Found problems: 2023
2011 Canadian Students Math Olympiad, 4
Circles $\Gamma_1$ and $\Gamma_2$ have centers $O_1$ and $O_2$ and intersect at $P$ and $Q$. A line through $P$ intersects $\Gamma_1$ and $\Gamma_2$ at $A$ and $B$, respectively, such that $AB$ is not perpendicular to $PQ$. Let $X$ be the point on $PQ$ such that $XA=XB$ and let $Y$ be the point within $AO_1 O_2 B$ such that $AYO_1$ and $BYO_2$ are similar. Prove that $2\angle{O_1 AY}=\angle{AXB}$.
[i]Author: Matthew Brennan[/i]
1998 Hungary-Israel Binational, 2
On the sides of a convex hexagon $ ABCDEF$ , equilateral triangles are constructd in its exterior. Prove that the third vertices of these six triangles are vertices of a regular hexagon if and only if the initial hexagon is [i]affine regular[/i]. (A hexagon is called affine regular if it is centrally symmetric and any two opposite sides are parallel to the diagonal determine by the remaining two vertices.)
2012 Paraguay Mathematical Olympiad, 5
Let $ABC$ be an equilateral triangle. Let $Q$ be a random point on $BC$, and let $P$ be the meeting point of $AQ$ and the circumscribed circle of $\triangle ABC$.
Prove that $\frac{1}{PQ}=\frac{1}{PB}+\frac{1}{PC}$.
2002 Tournament Of Towns, 5
An acute triangle was dissected by a straight cut into two pieces which are not necessarily triangles. Then one of the pieces were dissected by a straight cut into two pieces and so on. After a few dissections it turns out the pieces were all triangles. Is it possible they were all obtuse?
1994 Iran MO (2nd round), 2
The incircle of triangle $ABC$ meet the sides $AB, AC$ and $BC$ in $M,N$ and $P$, respectively. Prove that the orthocenter of triangle $MNP,$ the incenter and the circumcenter of triangle $ABC$ are collinear.
[asy]
import graph; size(300); real lsf = 0.5; pen dp = linewidth(0.7) + fontsize(10); defaultpen(dp); pen ds = black; pen ttttff = rgb(0.2,0.2,1); pen ffwwww = rgb(1,0.4,0.4); pen xdxdff = rgb(0.49,0.49,1);
draw((8,17.58)--(2.84,9.26)--(20.44,9.21)--cycle); draw((8,17.58)--(2.84,9.26),ttttff+linewidth(2pt)); draw((2.84,9.26)--(20.44,9.21),ttttff+linewidth(2pt)); draw((20.44,9.21)--(8,17.58),ttttff+linewidth(2pt)); draw(circle((9.04,12.66),3.43),blue+linewidth(1.2pt)+linetype("8pt 8pt")); draw((6.04,14.42)--(8.94,9.24),ffwwww+linewidth(1.2pt)); draw((8.94,9.24)--(11.12,15.48),ffwwww+linewidth(1.2pt)); draw((11.12,15.48)--(6.04,14.42),ffwwww+linewidth(1.2pt)); draw((8.94,9.24)--(7.81,14.79)); draw((11.12,15.48)--(6.95,12.79)); draw((6.04,14.42)--(10.12,12.6));
dot((8,17.58),ds); label("$A$", (8.11,18.05),NE*lsf); dot((2.84,9.26),ds); label("$B$", (2.11,8.85), NE*lsf); dot((20.44,9.21),ds); label("$C$", (20.56,8.52), NE*lsf); dot((9.04,12.66),ds); label("$O$", (8.94,12.13), NE*lsf); dot((6.04,14.42),ds); label("$M$", (5.32,14.52), NE*lsf); dot((11.12,15.48),ds); label("$N$", (11.4,15.9), NE*lsf); dot((8.94,9.24),ds); label("$P$", (8.91,8.58), NE*lsf); dot((7.81,14.79),ds); label("$D$", (7.81,15.14),NE*lsf); dot((6.95,12.79),ds); label("$F$", (6.64,12.07),NE*lsf); dot((10.12,12.6),ds); label("$G$", (10.41,12.35),NE*lsf); dot((8.07,13.52),ds); label("$H$", (8.11,13.88),NE*lsf); clip((-0.68,-0.96)--(-0.68,25.47)--(30.71,25.47)--(30.71,-0.96)--cycle);
[/asy]
2011 Switzerland - Final Round, 5
Let $\triangle{ABC}$ be a triangle with circumcircle $\tau$. The tangentlines to $\tau$ through $A$ and $B$ intersect at $T$. The circle through $A$, $B$ and $T$ intersects $BC$ and $AC$ again at $D$ and $E$, respectively; $CT$ and $BE$ intersect at $F$.
