$ be a parallelogram and $P$ be an arbitrary point of the circumcircle of $\Delta ABD$, different from the vertices. Line $PA$ intersects the line $CD$ at point $Q$. Let $O$ be the center of the circumcircle $\Delta PCQ$. Prove that $\angle ADO = 90^o$.

Let $AD$ be the inner angle bisector of the triangle $ABC$. The perpendicular on the side $BC$ at the point $D$ intersects the outer bisector of $\angle CAB$ at point $I$. The circle with center $I$ and radius $ID$ intersects the sides $AB$ and $AC$ at points $F$ and $E$ respectively. $A$-symmedian of $\Delta AFE$ intersects the circumcircle of $\Delta AFE$ again at point $X$. Prove that the circumcircles of $\Delta AFE$ and $\Delta BXC$ are tangent.

                                           2020 (14p)

In triangle $ABC$, bisectors are drawn $AA_1$ and $CC_1$. Prove that if the length of the perpendiculars drawn from the vertex $B$ on lines $AA1$ and $CC_1$ are equal, then $\vartriangle ABC$ is isosceles.

Inside the triangle $ABC$ is point $P$, such that $BP > AP$ and $BP > CP$. Prove that $\angle ABC$ is acute.

2020 Ukrainian Geometry Olympiad  VIII p3
Triangle $ABC$. Let $B_1$ and $C_1$ be such points, that $AB= BB_1,  AC=CC_1$ and $B_1, C_1$ lie on the circumscribed circle $\Gamma$ of $\vartriangle ABC$. Perpendiculars drawn from from points $B_1$ and $C_1$ on the lines $AB$ and $AC$ intersect $\Gamma$ at points $B_2$ and $C_2$ respectively, these points lie on smaller arcs $AB$ and $AC$ of circle $\Gamma$ respectively, Prove that $BB_2 \parallel CC_2$.

On the sides $AB$ and $AD$ of the square $ABCD$, the points $N$ and $P$ are selected respectively such that $NC=NP$. The point $Q$ is chosen on the segment $AN$ so that $\angle QPN = \angle NCB$. Prove that $2\angle BCQ = \angle AQP$.

The plane shows $2020$ straight lines in general position, that is, there are none three intersecting at one point but no two parallel. Let's say, that the drawn line $a$ detaches the drawn line $b$ if all intersection points of line $b$ with the other drawn lines lie in one half plane wrt to line $a$ (given the most straightforward $a$). Prove that you can be guaranteed find two drawn lines $a$ and $b$ that $a$ detaches $b$, but $b$ does not detach $a$.

grade IX

2020 Ukrainian Geometry Olympiad IX p1, VIII p2
Inside the triangle $ABC$ is point $P$, such that $BP > AP$ and $BP > CP$. Prove that $\angle ABC$ is acute.

2020 Ukrainian Geometry Olympiad IX p2, X p1, XI p1
Let $\Gamma$ be a circle and $P$ be a point outside, $PA$ and $PB$ be tangents to $\Gamma$ , $A, B \in  \Gamma$ . Point $K$ is an arbitrary point on the segment $AB$. The circumscirbed circle of $\vartriangle PKB$ intersects $\Gamma$ for the second time at point $T$, point $P'$ is symmetric to point $P$ wrt point $A$. Prove that $\angle PBT = \angle P'KA$.

2020 Ukrainian Geometry Olympiad IX p3, X p2
Let $H$ be the orthocenter of the acute-angled triangle $ABC$. Inside the segment $BC$ arbitrary point $D$ is selected. Let $P$ be such that $ADPH$ is a parallelogram. Prove that $\angle BCP<  \angle BHP$.

2020 Ukrainian Geometry Olympiad IX p4, VIII p5
The plane shows $2020$ straight lines in general position, that is, there are none three intersecting at one point but no two parallel. Let's say, that the drawn line $a$ detaches the drawn line $b$ if all intersection points of line $b$ with the other drawn lines lie in one half plane wrt to line $a$ (given the most straightforward $a$). Prove that you can be guaranteed find two drawn lines $a$ and $b$ that $a$ detaches $b$, but $b$ does not detach $a$.

Given a convex pentagon $ABCDE$, with $\angle BAC =  \angle ABE = \angle DEA - 90^o$, $\angle BCA =  \angle ADE$ and also $BC = ED$. Prove that $BCDE$ is parallelogram.

grade X

2020 Ukrainian Geometry Olympiad X p1, IX p2, XI p1
Let $\Gamma$ be a circle and $P$ be a point outside, $PA$ and $PB$ be tangents to $\Gamma$ , $A, B \in  \Gamma$ . Point $K$ is an arbitrary point on the segment $AB$. The circumscirbed circle of $\vartriangle PKB$ intersects $\Gamma$ for the second time at point $T$, point $P'$ is symmetric to point $P$ wrt point $A$. Prove that $\angle PBT = \angle P'KA$.

