geometry problems from final round of Kurchatov Olympiad (Russia) with aops links in the names
ОЛИМПИАДА «КУРЧАТОВ»
collected inside aops here
2013 - 2022
Baron Munchausen has a set of four triangles. He states that the first of the triangles is not equal to the second. And he also says that whichever of the four is lost, from the other three you can make one a larger triangle (no overlaps and holes). Could all the words of the baron be true?
$ABCD$ is trapezoid, $M$ is midpoint of base $AD$ , $E$ is intersection point of the diagonals.
It is known that $AD = 2BC = 6ME$. Prove that $AB^2+ CD^2= BC^2$.
There are $n$ different points on the plane. Prove that they can be denoted by the letters $A_1, A_2, ..., A_n$ so that the vectors $\overrightarrow{A_1A_2}, \overrightarrow{A_2A_3}, ..., \overrightarrow{A_{n-1}A_n}$ are all different.
A perpendicular bisector was drawn to the leg AB of an isosceles trapezoid $ABCD$. It crossed the segment $BC$ at point $E$. Find the angle $ABC$ if it is known that the lines $AE$ and $CD$ perpendicular.
Points $M$ and $N$ are marked on the sides of the angle with vertex $Q$, and point $E$ is marked inside the angle so that $QE = MN$, $\angle MQE = \angle QNM$, $\angle EQN + \angle QNE = \angle QMN$. Find $\angle MQN$.
The coordinate axes and the graph of the function $y=\frac{2}{x}$ are drawn on the plane for $x> 0$. The scale is not specified, but it is known to be the same on both axes. is the same. Unfortunately, a small piece of the graph near the beginning coordinates was accidentally erased (see fig.) Using a compass and a ruler , restore the point with abscissa $1$ on this graph.
The coordinate axes and the graph function of the funtion $y=\frac{2}{x}$ are drawn on the plane . The scale is not specified, but it is known to be the same on both axes. is the same. Using a compass and a ruler , restore the point at which the abscissa is positive and $2$ less than the ordinate.
Two disjoint circles are inscribed in an angle. They touch one side of the corner at points $K$ and $L$, the other at points $M$ and $N$ (see figure), $C$ is the midpoint of the segment $KL$, $A$ and $B$ are the points of intersection of the segments $CM$ and$ CN$ with the circles. Prove that
a) points $A, B, M$ and $N$ lie on the same circle.
b) points $A, B, K$ and $L$ lie on the same circle.
Divide an isosceles right triangle into two smaller triangles so that some median of one of these triangles is parallel to one of atltides of the second triangle.
Vasya, an invisible virus, is sitting on a square plate with a side of $1$ cm. He and the doctor Petya take turns. With the next $n$-th move, Petya draws with the vaccine like ink a segment $1$ micron long, and then Vasya must choose a direction and crawl in this straight line a direction of $1 / n$ micron distance (without going beyond the edge of the plate). If Vasya crawls through any of the points with the vaccine or touches it, he will die. Petya can act with any precision. Can Petya within a finite number of moves for sure kill the virus?
In a right-angled triangle $ABC$, angle bisector $AL$ was drawn and a a point $K$ was marked on the hypotenuse $AB$ such such that $AB = 3BK$. It turned out that the angle $ALK$ is right . Prove that $AL = BL$.
In the convex quadrilateral $ABCD$, the perpendiculars are drawn to sides $AB, BC$ and $CD$. Inside the quadrilateral, these perpendiculars do not cross . The intersection points of these perpendiculars split the $AD$ side into $4$ equal parts. Prove that $AD\parallel BC$.
On the median $CM$ of triangle $ABC$, a point $D$ is chosen such that $2CD = AB$. Line $BD$ meets side $AC$ at point $E$. Prove that if $DE = CE$, then angle $\angle BMC = 120^o.$
$ABCD$ is a cyclic quadrilateral, $AB> CD$, $BC> AD$. On sides $AB$ and $BC$ marked points $X$ and $Y$ so that $AX = CD$ and $AD = CY$. $M$ is the midpoint of $XY$. Prove that the angle $AMC$ is right.
