Ukrainian Geometry 2017

geometry problems
with aops links in the names
from Ukrainian Geometry Olympiad

(as far as I know it took place only in year 2017, 19 November 2017, as a thematic olympiad)

problems in pdf in English here
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.


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