Kyiv City MO Juniors Round2 2010+ geometry

2010.89.3

In the acute-angled triangle $ABC$ the angle$ \angle B = 30^o$, point $H$ is the intersection point of its altitudes. Denote by $O_1, O_2$ the centers of circles inscribed in triangles $ABH ,CBH$ respectively. Find the degree of the angle between the lines $AO_2$ and $CO_1$.

2011.8.3

On the sides $AD , BC$ of the square $ABCD$ the points $M, N$ are selected $N$, respectively, such that $AM = BN$. Point $X$ is the foot of the perpendicular from point $D$ on the line $AN$. Prove that the angle $MXC$ is right. (Mirchev Borislav)

2011.9.4

Let two circles be externally tangent at point $C$, with parallel diameters $A_1A_2, B_1B_2$ (i.e. the quadrilateral $A_1B_1B_2A_2$ is a trapezoid with bases $A_1A_2$ and $B_1B_2$ or parallelogram). Circle with the center on the common internal tangent to these two circles, passes through the intersection point of lines $A_1B_2$ and $A_2B_1$ as well intersects those lines at points $M, N$. Prove that the line $MN$ is perpendicular to the parallel diameters $A_1A_2, B_1B_2$. (Yuri Biletsky)

2012.7.3

In the triangle $ABC $ the median $BD$ is drawn, which is divided into three equal parts by the points $E $ and $F$ ($BE = EF = FD$). It is known that $AD = AF$ and $AB = 1$. Find the length of the segment $CE$.

2012.8.5

In the triangle $ABC$ on the sides $AB$ and $AC$ outward constructed equilateral triangles $ABD$ and $ACE$. The segments $CD$ and $BE$ intersect at point $F$. It turns out that point $A$ is the center of the circle inscribed in triangle $ DEF$. Find the angle $BAC$. (Rozhkova Maria)

2012.9.4

In an acute-angled triangle $ABC$, the point $O$ is the center of the circumcircle, and the point $H$ is the orthocenter. It is known that the lines $OH$ and $BC$ are parallel, and $BC = 4OH $. Find the value of the smallest angle of triangle $ ABC $. (Black Maxim)

2013.7.3

In the square $ABCD$ on the sides $AD$ and $DC$, the points $M$ and $N$ are selected so that $\angle BMA = \angle NMD = 60 { } ^ \circ $. Find the value of the angle $MBN$.

2013.8.3

Inside $\angle BAC = 45 {} ^ \circ$ the point $P$ is selected that the conditions $\angle APB = \angle APC = 45 {} ^ \circ $ are fulfilled. Let the points $M$ and $N$ be the projections of the point $P$ on the lines $AB$ and $AC$, respectively. Prove that $BC\parallel MN $. (Serdyuk Nazar)

2013.9.5

Given a triangle $ ABC $, $ AD $ is its angle bisector. Let $ E, F $ be the centers of the circles inscribed in the triangles $ ADC $ and $ ADB $, respectively. Denote by $ \omega $, the circle circumscribed around the triangle $ DEF $, and by $ Q $, the intersection point of $ BE $ and $ CF $, and $ H, J, K, M $ , respectively the second intersection point of the lines $ CE, CF, BE, BF $ with circle $ \omega $. Let $\omega_1, \omega_2 $ the circles be circumscribed around the triangles $ HQJ $ and $ KQM $ Prove that the intersection point of the circles $\omega_1, \omega_2 $ different from $ Q $ lies on the line $ AD $. (Kivva Bogdan)

2014.7.4

The median $BM$ is drawn in the triangle $ABC$. It is known that $\angle ABM = 40 {} ^ \circ$ and $\angle CBM = 70 {} ^ \circ $ Find the ratio $AB: BM$.

2014.89.3

Given a triangle $ABC$, on the side $BC$ which marked the point $E$ such that $BE \ge CE$. Construct on the sides $AB$ and $AC$ the points $D$ and $F$, respectively, such that $\angle DEF = 90 {} ^ \circ$ and the segment $BF$ is bisected by the segment $DE $. (Black Maxim)

2015.789.4

In the acute triangle $ABC$ the side $BC> AB$, and the angle bisector $BL = AB$. On the segment $BL$ there is a point $M$, for which $\angle AML = \angle BCA$. Prove that $AM = LC$.

2015.7.41

The equal segments $AB$ and $CD$ intersect at the point $O$ and divide it by the relation $AO: OB = CO: OD = 1: 2 $. The lines $AD$ and $BC$ intersect at the point $M$. Prove that $DM = MB$.

2015.8.41

On the sides $AB, \, \, BC, \, \, CA$ of the triangle $ABC$ the points ${{C} _ {1}}, \, \, {{A} _ { 1}},\, \, {{B} _ {1}}$ are selected respectively, that are different from the vertices. It turned out that $\Delta {{A} _ {1}} {{B} _ {1}} {{C} _ {1}}$ is equilateral, $\angle B{{C}_{1}}{{A}_{1}}=\angle {{C}_{1}}{{B}_{1}}A$ and $\angle B{{A}_{1}}{{C}_{1}}=\angle {{A}_{1}}{{B}_{1}}C$ . Is $ \Delta ABC$ equilateral?

