Olympiad problems

2013 Bosnia and Herzegovina Junior BMO TST, 3

Let $M$ and $N$ be touching points of incircle with sides $AB$ and $AC$ of triangle $ABC$, and $P$ intersection point of line $MN$ and angle bisector of $\angle ABC$. Prove that $\angle BPC =90 ^{\circ}$

2017 CCA Math Bonanza, L1.3

Tags: angle bisector , geometry , analytic geometry

Triangle $ABC$ has points $A$ at $\left(0,0\right)$, $B$ at $\left(9,12\right)$, and $C$ at $\left(-6,8\right)$ in the coordinate plane. Find the length of the angle bisector of $\angle{BAC}$ from $A$ to where it intersects $BC$. [i]2017 CCA Math Bonanza Lightning Round #1.3[/i]

2008 Bulgaria National Olympiad, 1

Tags: circumcircle , trigonometry , geometry , angle bisector

Let $ ABC$ be an acute triangle and $ CL$ be the angle bisector of $ \angle ACB$. The point $ P$ lies on the segment $CL$ such that $ \angle APB\equal{}\pi\minus{}\frac{_1}{^2}\angle ACB$. Let $ k_1$ and $ k_2$ be the circumcircles of the triangles $ APC$ and $ BPC$. $ BP\cap k_1\equal{}Q, AP\cap k_2\equal{}R$. The tangents to $ k_1$ at $ Q$ and $ k_2$ at $ B$ intersect at $ S$ and the tangents to $ k_1$ at $ A$ and $ k_2$ at $ R$ intersect at $ T$. Prove that $ AS\equal{}BT.$

2017 CHMMC (Fall), 7

Tags: angle bisector , geometry

Triangle $ABC$ has side lengths $AB=18$, $BC=36$, and $CA=24$. The circle $\Gamma$ passes through point $C$ and is tangent to segment $AB$ at point $A$. Let $X$, distinct from $C$, be the second intersection of $\Gamma$ with $BC$. Moreover, let $Y$ be the point on $\Gamma$ such that segment $AY$ is an angle bisector of $\angle XAC$. Suppose the length of segment $AY$ can be written in the form $AY=\frac{p\sqrt{r}}{q}$ where $p$, $q$, and $r$ are positive integers such that $gcd(p, q)=1$ and $r$ is square free. Find the value of $p+q+r$.

1991 Chile National Olympiad, 6

Tags: angle chasing , angle , angle bisector , geometry

Given a triangle with $ \triangle ABC $, with: $ \angle C = 36^o$ and $ \angle A = \angle B $. Consider the points $ D $ on $ BC $, $ E $ on $ AD $, $ F $ on $ BE $, $ G $ on $ DF $ and $ H $ on $ EG $, so that the rays $ AD, BE, DF, EG, FH $ bisect the angles $ A, B, D, E, F $ respectively. It is known that $ FH = 1 $. Calculate $ AC$.

2016 CCA Math Bonanza, T9

Tags: angle bisector , geometry

Let ABC be a triangle with $AB = 8$, $BC = 9$, and $CA = 10$. The line tangent to the circumcircle of $ABC$ at $A$ intersects the line $BC$ at $T$, and the circle centered at $T$ passing through $A$ intersects the line $AC$ for a second time at $S$. If the angle bisector of $\angle SBA$ intersects $SA$ at $P$, compute the length of segment $SP$. [i]2016 CCA Math Bonanza Team #9[/i]

2012 National Olympiad First Round, 17

Tags: symmetry , angle bisector , reflection , geometric transformation , geometry , trigonometry

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}$

2007 Iran Team Selection Test, 1

Tags: trigonometry , geometry , angle bisector , circumcircle

In triangle $ABC$, $M$ is midpoint of $AC$, and $D$ is a point on $BC$ such that $DB=DM$. We know that $2BC^{2}-AC^{2}=AB.AC$. Prove that \[BD.DC=\frac{AC^{2}.AB}{2(AB+AC)}\]

2023 Canadian Junior Mathematical Olympiad, 2

Tags: angle bisector , geometry , circumcircle

An acute triangle is a triangle that has all angles less that $90^{\circ}$ ($90^{\circ}$ is a Right Angle). Let $ABC$ be a right-angled triangle with $\angle ACB =90^{\circ}.$ Let $CD$ be the altitude from $C$ to $AB,$ and let $E$ be the intersection of the angle bisector of $\angle ACD$ with $AD.$ Let $EF$ be the altitude from $E$ to $BC.$ Prove that the circumcircle of $BEF$ passes through the midpoint of $CE.$

2008 Turkey Team Selection Test, 5

Tags: geometry , angle bisector , trigonometry

$ D$ is a point on the edge $ BC$ of triangle $ ABC$ such that $ AD\equal{}\frac{BD^2}{AB\plus{}AD}\equal{}\frac{CD^2}{AC\plus{}AD}$. $ E$ is a point such that $ D$ is on $ [AE]$ and $ CD\equal{}\frac{DE^2}{CD\plus{}CE}$. Prove that $ AE\equal{}AB\plus{}AC$.

