This website contains problems from math contests. Problems and corresponding tags were obtained from the Art of Problem Solving website.

Tags were heavily modified to better represent problems.

AND:
OR:
NO:

Found problems: 1065

1999 Baltic Way, 12

In a triangle $ABC$ it is given that $2AB=AC+BC$. Prove that the incentre of $\triangle ABC$, the circumcentre of $\triangle ABC$, and the midpoints of $AC$ and $BC$ are concyclic.

2012 Greece Team Selection Test, 2

Given is an acute triangle $ABC$ $\left(AB<AC<BC\right)$,inscribed in circle $c(O,R)$.The perpendicular bisector of the angle bisector $AD$ $\left(D\in BC\right)$ intersects $c$ at $K,L$ ($K$ lies on the small arc $\overarc{AB}$).The circle $c_1(K,KA)$ intersects $c$ at $T$ and the circle $c_2(L,LA)$ intersects $c$ at $S$.Prove that $\angle{BAT}=\angle{CAS}$. [hide=Diagram][asy]import graph; size(10cm); real labelscalefactor = 0.5; /* changes label-to-point distance */ pen dps = linewidth(0.7) + fontsize(10); defaultpen(dps); /* default pen style */ pen dotstyle = black; /* point style */ real xmin = -6.94236331697463, xmax = 15.849400903703716, ymin = -5.002235438802758, ymax = 7.893104843949444; /* image dimensions */ pen aqaqaq = rgb(0.6274509803921569,0.6274509803921569,0.6274509803921569); pen uququq = rgb(0.25098039215686274,0.25098039215686274,0.25098039215686274); pen qqqqtt = rgb(0.,0.,0.2); draw((1.8318261909633622,3.572783369254345)--(0.,0.)--(6.,0.)--cycle, aqaqaq); draw(arc((1.8318261909633622,3.572783369254345),0.6426249310341638,-117.14497824050169,-101.88970202103212)--(1.8318261909633622,3.572783369254345)--cycle, qqqqtt); draw(arc((1.8318261909633622,3.572783369254345),0.6426249310341638,-55.85706977865775,-40.60179355918817)--(1.8318261909633622,3.572783369254345)--cycle, qqqqtt); /* draw figures */ draw((1.8318261909633622,3.572783369254345)--(0.,0.), uququq); draw((0.,0.)--(6.,0.), uququq); draw((6.,0.)--(1.8318261909633622,3.572783369254345), uququq); draw(circle((3.,0.7178452373968209), 3.0846882800136055)); draw((2.5345020274407277,0.)--(1.8318261909633622,3.572783369254345)); draw(circle((-0.01850947366601585,1.3533783539547308), 2.889550258039566)); draw(circle((5.553011501106743,2.4491551634556963), 3.887127532933951)); draw((-0.01850947366601585,1.3533783539547308)--(5.553011501106743,2.4491551634556963), linetype("2 2")); draw((1.8318261909633622,3.572783369254345)--(0.7798408954511686,-1.423695174396108)); draw((1.8318261909633622,3.572783369254345)--(5.22015910454883,-1.4236951743961088)); /* dots and labels */ dot((1.8318261909633622,3.572783369254345),linewidth(3.pt) + dotstyle); label("$A$", (1.5831274347452782,3.951671933606579), NE * labelscalefactor); dot((0.,0.),linewidth(3.pt) + dotstyle); label("$B$", (-0.6,0.05), NE * labelscalefactor); dot((6.,0.),linewidth(3.pt) + dotstyle); label("$C$", (6.188606107156787,0.07450151636712989), NE * labelscalefactor); dot((2.5345020274407277,0.),linewidth(3.pt) + dotstyle); label("$D$", (2.3,-0.7), NE * labelscalefactor); dot((-0.01850947366601585,1.3533783539547308),linewidth(3.pt) + dotstyle); label("$K$", (-0.3447473583572136,1.6382221818835927), NE * labelscalefactor); dot((5.553011501106743,2.4491551634556963),linewidth(3.pt) + dotstyle); label("$L$", (5.631664500260511,2.580738747400365), NE * labelscalefactor); dot((0.7798408954511686,-1.423695174396108),linewidth(3.pt) + dotstyle); label("$T$", (0.5977692071595602,-1.960477431907719), NE * labelscalefactor); dot((5.22015910454883,-1.4236951743961088),linewidth(3.pt) + dotstyle); label("$S$", (5.160406217502124,-1.8747941077698307), NE * labelscalefactor); clip((xmin,ymin)--(xmin,ymax)--(xmax,ymax)--(xmax,ymin)--cycle); /* end of picture */[/asy][/hide]

2023 Sharygin Geometry Olympiad, 10

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.