2015.9.4

Circles ${{w} _ {1}}$ and ${{w} _ {2}}$ with centers ${{O} _ {1}}$ and ${{O} _ {2}}$ intersect at points $A$ and $B$, respectively. The line ${{O} _ {1}} {{O} _ {2}}$ intersects ${{w} _ {1}}$ at the point $Q$, which does not lie inside the circle ${{w} _ {2}}$, and ${{w} _ {2}}$ at the point $X$ lying inside the circle ${{w} _ {1} }$. Around the triangle ${{O} _ {1}} AX$ circumscribe a circle ${{w} _ {3}}$ intersecting the circle ${{w} _ {1}}$ for the second time in point $T$. The line $QT$ intersects the circle ${{w} _ {3}}$ at the point $K$, and the line $QB$ intersects ${{w} _ {2}}$ the second time at the point $H$. Prove that a) points $T, \, \, X, \, \, B$ lie on one line; b) points $K, \, \, X, \, \, H$ lie on one line. (Vadym Mitrofanov)

2016.7.3

In an acute triangle $ABC$, the bisector $AL$, the altitude $BH$, and the perpendicular bisector of the side $AB$ intersect at one point. Find the value of the angle $BAC$.

2016.8.1

In a right triangle, the point $O$ is the center of the circumcircle. Another circle of smaller radius centered at the point $O$ touches the larger leg and the altitude drawn from the top of the right angle. Find the acute angles of a right triangle and the ratio of the radii of the circumscribed and smaller circles.

2016.9.2

The bisector of the angle $BAC$of the acute triangle $ABC$ ( $AC \ne AB$) intersects its circumscribed circle for the second time at the point $W$. Let $O$ be the center of the circumscribed circle $\Delta ABC$. The line $AW$ intersects for the second time the circumcribed circles of triangles $OWB$ and $OWC$ at the points $N$ and $M$, respectively. Prove that $BN + MC = AW$. (Mitrofanov V., Hilko D.)

2017.7.4

On the sides $AD$ and $BC$ of a rectangle $ABCD$ select points $M, N$ and $P, Q$ respectively such that $AM = MN = ND = BP = PQ = QC$. On segment $QC$ selected point $X$, different from the ends of the segment. Prove that the perimeter of $\vartriangle ANX$ is more than the perimeter of $\vartriangle MDX$.

2017.7.41

Let $AC$ be the largest side of the triangle $ABC$. The point M is selected on the ray $AC$ ray, and point $N$ on ray $CA$ such that $CN = CB$ and$ AM = AB$ . a) Prove that $\vartriangle ABC$ is isosceles if we know that $BM = BN$. b) Will the statement remain true if $AC$ is not necessarily the largest side of triangle $ABC$?

2017.8.2

Triangle $ABC$ is right-angled and isosceles with a right angle at the vertex $C$. On rays $CB$ on vertex $B$ is selected point F, on rays $BA$ on vertex $A$ is selected point G so that $AG = BF.$ The ray $GD$ is drawn so that it intersects with ray $AC$ at point $D$ with $\angle FGD = 45^o$. Find $\angle FDG$. (Bogdan Rublev)

2017.9.1

Find the angles of the triangle $ABC$, if we know that its center $O$ of the circumscribed circle and the center $I_A$ of the exscribed circle (tangent to $BC$) are symmetric wrt $BC$. (Bogdan Rublev)

2018.8.3

In the triangle $ABC$ it is known that $\angle ACB> 90 {} ^ \circ$, $\angle CBA> 45 {} ^ \circ$. On the sides $AC$ and $AB$, respectively, there are points $P$ and $T$ such that $ABC$ and $PT = BC$. The points ${{P} _ {1}}$ and ${{T} _ {1}}$ on the sides $AC$ and $AB$ are such that $AP = C {{P} _ {1}}$ and $AT = B {{T} _ {1}}$. Prove that $\angle CBA- \angle {{P} _ {1}} {{T} _ {1}} A = 45 {} ^ \circ$. (Anton Trygub)

2018.8.31

On the sides $AB$, $BC$ and $CA$ of the isosceles triangle $ABC$ with the vertex at the point $B$ marked the points $M$, $D$ and $K$ respectively so that $AM = 2DC$ and $\angle AMD = \angle KDC$. Prove that $MD = KD$.

2018.9.1

Cut a right triangle with an angle of $30^o$ into three isosceles non-acute triangles, among which there are no congruent ones. (Maria Rozhkova)

2019.7.3

In the quadrilateral $ABCD$ it is known that $\angle ABD= \angle DBC$ and $AD= CD$. Let $DH$ be the altitude of $\vartriangle ABD$. Prove that $| BC - BH | = HA$. (Hilko Danilo)

2019.7.31

The teacher drew a coordinate plane on the board and marked some points on this plane. Unfortunately, Vasya's second-grader, who was on duty, erased almost the entire drawing, except for two points $A (1, 2)$ and $B (3,1)$. Will the excellent Andriyko be able to follow these two points to construct the beginning of the coordinate system point $O (0, 0)$? Point A on the board located above and to the left of point $B$.