2009 Bundeswettbewerb Mathematik, 3

Tags: circumcircle , geometric inequality , angle bisector , geometry

Given a triangle $ABC$ and a point $P$ on the side $AB$ . Let $Q$ be the intersection of the straight line $CP$ (different from $C$) with the circumcicle of the triangle. Prove the inequality $$\frac{\overline{PQ}}{\overline{CQ}} \le \left(\frac{\overline{AB}}{\overline{AC}+\overline{CB}}\right)^2$$ and that equality holds if and only if the $CP$ is bisector of the angle $ACB$. [img]https://cdn.artofproblemsolving.com/attachments/b/1/068fafd5564e77930160115a1cd409c4fdbf61.png[/img]

2012 Balkan MO Shortlist, G7

Tags: cyclic , rectangle , angle bisector , geometry , cyclic quadrilateral

$ABCD$ is a cyclic quadrilateral. The lines $AD$ and $BC$ meet at X, and the lines $AB$ and $CD$ meet at $Y$ . The line joining the midpoints $M$ and $N$ of the diagonals $AC$ and $BD$, respectively, meets the internal bisector of angle $AXB$ at $P$ and the external bisector of angle $BYC$ at $Q$. Prove that $PXQY$ is a rectangle

2018 Pan-African Shortlist, G6

Tags: geometry , angle bisector

Let $\Gamma$ be the circumcircle of an acute triangle $ABC$. The perpendicular line to $AB$ passing through $C$ cuts $AB$ in $D$ and $\Gamma$ again in $E$. The bisector of the angle $C$ cuts $AB$ in $F$ and $\Gamma$ again in $G$. The line $GD$ meets $\Gamma$ again at $H$ and the line $HF$ meets $\Gamma$ again at $I$. Prove that $AI = EB$.

1997 IMO Shortlist, 16

Tags: angle bisector , incenter , collinearity , trigonometry , circumcircle , geometry

In an acute-angled triangle $ ABC,$ let $ AD,BE$ be altitudes and $ AP,BQ$ internal bisectors. Denote by $ I$ and $ O$ the incenter and the circumcentre of the triangle, respectively. Prove that the points $ D, E,$ and $ I$ are collinear if and only if the points $ P, Q,$ and $ O$ are collinear.

2020 Israel Olympic Revenge, G

Tags: nine point center , angle bisector , apollonius circle , geometry

Let $ABC$ be an acute triangle with $AB\neq AC$. The angle bisector of $\angle BAC$ intersects with $BC$ at a point $D$. $BE,CF$ are the altitudes of the triangle and $Ap_1,Ap_2$ are the isodynamic points of triangle $ABC$.Let the $A$-median of $ABC$ intersect $EF$ at $T$. Show that the line connecting $T$ with the nine-point center of $ABC$ is perpendicular to $BC$ if and only if $\angle Ap_1DAp_2=90^\circ$.

1953 AMC 12/AHSME, 28

Tags: geometry , angle bisector , ratio

In triangle $ ABC$, sides $ a,b$ and $ c$ are opposite angles $ A,B$ and $ C$ respectively. $ AD$ bisects angle $ A$ and meets $ BC$ at $ D$. Then if $ x \equal{} \overline{CD}$ and $ y \equal{} \overline{BD}$ the correct proportion is: $ \textbf{(A)}\ \frac {x}{a} \equal{} \frac {a}{b \plus{} c} \qquad\textbf{(B)}\ \frac {x}{b} \equal{} \frac {a}{a \plus{} c} \qquad\textbf{(C)}\ \frac {y}{c} \equal{} \frac {c}{b \plus{} c} \\ \textbf{(D)}\ \frac {y}{c} \equal{} \frac {a}{b \plus{} c} \qquad\textbf{(E)}\ \frac {x}{y} \equal{} \frac {c}{b}$

