Found problems: 2023
1994 All-Russian Olympiad Regional Round, 10.7
In a convex pentagon $ ABCDE$ side $ AB$ is perpendicular to $ CD$ and side $ BC$ is perpendicular to $ DE$. Prove that if $ AB \equal{} AE \equal{} ED \equal{} 1$, then
$ BC \plus{} CD < 1$.
2007 Indonesia TST, 1
Let $ ABCD$ be a cyclic quadrilateral and $ O$ be the intersection of diagonal $ AC$ and $ BD$. The circumcircles of triangle $ ABO$ and the triangle $ CDO$ intersect at $ K$. Let $ L$ be a point such that the triangle $ BLC$ is similar to $ AKD$ (in that order). Prove that if $ BLCK$ is a convex quadrilateral, then it has an incircle.
1997 All-Russian Olympiad, 2
A circle centered at $O$ and inscribed in triangle $ABC$ meets sides $AC$;$AB$;$BC$ at $K$;$M$;$N$, respectively. The median $BB_1$ of the triangle meets $MN$ at $D$. Show that $O$;$D$;$K$ are collinear.
[i]M. Sonkin[/i]
1992 Baltic Way, 4
Is it possible to draw a hexagon with vertices in the knots of an integer lattice so that the squares of the lengths of the sides are six consecutive positive integers?
2009 Sharygin Geometry Olympiad, 20
Suppose $ H$ and $ O$ are the orthocenter and the circumcenter of acute triangle $ ABC$; $ AA_1$, $ BB_1$ and $ CC_1$ are the altitudes of the triangle. Point $ C_2$ is the reflection of $ C$ in $ A_1B_1$. Prove that $ H$, $ O$, $ C_1$ and $ C_2$ are concyclic.
2012 Tuymaada Olympiad, 2
Quadrilateral $ABCD$ is both cyclic and circumscribed. Its incircle touches its sides $AB$ and $CD$ at points $X$ and $Y$, respectively. The perpendiculars to $AB$ and $CD$ drawn at $A$ and $D$, respectively, meet at point $U$; those drawn at $X$ and $Y$ meet at point $V$, and finally, those drawn at $B$ and $C$ meet at point $W$. Prove that points $U$, $V$ and $W$ are collinear.
[i]Proposed by A. Golovanov[/i]
2016 All-Russian Olympiad, 2
Diagonals $AC,BD$ of cyclic quadrilateral $ABCD$ intersect at $P$.Point $Q$ is on$BC$ (between$B$ and $C$) such that $PQ \perp AC$.Prove that the line passes through the circumcenters of triangles $APD$ and $BQD$ is parallel to $AD$.(A.Kuznetsov)
1988 Romania Team Selection Test, 12
The four vertices of a square are the centers of four circles such that the sum of theirs areas equals the square's area. Take an arbitrary point in the interior of each circle. Prove that the four arbitrary points are the vertices of a convex quadrilateral.
