Ukraine (hidden)

geometry problems from Ukrainian Mathematical Olympiads
with aops links in the names

2009 - 2018
under construction


2009 Ukraine MO grade VIII P4
In the triangle $ABC$ given that $\angle ABC = 120^\circ .$ The bisector of $\angle B$ meet $AC$ at $M$ and external bisector of $\angle BCA$ meet $AB$ at $P.$ Segments $MP$ and $BC$ intersects at $K$. Prove that $\angle AKM = \angle KPC .$

2009 Ukraine MO grade VIII P6
In acute-angled triangle $ABC,$ let $M$ be the midpoint of $BC$ and let $K$ be a point on side $AB.$ We know that $AM$ meet $CK$ at $F.$ Prove that if $AK = KF$ then $AB = CF .$

2009 Ukraine MO grade IX P4, grade X P3
In triangle $ABC$ points $M, N$ are midpoints of $BC, CA$ respectively. Point $P$ is inside $ABC$ such that $\angle BAP = \angle PCA = \angle MAC .$ Prove that $\angle PNA = \angle AMB .$

2009 Ukraine MO grade IX P8
In the trapezoid $ABCD$ we know that $CD \perp BC, $ and $CD \perp AD .$ Circle $w$ with diameter $AB$ intersects $AD$ in points $A$ and $P,$ tangent from $P$ to $w$ intersects $CD$ at $M.$ The second tangent from $M$ to $w$ touches $w$ at $Q.$ Prove that midpoint of $CD$ lies on $BQ.$

2009 Ukraine MO grade X P8
Let $ABCD$ be a parallelogram with $\angle BAC = 45^\circ,$ and $AC > BD .$ Let $w_1$ and $w_2$ be two circles with diameters $AC$ and $DC,$ respectively. The circle $w_1$ intersects $AB$ at $E$ and the circle $w_2$ intersects $AC$ at $O$ and $C$, and $AD$ at $F.$ Find the ratio of areas of triangles $AOE$ and $COF$ if $AO = a,$ and $FO = b .$

2009 Ukraine MO grade XI P3
In triangle $ABC$ let $M$ and $N$ be midpoints of $BC$ and $AC,$ respectively. Point $P$ is inside $ABC$ such that $\angle BAP = \angle PBC = \angle PCA .$ Prove that if $\angle PNA = \angle AMB,$ then $ABC$ is isosceles triangle.


2010 Ukraine MO grade VIII P
2010 Ukraine MO grade VIII P
2010 Ukraine MO grade IX P
2010 Ukraine MO grade IX P
2010 Ukraine MO grade X P
2010 Ukraine MO grade X P
2010 Ukraine MO grade XI P
2010 Ukraine MO grade XI P

2011 Ukraine MO grade VIII P
2011 Ukraine MO grade VIII P
2011 Ukraine MO grade IX P
2011 Ukraine MO grade IX P
2011 Ukraine MO grade X P
2011 Ukraine MO grade X P
2011 Ukraine MO grade XI P
2011 Ukraine MO grade XI P

2012 Ukraine MO grade VIII P
2012 Ukraine MO grade VIII P
2012 Ukraine MO grade IX P
2012 Ukraine MO grade IX P
2012 Ukraine MO grade X P
2012 Ukraine MO grade X P
2012 Ukraine MO grade XI P
2012 Ukraine MO grade XI P

2013 Ukraine MO grade VIII P
2013 Ukraine MO grade VIII P
2013 Ukraine MO grade IX P
2013 Ukraine MO grade IX P
2013 Ukraine MO grade X P
2013 Ukraine MO grade X P
2013 Ukraine MO grade XI P
2013 Ukraine MO grade XI P

2014 Ukraine MO grade VIII P
2014 Ukraine MO grade VIII P
2014 Ukraine MO grade IX P
2014 Ukraine MO grade IX P

2014 Ukraine MO grade X P2
Let $M$ be the midpoint of the side $BC$ of $\triangle ABC$. On the side $AB$ and $AC$ the points $E$ and $F$ are chosen. Let $K$ be the point of the intersection of $BF$ and $CE$ and $L$ be chosen in a way that $CL\parallel AB$ and $BL\parallel CE$. Let $N$ be the point of intersection of $AM$ and $CL$. Show that $KN$ is parallel to $FL$.

2014 Ukraine MO grade X P8
Let $M$ be the midpoint of the internal bisector $AD$ of $\triangle ABC$.Circle $\omega_1$ with diameter $AC$ intersects $BM$ at $E$ and circle $\omega_2$ with diameter $AB$ intersects $CM$ at $F$.Show that $B,E,F,C$ are concyclic.

