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Serbia JBMO TST 2007-22 20p (-12,-13)

geometry problems from Serbian Junior Balkan Mathematical Olympiads Team Selection Tests (JBMO TST) with aops links

[not in JBMO Shortlist]
collected inside aops here 

2007-2022 (-12,-13)
[+ Training Tests 2004, 2006, 2007]

2004 Serbia JBMO Training Test p2
Let $ABCDEF$ be a regular hexagon with center $O$. Let $M, N$ be the midpoints of sides $[CD]$, respectively $[DE]$, and $L$ be the intersection of the lines $AM$ and $BN$. Prove that the area of the triangle $ALB$ is equal to the area of the quadrilateral $DMLN$, $\angle OLD  = 90^o$ and $\angle  ALO = \angle OLN = 60^o$

Let $E$ and $F$ lies on sides $AC$ and $AB$ of triangle $ABC$ respectively, such that $EF\parallel BC$. .Prove that the intersections of the circles with diameters $BE$ and $CF$ lay on the altitude from $A$.

2007 Serbia JBMO Training Test p2
The rectangles $BCKL$ and $ACPQ$ are constructed on the sides $[BC]$ and $[AC]$ of the acute triangle $ABC$ so that their areas are equal. Show that the points $C$, the center $O$ of the circle circumscribed around the triangle $ABC$, and the middle of the segment $[PK]$ are collinear

2007 Serbia JBMO TST 1.1
Inside a parallelogram $ABCD$ is considered a point $P$ such that $\angle ADP = \angle ABP$  and $\angle DCP = 30^o$ . Find the measure of the angle $\angle DAP$.

2007 Serbia JBMO TST 2.3
On the sides $AB, BC , CA$ of the acute triangle $ABC$ consider the points $D, E$, respectively  so that the lengths of the segments $[AE], [BF]$ ¸and $[CD]$ are at most $\sqrt3$ cm. Prove that the area of triangle $ABC$ is at most $\sqrt3$ cm$^2$

2008 Serbia JBMO TST 1.1
In triangle $ABC$,  $\angle A = 120^o$  , $AB = 3$ and $AC = 6$. The bisector of the angle $A$ intersects the side $BC$ at point $D$. Find the length of the segment  $[AD]$.

2008 Serbia JBMO TST 2.3
In triangle $ABC$ the side $[BC]$ is the shortest. On the sides $[AB]$ ,$[AC]$ are considered the points $D ,E$  so that $BD = CE = BC$. Show that the radius of the circumcircle of the triangle $ADE$ is equal to the distance between the center of the circumcircle of the triangle $ABC$ and the center of the incircle of the triangle $ABC$.

2009 Serbia JBMO TST 1.2
In isosceles right triangle $ ABC$ a circle is inscribed. Let $ CD$ be the hypotenuse height ($ D\in AB$), and let $ P$ be the intersection of inscribed circle and height $ CD$. In which ratio does the circle divide segment $ AP$?

2009 Serbia JBMO TST 2.3
Let $ ABCD$ be a convex quadrilateral, such that $ \angle CBD=2\cdot\angle ADB, \angle ABD=2\cdot\angle CDB$ and $ AB=CB$. Prove that quadrilateral $ ABCD$ is a kite.

2010 Serbia JBMO TST 1.1
Points $E$ and $F$ are the feet of the altitudes at the vertices $B$ and $C$ of the of the triangle $ABC$. The point $M$ is the foot of the perpendicular drawn from $F$ to $BC$, and$ N$ is the foot of the perpendicular from $B$ to $EF$. Show that the lines $AC$ and $MN$ are parallel.

2010 Serbia JBMO TST 2.4 (Croatian TST 2008)
Point $ M$ is taken on side $ BC$ of a triangle $ ABC$ such that the centroid $ T_c$ of triangle $ ABM$ lies on the circumcircle of $ \triangle ACM$ and the centroid $ T_b$ of $ \triangle ACM$ lies on the circumcircle of $ \triangle ABM$. Prove that the medians of the triangles $ ABM$ and $ ACM$ from $ M$ are of the same length.

2011 Serbia JBMO TST p3
Let $ABC$ be a right triangle $\angle ACB=90^{o}$ and $AC \ge BC$. Point $M$ is on $AC$ and $AM=BC$, $N$ is on $BC$ and $BN=MC$. Lines $AN$ and $BM$ intersect in $K$. Find $\angle AKM$.

