Found problems: 41
In a triangle $ABC$, points $H, L, K$ are chosen on the sides $AB, BC, AC$, respectively, so that $CH \perp AB$, $HL \parallel AC$ and $HK \parallel BC$. In the triangle $BHL$, let $P, Q$ be the feet of the heights from the vertices $B$ and $H$. In the triangle $AKH$, let $R, S$ be the feet of the heights from the vertices $A$ and $H$. Show that the four points $P, Q, R, S$ are collinear.
Given a nonequilateral triangle $ABC$, the vertices listed counterclockwise, find the locus of the centroids of the equilateral triangles $A'B'C'$ (the vertices listed counterclockwise) for which the triples of points $A,B', C'; A',B, C';$ and $A',B', C$ are collinear.
[i]Proposed by Poland.[/i]
Let $ABCD$ be a cyclic quadrilateral inscirbed in circle $O$. Let the tangent to $O$ at $A$ meet $BC$ at $S$, and the tangent to $O$ at $B$ meet $CD$ at $T$. Circle with $S$ as its center and passing $A$ meets $BC$ at $E$, and $AE$ meets $O$ again at $F(\ne A)$. The circle with $T$ as its center and passing $B$ meets $CD$ at $K$. Let $P = BK \cap AC$. Prove that $P,F,D$ are collinear if and only if $AB = AP$.
Let $ABC$ be a triangle such that $AB\ne AC$ and $\omega$ be the circumcircle of this triangle.
Let $I$ be the center of the inscribed circle of $ABC$ which touches $BC$ at $D$.
Let the circle with diameter $AI$ meets $\omega$ again at $K$.
If the line $AI$ intersects $\omega$ again at $M$, show that $K, D, M$ are collinear.
A triangle $ABC$ is given. The point $K$ is the base of the external bisector of angle $A$. The point $M$ is the midpoint of the arc $AC$ of the circumcircle. The point $N$ on the bisector of angle $C$ is such that $AN \parallel BM$. Prove that the points $M,N,K$ are collinear.
[i](Proposed by Ilya Bogdanov)[/i]
In triangle $ABC$ $AA_0$ and $BB_0$ are medians, $AA_1$ and $BB_1$ are altitudes. The circumcircles of triangles $CA_0B_0$ and $CA_1B_1$ meet again in point $M_c$. Points $M_a, M_b$ are defined similarly. Prove that points $M_a, M_b, M_c$ are collinear and lines $AM_a, BM_b, CM_c$ are parallel.
Given a cyclic quadrilateral $ABCD$, let it's diagonals intersect at the point $O$. Take the midpoints of $AD$ and $BC$ as $M$ and $N$ respectively. Take a point $S$ on the arc $AB$ not containing $C$ or $D$ such that $$\angle SMA=\angle SNB$$ Prove that if the diagonals of the quadrilateral made from the lines $SM$, $SN$, $AB$, and $CD$ intersect at the point $T$, then $S$, $O$, and $T$ are collinear.
Suppose $ABCD$ and $AEFG$ are rectangles such that the points $B,E,D,G$ are collinear (in this order). Let the lines $BC$ and $GF$ intersect at point $T$ and let the lines $DC$ and $EF$ intersect at point $H$. Prove that points $A, H$ and $T$ are collinear.
Let $ABC$ be an acute triangle such that $AB < AC$. Let $\omega$ be the circumcircle of $ABC$
and assume that the tangent to $\omega$ at $A$ intersects the line $BC$ at $D$. Let $\Omega$ be the circle with
center $D$ and radius $AD$. Denote by $E$ the second intersection point of $\omega$ and $\Omega$. Let $M$ be the
midpoint of $BC$. If the line $BE$ meets $\Omega$ again at $X$, and the line $CX$ meets $\Omega$ for the second
time at $Y$, show that $A, Y$, and $M$ are collinear.
[i]Proposed by Nikola Velov, North Macedonia[/i]
An acute $\triangle ABC$ with circumcircle $\Gamma$ is given. $I$ and $I_a$ are the incenter and $A-$excenter of $\triangle ABC$ respectively. The line $AI$ intersects $\Gamma$ again at $D$ and $A'$ is the antipode of $A$ with respect to $\Gamma$.
$X$ and $Y$ are point on $\Gamma$ such that $\angle IXD = \angle I_aYD = 90^\circ$. The tangents to $\Gamma$ at $X$ and $Y$ intersect in point $Z$. Prove that $A', D$ and $Z$ are collinear.
[i]Proposed by Nikola Velov[/i]
Let $ A_1A_2A_3$ be a non-isosceles triangle with incenter $ I.$ Let $ C_i,$ $ i \equal{} 1, 2, 3,$ be the smaller circle through $ I$ tangent to $ A_iA_{i\plus{}1}$ and $ A_iA_{i\plus{}2}$ (the addition of indices being mod 3). Let $ B_i, i \equal{} 1, 2, 3,$ be the second point of intersection of $ C_{i\plus{}1}$ and $ C_{i\plus{}2}.$ Prove that the circumcentres of the triangles $ A_1 B_1I,A_2B_2I,A_3B_3I$ are collinear.
In a trapezoid $ABCD$ with $AB<CD$ and $AB \parallel CD$, the diagonals intersect each other at $E$. Let $F$ be the midpoint of the arc $BC$ (not containing the point $E$) of the circumcircle of the triangle $EBC$. The lines $EF$ and $BC$ intersect at $G$. The circumcircle of the triangle $BFD$ intersects the ray $[DA$ at $H$ such that $A \in [HD]$. The circumcircle of the triangle $AHB$ intersects the lines $AC$ and $BD$ at $M$ and $N$, respectively. $BM$ intersects $GH$ at $P$, $GN$ intersects $AC$ at $Q$. Prove that the points $P, Q, D$ are collinear.