2014 Ukraine MO grade XI P
2014 Ukraine MO grade XI P

2015 Ukraine MO grade VIII P
2015 Ukraine MO grade VIII P
2015 Ukraine MO grade IX P
2015 Ukraine MO grade IX P
2015 Ukraine MO grade X P
2015 Ukraine MO grade X P

2015 Ukraine MO grade XI P4
In convex quadrilateral $ABCD$ with angles $ABC$ and $BCD$ equal to $120^{\circ}$, $O$ is intersection of diagonals, $M$ is midpoint of $BC$. $K$ is intersection of segments $MO$ and $AD$. It's known that $\angle BKC=60^{\circ}$. Prove that $\angle BKA=\angle CKD = 60^{\circ}$.

by Nazar Serdyuk

2015 Ukraine MO grade XI P

2016 Ukraine MO grade VIII P
2016 Ukraine MO grade VIII P
2016 Ukraine MO grade IX P
2016 Ukraine MO grade IX P
2016 Ukraine MO grade X P
2016 Ukraine MO grade X P
2016 Ukraine MO grade XI P

2016 Ukraine MO grade XI P7
Triangle $ABC$ is given. Circle $\omega$ with center $Q$ is tangent to side $BC$ and touches the circumcircle of triangle $ABC$ internally at point $A.$ Let $M$ be the midpoint of $BC$ and $N$ be the midpoint of arc $BAC$ of the circumcircle of triangle $ABC.$ Point $S$ is chosen on the segment $BC$ such that $\angle BAM=\angle SAC$. Prove that points $N,$ $Q$ and $S$ are collinear.

by M. Plotnikov

2017 Ukraine MO grade VIII P4
$\triangle ABC$. $\angle C = 90^\circ$. Inside $\triangle ABC$ there is point $K$, such that $\angle AKC = 90^\circ$ and $\angle CKB = 2 \angle CAB$. On segment $KB$ there is point $T$, such that $\angle KTC = \angle CAK$.  $P = AK \cap BC$.  Prove that  $\angle TPA = \angle ABC$.

2017 Ukraine MO grade VIII P

2017 Ukraine MO grade IX P4
$\omega$ - circle with diameter $AB$ and center $O$. $CD$ - chord perpendicular $AB$. $E$ - middle point of $OC$. Line $AE$ intersects $\omega$  at $F$ and $BC$ at $M$. $L = BC \cap DF$. $\odot (DLM)$ intersects $\omega$ at $K$.  Prove that $KM=MB$

2017 Ukraine MO grade IX P
2017 Ukraine MO grade X P
2017 Ukraine MO grade X P
2017 Ukraine MO grade XI P

2017 Ukraine MO grade XI P6
The quadrilateral $ABCD$ is inscribed in a circle $\omega$ with the center $O$ . Its diagonals
intersect at $H$ .  $O_1$ and $O_2$ are the centers of the $\odot (AHD)$ and $\odot (BHC)$ respectively. Line through $H$ intersects $\omega$ at $M_1$ and $M_2$. It also intersects $\odot (O_1HO) and \odot (O_2HO)$ at $N_1$ and $N_2$ respectively. $N_1$ and $N_2$ are lay inside of $ \omega$. Prove that $M_1N_1=M_2N_2$.



2018 Ukraine MO grade VIII P
2018 Ukraine MO grade VIII P
2018 Ukraine MO grade IX P
2018 Ukraine MO grade IX P
2018 Ukraine MO grade X P
2018 Ukraine MO grade X P

2018 Ukraine MO grade XI P2
In acute-angled triangle $ABC$, $AH$ is an altitude and $AM$ is a median. Points $X$ and $Y$ on lines $AB$ and $AC$ respectively are such that $AX=XC$ and $AY=YB$. Prove that the midpoint of $XY$ is equidistant from $H$ and $M$.

by Danylo Khilko

2018 Ukraine MO grade XI P8
Given an acute-angled triangle $ABC$, $AA_1$ and $CC_1$ are its angle bisectors, $I$ is its incenter, $M$ and $N$ are the midpoints of $AI$ and $CI$. Points $K$ and $L$ in the interior of triangles $AC_1I$ and $CA_1I$ respectively are such that $\angle AKI = \angle CLI = \angle AIC$, $\angle AKM = \angle ICA$, $\angle CLN = \angle IAC$. Prove that the circumradii of triangles $KIL$ and $ABC$ are equal.

by Anton Trygub

source: matholymp.org.ua/
mathedu.kharkiv.ua/examples/sefu/

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