To prepare mashed potatoes, Kolya, the chef, needs to get the specified amount of peeled potatoes as soon as possible. Not caring about saving peelings, he cuts cubes from spherical potatoes, with each stroke clearing one edge of the knife. Can he complete the task faster when the same frequency of knife strokes if you cut out any other polyhedra? (formally: is it true that of all polyhedra cut from of the given sphere, the largest ratio of volume to the number of faces is for the inscribed cube?)
If the width of the rectangle is increased by $3$ cm, and the length is reduced by $3$ cm, it's area will not change. And how will the area change if, instead, the original reduce the width of the rectangle by $4$ cm, and increase the length by $4$ cm?
The segments $KL$ and $MN$ intersect at point $T$. It is known that the triangle $KNT$ is equilateral and $KL = MT$. Prove that triangle $LMN$ is isosceles.
If the width of the rectangle is increased by $30\%$, and the length is decreased by $20\%$, it's perimeter will not change. And the perimeter will decrease or increase, and by what percentage, if instead of that the original rectangle has its width reduced by $20\%$, and the length increase by $30\%$?
The diagonals of parallelogram $ABCD$ meet at point $O$. In triangles $OAB, OBC, OCD$ medians $OM, OM', OM''$ and angle bisectors $OL, OL', OL''$ are drawn respectively. Prove that angles $MM'M''$ and $LL'L''$ are equal.
You are given an isosceles triangle $ABC$. On the lateral side $AB$, a point $M$ is marked such that $CM = AC$. Then, on the lateral side $BC$, a point $N$ was marked such that $BN = MN$, and the angle bisector $NH$ was drawn at triangle $CNM$. Prove that $H$ lies on the median $BK$ of the triangle $ABC$.
Given a triangle $ABC$. From point $P$ inside it, the perpendiculars $PA', PB', PC'$ are drawn on the sides $BC,CA,AB$ respectively. Then from point $P$ perpendiculars $PA'', PB''$ are drawn on the sides $B'C'$ and $C'A' $ respectively. Prove that $PA\cdot PA'\cdot PA''= PB\cdot PB'\cdot PB''$.
Two right-angled triangles have the same area and perimeter. Is it obligatory are these triangles congruent?
Two opposite vertices $M$ and $M '$ were chosen from the cube and flat sections $ABC$ and $A'B'C'$ cut off two triangular pyramids $MABC$ and $M'A'B'C '$. It turned out to be an octahedron (see fig.) Three distances turned out to be are pairwise different: between straight lines $AB$ and $A'B '$, between straight lines $BC$ and $B'C'$ and between straight lines $AC$ and $A'C'$. Prove that lines $AA ', BB'$ and $CC '$ have a common point.
All angles of a convex hexagon $ABCDEF$ are equal. Prove that $AB - DE = EF - BC = CD - FA$.
The diagonals of the trapezoid $ABCD$ ($AD \parallel BC$) meet at point $S$. It is known that that $AD \perp AC$ and $BS = 2CD$. Prove that $\angle CDB = 2\angle ADB$.
Points $X$ and $Y$ are chosen on sides $AB, AD$ of square $ABCD$ so that $AX =DY$. Lines $BC$ and $DX$ meet at point $P$, lines $CD$ and $BY$ at point $Q$. Prove that points $P, Q, A$ are collinear.
Let $A$ and $B$ be different points belonging to the line of intersection of the perpendicular planes $\pi_1$ and $\pi_2$. Point $C$ belongs to the plane $\pi_2$, but does not belong to $\pi_1$. Let $P$ denote the intersection point of the bisector of the angle $ACB$ with line $AB$ and through $\omega$ a circle with diameter $AB$ in the plane $\pi_1$. The plane $\pi_3$ containing $CP$ meets the circle $\omega$ at points $D$ and $E$. Prove that $CP$ is the bisector of the angle $DCE$.
Points $X$ and $Y$ are marked on the sides $AB, BC$ of triangle $ABC$, respectively, so that $AY = AB$ and $CX = CB$. The straight line passing through the vertex $A$ parallel to the side $BC$ intersects the straight line, passing through $C$ parallel to side $AB$, at point $D$. Prove that $DX = DY$.