2006 Sharygin Geometry Olympiad, 12

In the triangle $ABC$, the bisector of angle $A$ is equal to the half-sum of the height and median drawn from vertex $A$. Prove that if $\angle A$ is obtuse, then $AB = AC$.

2015 Auckland Mathematical Olympiad, 4

The bisector of angle $A$ in parallelogram $ABCD$ intersects side $BC$ at $M$ and the bisector of $\angle AMC$ passes through point $D$. Find angles of the parallelogram if it is known that $\angle MDC = 45^o$. [img]https://cdn.artofproblemsolving.com/attachments/e/7/7cfb22f0c26fe39aa3da3898e181ae013a0586.png[/img]

Kvant 2019, M2553

A circle centred at $I$ is tangent to the sides $BC, CA$, and $AB$ of an acute-angled triangle $ABC$ at $A_1, B_1$, and $C_1$, respectively. Let $K$ and $L$ be the incenters of the quadrilaterals $AB_1IC_1$ and $BA_1IC_1$, respectively. Let $CH$ be an altitude of triangle $ABC$. Let the internal angle bisectors of angles $AHC$ and $BHC$ meet the lines $A_1C_1$ and $B_1C_1$ at $P$ and $Q$, respectively. Prove that $Q$ is the orthocenter of the triangle $KLP$. Kolmogorov Cup 2018, Major League, Day 3, Problem 1; A. Zaslavsky

2018 Moscow Mathematical Olympiad, 5

We have a blue triangle. In every move, we divide the blue triangle by angle bisector to $2$ triangles and color one triangle in red. Prove, that after some moves we color more than half of the original triangle in red.

2019 India PRMO, 29

In a triangle $ABC$, the median $AD$ (with $D$ on $BC$) and the angle bisector $BE$ (with $E$ on $AC$) are perpedicular to each other. If $AD = 7$ and $BE = 9$, find the integer nearest to the area of triangle $ABC$.

2012 NIMO Problems, 5

In convex hexagon $ABCDEF$, $\angle A \cong \angle B$, $\angle C \cong \angle D$, and $\angle E \cong \angle F$. Prove that the perpendicular bisectors of $\overline{AB}$, $\overline{CD}$, and $\overline{EF}$ pass through a common point. [i]Proposed by Lewis Chen[/i]

2009 Sharygin Geometry Olympiad, 3

The cirumradius and the inradius of triangle $ABC$ are equal to $R$ and $r, O, I$ are the centers of respective circles. External bisector of angle $C$ intersect $AB$ in point $P$. Point $Q$ is the projection of $P$ to line $OI$. Find distance $OQ.$ (A.Zaslavsky, A.Akopjan)

2009 Serbia Team Selection Test, 1

Let $ \alpha$ and $ \beta$ be the angles of a non-isosceles triangle $ ABC$ at points $ A$ and $ B$, respectively. Let the bisectors of these angles intersect opposing sides of the triangle in $ D$ and $ E$, respectively. Prove that the acute angle between the lines $ DE$ and $ AB$ isn't greater than $ \frac{|\alpha\minus{}\beta|}3$.

2008 Sharygin Geometry Olympiad, 10

(A.Zaslavsky, 9--10) Quadrilateral $ ABCD$ is circumscribed arounda circle with center $ I$. Prove that the projections of points $ B$ and $ D$ to the lines $ IA$ and $ IC$ lie on a single circle.

2010 China Team Selection Test, 1

Let $ABCD$ be a convex quadrilateral with $A,B,C,D$ concyclic. Assume $\angle ADC$ is acute and $\frac{AB}{BC}=\frac{DA}{CD}$. Let $\Gamma$ be a circle through $A$ and $D$, tangent to $AB$, and let $E$ be a point on $\Gamma$ and inside $ABCD$. Prove that $AE\perp EC$ if and only if $\frac{AE}{AB}-\frac{ED}{AD}=1$.

2017 Ukrainian Geometry Olympiad, 4

Let $AD$ be the inner angle bisector of the triangle $ABC$. The perpendicular on the side $BC$ at the point $D$ intersects the outer bisector of $\angle CAB$ at point $I$. The circle with center $I$ and radius $ID$ intersects the sides $AB$ and $AC$ at points $F$ and $E$ respectively. $A$-symmedian of $\Delta AFE$ intersects the circumcircle of $\Delta AFE$ again at point $X$. Prove that the circumcircles of $\Delta AFE$ and $\Delta BXC$ are tangent.

