Found problems: 83
Ukrainian TYM Qualifying - geometry, 2020.13
In the triangle $ABC$ on the side $BC$, the points$ D$ and $E$ are chosen so that the angle $BAD$ is equal to the angle $EAC$. Let $I$ and $J$ be the centers of the inscribed circles of triangles $ABD$ and $AEC$ respectively, $F$ be the point of intersection of $BI$ and $EJ$, $G$ be the point of intersection of $DI$ and $CJ$. Prove that the points $I, J, F, G$ lie on one circle, the center of which belongs to the line $I_bI_c$, where $I_b$ and $I_c$ are the centers of the exscribed circles of the triangle $ABC$, which touch respectively sides $AC$ and $AB$.
Ukrainian TYM Qualifying - geometry, 2017.4
Specify at least one right triangle $ABC$ with integer sides, inside which you can specify a point $M$ such that the lengths of the segments $MA, MB, MC$ are integers. Are there many such triangles, none of which are are similar?
Ukrainian TYM Qualifying - geometry, IV.11
In the tetrahedron $ABCD$, the point $E$ is the projection of the point $D$ on the plane $(ABC)$. Prove that the following statements are equivalent:
a) $C = E$ or $CE \parallel AB$
b) For each point M belonging to the segment $CD$, the following equation is satisfied
$$S^2_{\vartriangle ABM}= \frac{CM^2}{CD^2}\cdot S^2_{\vartriangle ABD}+\left(1- \frac{CM^2}{CD^2} \right)S^2_{\vartriangle ABC}$$
where $S_{\vartriangle XYZ}$ means the area of triangle $XYZ$.
Ukrainian TYM Qualifying - geometry, VIII.2
Investigate the properties of the tetrahedron $ABCD$ for which there is equality
$$\frac{AD}{ \sin \alpha}=\frac{BD}{\sin \beta}=\frac{CD}{ \sin \gamma}$$
where $\alpha, \beta, \gamma$ are the values of the dihedral angles at the edges $AD, BD$ and $CD$, respectively.
Ukrainian TYM Qualifying - geometry, 2020.12
On the side $CD$ of the square $ABCD$, the point $F$ is chosen and the equal squares $DGFE$ and $AKEH$ are constructed ($E$ and $H$ lie inside the square). Let $M$ be the midpoint of $DF$, $J$ is the incenter of the triangle $CFH$. Prove that:
a) the points $D, K, H, J, F$ lie on the same circle;
b) the circles inscribed in triangles $CFH$ and $GMF$ have the same radii.
Ukrainian TYM Qualifying - geometry, 2015.20
What is the smallest value of the ratio of the lengths of the largest side of the triangle to the radius of its inscribed circle?
Ukrainian TYM Qualifying - geometry, 2015.24
The inscribed circle $\omega$ of the triangle $ABC$ touches its sides $BC, CA$, and $AB$ at the points $D, E$, and $F$, respectively. Let the points $X, Y$, and $Z$ of the circle $\omega$ be diametrically opposite to the points $D, E$, and $F$, respectively. Line $AX, BY$ and $CZ$ intersect the sides $BC, CA$ and $AB$ at the points $D', E'$ and $F'$, respectively. On the segments $AD', BE'$ and $CF'$ noted the points $X', Y'$ and $Z'$, respectively, so that $D'X'= AX$, $E'Y' = BY$, $F'Z' = CZ$. Prove that the points $X', Y'$ and $Z'$ coincide.
Ukrainian TYM Qualifying - geometry, II.1
Inside a right cylinder with a radius of the base $R$ are placed $k$ ($k\ge 3$) of equal balls, each of which touches the side surface and the lower base of the cylinder and, in addition, exactly two other balls. After that, another equal ball is placed inside the cylinder so that it touches the upper base of the cylinder and all other balls. Find the volume $V (R, k)$ of the cylinder.