An overlapping square and circle are drawn on the plane. Together they cover an area of $2018$ cm$^2$. The intersection area is $137$ cm$^2$ .The area of the circle is $1371$ cm$^2$ . What is the perimeter of the square?
In triangle $ABC$, point $D$ is chosen on the side $AC$, and point $E$ on the side $BC$ so that relations $CD = AB$, $BE = BD$, $AB \cdot AC = BC^2$. Find $\angle DEA$ if it is known that $\angle DBC = 40^o$.
In an acute-angled triangle $ABC$, a straight line $\ell$ is drawn through the vertex $A$, perpendicular to the median, starting from the vertex $A$. The extensions of the altitudes $BD$ and $CE$ of the triangle intersect the straight line $\ell$ at the points $M$ and $N$. Prove that $AM = AN$.
The inscribed circle of triangle $ABC$ touches sides $AB$ and $AC$ at points $D$ and $E$, respectively. Point $I_A$ is the center of the $A$-excircle of triangle $ABC$, and points $K$ and $L$ are the midpoints of the segments $DI_A$ and $EI_A $ , respectively. Lines $BK$ and $CL$ meet at point $F$ lying inside the angle $BAC$. Find $\angle BFC$ if $\angle BAC = 50^o$.
A tetrahedron $ABCD$ with acute-angled edges is inscribed in a sphere centered at $O$. The straight line passing through point $O$ perpendicular to the plane $ABC$ intersects the sphere at point $E$ such that $D$ and $E$ lie on opposite sides relative to plane $ABC$. Line $DE$ intersects plane $ABC$ at point $F$, lying inside the triangle $ABC$. It turned out that $ \angle ADE = \angle BDE$, $AF \ne BF$ and $\angle AFB = 80^o$. Find the value for $\angle ACB$.
A square with a side of $1$ m is cut into three rectangles with equal perimeters. What can these perimeters be equal to? Indicate all possible options and explain why there are no others.
Points $X$ and $Y$ are chosen on the sides $AB$ and $AC$ of triangle $ABC$, respectively, so that $\angle AYB = \angle AXC = 134^o$ . On ray $YB$, point $M$ was marked beyond point $B$, and on ray $XC$ beyond point $C$ point $N$. It turned out that $MB = AC$, $AB = CN$. Find $\angle MAN$.
Distances from a point $P$, which lies inside an equilateral triangle, to its vertices are $3, 4$ and $5$. Find the area of the triangle.
The circle $\omega$ centered at point $I$ is inscribed in a convex quadrilateral $ABCD$ and touches side $AB$ at point $M$, and side $CD$ at point $N$, while $\angle BAD + \angle ADC <180^o$. On the line $MN$, a point $K \ne M$ is chosen such that $AK = AM$. In what ratio does the straight line $DI$ divide the segment $KN$? Give all the possible answers and prove that there are no others.
In tetrahedron $ABCD$, the following equalities hold:
$\angle BAC + \angle BDC = \angle ABD + \angle ACD$, $\angle BAD + \angle BCD = \angle ABC + \angle ADC$. Prove that the center of the circumscribed sphere of the tetrahedron lies on the line connecting the midpoints of the edges $AB$ and $CD$.
The angle bisector $BL$ is drawn in right-angled triangle $ABC$ with right angle $A$. Point $E$ is selected on segment $BC$, and that segment $CL$ is point $D$ so that $\angle LDE = 90^o$, $AL = DE$. Prove that $AB = LD + BE$.
In a right-angled triangle $ABC$ with a right angle $A$, the altitude $AH$ is drawn. On the extension of the hypotenuse $BC$ beyond the point $C$, there is a point $X$ such that $HX = \frac{BH + CX}{3}$. Prove that $2\angle ABC = \angle AXC$.
In triangle $ABC$ with angles $\angle A = 35^o, \angle B = 20^o$ and $ \angle C = 125^o$, point $O$ is the center of the circumscribed circle . Prove that points $O, A, B, C$ are the vertices of a trapezoid.
