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Czech-Polish-Slovak Junior Match 2012-22 (CPSJ) 38p

geometry problems from Czech-Polish-Slovak Junior Mathematical Match (CPSJ)
with aops links in the names

aops post collection: here
collected inside aops here
2012  - 2022
it did not take place in 2020

Point $P$ lies inside the triangle $ABC$. Points $K, L, M$ are symmetrics of point $P$ wrt the midpoints of the sides $BC, CA, AB$. Prove that the straight $AK, BL, CM$ intersect at one point.

Different points $A, B, C, D$ lie on a circle with a center at the point $O$ at such way that $\angle AOB$ $= \angle BOC =$  $\angle COD =$ $60^o$. Point $P$ lies on the shorter arc $BC$ of this circle. Points $K, L, M$ are projections of  $P$ on lines $AO, BO, CO$ respectively . Show that
(a) the triangle $KLM$ is equilateral,
(b) the area of triangle $KLM$ does not depend on the choice of the position of point $P$ on the shorter arc $BC$

On the circle $k$, the points $A,B$ are given, while $AB$ is not the diameter of the circle $k$. Point $C$ moves along  the long arc $AB$  of circle $k$ so that the triangle $ABC$ is acute. Let $D,E$ be the feet of the altitudes from $A, B$ respectively. Let $F$ be the projection of point $D$ on line $AC$ and $G$  be the projection of point $E$ on line $BC$.
(a) Prove that the lines $AB$ and $FG$ are parallel.
(b) Determine the set of midpoints $S$ of segment $FG$ while along all allowable positions of point $C$.

CPSJ Match 2012 Team p4
A rhombus $ABCD$ is given with $\angle BAD = 60^o$ . Point $P$ lies inside the rhombus such that $BP = 1$, $DP = 2$, $CP = 3$. Determine the length of the segment $AP$.

The $ABCDE$ pentagon is inscribed in a circle and $AB = BC = CD$. Segments $AC$ and $BE$ intersect at $K$, and Segments $AD$ and $CE$ intersect at point $L$. Prove that $AK = KL$.

Point $M$ is the midpoint of the side $AB$ of an acute triangle $ABC$. Point $P$ lies on the segment $AB$, and points $S_1$ and $S_2$ are the centers of the circumcircles of $APC$ and $BPC$, respectively. Show that the midpoint of segment $S_1S_2$ lies on the perpendicular bisector of segment $CM$.

Let $ABCD$ be a convex quadrilateral with $\angle DAB =\angle ABC =\angle BCD > 90^o$. The circle circumscribed around the triangle $ABC$ intersects the sides $AD$ and $CD$ at points $K$ and $L$, respectively, different from any vertex of the quadrilateral $ABCD$ . Segments $AL$ and $CK$ intersect at point $P$. Prove that $\angle ADB =\angle PDC$.

There is a square $ABCD$ in the plane with $|AB|=a$. Determine the smallest possible radius value of three equal circles to cover a given square.

On the plane circles $k$ and $\ell$ are intersected at points $C$ and $D$, where circle $k$ passes through the center $L$ of circle $\ell$. The straight line passing through point $D$ intersects circles $k$ and $\ell$ for the second time at points $A$ and $B$ respectively in such a way that $D$ is the interior point of segment $AB$. Show that $AB = AC$.

Let $ABCD$ be a parallelogram with $\angle BAD<90^o$ and $AB> BC$ . The angle bisector of $BAD$ intersects line $CD$ at point $P$ and line $BC$ at point $Q$. Prove that the center of the circle circumscirbed around the triangle $CPQ$ is equidistant from points $B$ and $D$.

Point $M$ is the midpoint of the side $AB$ of an acute triangle $ABC$. Circle with  center $M$ passing through point $ C$, intersects lines $AC ,BC$ for the second time at points $P,Q$ respectively.  Point $R$ lies on segment $AB$ such that the triangles $APR$ and $BQR$ have equal areas. Prove that lines $PQ$ and $CR$ are perpendicular.

In the right triangle $ABC$ with shorter side $AC$ the hypotenuse $AB$ has  length $12$. Denote $T$ its centroid and $D$ the feet of altitude from the vertex $C$. Determine the size of its inner angle at the vertex $B$ for which the triangle $DTC$ has the greatest possible area.

