Found problems: 2023
1994 Iran MO (2nd round), 2
In the following diagram, $O$ is the center of the circle. If three angles $\alpha, \beta$ and $\gamma$ be equal, find $\alpha.$
[asy]
unitsize(40);
import graph; size(300); real lsf = 0.5; pen dp = linewidth(0.7) + fontsize(10); defaultpen(dp); pen ds = black; pen ttttff = rgb(0.2,0.2,1); pen ffttww = rgb(1,0.2,0.4); pen qqwuqq = rgb(0,0.39,0);
draw(circle((0,0),2.33),ttttff+linewidth(2.8pt)); draw((-1.95,-1.27)--(0.64,2.24),ffttww+linewidth(2pt)); draw((0.64,2.24)--(1.67,-1.63),ffttww+linewidth(2pt)); draw((-1.95,-1.27)--(1.06,0.67),ffttww+linewidth(2pt)); draw((1.67,-1.63)--(-0.6,0.56),ffttww+linewidth(2pt)); draw((-0.6,0.56)--(1.06,0.67),ffttww+linewidth(2pt)); pair parametricplot0_cus(real t){
return (0.6*cos(t)+0.64,0.6*sin(t)+2.24);
}
draw(graph(parametricplot0_cus,-2.2073069497794027,-1.3111498158746024)--(0.64,2.24)--cycle,qqwuqq); pair parametricplot1_cus(real t){
return (0.6*cos(t)+-0.6,0.6*sin(t)+0.56);
}
draw(graph(parametricplot1_cus,0.06654165390165974,0.9342857038103908)--(-0.6,0.56)--cycle,qqwuqq); pair parametricplot2_cus(real t){
return (0.6*cos(t)+-0.6,0.6*sin(t)+0.56);
}
draw(graph(parametricplot2_cus,-0.766242589858673,0.06654165390165967)--(-0.6,0.56)--cycle,qqwuqq);
dot((0,0),ds); label("$O$", (-0.2,-0.38), NE*lsf); dot((0.64,2.24),ds); label("$A$", (0.72,2.36), NE*lsf); dot((-1.95,-1.27),ds); label("$B$", (-2.2,-1.58), NE*lsf); dot((1.67,-1.63),ds); label("$C$", (1.78,-1.96), NE*lsf); dot((1.06,0.67),ds); label("$E$", (1.14,0.78), NE*lsf); dot((-0.6,0.56),ds); label("$D$", (-0.92,0.7), NE*lsf); label("$\alpha$", (0.48,1.38),NE*lsf); label("$\beta$", (-0.02,0.94),NE*lsf); label("$\gamma$", (0.04,0.22),NE*lsf); clip((-8.84,-9.24)--(-8.84,8)--(11.64,8)--(11.64,-9.24)--cycle);
[/asy]
1997 Turkey MO (2nd round), 2
Let $F$ be a point inside a convex pentagon $ABCDE$, and let $a_{1}$, $a_{2}$, $a_{3}$, $a_{4}$, $a_{5}$ denote the distances from $F$ to the lines $AB$, $BC$, $CD$, $DE$, $EA$, respectively. The points $F_{1}$, $F_{2}$, $F_{3}$, $F_{4}$, $F_{5}$ are chosen on the inner bisectors of the angles $A$, $B$, $C$, $D$, $E$ of the pentagon respectively, so that $AF_{1} = AF$ , $BF_{2} = BF$ , $CF_{3} = CF$ , $DF_{4} = DF$ and $EF_{5} = EF$ . If the distances from $F_{1}$, $F_{2}$, $F_{3}$, $F_{4}$, $F_{5}$ to the lines $EA$, $AB$, $BC$, $CD$, $DE$ are $b_{1}$, $b_{2}$, $b_{3}$, $b_{4}$, $b_{5}$, respectively.