2016 Iran MO (3rd Round), 3

Tags: angle bisector , geometry

Given triangle $\triangle ABC$ and let $D,E,F$ be the foot of angle bisectors of $A,B,C$ ,respectively. $M,N$ lie on $EF$ such that $AM=AN$. Let $H$ be the foot of $A$-altitude on $BC$. Points $K,L$ lie on $EF$ such that triangles $\triangle AKL, \triangle HMN$ are correspondingly similiar (with the given order of vertices) such that $AK \not\parallel HM$ and $AK \not\parallel HN$. Show that: $DK=DL$

2001 JBMO ShortLists, 11

Tags: law of cosines , trig identities , angle bisector , geometry , trigonometry

Consider a triangle $ABC$ with $AB=AC$, and $D$ the foot of the altitude from the vertex $A$. The point $E$ lies on the side $AB$ such that $\angle ACE= \angle ECB=18^{\circ}$. If $AD=3$, find the length of the segment $CE$.

2013 Denmark MO - Mohr Contest, 5

Tags: ratio , geometry , angle bisector , equal segments

The angle bisector of $A$ in triangle $ABC$ intersects $BC$ in the point $D$. The point $E$ lies on the side $AC$, and the lines $AD$ and $BE$ intersect in the point $F$. Furthermore, $\frac{|AF|}{|F D|}= 3$ and $\frac{|BF|}{|F E|}=\frac{5}{3}$. Prove that $|AB| = |AC|$. [img]https://1.bp.blogspot.com/-evofDCeJWPY/XzT9dmxXzVI/AAAAAAAAMVY/ZN87X3Cg8iMiULwvMhgFrXbdd_f1f-JWwCLcBGAsYHQ/s0/2013%2BMohr%2Bp5.png[/img]

2001 Tuymaada Olympiad, 3

Tags: concyclic , angle bisector , geometry , circumcircle

Let ABC be an acute isosceles triangle ($AB=BC$) inscribed in a circle with center $O$ . The line through the midpoint of the chord $AB$ and point $O$ intersects the line $AC$ at $L$ and the circle at the point $P$. Let the bisector of angle $BAC$ intersects the circle at point $K$. Lines $AB$ and $PK$ intersect at point $D$. Prove that the points $L,B,D$ and $P$ lie on the same circle.

2022 Israel TST, 3

Tags: circles , geometry , angle bisector , circumcircle

In triangle $ABC$, the angle bisectors are $BE$ and $CF$ (where $E, F$ are on the sides of the triangle), and their intersection point is $I$. Point $N$ lies on the circumcircle of $AEF$, and the angle $\angle IAN$ is right. The circumcircle of $AEF$ meets the line $NI$ a second time at the point $L$. Show that the circumcenter of $AIL$ lies on line $BC$.

1984 IMO Longlists, 48

Tags: angle bisector , geometry , incenter , analytic geometry

Let $ABC$ be a triangle with interior angle bisectors $AA_1, BB_1, CC_1$ and incenter $I$. If $\sigma[IA_1B] + \sigma[IB_1C] + \sigma[IC_1A] = \frac{1}{2}\sigma[ABC]$, where $\sigma[ABC]$ denotes the area of $ABC$, show that $ABC$ is isosceles.

1999 AIME Problems, 12

Tags: angle bisector , trigonometry , incenter , geometry

The inscribed circle of triangle $ABC$ is tangent to $\overline{AB}$ at $P,$ and its radius is 21. Given that $AP=23$ and $PB=27,$ find the perimeter of the triangle.

2007 Sharygin Geometry Olympiad, 4

Tags: angle bisector , symmetry , reflection , cyclic quadrilateral , geometry

A quadrilateral A$BCD$ is inscribed into a circle with center $O$. Points $C', D'$ are the reflections of the orthocenters of triangles $ABD$ and $ABC$ at point $O$. Lines $BD$ and $BD'$ are symmetric with respect to the bisector of angle $ABC$. Prove that lines $AC$ and $AC'$ are symmetric with respect to the bisector of angle $DAB$.

2008 Kazakhstan National Olympiad, 2

Tags: vector , power of a point , angle bisector , geometry , radical axis , incenter , circumcircle

Suppose that $ B_1$ is the midpoint of the arc $ AC$, containing $ B$, in the circumcircle of $ \triangle ABC$, and let $ I_b$ be the $ B$-excircle's center. Assume that the external angle bisector of $ \angle ABC$ intersects $ AC$ at $ B_2$. Prove that $ B_2I$ is perpendicular to $ B_1I_B$, where $ I$ is the incenter of $ \triangle ABC$.