Found problems: 3882
2015 India IMO Training Camp, 1
Let $ABC$ be a triangle in which $CA>BC>AB$. Let $H$ be its orthocentre and $O$ its circumcentre. Let $D$ and $E$ be respectively the midpoints of the arc $AB$ not containing $C$ and arc $AC$ not containing $B$. Let $D'$ and $E'$ be respectively the reflections of $D$ in $AB$ and $E$ in $AC$. Prove that $O, H, D', E'$ lie on a circle if and only if $A, D', E'$ are collinear.
2014 Saint Petersburg Mathematical Olympiad, 2
All angles of $ABC$ are in $(30,90)$. Circumcenter of $ABC$ is $O$ and circumradius is $R$. Point $K$ is projection of $O$ to angle bisector of $\angle B$, point $M$ is midpoint $AC$. It is known, that $2KM=R$. Find $\angle B$
1960 AMC 12/AHSME, 15
Triangle I is equilateral with side $A$, perimeter $P$, area $K$, and circumradius $R$ (radius of the circumscribed circle). Triangle II is equilateral with side $a$, perimeter $p$, area $k$, and circumradius $r$. If $A$ is different from $a$, then:
$ \textbf{(A)}\ P:p = R:r \text{ } \text{only sometimes} \qquad\textbf{(B)}\ P:p = R:r \text{ } \text{always}\qquad$
$\textbf{(C)}\ P:p = K:k \text{ } \text{only sometimes} \qquad\textbf{(D)}\ R:r = K:k \text{ } \text{always}\qquad$
$\textbf{(E)}\ R:r = K:k \text{ } \text{only sometimes} $
2019 Greece Junior Math Olympiad, 2
Let $ABCD$ be a quadrilateral inscribed in circle of center $O$. The perpendicular on the midpoint $E$ of side $BC$ intersects line $AB$ at point $Z$. The circumscribed circle of the triangle $CEZ$, intersects the side $AB$ for the second time at point $H$ and line $CD$ at point $G$ different than $D$. Line $EG$ intersects line $AD$ at point $K$ and line $CH$ at point $L$. Prove that the points $A,H,L,K$ are concyclic, e.g. lie on the same circle.
2007 AIME Problems, 3
Square $ABCD$ has side length $13$, and points $E$ and $F$ are exterior to the square such that $BE=DF=5$ and $AE=CF=12$. Find $EF^{2}$.
[asy]
size(200);
defaultpen(fontsize(10));
real x=22.61986495;
pair A=(0,26), B=(26,26), C=(26,0), D=origin, E=A+24*dir(x), F=C+24*dir(180+x);
draw(B--C--F--D--C^^D--A--E--B--A, linewidth(0.7));
dot(A^^B^^C^^D^^E^^F);
pair point=(13,13);
label("$A$", A, dir(point--A));
label("$B$", B, dir(point--B));
label("$C$", C, dir(point--C));
label("$D$", D, dir(point--D));
label("$E$", E, dir(point--E));
label("$F$", F, dir(point--F));[/asy]
2000 IMO Shortlist, 5
The tangents at $B$ and $A$ to the circumcircle of an acute angled triangle $ABC$ meet the tangent at $C$ at $T$ and $U$ respectively. $AT$ meets $BC$ at $P$, and $Q$ is the midpoint of $AP$; $BU$ meets $CA$ at $R$, and $S$ is the midpoint of $BR$. Prove that $\angle ABQ=\angle BAS$. Determine, in terms of ratios of side lengths, the triangles for which this angle is a maximum.
2010 Serbia National Math Olympiad, 2
In an acute-angled triangle $ABC$, $M$ is the midpoint of side $BC$, and $D, E$ and $F$ the feet of the altitudes from $A, B$ and $C$, respectively. Let $H$ be the orthocenter of $\Delta ABC$, $S$ the midpoint of $AH$, and $G$ the intersection of $FE$ and $AH$. If $N$ is the intersection of the median $AM$ and the circumcircle of $\Delta BCH$, prove that $\angle HMA = \angle GNS$.
[i]Proposed by Marko Djikic[/i]