Prove that $a_{1} + a_{2} + a_{3} + a_{4} + a_{5} \leq b_{1} + b_{2} + b_{3} + b_{4} + b_{5}$
2010 Sharygin Geometry Olympiad, 7
The line passing through the vertex $B$ of a triangle $ABC$ and perpendicular to its median $BM$ intersects the altitudes dropped from $A$ and $C$ (or their extensions) in points $K$ and $N.$ Points $O_1$ and $O_2$ are the circumcenters of the triangles $ABK$ and $CBN$ respectively. Prove that $O_1M=O_2M.$
2008 Mongolia Team Selection Test, 3
Let $ \Omega$ is circle with radius $ R$ and center $ O$. Let $ \omega$ is a circle inside of the $ \Omega$ with center $ I$ radius $ r$. $ X$ is variable point of $ \omega$ and tangent line of $ \omega$ pass through $ X$ intersect the circle $ \Omega$ at points $ A,B$. A line pass through $ X$ perpendicular with $ AI$ intersect $ \omega$ at $ Y$ distinct with $ X$.Let point $ C$ is symmetric to the point $ I$ with respect to the line $ XY$.Find the locus of circumcenter of triangle $ ABC$ when $ X$ varies on $ \omega$
1990 IMO Longlists, 6
Let $S, T$ be the circumcenter and centroid of triangle $ABC$, respectively. $M$ is a point in the plane of triangle $ABC$ such that $90^\circ \leq \angle SMT < 180^\circ$. $A_1, B_1, C_1$ are the intersections of $AM, BM, CM$ with the circumcircle of triangle $ABC$ respectively. Prove that $MA_1 + MB_1 + MC_1 \geq MA + MB + MC.$
2015 Turkey MO (2nd round), 5
In a cyclic quadrilateral $ABCD$ whose largest interior angle is $D$, lines $BC$ and $AD$ intersect at point $E$, while lines $AB$ and $CD$ intersect at point $F$. A point $P$ is taken in the interior of quadrilateral $ABCD$ for which $\angle EPD=\angle FPD=\angle BAD$. $O$ is the circumcenter of quadrilateral $ABCD$. Line $FO$ intersects the lines $AD$, $EP$, $BC$ at $X$, $Q$, $Y$, respectively. If $\angle DQX = \angle CQY$, show that $\angle AEB=90^\circ$.
2004 Regional Olympiad - Republic of Srpska, 2
Given an isosceles triangle $ABC$ with base $AB$, cirumcenter $O$, incenter $S$ and $\angle C<60^\circ$.
The circumcircle of $AOS$ intersects $AC$ at $D$. Prove that $SD\parallel BC$ and $AS\perp OD$.
1985 Iran MO (2nd round), 2
In the triangle $ABC$ the length of side $AB$, and height $AH$ are known. also we know that $\angle B = 2 \angle C.$ Plot this triangle.
2007 South africa National Olympiad, 4
Let $ ABC$ be a triangle and $ PQRS$ a square with $ P$ on $ AB$, $ Q$ on $ AC$, and $ R$ and $ S$ on $ BC$. Let $ H$ on $ BC$ such that $ AH$ is the altitude of the triangle from $ A$ to base $ BC$. Prove that:
(a) $ \frac{1}{AH} \plus{}\frac{1}{BC}\equal{}\frac{1}{PQ}$
(b) the area of $ ABC$ is twice the area of $ PQRS$ iff $ AH\equal{}BC$
2010 Iran MO (3rd Round), 3
[b]points in plane[/b]
set $A$ containing $n$ points in plane is given. a $copy$ of $A$ is a set of points that is made by using transformation, rotation, homogeneity or their combination on elements of $A$. we want to put $n$ $copies$ of $A$ in plane, such that every two copies have exactly one point in common and every three of them have no common elements.
a) prove that if no $4$ points of $A$ make a parallelogram, you can do this only using transformation. ($A$ doesn't have a parallelogram with angle $0$ and a parallelogram that it's two non-adjacent vertices are one!)
b) prove that you can always do this by using a combination of all these things.
time allowed for this question was 1 hour and 30 minutes
2009 Kazakhstan National Olympiad, 5
Quadrilateral $ABCD$ inscribed in circle with center $O$. Let lines $AD$ and $BC$ intersects at $M$, lines $AB$ and $CD$- at $N$, lines $AC$ and $BD$ -at $P$, lines $OP$ and $MN$ at $K$.
Proved that $ \angle AKP = \angle PKC$.
As I know, this problem was very short solution by polars, but in olympiad for this solution gives maximum 4 balls (in marking schemes written, that needs to prove all theorems about properties of polars)
2009 Vietnam Team Selection Test, 2
Let a circle $ (O)$ with diameter $ AB$. A point $ M$ move inside $ (O)$. Internal bisector of $ \widehat{AMB}$ cut $ (O)$ at $ N$, external bisector of $ \widehat{AMB}$ cut $ NA,NB$ at $ P,Q$. $ AM,BM$ cut circle with diameter $ NQ,NP$ at $ R,S$.
Prove that: median from $ N$ of triangle $ NRS$ pass over a fix point.
1999 Turkey MO (2nd round), 5
In an acute triangle $\vartriangle ABC$ with circumradius $R$, altitudes $\overline{AD},\overline{BE},\overline{CF}$ have lengths ${{h}_{1}},{{h}_{2}},{{h}_{3}}$, respectively. If ${{t}_{1}},{{t}_{2}},{{t}_{3}}$ are lengths of the tangents from $A,B,C$, respectively, to the circumcircle of triangle $\vartriangle DEF$, prove that
$\sum\limits_{i=1}^{3}{{{\left( \frac{t{}_{i}}{\sqrt{h{}_{i}}} \right)}^{2}}\le }\frac{3}{2}R$.
2008 Saint Petersburg Mathematical Olympiad, 2
Point $O$ is the center of the circle into which quadrilateral $ABCD$ is inscribed. If angles $AOC$ and $BAD$ are both equal to $110$ degrees and angle $ABC$ is greater than angle $ADC$, prove that $AB+AD>CD$.