[i]Laurentiu Panaitopol[/i]
2012 Canadian Mathematical Olympiad Qualification Repechage, 8
Suppose circles $\mathit{W}_1$ and $\mathit{W}2$, with centres $\mathit{O}_1$ and $\mathit{O}_2$ respectively, intersect at points $\mathit{M}$ and $\mathit{N}$. Let the tangent on $\mathit{W}_2$ at point $\mathit{N}$ intersect $\mathit{W}_1$ for the second time at $\mathit{B}_1$. Similarly, let the tangent on $\mathit{W}_1$ at point $\mathit{N}$ intersect $\mathit{W}_2$ for the second time at $\mathit{B}_2$. Let $\mathit{A}_1$ be a point on $\mathit{W}_1$ which is on arc $\mathit{B}_1\mathit{N}$ not containing $\mathit{M}$ and suppose line $\mathit{A}_1\mathit{N}$ intersects $\mathit{W}_2$ at point $\mathit{A}_2$. Denote the incentres of triangles $\mathit{B}_1\mathit{A}_1\mathit{N}$ and $\mathit{B}_2\mathit{A}_2\mathit{N}$ by $\mathit{I}_1$ and $\mathit{I}_2$, respectively.*
[asy]
/* Geogebra to Asymptote conversion, documentation at artofproblemsolving.com/Wiki, go to User:Azjps/geogebra */
import graph; size(10.1cm);
real labelscalefactor = 0.5; /* changes label-to-point distance */
pen dps = linewidth(0.7) + fontsize(10); defaultpen(dps); /* default pen style */
pen dotstyle = black; /* point style */
real xmin = -0.9748626324969808, xmax = 13.38440254515721, ymin = 0.5680051903627492, ymax = 10.99430986899034; /* image dimensions */
pair O_2 = (7.682929606970993,6.084708172218866), O_1 = (2.180000000000002,6.760000000000007), M = (4.560858774883258,8.585242858926296), B_2 = (10.07334553576748,9.291873850408265), A_2 = (11.49301008867042,4.866805580476367), B_1 = (2.113311869970955,9.759258690628950), A_1 = (0.2203184186713625,4.488514120712773);
/* draw figures */
draw(circle(O_2, 4.000000000000000));
draw(circle(O_1, 3.000000000000000));
draw((4.048892687647541,4.413249028538064)--B_2);
draw(B_2--A_2);
draw(A_2--(4.048892687647541,4.413249028538064));
draw((4.048892687647541,4.413249028538064)--B_1);
draw(B_1--A_1);
draw(A_1--(4.048892687647541,4.413249028538064));
/* dots and labels */
dot(O_2,dotstyle);
label("$O_2$", (7.788512439159622,6.243082420501817), NE * labelscalefactor);
dot(O_1,dotstyle);
label("$O_1$", (2.298205165350667,6.929370829727937), NE * labelscalefactor);
dot(M,dotstyle);
label("$M$", (4.383466101076183,8.935444641311980), NE * labelscalefactor);
dot((4.048892687647541,4.413249028538064),dotstyle);
label("$N$", (3.855551940133015,3.761885864068922), NE * labelscalefactor);
dot(B_2,dotstyle);
label("$B_2$", (10.19052187145104,9.463358802255147), NE * labelscalefactor);
dot(A_2,dotstyle);
label("$A_2$", (11.80066006232771,4.659339937672310), NE * labelscalefactor);
dot(B_1,dotstyle);
label("$B_1$", (1.981456668784765,10.09685579538695), NE * labelscalefactor);
dot(A_1,dotstyle);
label("$A_1$", (0.08096568938935705,3.973051528446190), NE * labelscalefactor);
clip((xmin,ymin)--(xmin,ymax)--(xmax,ymax)--(xmax,ymin)--cycle);
/* end of picture */[/asy]
Show that \[\angle\mathit{I}_1\mathit{MI}_2=\angle\mathit{O}_1\mathit{MO}_2.\]
*[size=80]Given a triangle ABC, the incentre of the triangle is defined to be the intersection of the angle bisectors of A, B, and C. To avoid cluttering, the incentre is omitted in the provided diagram. Note also that the diagram serves only as an aid and is not necessarily drawn to scale.[/size]
2008 Croatia Team Selection Test, 3
Point $ M$ is taken on side $ BC$ of a triangle $ ABC$ such that the centroid $ T_c$ of triangle $ ABM$ lies on the circumcircle of $ \triangle ACM$ and the centroid $ T_b$ of $ \triangle ACM$ lies on the circumcircle of $ \triangle ABM$. Prove that the medians of the triangles $ ABM$ and $ ACM$ from $ M$ are of the same length.
2012 Grigore Moisil Intercounty, 3
Let $ \Delta ABC$ be a triangle, with $ m(\angle A)=90^{\circ}$ and $ m(\angle B)=30^{\circ}.$
If $M$ is the middle of $[AB],$ $N$ is the middle of $[BC],$ and $P\in[BC],\ Q\in[MN],$ such that
\[\frac{PB}{PC}=4\cdot\frac{QM}{QN}+3,\]
prove that $ \Delta APQ$ is an equilateral triangle.
