Found problems: 410
Consider points $D,E$ and $F$ on sides $BC,AC$ and $AB$, respectively, of a triangle $ABC$, such that $AD, BE$ and $CF$ concurr at a point $G$. The parallel through $G$ to $BC$ cuts $DF$ and $DE$ at $H$ and $I$, respectively. Show that triangles $AHG$ and $AIG$ have the same areas.
Let $A_1$ be the center of the square inscribed in acute triangle $ABC$ with two vertices of the square on side $BC$. Thus one of the two remaining vertices of the square is on side $AB$ and the other is on $AC$. Points $B_1,\ C_1$ are defined in a similar way for inscribed squares with two vertices on sides $AC$ and $AB$, respectively. Prove that lines $AA_1,\ BB_1,\ CC_1$ are concurrent.
Let $ABC$ be a triangle and let $\omega$ be its incircle. Denote by $D_1$ and $E_1$ the points where $\omega$ is tangent to sides $BC$ and $AC$, respectively. Denote by $D_2$ and $E_2$ the points on sides $BC$ and $AC$, respectively, such that $CD_2=BD_1$ and $CE_2=AE_1$, and denote by $P$ the point of intersection of segments $AD_2$ and $BE_2$. Circle $\omega$ intersects segment $AD_2$ at two points, the closer of which to the vertex $A$ is denoted by $Q$. Prove that $AQ=D_2P$.
Let $P$ be a point inside an acute triangle $ABC$. Let the lines $BP$ and $CP$ intersect the sides $AC$ and $AB$ at $D$ and $E$, respectively. Let the circles with diameters $BD$ and $CE$ intersect at points $S$ and $T$. Prove that if the points $A$, $S$, and $T$ are colinear, then $P$ lies on a median of $\triangle ABC$.
Given a cyclic quadrilateral $ABCD$, let the diagonals $AC$ and $BD$ meet at $E$ and the lines $AD$ and $BC$ meet at $F$. The midpoints of $AB$ and $CD$ are $G$ and $H$, respectively. Show that $EF$ is tangent at $E$ to the circle through the points $E$, $G$ and $H$.
[i]Proposed by David Monk, United Kingdom[/i]
Let $ABC$ be a triangle and $X,Y,Z$ be the tangency points of its inscribed circle with the sides $BC, CA, AB$, respectively. Suppose that $C_1, C_2, C_3$ are circle with chords $YZ, ZX, XY$, respectively, such that $C_1$ and $C_2$ intersect on the line $CZ$ and that $C_1$ and $C_3$ intersect on the line $BY$. Suppose that $C_1$ intersects the chords $XY$ and $ZX$ at $J$ and $M$, respectively; that $C_2$ intersects the chords $YZ$ and $XY$ at $L$ and $I$, respectively; and that $C_3$ intersects the chords $YZ$ and $ZX$ at $K$ and $N$, respectively. Show that $I, J, K, L, M, N$ lie on the same circle.
.Let $CH$ be an altitude of right-angled triangle $ABC$ ($\angle C = 90^o$), $HA_1$, $HB_1$ be the bisectors of angles $CHB$, $AHC$ respectively, and $E, F$ be the midpoints of $HB_1$ and $HA_1$ respectively. Prove that the lines $AE$ and $BF$ meet on the bisector of angle $ACB$.
The diagonals of a trapezoid $ ABCD$ intersect at point $ P$. Point $ Q$ lies between the parallel lines $ BC$ and $ AD$ such that $ \angle AQD \equal{} \angle CQB$, and line $ CD$ separates points $ P$ and $ Q$. Prove that $ \angle BQP \equal{} \angle DAQ$.
[i]Author: Vyacheslav Yasinskiy, Ukraine[/i]
$P$ is a point in the interior of acute triangle $ABC$. $D,E,F$ are the reflections of $P$ across $BC,CA,AB$ respectively. Rays $AP,BP,CP$ meet the circumcircle of $\triangle ABC$ at $L,M,N$ respectively. Prove that the circumcircles of $\triangle PDL,\triangle PEM,\triangle PFN$ meet at a point $T$ different from $P$.
Let $ABC$ be a triangle with $AB=AC$ and let $D$ be the midpoint of $AC$. The angle bisector of $\angle BAC$ intersects the circle through $D,B$ and $C$ at the point $E$ inside the triangle $ABC$. The line $BD$ intersects the circle through $A,E$ and $B$ in two points $B$ and $F$. The lines $AF$ and $BE$ meet at a point $I$, and the lines $CI$ and $BD$ meet at a point $K$. Show that $I$ is the incentre of triangle $KAB$.
[i]Proposed by Jan Vonk, Belgium and Hojoo Lee, South Korea[/i]
In a triangle $ABC$, $\angle A$ is $60^\circ$. On sides $AB$ and $AC$ we make two equilateral triangles (outside the triangle $ABC$) $ABK$ and $ACL$. $CK$ and $AB$ intersect at $S$ , $AC$ and $BL$ intersect at $R$ , $BL$ and $CK$ intersect at $T$. Prove the radical centre of circumcircle of triangles $BSK, CLR$ and $BTC$ is on the median of vertex $A$ in triangle $ABC$.
[i]Proposed by Ali Zamani[/i]
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 $D$ be a point inside $\triangle ABC$ such that $m(\widehat{BAD})=20^{\circ}$, $m(\widehat{DAC})=80^{\circ}$, $m(\widehat{ACD})=20^{\circ}$, and $m(\widehat{DCB})=20^{\circ}$.
