Found problems: 254
Two circles that are not equal are tangent externally at point $R$. Suppose point $P$ is the intersection of the external common tangents of the two circles. Let $A$ and $B$ are two points on different circles so that $RA$ is perpendicular to $RB$. Show that the line $AB$ passes through $P$.
A triangle $ABC$ is inscribed in circle $(O)$. $P1, P2$ are two points in the plane of the triangle. $P_1A, P_1B, P_1C$ meet $(O)$ again at $A_1,B_1,C_1$ . $P_2A, P_2B, P_2C$ meet $(O)$ again at $A_2,B_2,C_2$.
a) $A_1A_2, B_1B_2, C_1C_2$ intersect $BC,CA,AB$ at $A_3,B_3,C_3$. Prove that three points $A_3,B_3,C_3$ are collinear.
b) $P$ is a point on the line $P_1P_2. A_1P,B_1P,C_1P$ meet (O) again at $A_4,B_4,C_4$. Prove that three lines $A_2A_4,B_2B_4,C_2C_4$ are concurrent.
Trần Quang Hùng
Let $ABC$ be an acute triangle. Suppose that circle $\Gamma_1$ has it's center on the side $AC$ and is tangent to the sides $AB$ and $BC$, and circle $\Gamma_2$ has it's center on the side $AB$ and is tangent to the sides $AC$ and $BC$. The circles $\Gamma_1$ and $ \Gamma_2$ intersect at two points $P$ and $Q$. Show that if $A, P, Q$ are collinear, then $AB = AC$.
A sphere $S$ lies within tetrahedron $ABCD$, touching faces $ABD, ACD$, and $BCD$, but having no point in common with plane $ABC$. Let $E$ be the point in the interior of the tetrahedron for which $S$ touches planes $ABE$, $ACE$, and $BCE$ as well. Suppose the line $DE$ meets face $ABC$ at $F$, and let $L$ be the point of $S$ nearest to plane $ABC$. Show that segment $FL$ passes through the centre of the inscribed sphere of tetrahedron $ABCE$.
KöMaL A.723. (April 2018), G. Kós
Let $\triangle ABC$ be a triangle and $H$ its orthocenter. We draw the circumference $\mathcal{C}_1$ that passes through $H$ and its tangent to $BC$ at $B$ and the circumference $\mathcal{C}_2$ that passes through $H$ and its tangent to $BC$ at $C$. If $\mathcal{C}_1$ cuts line $AB$ again at $X$ and $\mathcal{C}_2$ cuts line $AC$ again at $Y$. Prove that $X,Y$ and $H$ are collinear.
Let $ABC$ be a triangle such that $AB < AC$. Let the excircle opposite to A be tangent to the lines $AB, AC$, and $BC$ at points $D, E$, and $F$, respectively, and let $J$ be its centre. Let $P$ be a point on the side $BC$. The circumcircles of the triangles $BDP$ and $CEP$ intersect for the second time at $Q$. Let $R$ be the foot of the perpendicular from $A$ to the line $FJ$. Prove that the points $P, Q$, and $R$ are collinear.
(The [i]excircle[/i] of a triangle $ABC$ opposite to $A$ is the circle that is tangent to the line segment $BC$, to the ray $AB$ beyond $B$, and to the ray $AC$ beyond $C$.)
[i]Proposed by Bozhidar Dimitrov, Bulgaria[/i]
The $ AB, BC $ and $ CD $ segments of the polygon $ ABCD $ have the same length and are tangent to a circle $ S $, centered on the point $ O $. Let $ P $ be the point of tangency of $ BC $ with $ S $, and let $ Q $ be the intersection point of lines $ AC $ and $ BD $. Show that the point $ Q $ is collinear with the points $ P $ and $ O $.
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.
Points $A$, $B$ and $C$ lie on one side of the angle with the vertex at point $O$, and points $A'$, $B'$ and $C'$ lie on the other. It is known that$ B$ is the midpoint of the segment $AC$, $B'$ is the midpoint of the segment $A'C'$, and lines $AA'$, $BB'$ and $CC'$ are parallel (fig.). Prove that the centers of the circles circumscribed around the triangles $OAC$, $OA'C$ and $OBB'$ lie on the same straight line.
[img]https://cdn.artofproblemsolving.com/attachments/d/6/92831077781bc45f25e9f71077034f84753a59.png[/img]
Let $ABC$ be a triangle with $\angle BAC> 90 ^ \circ$ and $D$ a point on $BC$. Let $E$ and $F$be the reflections of the point $D$ about $AB$ and $AC$, respectively. Suppose that $BE$ and $CF$ intersect at $P$. Show that $AP$ passes through the circumcenter of triangle $ABC$.
Let $ABC$ be an acute triangle with incenter $O$, orthocenter $H$, altitude $AD. AO$ meets $BC$ at $E$. Line through $D$ parallel to $OH$ meet $AB,AC$ at $M,N$, respectively. Let $I$ be the midpoint of $AE$, and $DI$ intersect $AB,AC$ at $P,Q$ respectively. $MQ$ meets $NP$ at $T$. Prove that $D,O, T$ are collinear.
Trần Quang Hùng
Let $AD$ be the bisector of an acute-angled triangle $ABC$. The circle circumscribed around the triangle $ABD$ intersects the straight line perpendicular to $AD$ that passes through point $B$, at point $E$. Point $O$ is the center of the circumscribed circle of triangle $ABC$. Prove that the points $A, O, E$ lie on the same line.