Fresh translation.
2009 Sharygin Geometry Olympiad, 21
The opposite sidelines of quadrilateral $ ABCD$ intersect at points $ P$ and $ Q$. Two lines passing through these points meet the side of $ ABCD$ in four points which are the vertices of a parallelogram. Prove that the center of this parallelogram lies on the line passing through the midpoints of diagonals of $ ABCD$.
2014 Iran MO (3rd Round), 5
$X$ and $Y$ are two points lying on or on the extensions of side $BC$ of $\triangle{ABC}$ such that $\widehat{XAY} = 90$. Let $H$ be the orthocenter of $\triangle{ABC}$. Take $X'$ and $Y'$ as the intersection points of $(BH,AX)$ and $(CH,AY)$ respectively. Prove that circumcircle of $\triangle{CYY'}$,circumcircle of $\triangle{BXX'}$ and $X'Y'$ are concurrent.
2006 Vietnam Team Selection Test, 1
Given an acute angles triangle $ABC$, and $H$ is its orthocentre. The external bisector of the angle $\angle BHC$ meets the sides $AB$ and $AC$ at the points $D$ and $E$ respectively. The internal bisector of the angle $\angle BAC$ meets the circumcircle of the triangle $ADE$ again at the point $K$. Prove that $HK$ is through the midpoint of the side $BC$.
2008 Sharygin Geometry Olympiad, 7
(A.Zaslavsky, 8--9) Given a circle and a point $ O$ on it. Another circle with center $ O$ meets the first one at points $ P$ and $ Q$. The point $ C$ lies on the first circle, and the lines $ CP$, $ CQ$ meet the second circle for the second time at points $ A$ and $ B$. Prove that $ AB\equal{}PQ$.
2016 Croatia Team Selection Test, Problem 3
Let $ABC$ be an acute triangle with circumcenter $O$. Points $E$ and $F$ are chosen on segments $OB$ and $OC$ such that $BE = OF$. If $M$ is the midpoint of the arc $EOA$ and $N$ is the midpoint of the arc $AOF$, prove that $\sphericalangle ENO + \sphericalangle OMF = 2 \sphericalangle BAC$.
2017 Baltic Way, 15
Let $n \ge 3$ be an integer. What is the largest possible number of interior angles greater than $180^\circ$ in an $n$-gon in the plane, given that the $n$-gon does not intersect itself and all its sides have the same length?
2004 Iran Team Selection Test, 4
Let $ M,M'$ be two conjugates point in triangle $ ABC$ (in the sense that $ \angle MAB\equal{}\angle M'AC,\dots$). Let $ P,Q,R,P',Q',R'$ be foots of perpendiculars from $ M$ and $ M'$ to $ BC,CA,AB$. Let $ E\equal{}QR\cap Q'R'$, $ F\equal{}RP\cap R'P'$ and $ G\equal{}PQ\cap P'Q'$. Prove that the lines $ AG, BF, CE$ are parallel.
2008 Czech-Polish-Slovak Match, 2
$ABCDE$ is a regular pentagon. Determine the smallest value of the expression
\[\frac{|PA|+|PB|}{|PC|+|PD|+|PE|},\]
where $P$ is an arbitrary point lying in the plane of the pentagon $ABCDE$.
2007 Korea - Final Round, 1
Let $ O$ be the circumcenter of an acute triangle $ ABC$ and let $ k$ be the circle with center $ P$ that is tangent to $ O$ at $ A$ and tangent to side $ BC$ at $ D$. Circle $ k$ meets $ AB$ and $ AC$ again at $ E$ and $ F$ respectively. The lines $ OP$ and $ EP$ meet $ k$ again at $ I$ and $ G$. Lines $ BO$ and $ IG$ intersect at $ H$. Prove that $ \frac{{DF}^2}{AF}\equal{}GH$.
2024 Bundeswettbewerb Mathematik, 3
Let $ABC$ be a triangle. For a point $P$ in its interior, we draw the threee lines through $P$ parallel to the sides of the triangle. This partitions $ABC$ in three triangles and three quadrilaterals.
Let $V_A$ be the area of the quadrilateral which has $A$ as one vertex. Let $D_A$ be the area of the triangle which has a part of $BC$ as one of its sides. Define $V_B, D_B$ and $V_C, D_C$ similarly.
Determine all possible values of $\frac{D_A}{V_A}+\frac{D_B}{V_B}+\frac{D_C}{V_C}$, as $P$ varies in the interior of the triangle.
2007 Kyiv Mathematical Festival, 2
The point $D$ at the side $AB$ of triangle $ABC$ is given. Construct points $E,F$ at sides $BC, AC$ respectively such that the midpoints of $DE$ and $DF$ are collinear with $B$ and the midpoints of $DE$ and $EF$ are collinear with $C.$