1998 AMC 12/AHSME, 28

In triangle $ ABC$, angle $ C$ is a right angle and $ CB > CA$. Point $ D$ is located on $ \overline{BC}$ so that angle $ CAD$ is twice angle $ DAB$. If $ AC/AD \equal{} 2/3$, then $ CD/BD \equal{} m/n$, where $ m$ and $ n$ are relatively prime positive integers. Find $ m \plus{} n$. $ \textbf{(A)}\ 10\qquad \textbf{(B)}\ 14\qquad \textbf{(C)}\ 18\qquad \textbf{(D)}\ 22\qquad \textbf{(E)}\ 26$

1999 National Olympiad First Round, 13

Square $ BDEC$ with center $ F$ is constructed to the out of triangle $ ABC$ such that $ \angle A \equal{} 90{}^\circ$, $ \left|AB\right| \equal{} \sqrt {12}$, $ \left|AC\right| \equal{} 2$. If $ \left[AF\right]\bigcap \left[BC\right] \equal{} \left\{G\right\}$ , then $ \left|BG\right|$ will be $\textbf{(A)}\ 6 \minus{} 2\sqrt {3} \qquad\textbf{(B)}\ 2\sqrt {3} \minus{} 1 \qquad\textbf{(C)}\ 2 \plus{} \sqrt {3} \\ \qquad\textbf{(D)}\ 4 \minus{} \sqrt {3} \qquad\textbf{(E)}\ 5 \minus{} 2\sqrt {2}$

2004 IMO Shortlist, 1

1. Let $ABC$ be an acute-angled triangle with $AB\neq AC$. The circle with diameter $BC$ intersects the sides $AB$ and $AC$ at $M$ and $N$ respectively. Denote by $O$ the midpoint of the side $BC$. The bisectors of the angles $\angle BAC$ and $\angle MON$ intersect at $R$. Prove that the circumcircles of the triangles $BMR$ and $CNR$ have a common point lying on the side $BC$.

2009 AMC 12/AHSME, 16

Trapezoid $ ABCD$ has $ AD\parallel{}BC$, $ BD \equal{} 1$, $ \angle DBA \equal{} 23^{\circ}$, and $ \angle BDC \equal{} 46^{\circ}$. The ratio $ BC: AD$ is $ 9: 5$. What is $ CD$? $ \textbf{(A)}\ \frac {7}{9}\qquad \textbf{(B)}\ \frac {4}{5}\qquad \textbf{(C)}\ \frac {13}{15} \qquad \textbf{(D)}\ \frac {8}{9}\qquad \textbf{(E)}\ \frac {14}{15}$

2006 Iran MO (3rd Round), 1

Prove that in triangle $ABC$, radical center of its excircles lies on line $GI$, which $G$ is Centroid of triangle $ABC$, and $I$ is the incenter.

2001 Rioplatense Mathematical Olympiad, Level 3, 5

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: $\angle ANC + \angle ABC = 180$

2020-21 IOQM India, 22

In triangle $ABC$, let $P$ and $R$ be the feet of the perpendiculars from $A$ onto the external and internal bisectors of $\angle ABC$, respectively; and let $Q$ and $S$ be the feet of the perpendiculars from $A$ onto the internal and external bisectors of $\angle ACB$, respectively. If $PQ = 7, QR = 6$ and $RS = 8$, what is the area of triangle $ABC$?

2020 Saint Petersburg Mathematical Olympiad, 3.

$BB_1$ is the angle bisector of $\triangle ABC$, and $I$ is its incenter. The perpendicular bisector of segment $AC$ intersects the circumcircle of $\triangle AIC$ at $D$ and $E$. Point $F$ is on the segment $B_1C$ such that $AB_1=CF$.Prove that the four points $B, D, E$ and $F$ are concyclic.

1989 China Team Selection Test, 2

$AD$ 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 $\angle{A}$.

2008 Denmark MO - Mohr Contest, 4

In triangle $ABC$ we have $AB = 2, AC = 6$ and $\angle A = 120^o$ . The bisector of angle $A$ intersects the side BC at the point $D$. Determine the length of $AD$. The answer must be given as a fraction with integer numerator and denominator.

2002 Canada National Olympiad, 4

Let $\Gamma$ be a circle with radius $r$. Let $A$ and $B$ be distinct points on $\Gamma$ such that $AB < \sqrt{3}r$. Let the circle with centre $B$ and radius $AB$ meet $\Gamma$ again at $C$. Let $P$ be the point inside $\Gamma$ such that triangle $ABP$ is equilateral. Finally, let the line $CP$ meet $\Gamma$ again at $Q$. Prove that $PQ = r$.