Suppose $D$ is the midpoint of $BC$. Calculate the ratio $BF:BE$.
[i](Swiss Mathematical Olympiad 2011, Final round, problem 5)[/i]
2010 All-Russian Olympiad, 2
Into triangle $ABC$ gives point $K$ lies on bisector of $ \angle BAC$. Line $CK$ intersect circumcircle $ \omega$ of triangle $ABC$ at $M \neq C$. Circle $ \Omega$ passes through $A$, touch $CM$ at $K$ and intersect segment $AB$ at $P \neq A$ and $\omega $ at $Q \neq A$.
Prove, that $P$, $Q$, $M$ lies at one line.
1994 Baltic Way, 15
Does there exist a triangle such that the lengths of all its sides and altitudes are integers and its perimeter is equal to $1995$?
1985 IMO Longlists, 30
A plane rectangular grid is given and a “rational point” is defined as a point $(x, y)$ where $x$ and $y$ are both rational numbers. Let $A,B,A',B'$ be four distinct rational points. Let $P$ be a point such that $\frac{A'B'}{AB}=\frac{B'P}{BP} = \frac{PA'}{PA}.$ In other words, the triangles $ABP, A'B'P$ are directly or oppositely similar. Prove that $P$ is in general a rational point and find the exceptional positions of $A'$ and $B'$ relative to $A$ and $B$ such that there exists a $P$ that is not a rational point.
2014 Tajikistan Team Selection Test, 2
Let $M$be an interior point of triangle $ABC$. Let the line $AM$ intersect the circumcircle of the triangle $MBC$ for the second time at point $D$, the line $BM$ intersect the circumcircle of the triangle $MCA$ for the second time at point $E$, and the line $CM$ intersect the circumcircle of the triangle $MAB$ for the second time at point $F$. Prove that $\frac{AD}{MD} + \frac{BE}{ME} + \frac{CF}{MF} \geq \frac{9}{2}$.
[i]Proposed by Nairy Sedrakyan[/i]
1999 China National Olympiad, 1
Let $ABC$ be an acute triangle with $\angle C>\angle B$. Let $D$ be a point on $BC$ such that $\angle ADB$ is obtuse, and let $H$ be the orthocentre of triangle $ABD$. Suppose that $F$ is a point inside triangle $ABC$ that is on the circumcircle of triangle $ABD$. Prove that $F$ is the orthocenter of triangle $ABC$ if and only if $HD||CF$ and $H$ is on the circumcircle of triangle $ABC$.
2009 Sharygin Geometry Olympiad, 10
Let $ ABC$ be an acute triangle, $ CC_1$ its bisector, $ O$ its circumcenter. The perpendicular from $ C$ to $ AB$ meets line $ OC_1$ in a point lying on the circumcircle of $ AOB$. Determine angle $ C$.
2000 Iran MO (2nd round), 2
In a tetrahedron we know that sum of angles of all vertices is $180^\circ.$ (e.g. for vertex $A$, we have $\angle BAC + \angle CAD + \angle DAB=180^\circ.$)
Prove that faces of this tetrahedron are four congruent triangles.
2006 Junior Balkan Team Selection Tests - Moldova, 3
The convex polygon $A_{1}A_{2}\ldots A_{2006}$ has opposite sides parallel $(A_{1}A_{2}||A_{1004}A_{1005}, \ldots)$.
Prove that the diagonals $A_{1}A_{1004}, A_{2}A_{1005}, \ldots A_{1003}A_{2006}$ are concurrent if and only if opposite sides are equal.
2015 Turkey Team Selection Test, 4
Let $ABC$ be a triangle such that $|AB|=|AC|$ and let $D,E$ be points on the minor arcs $\overarc{AB}$ and $\overarc{AC}$ respectively. The lines $AD$ and $BC$ intersect at $F$ and the line $AE$ intersects the circumcircle of $\triangle FDE$ a second time at $G$. Prove that the line $AC$ is tangent to the circumcircle of $\triangle ECG$.