Ukrainian TYM Qualifying - geometry, X.12
Inside the convex polygon $A_1A_2...A_n$ , there is a point $M$ such that $\sum_{k=1}^n \overrightarrow {A_kM} = \overrightarrow{0}$. Prove that $nP\ge 4d$, where $P$ is the perimeter of the polygon, and $d=\sum_{k=1}^n |\overrightarrow {A_kM}|$ . Investigate the question of the achievement of equality in this inequality.
Ukrainian TYM Qualifying - geometry, 2019.10
At the altitude $AH_1$ of an acute non-isosceles triangle $ABC$ chose a point $X$ , from which draw the perpendiculars $XN$ and $XM$ on the sides $AB$ and $AC$ respectively. It turned out that $H_1A$ is the angle bisector $MH_1N$. Prove that $X$ is the point of intersection of the altitudes of the triangle $ABC$.
Ukrainian TYM Qualifying - geometry, 2017.1
In an isosceles trapezoid $ABCD$ with bases $AD$ and $BC$, diagonals intersect at point $P$, and lines $AB$ and $CD$ intersect at point $Q$. $O_1$ and $O_2$ are the centers of the circles circumscribed around the triangles $ABP$ and $CDP$, $r$ is the radius of these circles. Construct the trapezoid ABCD given the segments $O_1O_2$, $PQ$ and radius $r$.
Ukrainian TYM Qualifying - geometry, 2012.11
Let $E$ be an arbitrary point on the side $BC$ of the square $ABCD$. Prove that the inscribed circles of triangles $ABE$, $CDE$, $ADE$ have a common tangent.
Ukrainian TYM Qualifying - geometry, II.2
Is it true that when all the faces of a tetrahedron have the same area, they are congruent triangles?
Ukrainian TYM Qualifying - geometry, 2013.15
Inside the acute-angled triangle $ABC$, mark the point $O$ so that $\angle AOB=90^o$, a point $M$ on the side $BC$ such that $\angle COM=90^o$, and a point $N$ on the segment $BO$ such that $\angle OMN = 90^o$. Let $P$ be the point of intersection of the lines $AM$ and $CN$, and let $Q$ be a point on the side $AB$ that such $\angle POQ = 90^o$. Prove that the lines $AN, CO$ and $MQ$ intersect at one point.
Ukrainian TYM Qualifying - geometry, XI.4
Chords $AB$ and $CD$, which do not intersect, are drawn in a circle. On the chord $AB$ or on its extension is taken the point $E$. Using a compass and construct the point $F$ on the arc $AB$ , such that $\frac{PE}{EQ} = \frac{m}{n}$, where $m,n$ are given natural numbers, $P$ is the point of intersection of the chord $AB$ with the chord $FC$, $Q$ is the point of intersection of the chord $AB$ with the chord $FD$. Consider cases where $E\in PQ$ and $E \notin PQ$.
Ukrainian TYM Qualifying - geometry, 2012.2
The triangle $ABC$ is drawn on the board such that $AB + AC = 2BC$. The bisectors $AL_1, BL_2, CL_3$ were drawn in this triangle, after which everything except the points $L_1, L_2, L_3$ was erased. Use a compass and a ruler to reconstruct triangle $ABC$.
Ukrainian TYM Qualifying - geometry, IV.8
Prove that in an arbitrary convex hexagon there is a diagonal that cuts off from it a triangle whose area does not exceed $\frac16$ of the area of the hexagon. What are the properties of a convex hexagon, each diagonal of which is cut off from it is a triangle whose area is not less than $\frac16$ the area of the hexagon?
Ukrainian TYM Qualifying - geometry, 2016.14
Using only a compass and a ruler, reconstruct triangle $ABC$ given the following three points: point $M$ the intersection of its medians, point $I$ is the center of its inscribed circle and the point $Q_a$ is touch point of the inscribed circle to side $BC$.