Let $ABC$ ne a right triangle with $\angle ACB=90^o$. Let $E, F$ be respecitvely the midpoints of the $BC, AC$ and $CD$ be it's altitude. Next, let $P$ be the intersection of the internal angle bisector from $A$ and the line $EF$. Prove that $P$ is the center of the circle inscribed in the triangle $CDE$ .

Let $I$ be the center of the circle of the inscribed triangle $ABC$ and $M$ be the center of its side $BC$. If $|AI| = |MI|$, prove that there are two of the sides of triangle $ABC$, of which one is twice of the other.

Different points $A$ and $D$ are on the same side of the line $BC$, with $|AB| = | BC|= |CD|$ and lines $AD$ and $BC$ are perpendicular. Let $E$ be the intersection point of lines $AD$ and $BC$. Prove that $||BE| - |CE|| < |AD| \sqrt3$

CPSJ Match 2016 Individual p1
Let $AB$ be a given segment and $M$ be its midpoint. We consider the set of right-angled triangles $ABC$ with hypotenuses $AB$. Denote by $D$ the foot of the altitude from $C$. Let $K$ and $L$ be feet of perpendiculars from $D$ to the legs $BC$ and $AC$, respectively. Determine the largest possible area of the quadrilateral $MKCL$.
Czech Republic
We are given an acute-angled triangle $ABC$ with $AB < AC < BC$. Points $K$ and $L$ are chosen on segments $AC$ and $BC$, respectively, so that $AB = CK = CL$. Perpendicular bisectors of segments $AK$ and $BL$ intersect the line $AB$ at points $P$ and $Q$, respectively. Segments $KP$ and $LQ$ intersect at point $M$. Prove that $AK + KM = BL + LM$.
Poland

Let $ABC$ be a right-angled triangle with hypotenuse $AB$. Denote by $D$ the foot of the altitude from $C$. Let $Q, R$, and $P$ be the midpoints of the segments $AD, BD$, and $CD$, respectively. Prove that $\angle AP B + \angle QCR = 180^o$.
Czech Republic
CPSJ Match 2016 Team p5
Let $ABC$ be a triangle with $AB : AC : BC =5:5:6$. Denote by $M$ the midpoint of $BC$ and by $N$ the point on the segment $BC$ such that $BN = 5 \cdot CN$. Prove that the circumcenter of triangle $ABN$ is the midpoint of the segment connecting the incenters of triangles $ABC$ and $ABM$.
Slovakia

Given is the triangle $ABC$, with $| AB | + | AC | = 3 \cdot | BC | $. Let's denote $D, E$ also points that $BCDA$ and $CBEA$ are parallelograms. On the sides $AC$ and $AB$ sides, $F$ and $G$ are selected respectively so that $| AF | = | AG | = | BC |$. Prove that the lines $DF$ and $EG$ intersect at the line segment $BC$

Given is a right triangle $ABC$ with perimeter $2$, with $\angle B=90^o$ . Point $S$ is the center of the excircle to the side $AB$ of the triangle and $H$ is the intersection of the heights of the triangle $ABS$ . Determine the smallest possible length of the segment $HS $.

CPSJ Match 2017 Team p2
Decide if exists a convex hexagon with all sides longer than $1$ and all nine of its diagonals are less than $2$ in length.

Bolek draw a trapezoid $ABCD$ trapezoid ($AB // CD$) on the board, with its midsegment line $EF$ in it. Point intersection of his diagonal $AC, BD$ denote by $P,$ and his rectangular projection on straight $AB$ denote by $Q$. Lolek, wanting to tease Bolek, blotted from the board everything except segments $EF$ and $PQ$. When Bolek saw it, wanted to complete the drawing and draw the original trapezoid, but did not know how to do it. Can you help Bolek?

CPSJ Match 2018 Individual p2
A convex hexagon $ABCDEF$ is given whose sides $AB$ and $DE$ are parallel. Each of the diagonals $AD, BE, CF$ divides this hexagon into two quadrilaterals of equal perimeters. Show that these three diagonals intersect at one point.

CPSJ Match 2018 Individual p5
An acute triangle $ABC$ is given in which $AB <AC$. Point $E$ lies on the $AC$ side of the triangle, with $AB = AE$. The segment $AD$ is the diameter of the circumcircle of the triangle $ABC$, and point $S$ is the center of this arc $BC$ of this circle to which point $A$ does not belong. Point $F$ is symmetric of point $D$ wrt $S$. Prove that lines $F E$ and $AC$ are perpendicular.