2005 India IMO Training Camp, 1
For a given triangle ABC, let X be a variable point on the line BC such that the point C lies between the points B and X. Prove that the radical axis of the incircles of the triangles ABX and ACX passes through a point independent of X.
This is a slight extension of the [url=http://www.mathlinks.ro/Forum/viewtopic.php?t=41033]IMO Shortlist 2004 geometry problem 7[/url] and can be found, together with the proposed solution, among the files uploaded at http://www.mathlinks.ro/Forum/viewtopic.php?t=15622 . Note that the problem was proposed by Russia. I could not find the names of the authors, but I have two particular persons under suspicion. Maybe somebody could shade some light on this...
Darij
2011 Romania Team Selection Test, 4
Let $ABCDEF$ be a convex hexagon of area $1$, whose opposite sides are parallel. The lines $AB$, $CD$ and $EF$ meet in pairs to determine the vertices of a triangle. Similarly, the lines $BC$, $DE$ and $FA$ meet in pairs to determine the vertices of another triangle. Show that the area of at least one of these two triangles is at least $3/2$.
1993 Vietnam National Olympiad, 1
The tetrahedron $ABCD$ has its vertices on the fixed sphere $S$. Prove that $AB^{2}+AC^{2}+AD^{2}-BC^{2}-BD^{2}-CD^{2}$ is minimum iff $AB\perp AC,AC\perp AD,AD\perp AB$.
2014 Turkey Team Selection Test, 1
Let $P$ be a point inside the acute triangle $ABC$ with $m(\widehat{PAC})=m(\widehat{PCB})$. $D$ is the midpoint of the segment $PC$. $AP$ and $BC$ intersect at $E$, and $BP$ and $DE$ intersect at $Q$. Prove that $\sin\widehat{BCQ}=\sin\widehat{BAP}$.
1997 Irish Math Olympiad, 2
For a point $ M$ inside an equilateral triangle $ ABC$, let $ D,E,F$ be the feet of the perpendiculars from $ M$ onto $ BC,CA,AB$, respectively. Find the locus of all such points $ M$ for which $ \angle FDE$ is a right angle.
2004 Iran MO (3rd Round), 10
$f:\mathbb{R}^2 \to \mathbb{R}^2$ is injective and surjective. Distance of $X$ and $Y$ is not less than distance of $f(X)$ and $f(Y)$. Prove for $A$ in plane:
\[ S(A) \geq S(f(A))\]
where $S(A)$ is area of $A$
2007 All-Russian Olympiad Regional Round, 9.4
Two triangles have equal longest sides and equal smallest angles. A new triangle is constructed, such that its sides are the sum of the longest sides, the sum of the shortest sides, and the sum of the middle sides of the initial triangles. Prove that the area of the new triangle is at least twice as much as the sum of the areas of the initial ones.
2004 Baltic Way, 17
Consider a rectangle with sidelengths 3 and 4, pick an arbitrary inner point on each side of this rectangle. Let $x, y, z$ and $u$ denote the side lengths of the quadrilateral spanned by these four points. Prove that $25 \leq x^2+y^2+z^2+u^2 \leq 50$.
2005 Junior Balkan Team Selection Tests - Romania, 16
Let $AB$ and $BC$ be two consecutive sides of a regular polygon with 9 vertices inscribed in a circle of center $O$. Let $M$ be the midpoint of $AB$ and $N$ be the midpoint of the radius perpendicular to $BC$. Find the measure of the angle $\angle OMN$.
2014 Bosnia And Herzegovina - Regional Olympiad, 3
Let $ABCD$ be a parallelogram. Let $M$ be a point on the side $AB$ and $N$ be a point on the side $BC$ such that the segments $AM$ and $CN$ have equal lengths and are non-zero. The lines $AN$ and $CM$ meet at $Q$.
Prove that the line $DQ$ is the bisector of the angle $\measuredangle ADC$.
[i]Alternative formulation.[/i] Let $ABCD$ be a parallelogram. Let $M$ and $N$ be points on the sides $AB$ and $BC$, respectively, such that $AM=CN\neq 0$. The lines $AN$ and $CM$ intersect at a point $Q$.
Prove that the point $Q$ lies on the bisector of the angle $\measuredangle ADC$.