geometry problems from (final round of) AESC MSU Internet Olympiad (Kolmogorov Boarding School)
with aops links in the names
Интернет-олимпиада СУНЦ МГУ
collected inside aops here
2014 - 2022
In convex quadrilateral $ABCD$, the diagonals meet at point $E$. It is known that $AB = EC$, $AD =BE$ and $\angle AED = \angle BAD$. Prove that $BC> AD$.
There is a regular $20$-gon $A_1... A_{20}$. A square $A_1A_2XY$ and a regular pentagon $A_3A_4CDE$. Prove that $A_3, E$ and $X$ are collinear.
From right-angled triangle $ABC$ with hypotenuse $AB$, you know the positions of points $A$ and $C$ as well as some point $L$ lying on the bisector of angle $B$. Construct the triangle $ABC$ using a compass and ruler.
In trapezoid $ABCD$ with a smaller base $BC$, a straight line parallel to $CD$ is drawn through point $B$. It intersects the diagonal $AC$ at point $E$. Compare the areas of triangles $ABC$ and $DEC$.
Consider the right angle $ABC$ on the plane. For every $XY$ segment for which $X$ lies on the ray $BA$, and $Y$ lies on the ray $BC$, consider a point $Z$ on the segment $XY$ such that $XZ = 2ZY$. Find the geometric locus of such points $Z$.
In convex hexagon $ABCDEF$, all interior angles are equal. It is known that $AB = 2$, $CD = 5$, $DE = 7$, $EF = 1$. Find $BC$ and $AF$.
Point $D$ lies on the median $AM$ of triangle $ABC$. A straight line parallel $AB$ was drawn through point $D$, and a straight line parallel to AM was drawn through point $C$. These lines intersect at a point $X$. Prove that $BD = AX$.
The convex pentagon $ABCDE$ is such that $\angle DAC = \angle DBE$, $\angle ACE = \angle BEC$. Prove that if $AC <BE$, then $AD <BD$.
Point $D$ is chosen in triangle $ABC$ on side $AC$. It is known that $\angle CBD - \angle ABD = 60^o$, $ \angle BDC = 30^o$ and $AB \cdot BC = BD^2$. Find the angles of triangle $ABC$.
Point $D$ is chosen on side $BC$ of triangle $ABC$. In addition $\angle BAC: \angle ADC: \angle BCA = 3: 2: 1$. Prove that $AB + AD = BC$.
The point $X$ is marked on the side $BC$ of an equilateral triangle $ABC$, and point $Y$ on the extension of side $AC$ beyond point $C$ such that $AX = XY$. Prove that $BX = CY$.
On the $BC$ side of isosceles $\vartriangle ABC$ ($AB = BC$), we took points $N$ and $M$ ($N$ closer to $B$ than $M$) such that $NM = AM$, $\angle BAN = \angle MAC$. Find the angle $\angle NAC$.
Point $K$ is selected on side $BC$ of square $ABCD$. Line $AK$ intersects the extension of side $CD$ at point $L$, on ray $AB$ beyond point $B$, point $M$ such that $CM \perp LK$, line $MK$ meets $CD$ at point $N$. Prove that the circumcircle $\vartriangle ANL$ is tangent to the line $AD$.
A square sheet of $ABCD$ paper was dropped onto a square tile so that the corner $A$ of the sheet fell into the corner of the tile, side $BC$ went through the corner of the tile $E$, and the corner $C$ went to the edge of the tile. Find the angle $BAE$.
In the quadrilateral $ABCD$, the three angles are known: $\angle A = 70^o$, $\angle C = 129^o$, $\angle D = 102^o$. Compare $AB$ with $AD + CD$.
Diameter $AB$ of circle $\omega$ with center $O$ divides it into two semicircles. Points $C$ and $D$ are selected on one of the semicircles, and on the other, point $F$. Rays $CD$ and $AB$ intersect at point $E$. It is known that $DE = OF$ and $\angle COA = 78^o$. Find $\angle CFD$.