The circle inscribed in a triangle $ABC$ touches the sides $BC,CA,AB$ in $D,E, F$, respectively, and $X, Y,Z$ are the midpoints of $EF, FD,DE$, respectively. Prove that the centers of the inscribed circle and of the circles around $XYZ$ and $ABC$ are collinear.
Let $\triangle ABC$ be a triangle with circumcenter $O$. The perpendicular bisectors of the segments $OA,OB$ and $OC$ intersect the lines $BC,CA$ and $AB$ at $D,E$ and $F$, respectively. Prove that $D,E,F$ are collinear.
A bicentric quadrilateral is one that is both inscribable in and circumscribable about a circle, i.e. both the incircle and circumcircle exists. Show that for such a quadrilateral, the centers of the two associated circles are collinear with the point of intersection of the diagonals.
The circumcenter of an acute-triangle $ABC$ with $|AB|<|BC|$ is $O$, $D$ and $E$ are midpoints of $|AB|$ and $|AC|$, respectively. $OE$ intersects $BC$ at $K$, the circumcircle of $OKB$ intersects $OD$ second time at $L$. $F$ is the foot of altitude from $A$ to line $KL$. Show that the point $F$ lies on the line $DE$
An ellipse $\Gamma_1$ with foci at the midpoints of sides $AB$ and $AC$ of a triangle $ABC$ passes through $A$, and an ellipse $\Gamma_2$ with foci at the midpoints of $AC$ and $BC$ passes through $C$. Prove that the common points of these ellipses and the orthocenter of triangle $ABC$ are collinear.
Let $\Gamma$ be the circumcircle of a triangle $ABC$ with $AB <BC$, and let $M$ be the midpoint from the side $AC$ . The median of side $AC$ cuts $\Gamma$ at points $X$ and $Y$ ($X$ in the arc $ABC$). The circumcircle of the triangle $BMY$ cuts the line $AB$ at $P$ ($P \ne B$) and the line $BC$ in $Q$ ($Q \ne B$). The circumcircles of the triangles $PBC$ and $QBA$ are cut in $R$ ($R \ne B$). Prove that points $X, B$ and $R$ are collinear.
Given trapezoid $ABCD$ with bases $AB \parallel CD$ ($AB < CD$). Let $O$ be the intersection of $AC$ and $BD$. Two straight lines from $D$ and $C$ are perpendicular to $AC$ and $BD$ intersect at $E$ , i.e. $CE \perp BD$ and $DE \perp AC$ . By analogy, $AF \perp BD$ and $BF \perp AC$ . Are three points $E , O, F$ located on the same line?
One of the excircles of triangle $ABC$ is tangent to the side $AB$ and to the extensions of sides $CA$ and $CB$ at points $C_1$, $B_1$ and $A_1$ respectively. Another excircle is tangent to side $AB$ and to the extensions of sides $BA$ and $BC$ at points $B_2$, $C_2$ and $A_2$ respectively. Line $A_1B_1$ and $A_2B_2$ intersect at point $P$,. lines $A_1C_1$ and $A_2C_2$ intersect at point $Q$. Prove that the points $A$, $P$, $Q$ are collinear
[I]Proposed by S. Berlov[/i]
In an acute triangle $\triangle ABC$, let there be point $D$ on segment $AC, E$ on segment $AB$ such that $\angle ADE = \angle ABC$. Let the bisector of $\angle A$ hit $BC$ at $K$. Let the foot of the perpendicular from $K$ to $DE$ be $P$, and the foot of the perpendicular from $A$ to $DE$ be $L$. Let $Q$ be the midpoint of $AL$. If the incenter of $\triangle ABC$ lies on the circumcircle of $\triangle ADE$, prove that $P,Q$ and the incenter of $\triangle ADE$ are collinear.
In an acute non-isosceles triangle $ABC$, the inscribed circle touches side $BC$ at point $T, Q$ is the midpoint of altitude $AK$, $P$ is the orthocenter of the triangle formed by the bisectors of angles $B$ and $C$ and line $AK$. Prove that the points $P, Q$ and $T$ lie on the same line.
(D. Prokopenko)
In a triangle $ABC$ with $AB < BC < AC$, points $A' ,B' ,C'$ are such that $AA' \perp BC$ and $AA' = BC, BB' \perp CA$ and $BB'=CA$, and $CC' \perp AB$ and $CC'= AB$, as shown on the picture. Suppose that $\angle AC'B$ is a right angle. Prove that the points $A',B' ,C' $ are collinear.
Given a nonequilateral triangle $ABC$, the vertices listed counterclockwise, find the locus of the centroids of the equilateral triangles $A'B'C'$ (the vertices listed counterclockwise) for which the triples of points $A,B', C'; A',B, C';$ and $A',B', C$ are collinear.
[i]Proposed by Poland.[/i]
A convex quadrilateral $ABCD$ is such that $\angle B = \angle D$ and are both acute angles. $E$ is
on $AB$ such that $CB = CE$ and $F$ is on $AD$ such that $CF = CD$. If the circumcenter of $CEF$ is
$O_1$ and the circumcenter of $ABD$ is $O_2$. Prove that $C,O_1,O_2$ are collinear.
[i]Proposed by Kapil Pause[/i]