$m(\widehat{ABD})= ?$
$ \textbf{(A)}\ 5^{\circ} \qquad \textbf{(B)}\ 10^{\circ} \qquad \textbf{(C)}\ 15^{\circ} \qquad \textbf{(D)}\ 20^{\circ} \qquad \textbf{(E)}\ 25^{\circ}$
There is given a convex quadrilateral $ ABCD$. Prove that there exists a point $ P$ inside the quadrilateral such that
\[
\angle PAB \plus{} \angle PDC \equal{} \angle PBC \plus{} \angle PAD \equal{} \angle PCD \plus{} \angle PBA \equal{} \angle PDA \plus{} \angle PCB = 90^{\circ}
\]
if and only if the diagonals $ AC$ and $ BD$ are perpendicular.
[i]Proposed by Dusan Djukic, Serbia[/i]
The incircle of triangle $ABC$ is tangent to sides $BC$, $AC$, and $AB$ at points $D$, $E$ and $F$, respectively. Let $A'$ denote the point of the incircle for which circle $(A'BC)$ is tangent to the incircle. Define points $B'$ and $C'$ similarly. Prove that lines $A'D$, $BE'$, and $CF'$ are concurrent.
[i]Proposed by Áron Bán-Szabó, Budapest[/i]
Let $ ABC$ be an acute triangle with orthocenter $ H$ and let $ X$ be an arbitrary point in its plane. The circle with diameter $ HX$ intersects the lines $ AH$ and $ AX$ at $ A_{1}$ and $ A_{2}$, respectively. Similarly, define $ B_{1}$, $ B_{2}$, $ C_{1}$, $ C_{2}$. Prove that the lines $ A_{1}A_{2}$, $ B_{1}B_{2}$, $ C_{1}C_{2}$ are concurrent.
[hide][i]Remark[/i]. The triangle obviously doesn't need to be acute.[/hide]
Let $ABCD$ be a parallelogram. The circle with diameter $AC$ intersects the line $BD$ at points $P$ and $Q$. The perpendicular to the line $AC$ passing through the point $C$ intersects the lines $AB$ and $AD$ at points $X$ and $Y$, respectively. Prove that the points $P,Q,X$ and $Y$ lie on the same circle.
Given a triangle $ABC$ satisfying $AC+BC=3\cdot AB$. The incircle of triangle $ABC$ has center $I$ and touches the sides $BC$ and $CA$ at the points $D$ and $E$, respectively. Let $K$ and $L$ be the reflections of the points $D$ and $E$ with respect to $I$. Prove that the points $A$, $B$, $K$, $L$ lie on one circle.
[i]Proposed by Dimitris Kontogiannis, Greece[/i]
Let $ABC$ be a triangle, and $M$ an interior point such that $\angle MAB=10^\circ$, $\angle MBA=20^\circ$, $\angle MAC=40^\circ$ and $\angle MCA=30^\circ$. Prove that the triangle is isosceles.
Let $ABC$ be a triangle with $AB=AC$ and let $D$ be the midpoint of $AC$. The angle bisector of $\angle BAC$ intersects the circle through $D,B$ and $C$ at the point $E$ inside the triangle $ABC$. The line $BD$ intersects the circle through $A,E$ and $B$ in two points $B$ and $F$. The lines $AF$ and $BE$ meet at a point $I$, and the lines $CI$ and $BD$ meet at a point $K$. Show that $I$ is the incentre of triangle $KAB$.
[i]Proposed by Jan Vonk, Belgium and Hojoo Lee, South Korea[/i]
Point $D$ is chosen on the Euler line of triangle $ABC$ and it is inside of the triangle. Points $E,F$ are were the line $BD,CD$ intersect with $AC,AB$ respectively. Point $X$ is on the line $AD$ such that $\angle EXF =180 - \angle A$, also $A,X$ are on the same side of $EF$. If $P$ is the second intersection of circumcircles of $CXF,BXE$ then prove the lines $XP,EF$ meet on the altitude of $A$
Proposed by [i]Alireza Danaie[/i]
Let $ABC$ be an acute triangle with incircle $\omega$, incenter $I$, and $A$-excircle $\omega_{a}$. Let $\omega$ and $\omega_{a}$ meet $BC$ at $X$ and $Y$, respectively. Let $Z$ be the intersection point of $AY$ and $\omega$ which is closer to $A$. The point $H$ is the foot of the altitude from $A$. Show that $HZ$, $IY$ and $AX$ are concurrent.
[i]Proposed by Nikola Velov[/i]
Let $ABC$ be an acute angled triangle satisfying the conditions $AB>BC$ and $AC>BC$. Denote by $O$ and $H$ the circumcentre and orthocentre, respectively, of the triangle $ABC.$ Suppose that the circumcircle of the triangle $AHC$ intersects the line $AB$ at $M$ different from $A$, and the circumcircle of the triangle $AHB$ intersects the line $AC$ at $N$ different from $A.$ Prove that the circumcentre of the triangle $MNH$ lies on the line $OH$.
In triangle ABC, let $P$ and $Q$ be two interior points such that $\angle ABP = \angle QBC$ and $\angle ACP = \angle QCB$. Point $D$ lies on segment $BC$. Prove that $\angle APB + \angle DPC = 180^\circ$ if and only if $\angle AQC + \angle DQB = 180^\circ$.
There is given a convex quadrilateral $ ABCD$. Prove that there exists a point $ P$ inside the quadrilateral such that
\[
\angle PAB \plus{} \angle PDC \equal{} \angle PBC \plus{} \angle PAD \equal{} \angle PCD \plus{} \angle PBA \equal{} \angle PDA \plus{} \angle PCB = 90^{\circ}
\]
if and only if the diagonals $ AC$ and $ BD$ are perpendicular.
[i]Proposed by Dusan Djukic, Serbia[/i]