2012 -13 missing

2014 Serbia JBMO TST  p3
Consider parallelogram $ABCD$, with acute angle at $A$, $AC$ and $BD$ intersect at $E$. Circumscribed circle of triangle $ACD$ intersects $AB$, $BC$ and $BD$ at $K$, $L$ and $P$ (in that order). Then, circumscribed circle of triangle $CEL$ intersects $BD$ at $M$. Prove that: $KD\cdot KM=KL \cdot PC$

2015 Serbia JBMO TST p4
The diagonals $AD$, $BE$, $CF$ of cyclic hexagon $ABCDEF$ intersect in $S$ and $AB$ is parallel to $CF$ and lines $DE$ and $CF$ intersect each other in $M$. Let $N$ be a point such that $M$ is the midpoint of $SN$. Prove that circumcircle of $\triangle ADN$ is passing through midpoint of segment $CF$.

2016 Serbia JBMO TST p1
Let rightangled $\triangle ABC$ be given with right angle at vertex $C$. Let $D$ be foot of altitude from $C$ and let $k$ be circle that touches $BD$ at $E$, $CD$ at $F$ and circumcircle of $\triangle ABC$ at $G$.
a) Prove that points $A$, $F$ and $G$ are collinear.
b) Express radius of circle $k$ in terms of sides of $\triangle ABC$.

2017 Serbia JBMO TST p4
In the acute non-isosceles triangle $ABC$, $\angle C= 60^o$.Let $A'$ and $B'$ be the  feet in the heights of A and $B$, respectively, and $G$ the centroid  of the triangle. Rays  $(A'G$ and $(B'G$ intersect the circumcircle of the triangle at points $M,N$ respectively .Prove that $MN = AB$.

2018 Serbia JBMO TST  p1
Let $AD$ be an internal angle bisector in triangle $\Delta ABC$. An arbitrary point $M$ is chosen on the closed segment $AD$. A parallel to $BC$ through $M$ cuts $AB$ at $N$. Let $AD, CM$ cut circumcircle of $\Delta ABC$ at $K, L$, respectively. Prove that $K,N,L$ are collinear.

2019 Serbia JBMO TST  p3
Congruent circles $k_{1}$ and $k_{2}$ intersect in the points $A$ and $B$. Let $P$ be a variable point of arc $AB$ of circle $k_{2}$ which is inside $k_{1}$ and let $AP$ intersect $k_{1}$ once more in point $C$, and the ray $CB$ intersects $k_{2}$ once more in $D$. Let the angle bisector of $\angle CAD$ intersect $k_{1}$ in $E$, and the circle $k_{2}$ in $F$. Ray $FB$ intersects $k_{1}$ in $Q$. If $X$ is one of the intersection points of circumscribed circles of triangles $CDP$ and $EQF$, prove that the triangle $CFX$ is equilateral.

Given is triangle $ABC$ with arbitrary point $D$ on $AB$ and central of inscribed circle $I$. The perpendicular bisector of $AB$ intersects $AI$ and $BI$ at $P$ and $Q$, respectively. The circle $(ADP)$ intersects $CA$ at $E$, and the circle $(BDQ)$ intersects $BC$ at $F$ and $(ADP)$ intersects $(BDQ)$ at $K$. Prove that $E, F, K, I$ lie on one circle.

On sides $AB$ and $AC$ of an acute triangle $\Delta ABC$, with orthocenter $H$ and circumcenter $O$, are given points $P$ and $Q$ respectively such that $APHQ$ is a parallelogram. Prove the following equality: $\frac{PB\cdot PQ}{QA\cdot QO}=2$

Let $I$ be the incenter, $A_1$ and $B_1$ midpoints of sides $BC$ and $AC$ of a triangle $\Delta ABC$. Denote by $M$ and $N$ the midpoints of the arcs $AC$ and $BC$ of circumcircle of $\Delta ABC$ which do contain the other vertex of the triangle. If points $M$, $I$ and $N$ are collinear prove that: $\angle AIB_1=\angle BIA_1=90^{\circ}$


source:  https://pregatirematematicaolimpiadejuniori.wordpress.com/

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