Found problems: 1065
Let be given points $A,B$ on a circle and let $P$ be a variable point on that circle. Let point $M$ be determined by $P$ as the point that is either on segment $PA$ with $AM=MP+PB$ or on segment $PB$ with $AP+MP=PB$. Find the locus of points $M$.
Let $ABC$ be a triangle and $D$ be a point on $BC$ such that $AB+BD=AC+CD$. The line $AD$ intersects the incircle of triangle $ABC$ at $X$ and $Y$ where $X$ is closer to $A$ than $Y$ i. Suppose $BC$ is tangent to the incircle at $E$, prove that:
1) $EY$ is perpendicular to $AD$;
2) $XD=2IM$ where $I$ is the incentre and $M$ is the midpoint of $BC$.
Let $ABC$ be a triangle with $AB = 16$, $AC = 10$, $BC = 18$. Let $D$ be a point on $AB$ such that $4AD = AB$ and let E be the foot of the angle bisector from $B$ onto $AC$. Let $P$ be the intersection of $CD$ and $BE$. Find the area of the quadrilateral $ADPE$.
In a convex quadrilateral $ABCD$ we have $AB\cdot CD=BC\cdot DA$ and $2\angle A+\angle C=180^\circ$. Point $P$ lies on the circumcircle of triangle $ABD$ and is the midpoint of the arc $BD$ not containing $A$. It is known that the point $P$ lies inside the quadrilateral $ABCD$. Prove that $\angle BCA=\angle DCP$
[i]Proposed by S. Berlov[/i]
In a triangle $ ABC$, the angle bisector at $ B$ intersects $ AC$ at $ D$ and the circumcircle again at $ E$. The circumcircle of the triangle $ DAE$ meets the segment $ AB$ again at $ F$. Prove that the triangles $ DBC$ and $ DBF$ are congruent.
If P is a point inside a convex quadrilateral $ABCD$, let the angle bisectors of $\angle APB$, $\angle BPC$, $\angle CPD$ and $\angle DPA$ meet $AB$, $BC$, $CD$ and $DA$ at $K$, $L$, $M$ and $N$ respectively.
(a) Find a point $P$ such that $KLMN$ is a parallelogram.
(b) Find the locus of all such points $P$.
(S Tokarev)
Let ${I}$ be the incenter of $\triangle {ABC}$ and ${BX}$, ${CY}$ are its two angle bisectors. ${M}$ is the midpoint of arc $\overset{\frown}{BAC}$. It is known that $MXIY$ are concyclic. Prove that the area of quadrilateral $MBIC$ is equal to that of pentagon $BXIYC$.
[i]Proposed by Dominik Burek - Poland[/i]
Altitudes $BE$ and $CF$ of an acute-angled triangle $ABC$ meet at point $H$. The perpendicular from $H$ to $EF$ meets the line $\ell$ passing through $A$ and parallel to $BC$ at point $P$. The bisectors of two angles between $\ell$ and $HP$ meet $BC$ at points $S$ and $T$. Prove that the circumcircles of triangles $ABC$ and $PST$ are tangent.
A line in the plane of a triangle $ABC$ intersects the sides $AB$ and $AC$ respectively at points $X$ and $Y$ such that $BX = CY$ . Find the locus of the center of the circumcircle of triangle $XAY .$
Let $ABC$ be a acute-angled triangle with centroid $G$, the angle bisector of $\angle ABC$ intersects $AC$ in $D$. Let $P$ and $Q$ be points in $BD$ where $\angle PBA = \angle PAB$ and $\angle QBC = \angle QCB$. Let $M$ be the midpoint of $QP$, let $N$ be a point in the line $GM$ such that $GN = 2GM$(where $G$ is the segment $MN$), prove that:
$\angle ANC + \angle ABC = 180$
Let $C$ be a point on the extension of the diameter $AB$ of a circle. A line through $C$ is tangent to the circle at point $N$. The bisector of $\angle ACN$ meets the lines $AN$ and $BN$ at $P$ and $Q$ respectively. Prove that $PN = QN$.
Triangle $ABC$ has lengths $AB=20$, $AC=14$, $BC=22$. The median from $B$ intersects $AC$ at $M$ and the angle bisector from $C$ intersects $AB$ at $N$ and the median from $B$ at $P$. Let $\dfrac pq=\dfrac{[AMPN]}{[ABC]}$ for positive integers $p$, $q$ coprime. Note that $[ABC]$ denotes the area of triangle $ABC$. Find $p+q$.
Let $ABCD$ be a square of side length $1$, and let $P$ be a variable point on $\overline{CD}$. Denote by $Q$ the intersection point of the angle bisector of $\angle APB$ with $\overline{AB}$. The set of possible locations for $Q$ as $P$ varies along $\overline{CD}$ is a line segment; what is the length of this segment?
Point $X$ is chosen inside the non trapezoid quadrilateral $ABCD$ such that $\angle AXD +\angle BXC=180$.
Suppose the angle bisector of $\angle ABX$ meets the $D$-altitude of triangle $ADX$ in $K$, and the angle bisector of $\angle DCX$ meets the $A$-altitude of triangle $ADX$ in $L$.We know $BK \perp CX$ and $CL \perp BX$. If the circumcenter of $ADX$ is on the line $KL$ prove that $KL \perp AD$.