Let $P$ be the point of intersection of the diagonals of the inscribed quadrilateral $ABCD$. It turned out that the centers of the circumscribed circles of triangles $APB$ and $CPD$ lie on the circumcircle of the quadrilateral $ABCD$. Prove that $AC + BD = 2 (BC + AD)$.
In a convex hexagon $ABCDEF$, all angles of which are obtuse, $\angle A = \angle B$, $\angle C = \angle D$, $\angle E =\angle F$. Prove that the perpendicular bisectors of its sides $AB, CD, EF$ intersect at one point.
Vasya started saving money for a new spinner. All accumulated, he puts the money in a rectangular box. So far he saved up $8$ dimes. At the bottom of the box, they lie, as shown in the figure (adjacent dimes touch). Prove that the centers of three the top coins lie on one straight line.
Point $I$ is the center of the inscribed circle of triangle $ABC$. Line $AI$ intersects the circumscribed circle around the triangle at point $D$. It is known that $\angle BIC = 120^o$, $ID = \frac{7}{\sqrt{3}}$ and $BC- AB = 2$. Find $AI$.
The diagonals of a convex quadrilateral $ABCD$ meet at point $O$, Points $K, L, M, N$ are the midpoints of its sides $AB, BC, CD, DA$, respectively. Prove that $\angle ANO = \angle BLO$ if and only if $\angle BKO = \angle CMO$.
Outside the regular pentagon $ABCDE$, a point $X$ is chosen such that $\angle XAE = 90^o$ and the length of the segment $CX$ equal to sidelength of the pentagon. Find the angle $CXA$.
Point $X$ is selected inside the regular pentagon $ABCDE$ such that $\angle XAE = 90^o$ and the length of the segment $CX$ is equal to the side length of the pentagon. Find the angle $CXD$.
In a triangle $ABC$ with an obtuse angle $B$, the bisectors of the internal and external angle $BAC$ are drawn. Point $O$ is the center of the circumscribed circle of triangle $ABC$. Distances from point $O$ to both angle bisectors, as well as up to line $BC$, are equal. Find $\angle ABC$.
A common external tangent $\ell$ is drawn to disjoint circles $\omega_1$ and $\omega_2$. Through the points of tangency $\ell$ with the circles, a circle $\omega_3$ is drawn, intersecting $\omega_1$ and $\omega_2$ for the second time at points $A$ and $B$, respectively. Through points $A$ and $B$ tangents to $\omega_1$ and $\omega_2$, respectively, are drawn, which intersect at point $X$, and points $A, B, X$ lie in the same half-plane wrt $\ell$, and point $X$ lies outside $\omega_3$, . Prove that $AX = BX$.
On the plane there are four triangles with vertices at the vertices of cells with a common base $M N$.
Find the sum of the angles of these triangles at the vertices $A, B, C, D$.
You are given a parallelogram $ABCD$ with an obtuse angle $A$. The line passing through point $C$ perpendicular to $BC$, intersects the line passing through the point $D$ parallel to $AC$ at the point $E$. Prove that $EC$ is the bisector of angle $BED$.
Given a triangle $ABC$. Three parallel lines $AX, BY$ and $CZ$ meet lines $BC, AC$ and $AB$ in points $X, Y$ and $Z$, respectively. Points $P, Q, R$ were chosen on the segments $AX, BY, CZ$ so that$$AP: P X = BQ: QY = CR: RZ = 1: k.$$For what values of $k$ do the points $P, Q, R$ lie on one straight line?
Point $H$ is the intersection point of the altitudes of the triangle $ABC$, points $A', B'$ and $C'$ are the midpoints of the sides $BC, CA$ and $AB$ respectively. The circle $\omega$ centered at the point $H$ meets the line $B'C'$ at the points $X_1$ and $X_2$, the line $C'A' $ at points $Y_1$ and $Y_2$, and the line $A'B'$ at points $Z_1$ and $Z_2$. Prove that $AX_1 = AX_2 = BY_1 = BY_2 = CZ_1 = CZ_2$.
