geometry problems from Innopolis University Open Olympiad in Mathematics (Russia) with aops links in the names
Олимпиады Университета Иннополис Open по математике
collected inside aops here
it started in 2015
2015 - 2022
In quadrilateral ABCD, BD = CD and \angle CBD = \angle CAD, point E is the base of the perpendicular from D to line AB. Prove that the length of AE equals to half of the sum of the lengths AB and AC.
In a trapezoid, the angles with a larger base are 25^o and 75^o degrees, average length of the bases is 27 cm, and the segment connecting the midpoints of the bases is 9 cm. Find the length of a segment passing through the point of intersection of the diagonals of the trapezoid parallel to its bases with ends on the sides.
In parallelogram ABCD, the lengths of the sides AB and BC are 20 cm and 30 cm, respectively. \cos \angle DAB =\frac35 . Inside ABCD, a point M is chosen such that MC = 6\sqrt{10} cm, and the distance from M to line AD is 10 cm. Find the length of AM.
The chord CD of the circle centered at the point O is perpendicular to its diameter AB, the chord AE intersects the radius OC at the point M, and the chord DE intersects the chord BC at the point N. Prove that the line MN is parallel to the line AB.
The sides of a convex quadrilateral in some order equal to 6, 7, 8, 9. It is known that this quadrilateral can be inscribed in a circle and can be circumscribed around a circle. Find the area of the quadrilateral.
The two sides of the quadrilateral ABCD are parallel. Let M and N be the midpoints of sides BC and CD, respectively, and P be the intersection point of AN and DM. Prove that if AP = 4PN then ABCD is a parallelogram.
Point M is the midpoint of side AB of the convex quadrilateral ABCD. It turned out that \angle BCA = \angle ACD and \angle CDA = \angle BAC. Prove that \angle BCM = \angle ABD.
The tangent at point A to the circumscribed circle of triangle ABC intersects line BC at point K. On the perpendicular to the segment BC at point B, point L is taken such that AL = BL. On the perpendicular to the segment BC at point C, point M is taken such that AM = CM. Prove that K, L and M are collinear.
The medians AM_a,BM_b and CM_c of the triangle ABC intersect at the point M. The circle \omega_a passes through the midpoint of the segment AM and touches the side BC at the point M_a. Similarly, the circle \omega_b passes through the midpoint of the segment BM and touches the side CA at the point M_b. Let X and Y be the points of intersection of the circles \omega_a and \omega_b. Prove that points X, Y and M_c are collinear.
At the altitude AH of an acute-angled triangle ABC, point K is marked, and point L is marked on the side BC. It turned out that \frac{AK}{HK} = \frac{BL}{CL}. Point P is the base of the perpendicular drawn from point B on line AL. Prove that line KL is tangent to the circumcircle of triangle CLP.
Let h be the length of the greatest altitude in a triangle, R the radius of the circumscribed circle, m_a, m_b and m_c the lengths of the medians triangle. Prove the inequality m_a + m_b + m_c \le 3R + h.
Points A, B and C are marked on the straight line \ell. Point B lies between points A and C and AB <BC. Two circles \omega_1 and \omega_2, whose radii are greater than AB, but less than BC, lie on opposite sides of the straight line \ell and touch it at point B. Let K be the point of intersection of the tangent to \omega_1 drawn from point A, and tangent to \omega_2, drawn from point C, L be the intersection point of tangent to \omega_2, drawn from point A, and tangent to \omega_1 drawn from point C. Prove that the quadrilateral AKCL is tangential.
Three segments L_1, L_2 and L_3 are given. Describe how to construct such a segment L using a compass and a ruler, such that the ratio of the lengths of the segments L and L_ 1 is equal to the ratio of the volumes of cubes with sides L_2 and L_3 respectively.
Three segments L_1, L_2 and L_3 are given, the lengths of which are prime numbers. Describe how to construct such a segment L using a compass and a ruler, such that its length is equal to the volume of a rectangular parallelepiped with sides L_1, L_2 and L_3.
A circle is, as you know, a set of points on a plane that are distant from a given point (center) by a fixed distance (radius of the circle), and the number \pi is the ratio of the length circumference to the length of its diameter. This definition assumes by default that it is about Euclidean distance / length, which is calculated on a two-dimensional coordinate plane (XOY) for a segment with ends at points (x_1, y_1) and (x_2, y_2) by the formula \sqrt{(x_1-x_2)^2 + (y_1- y_2)^2}. However, the Euclidean definition of length is not the only one possible. For example, Manhattan the length of the segment with ends at points (x_1, y_1) and (x_2, y_2) is calculated by the formula | x_1-x_2| + | y_1 - y_2|.
(The name "Manhattan Distance" is related to the street layout of Manhattan, which is a rectangular grid of streets: “To the north from the south there are avenues, to the west from east - streets "V. V. Mayakovsky," Broadway ".)
What is the ratio of the Manhattan length of the Euclidean circle to the Manhattan length of it diameter?
There is a triangle ABC.
a) Prove that there is a unique set of such three circles \omega_A, \omega_B and \omega_C, which lie inside the triangle, touch each other in pairs, and also each of them touches the sides of the corresponding angle: \omega_A touches the sides of AB and AC, \omega_B touches the sides BA and BC, \omega_C touches the sides CA and CB.
b) Let’s denote the tangency point of the circles \omega_A and \omega_B as T_{AC}. Points T_{AB} and T_{BC} are defined similarly.
A designer wants to make a stained glass chandelier made of colored glass, in which the sides of the triangle ??? are a strong light (its weight is zero) contour, into which a massive flat disc weighing 1 kg is inscribed, and balancing weights are added at the vertices A,B and C of a triangle so that the suspension point of the chandelier is at the intersection of the segments AT_{BC}, BT_{AC} and CT_{AB} (other parts of the chandelier are of zero weight). Prove that such a chandelier project is feasible and determine the balancing weights at the vertices (that is, such that the chandelier hangs horizontally only at the suspension point) if the radii of the circles \omega are r_A,r_B,r_C and the radius of the inscribed disk of the triangle is equal to r.
A regular hexagon ABCDEF and a regular triangle APQ have no common interior points. It is known that PB <QB, and the point M is the midpoint of the segment PB. Find the angle between lines AM and QF.
Point M is the midpoint of side AB of a triangle ABC. The point K is chosen on the segment AB in such way that \angle BCK=\angle ACM . Points P and Q on sides BC and AC are such that KP \parallel AC and KQ \parallel BC. Prove that quadrilateral BPQA can be inscribed in a circle.
Oleg and Oliver ride bicycles with the same angular speeds with Oliver riding along a circular path A and Oleg riding along a circular path T of half the radius of A. They begin from the two closest points of the circles with Oleg’s circle lying inside Oliver’s circle. Two assistants also move along the circle T holding the screen (i.e., a chord with ends at the points where the assistants are located) such that the distance from each of the assistants to Oleg is always equal to distance from Oleg to Oliver. Prove that the screen touches some fixed circle throughout all the ride.
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