[b]Author: MARIN BANCOȘ[/b]
[b]Regional Mathematical Contest GRIGORE MOISIL, Romania, Baia Mare, 24.03.2012, 7th grade[/b]
2008 Sharygin Geometry Olympiad, 22
(A.Khachaturyan, 10--11) a) All vertices of a pyramid lie on the facets of a cube
but not on its edges, and each facet contains at least one vertex. What is the
maximum possible number of the vertices of the pyramid?
b) All vertices of a pyramid lie in the facet planes of a cube but not on the lines
including its edges, and each facet plane contains at least one vertex. What is the
maximum possible number of the vertices of the pyramid?
2009 Canadian Mathematical Olympiad Qualification Repechage, 7
A rectangular sheet of paper is folded so that two diagonally opposite corners come together. If the crease formed is the same length as the longer side of the sheet, what is the ratio of the longer side of the sheet to the shorter side?
2007 Cuba MO, 9
Let $O$ be the circumcircle of $\triangle ABC$, with $AC=BC$ end let $D=AO\cap BC$. If $BD$ and $CD$ are integer numbers and $AO-CD$ is prime, determine such three numbers.
2013 Saint Petersburg Mathematical Olympiad, 3
Let $M$ and $N$ are midpoint of edges $AB$ and $CD$ of the tetrahedron $ABCD$, $AN=DM$ and $CM=BN$. Prove that $AC=BD$.
S. Berlov
2006 QEDMO 2nd, 7
Let $H$ be the orthocenter of a triangle $ABC$, and let $D$ be the midpoint of the segment $AH$.
The altitude $BH$ of triangle $ABC$ intersects the perpendicular to the line $AB$ through the point $A$ at the point $M$.
The altitude $CH$ of triangle $ABC$ intersects the perpendicular to the line $CA$ through the point $A$ at the point $N$.
The perpendicular bisector of the segment $AB$ intersects the perpendicular to the line $BC$ through the point $B$ at the point $U$.
The perpendicular bisector of the segment $CA$ intersects the perpendicular to the line $BC$ through the point $C$ at the point $V$.
Finally, let $E$ be the midpoint of the side $BC$ of triangle $ABC$.
Prove that the points $D$, $M$, $N$, $U$, $V$ all lie on one and the same perpendicular to the line $AE$.
[i]Extensions.[/i] In other words, we have to show that the points $M$, $N$, $U$, $V$ lie on the perpendicular to the line $AE$ through the point $D$. Additionally, one can find two more points on this perpendicular:
[b](a)[/b] The nine-point circle of triangle $ABC$ is known to pass through the midpoint $E$ of its side $BC$. Let $D^{\prime}$ be the point where this nine-point circle intersects the line $AE$ apart from $E$. Then, the point $D^{\prime}$ lies on the perpendicular to the line $AE$ through the point $D$.
[b](b)[/b] Let the tangent to the circumcircle of triangle $ABC$ at the point $A$ intersect the line $BC$ at a point $X$. Then, the point $X$ lies on the perpendicular to the line $AE$ through the point $D$.
[i]Comment.[/i] The actual problem was created by Victor Thébault around 1950 (cf. Hyacinthos messages #1102 and #1551). The extension [b](a)[/b] initially was a (pretty trivial) lemma in Thébault's solution of the problem. Extension [b](b)[/b] is rather new; in the form "prove that $X\in UV$", it was [url=http://www.mathlinks.ro/Forum/viewtopic.php?t=3659]proposed by Valentin Vornicu for the Balkan MO 2003[/url], however it circulated in the Hyacinthos newsgroup before (Hyacinthos messages #7240 and #7242), where different solutions of the problem were discussed as well. Hereby, "Hyacinthos" always refers to the triangle geometry newsgroup "Hyacinthos", which can be found at http://groups.yahoo.com/group/Hyacinthos .
I proposed the problem for the QEDMO math fight wishing to draw some attention to it. It has a rather short and elementary solution, by the way (without using radical axes or inversion like the standard solutions).