Triangle $ABC$ is inscribed in a circle with center $O$. It's altitudes are $CC_1$ and $BB_1$ meet at point $H$. It is known that $AC_1=CC_1$. Prove that $OC_1HB_1$ is parallelogram.
Given a right-angled triangle $ABC$ with right angle $C$, $AA_1$ and $BB_1$ , it's angle bisectors. Points $A_2$ and $B_2$ are the feet of the perpendiculars drawn from $A_1$ and $B_1$ on $AB$. Prove that the center of the incircle of triangle $ABC$ coincides with the center of the circumcircle of triangle $A_2B_2C$.
On the extension of the altitude $BH$ of an obtuse ($\angle B> \frac{\pi}{2}$) of triangle $ABC$ mark
the point $K$ such that $\angle BKA = \angle BCA$. Let the altitude $KF$ of the triangle $BKA$
intersect the line $CA$ at point $R$. Prove that triangles $KHA$ and $BHR$ are similar.
Let $ABCD$ be an isosceles trapezoid (not a square) with a large base $AD$ and a smaller $BC$ such that the diagonal $BD$ is the bisector of the angle $ADC$, and triangle $ADB$ is isosceles. Find the ratio of the radii of the circle inscribed in triangle $ABD$, and the circle inscribed in triangle $BCD$.
An equilateral triangle with sides equal to $n$ is divided into regular triangles with sides $1$. Semyon drew a broken line, the links of which run along the segments of the partition. This broken line passed through all the vertices of the partition exactly once. What is the smallest number of links could such a broken line have?
For example, the polyline in the picture satisfies the conditions and has $10$ links.
Given triangle $ABC$. Point $B_1$ is the foot of the perpendicular from point $B$ on the bisector of angle $BMA$. Point $C_1$ is the foot of the perpendicular from point $C$ to the bisector of angle $CMA$. Lines $AM$ and $B_1C_1$ intersect at point $X$. Find the ratio $B_1X : XC_1$.
Given an obtuse triangle $ABC$ with an obtuse angle $B$. It is known that $AC > BC > AB$. Circle $\omega$ is circumscribed around $ABC$, having $BD$ and $CE$ as its diameters. Circle with center at point $A$ and radius $AD$ intersects ray $AB$ at a point $L$. Circle with center at point $A$ and radius $AE$ intersects ray $AC$ at point $K$. Lines $DL$ and $EK$ intersect at a point $X$. Prove that $X$ lies on the circle $\omega$.
The circles $\omega_1$ and $\omega_2$ intersect at points $A$ and $D$. The line $\ell_1$ intersects the circle $\omega_1$ at points $A$ and $B$ and touches the circle $\omega_2$. The line $\ell_2$ intersects the circle $\omega_2$ at points $A$ and $C$ and touches the circle $\omega_1$. Prove that $BC$ is a common tangent of circles if and only if the radii of $\omega_1$ and $\omega_2$ are the same and equal to the length of segment $AD$.
Prove that the area of any convex polygon is strictly less than the square of its perimeter divided by $8$.
The circles $\omega_1$ and $\omega_2$ intersect at points $A$ and $D$. The line $\ell_1$ intersects the circle $\omega_1$ at points $A$ and $B$ and touches the circle $\omega_2$. The line $\ell_2$ intersects the circle $\omega_2$ at points $A$ and $C$ and touches the circle $\omega_1$. Prove that point $B$, point $C$ and point $D$ lie on the same line if and only if $\angle BAC = 90^o$.
Roma and Margarita are playing a game. At the beginning of the game, they must sequentially draw on the board obtuse-angled triangle, and Margarita draws a triangle that is not similar to Roma's triangle . On each turn, the player chooses any of the two triangles drawn on the board, writes in it a circle and constructs a new triangle with vertices at the touchpoints of this circle of the original triangle. After that, the original triangle and the circle are erased, and two triangles remain on the board. Players play in turns. The first person to draw a triangle with all angles different from the correct angles, by less than $1$ degree. Roma goes first. Which of them is guaranteed to win, no matter how played by an opponent, and why?
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