Proposed by [i]Alireza Dadgarnia[/i]
$BB_{1}$ is a bisector of an acute triangle $ABC$. A perpendicular from $B_{1}$ to $BC$ meets a smaller arc $BC$ of a circumcircle of $ABC$ in a point $K$. A perpendicular from $B$ to $AK$ meets $AC$ in a point $L$. $BB_{1}$ meets arc $AC$ in $T$. Prove that $K$, $L$, $T$ are collinear.
[i]V. Astakhov[/i]
Let [i]O[/i] be the center of circle[i] W[/i]. Two equal chords [i]AB[/i] and [i]CD [/i]of[i] W [/i]intersect at [i]L [/i]such that [i]AL>LB [/i]and [i]DL>LC[/i]. Let [i]M [/i]and[i] N [/i]be points on [i]AL[/i] and [i]DL[/i] respectively such that ([i]ALC[/i])=2*([i]MON[/i]). Prove that the chord of [i]W[/i] passing through [i]M [/i]and [i]N[/i] is equal to [i]AB[/i] and [i]CD[/i].
Let $ABCD$ be a rectangle and denote by $M$ and $N$ the midpoints of $AD$ and $BC$ respectively. The point $P$ is on $(CD$ such that $D\in (CP)$, and $PM$ intersects $AC$ in $Q$. Prove that $m(\angle{MNQ})=m(\angle{MNP})$.
In $\triangle ABC$, shown on the right, let $r$ denote the radius of the inscribed circle, and let $r_A$, $r_B$, and $r_C$ denote the radii of the smaller circles tangent to the inscribed circle and to the sides emanating from $A$, $B$, and $C$, respectively. Prove that
$r \leq r_A + r_B + r_C$
Let $\vartriangle ABC$ be an isosceles triangle with $|AB| = |AC|$. Let $D, E$ and $F$ be points on line segments $BC, CA$ and $AB$, respectively, such that $|BF| = |BE|$ and such that $ED$ is the internal angle bisector of $\angle BEC$. Prove that $|BD|= |EF|$ if and only if $|AF| = |EC|$.
Let $ABCD$ be a cyclic quadrilateral. Let $P$, $Q$, $R$ be the feet of the perpendiculars from $D$ to the lines $BC$, $CA$, $AB$, respectively. Show that $PQ=QR$ if and only if the bisectors of $\angle ABC$ and $\angle ADC$ are concurrent with $AC$.
Triangle $ABC$ is isosceles, with $AB=AC$ and altitude $AM=11.$ Suppose that there is a point $D$ on $\overline{AM}$ with $AD=10$ and $\angle BDC=3\angle BAC.$ Then the perimeter of $\triangle ABC$ may be written in the form $a+\sqrt{b},$ where $a$ and $b$ are integers. Find $a+b.$
[asy] import graph; size(7cm); real lsf=0.5; pen dps=linewidth(0.7)+fontsize(10); defaultpen(dps); pen ds=black; real xmin=-1.55,xmax=7.95,ymin=-4.41,ymax=5.3; draw((1,3)--(0,0)); draw((0,0)--(2,0)); draw((2,0)--(1,3)); draw((1,3)--(1,0)); draw((1,0.7)--(0,0)); draw((1,0.7)--(2,0)); label("$11$",(0.75,1.63),SE*lsf); dot((1,3),ds); label("$A$",(0.96,3.14),NE*lsf); dot((0,0),ds); label("$B$",(-0.15,-0.18),NE*lsf); dot((2,0),ds); label("$C$",(2.06,-0.18),NE*lsf); dot((1,0),ds); label("$M$",(0.97,-0.27),NE*lsf); dot((1,0.7),ds); label("$D$",(1.05,0.77),NE*lsf); clip((xmin,ymin)--(xmin,ymax)--(xmax,ymax)--(xmax,ymin)--cycle); [/asy]
$ABC$ is a right-angled triangle, with $\angle ABC = 90^{\circ}$. $B'$ is the reflection of $B$ over $AC$. $M$ is the midpoint of $AC$. We choose $D$ on $\overrightarrow{BM}$, such that $BD = AC$. Prove that $B'C$ is the angle bisector of $\angle MB'D$.
NOTE: An important condition not mentioned in the original problem is $AB<BC$. Otherwise, $\angle MB'D$ is not defined or $B'C$ is the external bisector.
$AD$ is the altitude on side $BC$ of triangle $ABC$. If $BC+AD-AB-AC = 0$, find the range of $\angle BAC$.
[i]Alternative formulation.[/i] Let $AD$ be the altitude of triangle $ABC$ to the side $BC$. If $BC+AD=AB+AC$, then find the range of $\angle{A}$.
Let $ABC$ be a triangle, and let $D$ and $E$ be the orthogonal projections of $A$ onto the internal bisectors from $B$ and $C$. Prove that $DE$ is parallel to $BC$.
Let $X$ be an arbitrary point on side $BC$ of triangle $ABC$. Triangle $T$ is formed by the angle bisectors of the angles $\angle ABC$, $\angle ACB$ and $\angle AXC$. Prove that the circle circumscribed around the triangle $T$, passes through the vertex $A$.
(Dmytro Prokopenko)