Found problems: 1065
The side lengths of $\triangle ABC$ are integers with no common factor greater than $1$. Given that $\angle B = 2 \angle C$ and $AB < 600$, compute the sum of all possible values of $AB$.
[i]Proposed by Eugene Chen[/i]
In acute $\triangle ABC$ with circumcircle $\Gamma$ and incentre $I$, the incircle touches side $AB$ at $F$. The external angle bisector of $\angle ACB$ meets ray $AB$ at $L$. Point $K$ lies on the arc $CB$ of $\Gamma$ not containing $A$, such that $\angle CKI=\angle IKL$. Ray $KI$ meets $\Gamma$ again at $D\ne K$. Prove that $\angle ACF =\angle DCB$.
Let $BE$ and $CF$ be the bisectors of $\angle B$ and $\angle C$ of a triangle $ABC$ whose incentre is $I$. Suppose $EF$, extended, meets the circumcircle of $ABC$ in $M,N$. Show that the circumradius of $MIN$ is twice that of $ABC$.
$BB_1$ 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.
Let $ ABC$ be a triangle, and let the angle bisectors of its angles $ CAB$ and $ ABC$ meet the sides $ BC$ and $ CA$ at the points $ D$ and $ F$, respectively. The lines $ AD$ and $ BF$ meet the line through the point $ C$ parallel to $ AB$ at the points $ E$ and $ G$ respectively, and we have $ FG \equal{} DE$. Prove that $ CA \equal{} CB$.
[i]Original formulation:[/i]
Let $ ABC$ be a triangle and $ L$ the line through $ C$ parallel to the side $ AB.$ Let the internal bisector of the angle at $ A$ meet the side $ BC$ at $ D$ and the line $ L$ at $ E$ and let the internal bisector of the angle at $ B$ meet the side $ AC$ at $ F$ and the line $ L$ at $ G.$ If $ GF \equal{} DE,$ prove that $ AC \equal{} BC.$
Prove that in a triangle the following inequality holds: $$s\sqrt3 \ge \ell_a + \ell_b + \ell_c$$ where $\ell_a$ is the length of the angle bisector from angle $A$, and $s$ is the semiperimeter of the triangle
In $\triangle ABC$, $AB<AC$. $D$ and $P$ are the feet of the internal and external angle bisectors of $\angle BAC$, respectively. $M$ is the midpoint of segment $BC$, and $\omega$ is the circumcircle of $\triangle APD$. Suppose $Q$ is on the minor arc $AD$ of $\omega$ such that $MQ$ is tangent to $\omega$. $QB$ meets $\omega$ again at $R$, and the line through $R$ perpendicular to $BC$ meets $PQ$ at $S$. Prove $SD$ is tangent to the circumcircle of $\triangle QDM$.
[i]Proposed by Ray Li[/i]
Let $\Gamma_1$ and $\Gamma_2$ be two circles that tangents each other at point $N$, with $\Gamma_2$ located inside $\Gamma_1$. Let $A, B, C$ be distinct points on $\Gamma_1$ such that $AB$ and $AC$ tangents $\Gamma_2$ at $D$ and $E$, respectively. Line $ND$ cuts $\Gamma_1$ again at $K$, and line $CK$ intersects line $DE$ at $I$.
(i) Prove that $CK$ is the angle bisector of $\angle ACB$.
(ii) Prove that $IECN$ and $IBDN$ are cyclic quadrilaterals.
In a triangle $ABC$ with a right angle at $C$, the angle bisector $AL$ (where $L$ is on segment $BC$) intersects the altitude $CH$ at point $K$. The bisector of angle $BCH$ intersects segment $AB$ at point $M$. Prove that $CK=ML$
Let $ M,N $ be points on the segments $ AB,AC, $ respectively, of the triangle $ ABC. $ Also, let $ P,Q, $ be the midpoints of the segments $ MN,BC, $ respectively. Knowing that $ PQ $ is parallel to the bisector of $ \angle BAC , $ show that $
BM=CN. $
[i]Gheorghe Duță[/i]
In triangle $ABC$ the bisector of the angle at $B$ meets $AC$ at $D$ and the bisector of the angle at $C$ meets $AB$ at $E$. These bisectors intersect at $O$ and the lengths of $OD$ and $OE$ are equal. Prove that either $\angle BAC = 60^o$ or triangle $ABC$ is isosceles.
