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AESC MSU Internet MO 2014-22 VII-X (Russia) 40p

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 thatAP: 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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