Darij
2007 China Team Selection Test, 2
Let $ I$ be the incenter of triangle $ ABC.$ Let $ M,N$ be the midpoints of $ AB,AC,$ respectively. Points $ D,E$ lie on $ AB,AC$ respectively such that $ BD\equal{}CE\equal{}BC.$ The line perpendicular to $ IM$ through $ D$ intersects the line perpendicular to $ IN$ through $ E$ at $ P.$ Prove that $ AP\perp BC.$
2010 Regional Competition For Advanced Students, 3
Let $\triangle ABC$ be a triangle and let $D$ be a point on side $\overline{BC}$. Let $U$ and $V$ be the circumcenters of triangles $\triangle ABD$ and $\triangle ADC$, respectively. Show, that $\triangle ABC$ and $\triangle AUV$ are similar.
[i](41th Austrian Mathematical Olympiad, regional competition, problem 3)[/i]
2005 QEDMO 1st, 5 (G1)
Let $ABC$ be a triangle, and let $C^{\prime}$ and $A^{\prime}$ be the feet of its altitudes issuing from the vertices $C$ and $A$, respectively. Denote by $P$ the midpoint of the segment $C^{\prime}A^{\prime}$. The circumcircles of triangles $AC^{\prime}P$ and $CA^{\prime}P$ have a common point apart from $P$; denote this common point by $Q$. Prove that:
[b](a)[/b] The point $Q$ lies on the circumcircle of the triangle $ABC$.
[b](b)[/b] The line $PQ$ passes through the point $B$.
[b](c)[/b] We have $\frac{AQ}{CQ}=\frac{AB}{CB}$.
Darij
2013 Turkey Junior National Olympiad, 3
Let $ABC$ be a triangle such that $AC>AB.$ A circle tangent to the sides $AB$ and $AC$ at $D$ and $E$ respectively, intersects the circumcircle of $ABC$ at $K$ and $L$. Let $X$ and $Y$ be points on the sides $AB$ and $AC$ respectively, satisfying
\[ \frac{AX}{AB}=\frac{CE}{BD+CE} \quad \text{and} \quad \frac{AY}{AC}=\frac{BD}{BD+CE} \]
Show that the lines $XY, BC$ and $KL$ are concurrent.
2007 Tournament Of Towns, 3
$D$ is the midpoint of the side $BC$ of triangle $ABC$. $E$ and $F$ are points on $CA$ and $AB$ respectively, such that $BE$ is perpendicular to $CA$ and $CF$ is perpendicular to $AB$. If $DEF$ is an equilateral triangle, does it follow that $ABC$ is also equilateral?
2014 Contests, 3
Let $D, E, F$ be points on the sides $BC, CA, AB$ of a triangle $ABC$, respectively such that the lines $AD, BE, CF$ are concurrent at the point $P$. Let a line $\ell$ through $A$ intersect the rays $[DE$ and $[DF$ at the points $Q$ and $R$, respectively. Let $M$ and $N$ be points on the rays $[DB$ and $[DC$, respectively such that the equation
\[ \frac{QN^2}{DN}+\frac{RM^2}{DM}=\frac{(DQ+DR)^2-2\cdot RQ^2+2\cdot DM\cdot DN}{MN} \]
holds. Show that the lines $AD$ and $BC$ are perpendicular to each other.
2008 Moldova Team Selection Test, 3
Let $ \omega$ be the circumcircle of $ ABC$ and let $ D$ be a fixed point on $ BC$, $ D\neq B$, $ D\neq C$. Let $ X$ be a variable point on $ (BC)$, $ X\neq D$. Let $ Y$ be the second intersection point of $ AX$ and $ \omega$. Prove that the circumcircle of $ XYD$ passes through a fixed point.
2000 JBMO ShortLists, 17
A triangle $ABC$ is given. Find all the pairs of points $X,Y$ so that $X$ is on the sides of the triangle, $Y$ is inside the triangle, and four non-intersecting segments from the set $\{XY, AX, AY, BX,BY, CX, CY\}$ divide the triangle $ABC$ into four triangles with equal areas.
2009 District Round (Round II), 2
in a right-angled triangle $ABC$ with $\angle C=90$,$a,b,c$ are the corresponding sides.Circles $K.L$ have their centers on $a,b$ and are tangent to $b,c$;$a,c$ respectively,with radii $r,t$.find the greatest real number $p$ such that the inequality
$\frac{1}{r}+\frac{1}{t}\ge p(\frac{1}{a}+\frac{1}{b})$
always holds.