2019.8.4

In the triangle $ABC$ it is known that$\angle A = 75^o, \angle C = 45^o$. On the ray $BC$ beyond the point $C$ the point $T$ is taken so that $BC = CT$. Let $M$ be the midpoint of the segment $AT$. Find the measure of the $\angle BMC$. (Anton Trygub)

2019.8.41

Through the vertices $A, B$ of the parallelogram $ABCD$ passes a circle that intersects for the second time diagonals $BD$ and $AC$ at points $X$ and $Y$, respectively. The circumsccribed circle of $\vartriangle ADX$ intersects diagonal $AC$ for the second time at the point $Z$. Prove that $AY = CZ$.

2019.9.3

The equilateral triangle $ABC$ is inscribed in the circle $w$. Points $F$ and $E$ on the sides $AB$ and $AC$, respectively, are chosen such that $\angle ABE+ \angle ACF = 60^o$. The circumscribed circle of $\vartriangle AFE$ intersects the circle $w$ at the point $D$ for the second time. The rays $DE$ and $DF$ intersect the line $BC$ at the points $X$ and $Y$, respectively. Prove that the center of the inscribed circle of $\vartriangle DXY$ does not depend on the choice of points $F$ and $E$. (Hilko Danilo)

2019.9.31

A circle $k$ of radius $r$ is inscribed in $\vartriangle ABC$, tangent to the circle $k$, which are parallel respectively to the sides $AB, BC$ and $CA$ intersect the other sides of $\vartriangle ABC$ at points $M, N; P, Q$ and $L, T$ ($P, T \in AB$, $L, N \in BC$ and $M, Q\in AC$). Denote by $r_1,r_2,r_3$ the radii of inscribed circles in triangles $MNC, PQA$ and $LTB$. Prove that $r_1+r_2+r_3=r$.

2020.8.2

Given a convex quadrilateral $ABCD$, in which $\angle CBD = 90^o$, $\angle BCD =\angle CAD$ and $AD= 2BC$. Prove that $CA =CD$. (Anton Trygub)

2020.9.2

In the acute-angled triangle $ABC$ is drawn the altitude $CH$. A ray beginning at point $C$ that lies inside the $\angle BCA$ and intersects for second time the circles circumscribed circles of $\vartriangle BCH$ and $\vartriangle ABC$ at points $X$ and $Y$ respectively. It turned out that $2CX = CY$. Prove that the line $HX$ bisects the segment $AC$. (Hilko Danilo)

2021.7.4

The sides of the triangle $ABC$ are extended in both directions and on these extensions $6$ equal segments $AA_1 , AA_2, BB_1,BB_2, CC_1, CC_2$ are drawn (fig.). It turned out that all $6$ points $A_1,A_2,B_1,B_2,C_1, C_2$ lie on the same circle, is $\vartriangle ABC$ necessarily equilateral? (Bogdan Rublev)

2021.7.41

Point $C$ lies inside the right angle $AOB$. Prove that the perimeter of triangle $ABC$ is greater than $2 OC$.

2021.8.2

In a triangle $ABC$, $\angle B=90^o$ and $\angle A=60^o$, $I$ is the point of intersection of its angle bisectors. A line passing through the point $I$ parallel to the line $AC$, intersects the sides $AB$ and $BC$ at the points $P$ and $T$ respectively. Prove that $3PI+IT=AC$ . (Anton Trygub)

2021.9.2

In an acute triangle $AB$ the heights $ BE$ and $CF$ intersect at the orthocenter $H$, and $M$ is the midpoint of $BC$. The line $EF$ intersects the lines $MH$ and $BC$ at the points $P$ and $T$ , respectively. $AP$ intersects the cirumcscribed circle of $\vartriangle ABC$ for second time at the point $Q$ . Prove that $\angle AQT= 90^o$. (Fedir Yudin)

2022.7.3

In triangle $ABC$ the median $BM$ is equal to half of the side $BC$. Show that $\angle ABM = \angle BCA + \angle BAC$. (Proposed by Anton Trygub)

2022.8.4

Points $D, E, F$ are selected on sides $BC, CA, AB$ correspondingly of triangle $ABC$ with $\angle C = 90^\circ$ such that $\angle DAB = \angle CBE$ and $\angle BEC = \angle AEF$. Show that $DB = DF$. (Proposed by Mykhailo Shtandenko)

2022.9.4

Let $\omega$ denote the circumscribed circle of triangle $ABC$, $I$ be its incenter, and $K$ be any point on arc $AC$ of $\omega$ not containing $B$. Point $P$ is symmetric to $I$ with respect to point $K$. Point $T$ on arc $AC$ of $\omega$ containing point $B$ is such that $\angle KCT = \angle PCI$. Show that the bisectors of angles $AKC$ and $ATC$ meet on line $CI$. (Proposed by Anton Trygub)