Found problems: 509
1955 Moscow Mathematical Olympiad, 293
Consider a quadrilateral $ABCD$ and points $K, L, M, N$ on sides $AB, BC, CD$ and $AD$, respectively, such that $KB = BL = a, MD = DN = b$ and $KL \nparallel MN$. Find the set of all the intersection points of $KL$ with $MN$ as $a$ and $b$ vary.
2020 Novosibirsk Oral Olympiad in Geometry, 5
Point $P$ is chosen inside triangle $ABC$ so that $\angle APC+\angle ABC=180^o$ and $BC=AP.$ On the side $AB$, a point $K$ is chosen such that $AK = KB + PC$. Prove that $CK \perp AB$.
Kyiv City MO Seniors 2003+ geometry, 2005.10.4
In a right triangle $ABC $ with a right angle $\angle C $, n the sides $AC$ and $AB$, the points $M$ and $N$ are selected, respectively, that $CM = MN$ and $\angle MNB = \angle CBM$. Let the point $K$ be the projection of the point $C $ on the segment $MB $. Prove that the line $NK$ passes through the midpoint of the segment $BC$.
(Alex Klurman)
2020 Ukrainian Geometry Olympiad - December, 4
In an isosceles triangle $ABC$ with an angle $\angle A= 20^o$ and base $BC=12$ point $E$ on the side $AC$ is chosen such that $\angle ABE= 30^o$ , and point $F$ on the side $AB$ such that $EF = FC$ . Find the length of $FC$.
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 $|EX| = |EY|$.
2003 Switzerland Team Selection Test, 2
In an acute-angled triangle $ABC, E$ and $F$ are the feet of the altitudes from $B$ and $C$, and $G$ and $H$ are the projections of $B$ and $C$ on $EF$, respectively. Prove that $HE = FG$.
1995 All-Russian Olympiad Regional Round, 11.7
Circles $S_1$ and $S_2$ with centers $O_1$ and $O_2$ respectively intersect at $A$ and $B$. Ray $O_1B$ meets $S_2$ again at $F$, and ray $O_2B$ meets $ S_1$ again at $E$. The line through $B$ parallel to $ EF$ intersects $S_1$ and $S_2$ again at $M$ and $N$, respectively. Prove that $MN = AE +AF$.
1992 All Soviet Union Mathematical Olympiad, 571
$ABCD$ is a parallelogram. The excircle of $ABC$ opposite $A$ has center $E$ and touches the line $AB$ at $X$. The excircle of $ADC$ opposite $A$ has center $F$ and touches the line $AD$ at $Y$. The line $FC$ meets the line$ AB$ at $W$, and the line $EC$ meets the line $AD$ at $Z$. Show that $WX = YZ$.
2004 Mexico National Olympiad, 3
Let $Z$ and $Y$ be the tangency points of the incircle of the triangle $ABC$ with the sides $AB$ and $CA$, respectively. The parallel line to $Y Z$ through the midpoint $M$ of $BC$, meets $CA$ in $N$. Let $L$ be the point in $CA$ such that $NL = AB$ (and $L$ on the same side of $N$ than $A$). The line $ML$ meets $AB$ in $K$. Prove that $KA = NC$.
2010 Contests, 1a
The point $P$ lies on the edge $AB$ of a quadrilateral $ABCD$.
The angles $BAD, ABC$ and $CPD$ are right, and $AB = BC + AD$.
Show that $BC = BP$ or $AD = BP$.
2003 Junior Balkan Team Selection Tests - Moldova, 7
The triangle $ABC$ is isosceles with $AB=BC$. The point F on the side $[BC]$ and the point $D$ on the side $AC$ are the feets of the the internals bisectors drawn from $A$ and altitude drawn from $B$ respectively so that $AF=2BD$. Fine the measure of the angle $ABC$.
2023 Denmark MO - Mohr Contest, 4
In the $9$-gon $ABCDEFGHI$, all sides have equal lengths and all angles are equal. Prove that $|AB| + |AC| = |AE|$.
[img]https://cdn.artofproblemsolving.com/attachments/6/2/8c82e8a87bf8a557baaf6ac72b3d18d2ba3965.png[/img]
Champions Tournament Seniors - geometry, 2007.3
Given a triangle $ABC$. Point $M$ moves along the side $BA$ and point $N$ moves along the side $AC$ beyond point $C$ such that $BM=CN$. Find the geometric locus of the centers of the circles circumscribed around the triangle $AMN$.
