Found problems: 509
2013 Sharygin Geometry Olympiad, 1
Let $ABCDE$ be a pentagon with right angles at vertices $B$ and $E$ and such that $AB = AE$ and $BC = CD = DE$. The diagonals $BD$ and $CE$ meet at point $F$. Prove that $FA = AB$.
2019 Saudi Arabia JBMO TST, 2
In triangle $ABC$ point $M$ is the midpoint of side $AB$, and point $D$ is the foot of altitude $CD$.
Prove that $\angle A = 2\angle B$ if and only if $AC = 2MD$
2007 Bulgarian Autumn Math Competition, Problem 11.3
In $\triangle ABC$ we have that $CC_{1}$ is an angle bisector. The points $P\in C_{1}B$, $Q\in BC$, $R\in AC$, $S\in AC_{1}$ satisfy $C_{1}P=PQ=QC$ and $CR=RS=SC_{1}$. Prove that $CC_{1}$ bisects $\angle SCP$.
1955 Moscow Mathematical Olympiad, 289
Consider an equilateral triangle $\vartriangle ABC$ and points $D$ and $E$ on the sides $AB$ and $BC$csuch that $AE = CD$. Find the locus of intersection points of $AE$ with $CD$ as points $D$ and $E$ vary.
2010 Saudi Arabia BMO TST, 2
Quadrilateral $ABCD$ with perpendicular diagonals $AC$ and $BD$ is inscribed in a circle. Altitude $DE$ in triangle $ABD$ intersects diagonal $AC$ in $F$. Prove that $FB = BC$
Kyiv City MO Seniors Round2 2010+ geometry, 2019.10.3.1
Let $ABCDE$ be a regular pentagon with center $M$. Point $P \ne M$ is selected on segment $MD$. The circumscribed circle of triangle $ABP$ intersects the line $AE$ for second time at point $Q$, and a line that is perpendicular to the $CD$ and passes through $P$, for second time at the point $R$. Prove that $AR = QR$.
2012 Belarus Team Selection Test, 2
$A, B, C, D, E$ are five points on the same circle, so that $ABCDE$ is convex and we have $AB = BC$ and $CD = DE$. Suppose that the lines $(AD)$ and $(BE)$ intersect at $P$, and that the line $(BD)$ meets line $(CA)$ at $Q$ and line $(CE)$ at $T$. Prove that the triangle $PQT$ is isosceles.
(I. Voronovich)
Kyiv City MO Seniors 2003+ geometry, 2008.11.4
In the tetrahedron $SABC $ at the height $SH$ the following point $O$ is chosen, such that: $$\angle AOS + \alpha = \angle BOS + \beta = \angle COS + \gamma = 180^o, $$ where $\alpha, \beta, \gamma$ are dihedral angles at the edges $BC, AC, AB $, respectively, at this point $H$ lies inside the base $ABC$. Let ${{A} _ {1}}, \, {{B} _ {1}}, \, {{C} _ {1}} $be the points of intersection of lines and planes: ${{A} _ {1}} = AO \cap SBC $, ${{B} _ {1}} = BO \cap SAC $, ${{C} _ {1}} = CO \cap SBA$ . Prove that if the planes $ABC $ and ${{A} _ {1}} {{B} _ {1}} {{C} _ {1}} $ are parallel, then $SA = SB = SC $.
(Alexey Klurman)
Cono Sur Shortlist - geometry, 1993.6
Consider in the interior of an equilateral triangle $ABC$ points $D, E$ and $F$ such that$ D$ belongs to segment $BE$, $E$ belongs to segment $CF$ and$ F$ to segment $AD$. If $AD=BE = CF$ then $DEF$ is equilateral.
Kharkiv City MO Seniors - geometry, 2018.10.4
On the sides $AB, AC ,BC$ of the triangle $ABC$, the points $M, N, K$ are selected, respectively, such that $AM = AN$ and $BM = BK$. The circle circumscribed around the triangle $MNK$ intersects the segments $AB$ and $BC$ for the second time at points $P$ and $Q$, respectively. Lines $MN$ and $PQ$ intersect at point $T$. Prove that the line $CT$ bisects the segment $MP$.
2019 Tournament Of Towns, 2
Let $ABC$ be an acute triangle. Suppose the points $A',B',C'$ lie on its sides $BC,AC,AB$ respectively and the segments $AA',BB',CC'$ intersect in a common point $P$ inside the triangle. For each of those segments let us consider the circle such that the segment is its diameter, and the chord of this circle that contains the point $P$ and is perpendicular to this diameter. All three these chords occurred to have the same length. Prove that $P$ is the orthocenter of the triangle $ABC$.
(Grigory Galperin)
2013 Junior Balkan Team Selection Tests - Romania, 4
In the acute-angled triangle $ABC$, with $AB \ne AC$, $D$ is the foot of the angle bisector of angle $A$, and $E, F$ are the feet of the altitudes from $B$ and $C$, respectively. The circumcircles of triangles $DBF$ and $DCE$ intersect for the second time at $M$. Prove that $ME = MF$.
Leonard Giugiuc
2007 Oral Moscow Geometry Olympiad, 4
The midpoints of the opposite sides of the hexagon are connected by segments. It turned out that the points of pairwise intersection of these segments form an equilateral triangle. Prove that the drawn segments are equal.