In triangle $ABC$ let $I_a$ be the $A$-excenter. Let $\omega$ be an arbitrary circle that passes through $A,I_a$ and intersects the extensions of sides $AB,AC$ (extended from $B,C$) at $X,Y$ respectively. Let $S,T$ be points on segments $I_aB,I_aC$ respectively such that $\angle AXI_a=\angle BTI_a$ and $\angle AYI_a=\angle CSI_a$.Lines $BT,CS$ intersect at $K$. Lines $KI_a,TS$ intersect at $Z$.
Prove that $X,Y,Z$ are collinear.
[i]Proposed by Hooman Fattahi[/i]
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)
$OA, OB, OC$ are three chords of a circle. The circles with diameters $OA, OB$ meet again at $Z$, the circles with diameters $OB, OC$ meet again at $X$, and the circles with diameters $OC, OA$ meet again at $Y$. Show that $X, Y, Z$ are collinear.
In a triangle $ABC, P$ and $P'$ are points on side $BC, Q$ on side $CA$, and $R $ on side $AB$, such that $\frac{AR}{RB}=\frac{BP}{PC}=\frac{CQ}{QA}=\frac{CP'}{P'B}$ . Let $G$ be the centroid of triangle $ABC$ and $K$ be the intersection point of $AP'$ and $RQ$. Prove that points $P,G,K$ are collinear.
Altitudes $AD$ and $BE$ of an acute triangle $ABC$ intersect at $H$. Let $P \ne E$ be the point of tangency of the circle with radius $HE$ centred at $H$ with its tangent line going through point $C$, and let $Q \ne E$ be the point of tangency of the circle with radius $BE$ centred at $B$ with its tangent line going through $C$. Prove that the points $D, P$ and $Q$ are collinear.
Let $ABCDEF$ be a cyclic hexagon, points $K, L, M, N$ be the common points of lines $AB$ and $CD$, $AC$ and $BD$, $AF$ and $DE$, $AE$ and $DF$ respectively. Prove that if three of these points are collinear then the fourth point lies on the same line.
In the convex quadrilateral of the $ABCD$, the diagonals of $AC$ and $BD$ are mutually perpendicular and intersect at point $E$. On the side of $AD$, a point $P$ is chosen such that $PE = EC$. The circumscribed circle of the triangle $BCD$ intersects the segment $AD$ at the point $Q$. The circle passing through point $A$ and tangent to the line $EP$ at point $P$ intersects the segment $AC$ at point $R$. It turns out that points $B, Q, R$ are collinear. Prove that $\angle BCD = 90^o$.
Given a square $ABCD$, point $E$ is the midpoint of $AD$. Let $F$ be the foot of the perpendicular drawn from point $B$ on $EC$. Point $K$ on $AB$ is such that $\angle DFK = 90^o$. The point $N$ on the $CE$ is such that $\angle NKB = 90^o$. Prove that the point $N$ lies on the segment $BD$.
(Matvii Kurskyi)
[img]https://cdn.artofproblemsolving.com/attachments/4/2/d42b8c8117ec1d5e5c5b981904779b156fce93.png[/img]
Let $A, B, C$ and $D$ be points on the circle $\omega$ such that $ABCD$ is a convex quadrilateral. Suppose that $AB$ and $CD$ intersect at a point $E$ such that $A$ is between $B$ and $E$ and that $BD$ and $AC$ intersect at a point $F$. Let $X \ne D$ be the point on $\omega$ such that $DX$ and $EF$ are parallel. Let $Y$ be the reflection of $D$ through $EF$ and suppose that $Y$ is inside the circle $\omega$.
Show that $A, X$, and $Y$ are collinear.
$ABCD$ has $AB = CD$, but $AB$ not parallel to $CD$, and $AD$ parallel to $BC$. The triangle is $ABC$ is rotated about $C$ to $A'B'C$. Show that the midpoints of $BC, B'C$ and $A'D$ are collinear.
On the sides $(AB)$ and $(AC)$ of the triangle $ABC$ are considered the points $M$ and $N$ respectively so that $ \angle ABC =\angle ANM$. Point $D$ is symmetric of point $A$ with respect to $B$, and $P$ and $Q$ are the midpoints of the segments $[MN]$ and $[CD]$, respectively. Prove that the points $A, P$ and $Q$ are collinear if and only if $AC = AB \sqrt {2}$
Let $ABC$ be an isosceles right triangle with hypotenuse $AB$, $D$ be some point in the plane such that $2CD = AB$ and point $C$ inside the triangle $ABD$. We construct two rays with a start in $C$, intersecting $AD$ and $BD$ and perpendicular to them. On the first one, intersecting $AD$, we will plot the segment $CK = AD$, and on the second one - $CM = BD$. Prove that points $M$, $D$ and $K$ lie on the same line.
Point $ O $ is the center of the circumscribed circle of an acute triangle $ Abc $. A certain circle passes through the points $ B $ and $ C $ and intersects sides $ AB $ and $ AC $ of a triangle. On its arc lying inside the triangle, points $ D $ and $ E $ are chosen so that the segments $ BD $ and $ CE $ pass through the point $ O $. Perpendicular $ DD_1 $ to $ AB $ side and perpendicular $ EE_1 $ to $ AC $ side intersect at $ M $. Prove that the points $ A $, $ M $ and $ O $ lie on the same straight line.