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Minsk City Internet MO 2014-22 VI-IX (Belarus) 19p

geometry problems from  Minsk City Internet Olympiad in Mathematics (Belarus)  with aops links in the names

collected inside aops here

2014 - 2022


The large triangle is divided by three bold line segments into $4$ triangles and $3$ quadrangles. Sum of perimeters of quadrangles is $25$. The sum of the perimeters of four of triangles is $20$. The perimeter of the original triangle is $19$. Find the sum of the lengths of the bold line segments.
In an acute-angled triangle $ABC$ on the sides $AC$ and $AB$ such points $K$ and $L$ are marked respectively, that the straight line $KL$ is parallel to $BC$ and at the same time $KL = KC$. On the side $BC$ lies point $M$ so that $\angle KMC = \angle BAC$. Prove that $KM = AL$.

Square $100 \times 100$ centimeters divided into $9$ rectangles by two vertical and two horizontal lines. The inner rectangle has size $45 \times 30$ centimeters, and the sides of the remaining rectangles cannot be expressed as integer numbers of centimeters. Find the sum of the areas of the four corner rectangles.

Points $A, B, C$ and $D$ are marked on the straight line. It is known that $AC = 3, BD = 4$. What can the distance between the midpoints of the segments $AB$ and $CD$ be equal to? Please indicate all possibilities.

Given a triangle $ABC$. Find a point $D$ on the side$ AC$ so that the perimeter of the triangle $ABD$ equals the length of the side $BC$. For which ratio, is this possible?

Point $O$, lying inside a regular $2n$-gon, is connected to its vertices. The resulting triangles are colored red and blue one by one. Prove that the sum of the areas of the red triangles is equal to the sum of the areas of the blue ones.

There are $9$ segments, and it is known that the length of each is an integer in centimeters. The two shortest segments are $1$ centimeter, the longest is $32$ centimeters. Prove that among the segments you can choose three such that the length of one of they are less than the sum of the other two.

Points $A, B, C, D$ are given so that segments $AC$ and $BD$ intersect at point $E$. Segment $AE$ is $1$ cm shorter than line $AB$, $AE = DC$, $AD = BE$, $\angle ADC = \angle DEC$. Find the length of segment $EC$.

Given a right-angled triangle $ABC$ with a right angle $C$. On straight line $AB$ on both sides of the hypotenuse noted such points $K$ and $M$ such that $AK = AC$ and $BM = BC$. Find the angle $KCM$.

The altitude $BH$ is drawn in an acute-angled triangle ABC. On segment $BC$ lies point $D$, and on the extension of segment $AB$ beyond point $B$ lies point $E$, such that $AD = DC$ and $AE = EC$. Lines $AD$ and $CE$ intersect line $BH$ at points $D_1$ and $E_1$, respectively. Prove that $2DE = D_1E_1$.

In an acute-angled triangle $ABC$: $\angle A = 30^o$, $BB_1$ and $CC_1$ are altitudes, $B_2$ and $C_2$ are midpoints of sides $AC$ and $AB$, respectively. At what angle do the straight lines $B_1C_2$ and $C_1B_2$ intersect?

In an isosceles triangle $ABC$, the angle $BAC$ is $120^o$. Point $M$ is the midpoint of side $AB$. Point $P$ is symmetric to point $M$ wrt side $BC$. The segments AP and BC meet at point $Q$. Lines $QM$ and $AC$ intersect at point $R$. What is the ratio of the segments $MR: AP$ ?

The convex quadrilateral $ABCD$ satisfies the condition $AD = AB + CD$. Bisectors of angles $BAD$ and $ADC$ intersect at point $P$, which lies inside the quadrilateral $ABCD$. Prove that $BP = CP$.

Outside an equilateral triangle $ABC$, points $S$ and $T$ are marked such that $\angle SAB =\angle TCA = 45^o$ and $\angle  SBA = \angle TAC = 15^o$.
a) Prove that $ST = AB$.
b) Find the angle between straight lines $ST$ and $AB$

Points $K, L$ and $M$ are located on sides $AB, BC$ and $CD$ of square $ABCD$ respectively so that triangle $KLM$ is isosceles right-angled triangle with a right angle at the vertex $L$. From point $D$ draw a straight line parallel to $LM$, which intersects the segment $AL$ at point $P$. In what angle do lines $KP$ and $DL$ intersect?

All sides and diagonals of the hexagon were colored red or green. Prove that some three vertices of this hexagon with connecting them segments (sides or diagonals) form a triangle with sides of one colors. Assume that all the diagonals of the hexagon are entirely inside it.

In a right-angled triangle $ABC$, the $AC$ leg is larger than the leg $BC$ , but less than doubled leg $BC$. Point $M$ is marked on the leg $AC$ so that $AM = BC$, and point N, on the leg $BC$ such that $BN = MC$. Find the angle between lines $AN$ and $BM$.

In triangles $ABC$ and $A_1B_1C_1$:  $\angle A=  \angle A_1$, altitudes drawn from vertices $B$ and $B_1$ are equal , the medians drawn from the vertices $C$ and $C_1$ are also equal. Is it obligatory that these triangles are congruent?

In a convex quadrilateral $ABCD$, the angles at vertices $A, B, C$ are equal. Point $E$ is marked on side $AB$. Prove that if $AD=CD=BE$, then $CE$ is the bisector of angle $BCD$.


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