Ukrainian TYM Qualifying - geometry, 2019.8
Hannusya, Petrus and Mykolka drew independently one isosceles triangle $ABC$, all angles of which are measured as a integer number of degrees. It turned out that the bases $AC$ of these triangles are equals and for each of them on the ray $BC$ there is a point $E$ such that $BE=AC$, and the angle $AEC$ is also measured by an integer number of degrees. Is it in necessary that:
a) all three drawn triangles are equal to each other?
b) among them there are at least two equal triangles?
Ukrainian TYM Qualifying - geometry, XII.17
Given a triangle $ABC$, inside which the point $M$ is marked. On the sides $BC,CA$ and $AB$ the following points $A_1,B_1$ and $C_1$ are chosen, respectively, that $MA_1 \parallel CA$, $MB_1 \parallel AB$, $MC_1 \parallel BC$. Let S be the area of triangle $ABC, Q_M$ be the area of the triangle $A_1 B_1 C_1$.
a) Prove that if the triangle $ABC$ is acute, and M is the point of intersection of its altitudes , then $3Q_M \le S$. Is there such a number $k> 0$ that for any acute-angled triangle $ABC$ and the point $M$ of intersection of its altitudes, such thatthe inequality $Q_M> k S$ holds?
b) For cases where the point $M$ is the point of intersection of the medians, the center of the inscribed circle, the center of the circumcircle, find the largest $k_1> 0$ and the smallest $k_2> 0$ such that for an arbitrary triangle $ABC$, holds the inequality $k_1S \le Q_M\le k_2S$ (for the center of the circumscribed circle, only acute-angled triangles $ABC$ are considered).
Ukrainian TYM Qualifying - geometry, 2017.3
The altitude $AH, BT$, and $CR$ are drawn in the non isosceles triangle $ABC$. On the side $BC$ mark the point $P$; points $X$ and $Y$ are projections of $P$ on $AB$ and $AC$. Two common external tangents to the circumscribed circles of triangles $XBH$ and $HCY$ intersect at point $Q$. The lines $RT$ and $BC$ intersect at point $K$.
a). Prove that the point $Q$ lies on a fixed line independent of choice$ P$.
b). Prove that $KQ = QH$.
Ukrainian TYM Qualifying - geometry, I.13
A candle and a man are placed in a dihedral mirror angle. How many reflections can the man see ?
Ukrainian TYM Qualifying - geometry, 2015.21
Let $CH$ be the altitude of the triangle $ABC$ drawn on the board, in which $\angle C = 90^o$, $CA \ne CB$. The mathematics teacher drew the perpendicular bisectors of segments$ CA$ and $CB$, which cut the line CH at points $K$ and $M$, respectively, and then erased the drawing, leaving only the points $C, K$ and $M$ on the board. Restore triangle $ABC$, using only a compass and a ruler.
Ukrainian TYM Qualifying - geometry, 2014.22
In $\vartriangle ABC$ on the sides $BC, CA, AB$ mark feet of altitudes $H_1, H_2, H_3$ and the midpoint of sides $M_1, M_3, M_3$. Let $H$ be orthocenter $\vartriangle ABC$. Suppose that $X_2, X_3$ are points symmetric to $H_1$ wrt $BH_2$ and $CH_3$. Lines $M_3X_2$ and $M_2X_3$ intersect at point $X$. Similarly, $Y_3,Y_1$ are points symmetric to $H_2$ wrt $C_3H$ and $AH_1$.Lines $M_1Y_3$ and $M_3Y_1$ intersect at point $Y.$ Finally, $Z_1,Z_2$ are points symmetric to $H_3$ wrt $AH_1$ and $BH_2$. Lines $M_1Z_2$ and $M_2Z_1$ intersect at the point $Z$ Prove that $H$ is the incenter $\vartriangle XYZ$ .
Ukrainian TYM Qualifying - geometry, 2014.8
In the triangle $ABC$ on the ray $BA$ mark the point $K$ so that $\angle BCA= \angle KCA$ , and on the median $BM$ mark the point $T$ so that $\angle CTK=90^o$ . Prove that $\angle MTC=\angle MCB$ .