2018 Romania National Olympiad, 4
In the rectangular parallelepiped $ABCDA'B'C'D'$ we denote by $M$ the center of the face $ABB'A'$. We denote by $M_1$ and $M_2$ the projections of $M$ on the lines $B'C$ and $AD'$ respectively. Prove that:
a) $MM_1 = MM_2$
b) if $(MM_1M_2) \cap (ABC) = d$, then $d \parallel AD$;
c) $\angle (MM_1M_2), (A B C)= 45^ o \Leftrightarrow \frac{BC}{AB}=\frac{BB'}{BC}+\frac{BC}{BB'}$.
Ukraine Correspondence MO - geometry, 2018.6
Let $AD$ and $AE$ be the altitude and median of triangle $ABC$, in with $\angle B = 2\angle C$. Prove that $AB = 2DE$.
2016 Junior Balkan Team Selection Tests - Moldova, 7
Let $ABCD$ ba a square and let point $E$ be the midpoint of side $AD$. Points $G$ and $F$ are located on the segment $(BE)$ such that the lines $AG$ and $CF$ are perpendicular on the line $BE$. Prove that $DF= CG$.
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)
2002 Croatia Team Selection Test, 2
A quadrilateral $ABCD$ is circumscribed about a circle. Lines $AC$ and $DC$ meet at point $E$ and lines $DA$ and $BC$ meet at $F$, where $B$ is between $A$ and $E$ and between $C$ and $F$. Let $I_1, I_2$ and $I_3$ be the incenters of triangles $AFB, BEC$ and $ABC$, respectively. The line $I_1I_3$ intersects $EA$ at $K$ and $ED$ at $L$, whereas the line $I_2I_3$ intersects $FC$ at $M$ and $FD$ at $N$. Prove that $EK = EL$ if and only if $FM = FN$
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.
2003 Olympic Revenge, 3
Let $ABC$ be a triangle with $\angle BAC =60^\circ$. $A'$ is the symmetric point of $A$ wrt $\overline{BC}$. $D$ is the point in $\overline{AC}$ such that $\overline{AB}=\overline{AD}$. $H$ is the orthocenter of triangle $ABC$. $l$ is the external angle bisector of $\angle BAC$. $\{M\}=\overline{A'D}\cap l$,$\{N\}=\overline{CH} \cap l$. Show that $\overline{AM}=\overline{AN}$.
2017 Saudi Arabia JBMO TST, 4
Let $ABC$ be an acute, non isosceles triangle and $(O)$ be its circumcircle (with center $O$). Denote by $G$ the centroid of the triangle $ABC$, by $H$ the foot of the altitude from $A$ onto the side $BC$ and by $I$ the midpoint of $AH$. The line $IG$ intersects $BC$ at $K$.
1. Prove that $CK = BH$.
2. The ray $(GH$ intersects $(O)$ at L. Denote by $T$ the circumcenter of the triangle $BHL$. Prove that $AO$ and $BT$ intersect on the circle $(O)$.
2010 China Northern MO, 2
From a point $P$ exterior of circle $\odot O$, we draw tangents $PA$, $PB$ and the secant $PCD$ . The line passing through point $C$ parallel to $PA$ intersects chords $AB$, $AD$ at points $E$, $F$ respectively. Prove that $CE = EF$.
[img]https://cdn.artofproblemsolving.com/attachments/8/c/bf15595bc341b917df30b3aa93067887317c65.png[/img]
2014 Denmark MO - Mohr Contest, 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]
2021 Portugal MO, 2
Let $ABC$ be a triangle such that $AB = AC$. Let $D$ be a point in $[BC]$ and $E$ a point in $[AD]$ such that
$\angle BE D = \angle BAC = 2 \angle DEC$. Shows that $DB = 2CD$.
[img]https://cdn.artofproblemsolving.com/attachments/d/5/677e19d8e68a89134e17a4ab6051e41f283486.png[/img]
Durer Math Competition CD Finals - geometry, 2017.D+5
The inscribed circle of the triangle $ABC$ touches the sides $BC, CA, AB$ at points $A_1, B_1, C_1$ respectively. The points $P_b, Q_b, R_b$ are the points of the segments $BC_1, C_1A_1, A_1B$, respectively, such that $BP_bQ_bR_b$ is parallelogram. In the same way, the points $P_c, Q_c, R_c$ are the points of the sections $CB_1, B_1A_1, A_1C$, respectively such that $CP_cQ_cR_c$ is a parallelogram. The intersection of the lines $P_bR_b$ and $P_cR_c$ is $T$. Show that $TQ_b = TQ_c$.