(M. Volchkevich)
Denmark (Mohr) - geometry, 2014.3
The points $C$ and $D$ lie on a halfline from the midpoint $M$ of a segment $AB$, so that $|AC| = |BD|$. Prove that the angles $u = \angle ACM$ and $v = \angle BDM$ are equal.
[img]https://1.bp.blogspot.com/-tQEJ1VBCa8U/XzT7IhwlZHI/AAAAAAAAMVI/xpRdlj5Rl64VUt_tCRsQ1UxIsv_SGrMlACLcBGAsYHQ/s0/2014%2BMohr%2Bp3.png[/img]
Kyiv City MO Juniors 2003+ geometry, 2016.8.51
In the quadrilateral $ABCD$, shown in fig. , the equations are true: $\angle ABC = \angle BCD$ and $2AB = CD$. On the side $BC$, a point $X$ is selected such that $\angle BAX = \angle CDA$. Prove that $AX = AD$.
[img]https://cdn.artofproblemsolving.com/attachments/2/9/0884eb311d1e40300c1e5980fd53eaadfa7a25.png[/img]
Geometry Mathley 2011-12, 14.3
Let $ABC$ be a triangle inscribed in circle $(I)$ that is tangent to the sides $BC,CA,AB$ at points $D,E, F$ respectively. Assume that $L$ is the intersection of $BE$ and $CF,G$ is the centroid of triangle $DEF,K$ is the symmetric point of $L$ about $G$. If $DK$ meets $EF$ at $P, Q$ is on $EF$ such that $QF = PE$, prove that $\angle DGE + \angle FGQ = 180^o$.
Nguyễn Minh Hà
Kyiv City MO Juniors Round2 2010+ geometry, 2020.8.2
Given a convex quadrilateral $ABCD$, in which $\angle CBD = 90^o$, $\angle BCD =\angle CAD$ and $AD= 2BC$. Prove that $CA =CD$.
(Anton Trygub)
2021 Dutch IMO TST, 1
Let $\Gamma$ be the circumscribed circle of a triangle $ABC$ and let $D$ be a point at line segment $BC$. The circle passing through $B$ and $D$ tangent to $\Gamma$ and the circle passing through $C $and $D$ tangent to $\Gamma$ intersect at a point $E \ne D$. The line $DE$ intersects $\Gamma$ at two points $X$ and $Y$ . Prove that $|EX| = |EY|$.
Novosibirsk Oral Geo Oly VII, 2021.6
Inside the equilateral triangle $ABC$, points $P$ and $Q$ are chosen so that the quadrilateral $APQC$ is convex, $AP = PQ = QC$ and $\angle PBQ = 30^o$. Prove that $AQ = BP$.
2011 Sharygin Geometry Olympiad, 5
A line passing through vertex $A$ of regular triangle $ABC$ doesn’t intersect segment $BC$. Points $M$ and $N$ lie on this line, and $AM = AN = AB$ (point $B$ lies inside angle $MAC$). Prove that the quadrilateral formed by lines $AB, AC, BN, CM$ is cyclic.
Ukrainian TYM Qualifying - geometry, XI.6
Prove that there exists a point $K$ in the plane of $\vartriangle ABC$ such that $$AK^2 - BC^2 = BK^2 - AC^2 = CK^2 - AB^2.$$ Let $Q, N, T$ be the points of intersection of the medians of the triangles $BKC, CKA, AKB$, respectively. Prove that the segments $AQ, BN$ and $CT$ are equal and have a common point.
Denmark (Mohr) - geometry, 2019.5
In the figure below the triangles $BCD, CAE$ and $ABF$ are equilateral, and the triangle $ABC$ is right-angled with $\angle A = 90^o$. Prove that $|AD| = |EF|$.
[img]https://1.bp.blogspot.com/-QMMhRdej1x8/XzP18QbsXOI/AAAAAAAAMUI/n53OsE8rwZcjB_zpKUXWXq6bg3o8GUfSwCLcBGAsYHQ/s0/2019%2Bmohr%2Bp5.png[/img]
2018 Estonia Team Selection Test, 7
Let $AD$ be the altitude $ABC$ of an acute triangle. On the line $AD$ are chosen different points $E$ and $F$ so that $|DE |= |DF|$ and point $E$ is in the interior of triangle $ABC$. The circumcircle of triangle $BEF$ intersects $BC$ and $BA$ for second time at points $K$ and $M$ respectively. The circumcircle of the triangle $CEF$ intersects the $CB$ and $CA$ for the second time at points $L$ and $N$ respectively. Prove that the lines $AD, KM$ and $LN$ intersect at one point.
Kyiv City MO Juniors 2003+ geometry, 2016.9.5
On the sides $BC$ and $AB$ of the triangle $ABC$ the points ${{A} _ {1}}$ and ${{C} _ {1}} $ are selected accordingly so that the segments $A {{A} _ {1}}$ and $C {{C} _ {1}}$ are equal and perpendicular. Prove that if $\angle ABC = 45 {} ^ \circ$, then $AC = A {{A} _ {1}} $.
(Gogolev Andrew)
Cono Sur Shortlist - geometry, 2018.G3
Consider the pentagon $ABCDE$ such that $AB = AE = x$, $AC = AD = y$, $\angle BAE = 90^o$ and $\angle ACB = \angle ADE = 135^o$. It is known that $C$ and $D$ are inside the triangle $BAE$. Determine the length of $CD$ in terms of $x$ and $y$.