Let $\triangle ABC$ be a triangle such that $\angle C$ and $\angle B$ are acute. Let $D$ be a variable point on $BC$ such that $D\neq B, C$ and $AD$ is not perpendicular to $BC.$ Let $d$ be the line passing through $D$ and perpendicular to $BC.$ Assume $d \cap AB= E, d \cap AC =F.$ If $M, N, P$ are the incentres of $\triangle AEF, \triangle BDE,\triangle CDF.$ Prove that $A, M, N, P$ are concyclic if and only if $d$ passes through the incentre of $\triangle ABC.$
In a triangle $ABC$, let $D$ be the midpoint of the side $BC$, and let $E$ be a point on the side $AC$. The lines $BE$ and $AD$ meet at a point $F$.
Prove: If $\frac{BF}{FE}=\frac{BC}{AB}+1$, then the line $BE$ bisects the angle $ABC$.
Angle bisectors $AA', BB'$and $CC'$ are drawn in triangle $ABC$ with angle $\angle B= 120^o$. Find $\angle A'B'C'$.
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 $\angle{BAT}=\angle{CAS}$.
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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 $|AF| = |EC|$.
Inside triangle $ABC$, the point $P$ is chosen such that $\angle PAB = \angle PCB =\frac14 (\angle A+ \angle C)$. Let $BL$ be the bisector of $\vartriangle ABC$. Line $PL$ intersects the circumcircle of $\vartriangle APC$ at point $Q$. Prove that the line $QB$ is the bisector of $\angle AQC$.
Angle bisectors $AA', BB'$and $CC'$ are drawn in triangle $ABC$ with angle $\angle B= 120^o$. Find $\angle A'B'C'$.
Let $ ABC$ be a triangle and let $ P$ be a point on the angle bisector $ AD$, with $ D$ on $ BC$. Let $ E$, $ F$ and $ G$ be the intersections of $ AP$, $ BP$ and $ CP$ with the circumcircle of the triangle, respectively. Let $ H$ be the intersection of $ EF$ and $ AC$, and let $ I$ be the intersection of $ EG$ and $ AB$. Determine the geometric place of the intersection of $ BH$ and $ CI$ when $ P$ varies.
Let triangle $ABC$ be an isosceles triangle with $AB = AC$. Suppose that the angle bisector of its angle $\angle B$ meets the side $AC$ at a point $D$ and that $BC = BD+AD$.
Determine $\angle A$.
Two straight lines are given on a plane, intersecting at point $O$ at an angle $a$. Let $A$, $B$ and $C $ be three points on one of the lines, located on one side of$ O$ and following in the indicated order, $M$ be an arbitrary point on another line, different from $O$, Let $\angle AMB=\gamma$, $\angle BMC = \phi$. Consider the function $F(M) = ctg \gamma + ctg \phi$ . Prove that$ F(M)$ takes the smallest value on each of the rays into which $O$ divides the second straight line. (Each has its own.) Let us denote one of these smallest values by $q$, and the other by $p$. Prove that the exprseeion $\frac{p}{q}$ is independent of choice of points $A$, $B$ and $C$. Express this relationship in terms of $a$.
Consider a triangle $ABC$ and a point $D$ on its side $\overline{AB}$. Let $I$ be a point inside $\triangle ABC$ on the angle bisector of $ACB$. The second intersections of lines $AI$ and $CI$ with circle $ACD$ are $P$ and $Q$, respectively. Similarly, the second intersection of lines $BI$ and $CI$ with circle $BCD$ are $R$ and $S$, respectively. Show that if $P\neq Q$ and $R\neq S$, then lines $AB$, $PQ$ and $RS$ pass through a point or are parallel.
Point $A$ lies inside an angle with vertex $M$. A ray issuing from point $A$ is reflected in one side of the angle at point $B$, then in the other side at point $C$ and then returns back to point $A$ (the ordinary rule of reflection holds). Prove that the center of the circle circumscribed about triangle $\triangle BCM$ lies on line $AM$.
Let $ABC$ be a triangle with $\angle C = 90^\circ$ and $CA \neq CB$. Let $CH$ be an altitude and $CL$ be an interior angle bisector. Show that for $X \neq C$ on the line $CL$, we have $\angle XAC \neq \angle XBC$. Also show that for $Y \neq C$ on the line $CH$ we have $\angle YAC \neq \angle YBC$.
[i]Bulgaria[/i]
In the triangle $ABC$ we have $|AB|<|AC|$. The bisectors of the angles at $B$ and $C$ meet $AC$ and $AB$ at $D$ and $E$ respectively. $BD$ and $CE$ intersect at the incenter $I$ of $\triangle ABC$.
Prove that $\angle BAC=60^\circ$ if and only if $|IE|=|ID|$