2020 Ukrainian Geometry Olympiad X p2, IX p3
Let $H$ be the orthocenter of the acute-angled triangle $ABC$. Inside the segment $BC$ arbitrary point $D$ is selected. Let $P$ be such that $ADPH$ is a parallelogram. Prove that $\angle BCP<  \angle BHP$.

2020 Ukrainian Geometry Olympiad X p3
The circles $\omega_1$ and $\omega_2$ intersect at points $A$ and $B$, point $M$ is the midpoint of $AB$. On line $AB$ select points $S_1$ and $S_2$. Let $S_1X_1$ and $S_1Y_1$ be tangents drawn from $S_1$ to circle $\omega_1$, similarly $S_2X_2$ and $S_2Y_2$ are tangents drawn from $S_2$ to circles $\omega_2$. Prove that if the point $M$ lies on the line $X_1X_2$, then it also lies on the line $Y_1Y_2$.

2020 Ukrainian Geometry Olympiad X p4
Inside triangle $ABC$, the point $P$ is chosen such that $\angle PAB = \angle  PCB =\frac14 (\angle A+ \angle C)$. Let $BL$ be the bisector of $\vartriangle ABC$. Line $PL$ intersects the circumcircle of $\vartriangle APC$ at point $Q$. Prove that the line $QB$ is the bisector of $\angle AQC$.

2020 Ukrainian Geometry Olympiad X p5, XI p4
On the plane painted $101$ points in brown and another $101$ points in green so that there are no three lying on one line. It turns out that the sum of the lengths of all $5050$ segments with brown ends equals the length of all $5050$ segments with green ends equal to $1$, and the sum of the lengths of all $10201$ segments with multicolored equals $400$. Prove that it is possible to draw a straight line so that all brown points are on one side relative to it and all green points are on the other.

grade XI

2020 Ukrainian Geometry Olympiad XI p1, IX p2, X p1
Let $\Gamma$ be a circle and $P$ be a point outside, $PA$ and $PB$ be tangents to $\Gamma$ , $A, B \in  \Gamma$ . Point $K$ is an arbitrary point on the segment $AB$. The circumscirbed circle of $\vartriangle PKB$ intersects $\Gamma$ for the second time at point $T$, point $P'$ is symmetric to point $P$ wrt point $A$. Prove that $\angle PBT = \angle P'KA$.

2020 Ukrainian Geometry Olympiad XI p2
Let $ABC$ be an isosceles triangle with $AB=AC$. Circle $\Gamma$ lies outside $ABC$ and touches line $AC$ at point $C$. The point $D$ is chosen on circle $\Gamma$ such that the circumscribed circle of the triangle $ABD$ touches externally circle $\Gamma$. The segment $AD$ intersects circle $\Gamma$ at a point $E$ other than $D$. Prove that $BE$ is tangent to circle $\Gamma$ .

2020 Ukrainian Geometry Olympiad XI p3
The angle $POQ$ is given ($OP$ and $OQ$ are rays). Let $M$ and $N$ be points inside the angle $POQ$ such that $\angle POM = \angle QON$ and $\angle POM < \angle PON$. Consider two circles: one touches the rays $OP$ and $ON$, the other touches the rays $OM$ and $OQ$. Denote by $B$ and $C$ the points of their intersection. Prove that $\angle POC = \angle QOB$.

2020 Ukrainian Geometry Olympiad XI p4 , X p5
On the plane painted $101$ points in brown and another $101$ points in green so that there are no three lying on one line. It turns out that the sum of the lengths of all $5050$ segments with brown ends equals the length of all $5050$ segments with green ends equal to $1$, and the sum of the lengths of all $10201$ segments with multicolored equals $400$. Prove that it is possible to draw a straight line so that all brown points are on one side relative to it and all green points are on the other.

2020 Ukrainian Geometry Olympiad XI p5
Inside the convex quadrilateral $ABCD$ there is a point $M$ such that $\angle AMB = \angle ADM +  \angle BCM$ and $\angle AMD = \angle ABM +   \angle DCM$. Prove that $AM \cdot  CM + BM  \cdot DM \ge \sqrt{AB \cdot BC\cdot CD \cdot DA}$

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