Point $H$ is the orthocenter of an acute-angled triangle $ABC$. Point $G$ is such that the quadrilateral $ABGH$ is a parallelogram. Let $I$ be a point on the line $GH$ such that $AC$ divides $HI$ in half. Line $AC$ meets the circumcircle of triangle $GCI$ at points $C$ and $J$. Prove that $IJ = AH$.
Let $SABCD$ be a regular rectangular pyramid with base $ABCD$. On the segment $AC$, a point $M$ was found such that $SM = MB$ and the planes $SBM$ and $SAB$ are perpendicular. Find the ratio $AM: AC$.
The altitude $AD$ is drawn in an acute-angled triangle $ABC$. It turned out that $AB + BD = DC$. Prove that $\angle B = 2\angle C$.
On the side $BC$ of an acute-angled triangle $ABC$, mark the point $D$ such that $AB + BD = DC$. Prove that $\angle ADC = 90^o$ if it is known that $\angle B = 2\angle C$.
A point $M$ is marked on the side $AC$ of triangle $ABC$ such that $AM = AB + MC$. Prove that the perpendicular to $AC$ passing through $M$ is divides the arc $BC$ of the circumcircle $ABC$ in half.
Given a rectangular trapezoid $ABCD$ with a right angle $A$ ($BC \parallel AD$). It is known that $BC = 1$, $AD = 4$. Point X is marked on the side $AB$, and point $Y$ on side $CD$such that $XY = 2$, $XY \perp CD$. Prove that the circumscribed circle of the triangle $XCD$ touches $AB$.
The diagonals of the trapezoid $ABCD$ ($AD \parallel BC$) meet at point $O$. On $AB$ mark a point $E$ such that line $EO$ is parallel to the base of the trapezoid. It turned out that $EO$ is the bisector of the angle $CED$ . Prove that the trapezoid is right.
The figure shows three squares. Find the marked angles, if the other two angles in the figure are known.
In triangle $ABC$, the angle at vertex $B$ is $120^o$, point $M$ is the midpoint of side $AC$. On sides $AB$ and $BC$, points $E$ and $F$ are selected, respectively, such that $AE = EF = FC$. Find $\angle EMF$.
Given a triangle $ABC$ such that $\angle BAC = 2\angle BCA$. Point $L$ lies on side $BC$ is that $\angle BAL = \angle CAL$. Point $M$ is the midpoint of side $AC$. Point $H$ lies on segment $AL$ is such that $MH \perp AL$. There is a point $K$ on the side of $BC$ such that triangle $KMH$ is equilateral. Prove that points $B, H$ and $M$ lie on the same line.
Diagonals of convex quadrilateral $ABCD$ intersect at point $O$. Points $P$ and $Q$ are the midpoints of segments $AC$ and $BD$, respectively. On the segments $OA$, $OB$, $OC$, $OD$ points $A_1$, $B_1$, $C_1$, $D_1$ respectively such that $AA_1=CC_1$, $BB_1=DD_1$.
$\bullet$ The circumscribed circles of triangles $AOB$ and $COD$ intersect at points $K$ and $O$.
$\bullet$ The circumscribed circles of triangles $A_1OB_1$ and $C_1OD_1$ intersect at points $M$ and $O$.
Prove that the points $K, M, P, Q$ lie on the same circle.
Given an acute-angled non-isosceles triangle $ABC$, point $O$ is the center of its circumscribed circle. The extension of altitude $BH$ of triangle $ABC$ intersects it's circumscribed circle at point $N$. On the sides $AB$ and $BC$ the points $X$ and $Y$ are marked, respectively, such that $OX \parallel AN$ and $OY\parallel CN$ . The circumscribed circle of the triangle $XBY$ intersects the segment $BH$ at the point $Z$. Prove that $XY \parallel OZ$.
A point $P$ inside an acute triangle $ABC$ is such that $\angle BAP =\angle CAP$. Point $M$ is the midpoint of side $BC$. The line $MP$ intersects the circumscribed circles of triangles $ABP$ and $ACP$ at points $D$ and $E$, respectively (point $P$ lies between the points $M$ and $E$, the point $E$ lies between the points $P$ and $D$). It turned out that $DE = MP$. Prove that $BC = 2BP$.
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