Given a right triangle $ABC$ with the hypotenuse $AB$. Let $K$ be any interior point of triangle $ABC$ and points $L, M$ are symmetric of point $K$ wrt lines $BC, AC$ respectively. Specify all possible values for $S_{ABLM} / S_{ABC}$, where $S_{XY ... Z}$ indicates the area of the polygon $XY...Z$ .

A line passing through the center $M$ of the equilateral triangle $ABC$ intersects sides $BC$ and $CA$, respectively, in points $D$ and $E$. Circumcircles of triangle $AEM$ and $BDM$ intersects, besides point $M$, also at point $P$. Prove that the center of circumcircle of triangle $DEP$ lies on the perpendicular bisector of the segment $AB$.

Let $ABC$ be a triangle with centroid $T$. Denote by $M$ the midpoint of $BC$. Let $D$ be a point on the ray opposite to the ray $BA$ such that $AB = BD$. Similarly, let $E$ be a point on the ray opposite to the ray $CA$ such that $AC = CE$. The segments $T D$ and $T E$ intersect the side $BC$ in $P$ and $Q$, respectively. Show that the points $P, Q$ and $M$ split the segment $BC$ into four parts of equal length.

CPSJ Match 2019 Individual p4
Let $k$ be a circle with diameter $AB$. A point $C$ is chosen inside the segment $AB$ and a point $D$ is chosen on $k$ such that $BCD$ is an acute-angled triangle, with circumcentre denoted by $O$. Let $E$ be the intersection of the circle $k$ and the line $BO$ (different from $B$). Show that the triangles $BCD$ and $ECA$ are similar.

Let $ABCD$ be a convex quadrilateral with perpendicular diagonals, such that $\angle BAC = \angle ADB$, $\angle CBD = \angle DCA$, $AB = 15$, $CD = 8$. Show that $ABCD$ is cyclic and find the distance between its circumcentre and the point of intersection of its diagonals.

Given is a cyclic quadrilateral $ABCD$. Points $K, L, M, N$ lying on sides $AB, BC, CD, DA$, respectively, satisfy $\angle ADK=\angle BCK$, $\angle BAL=\angle CDL$, $\angle CBM =\angle DAM$, $\angle DCN =\angle ABN$. Prove that lines $KM$ and $LN$ are perpendicular.

An acute triangle $ABC$ is given. Let us denote by $D$ and $E$ the orthogonal projections, respectively of points $ B$ and $C$ on the bisector of the external angle $BAC$. Let $F$ be the point of intersection of the lines $BE$ and $CD$. Show that the lines $AF$ and $DE$ are perpendicular.

A regular heptagon $ABCDEFG$ is given. The lines $AB$ and $CE$ intersect at $ P$. Find the measure of the angle $\angle PDG$.

Consider a trapezoid $ABCD$ with bases $AB$ and $CD$ satisfying $| AB | > | CD |$. Let $M$ be the midpoint of $AB$. Let the point $P$ lie inside $ABCD$ such that $| AD | = | PC |$ and $| BC | = | PD |$. Prove that if $| \angle CMD | = 90^o$, then the quadrilaterals $AMPD$ and $BMPC$ have the same area.

Given is a convex pentagon $ABCDE$ in which $\angle A = 60^o$, $\angle  B = 100^o$, $\angle  C = 140^o$. Show that this pentagon can be placed in a circle with a radius of $\frac23 AD$.

The points $D, E, F$ lie respectively on the sides $BC$, $CA$, $AB$ of the triangle ABC such that $F B = BD$, $DC = CE$, and the lines $EF$ and $BC$ are parallel. Tangent to the circumscribed circle of triangle $DEF$ at point $F$ intersects line $AD$ at point $P$. Perpendicular bisector of segment $EF$ intersects the segment $AC$ at $Q$. Show that the lines $P Q$ and $BC$ are parallel.

Given a regular nonagon $A_1A_2A_3A_4A_5A_6A_7A_8A_9$ with side length $1$. Diagonals $A_3A_7$ and $A_4A_8$ intersect at point $P$. Find the length of segment $P A_1$.



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