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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