Olympiad problems

1986 China Team Selection Test, 1

Given a square $ABCD$ whose side length is $1$, $P$ and $Q$ are points on the sides $AB$ and $AD$. If the perimeter of $APQ$ is $2$ find the angle $PCQ$.

III Soros Olympiad 1996 - 97 (Russia), 9.3

Tags: angle bisector , geometry

In triangle $ABC$, sides $CB$ and $CA$ are equal to $a$ and $b$, respectively. The bisector of the angle $ACB$ intersects the side $AB$ at the point $K$ and the circumscribed circle of the triangle at point $M$. The circumscribed circle of the triangle $AMK$ intersects for second time the straight line $CA$ at the point $P$. Find the length of the segment $AP$.

2004 IMO Shortlist, 5

Tags: angle bisector , geometry

Let $A_1A_2A_3\ldots A_n$ be a regular $n$-gon. Let $B_1$ and $B_{n-1}$ be the midpoints of its sides $A_1A_2$ and $A_{n-1}A_n$. Also, for every $i\in\left\{2,3,4,\ldots ,n-2\right\}$. Let $S$ be the point of intersection of the lines $A_1A_{i+1}$ and $A_nA_i$, and let $B_i$ be the point of intersection of the angle bisector bisector of the angle $\measuredangle A_iSA_{i+1}$ with the segment $A_iA_{i+1}$. Prove that $\sum_{i=1}^{n-1} \measuredangle A_1B_iA_n=180^{\circ}$. [i]Proposed by Dusan Dukic, Serbia and Montenegro[/i]

2009 India Regional Mathematical Olympiad, 1

Tags: trigonometry , angle bisector , perpendicular bisector , geometry

Let $ ABC$ be a triangle in which $ AB \equal{} AC$ and let $ I$ be its in-centre. Suppose $ BC \equal{} AB \plus{} AI$. Find $ \angle{BAC}$

2014 Saudi Arabia GMO TST, 4

Tags: geometry , angle bisector , concyclic

Let $ABC$ be a triangle, $D$ the midpoint of side $BC$ and $E$ the intersection point of the bisector of angle $\angle BAC$ with side $BC$. The perpendicular bisector of $AE$ intersects the bisectors of angles $\angle CBA$ and $\angle CDA$ at $M$ and $N$, respectively. The bisectors of angles $\angle CBA$ and $\angle CDA$ intersect at $P$ . Prove that points $A, M, N, P$ are concyclic.

2006 Bulgaria Team Selection Test, 1

Tags: symmetry , power of a point , angle bisector , parallelogram , circumcircle , geometry

[b]Problem 1.[/b] Points $D$ and $E$ are chosen on the sides $AB$ and $AC$, respectively, of a triangle $\triangle ABC$ such that $DE\parallel BC$. The circumcircle $k$ of triangle $\triangle ADE$ intersects the lines $BE$ and $CD$ at the points $M$ and $N$ (different from $E$ and $D$). The lines $AM$ and $AN$ intersect the side $BC$ at points $P$ and $Q$ such that $BC=2\cdot PQ$ and the point $P$ lies between $B$ and $Q$. Prove that the circle $k$ passes through the point of intersection of the side $BC$ and the angle bisector of $\angle BAC$. [i]Nikolai Nikolov[/i]

Swiss NMO - geometry, 2020.7

Tags: geometry , trapezoid , angle bisector , cyclic quadrilateral

Let $ABCD$ be an isosceles trapezoid with bases $AD> BC$. Let $X$ be the intersection of the bisectors of $\angle BAC$ and $BC$. Let $E$ be the intersection of$ DB$ with the parallel to the bisector of $\angle CBD$ through $X$ and let $F$ be the intersection of $DC$ with the parallel to the bisector of $\angle DCB$ through $X$. Show that quadrilateral $AEFD$ is cyclic.

2008 AMC 12/AHSME, 20

Tags: analytic geometry , ratio , inradius , incenter , angle bisector , geometry , trigonometry

Triangle $ ABC$ has $ AC\equal{}3$, $ BC\equal{}4$, and $ AB\equal{}5$. Point $ D$ is on $ \overline{AB}$, and $ \overline{CD}$ bisects the right angle. The inscribed circles of $ \triangle ADC$ and $ \triangle BCD$ have radii $ r_a$ and $ r_b$, respectively. What is $ r_a/r_b$? $ \textbf{(A)}\ \frac{1}{28}\left(10\minus{}\sqrt{2}\right) \qquad \textbf{(B)}\ \frac{3}{56}\left(10\minus{}\sqrt{2}\right) \qquad \textbf{(C)}\ \frac{1}{14}\left(10\minus{}\sqrt{2}\right) \qquad \textbf{(D)}\ \frac{5}{56}\left(10\minus{}\sqrt{2}\right) \\ \textbf{(E)}\ \frac{3}{28}\left(10\minus{}\sqrt{2}\right)$

2011 Mediterranean Mathematics Olympiad, 4

Tags: trigonometry , incenter , angle bisector , geometry

Let $D$ be the foot of the internal bisector of the angle $\angle A$ of the triangle $ABC$. The straight line which joins the incenters of the triangles $ABD$ and $ACD$ cut $AB$ and $AC$ at $M$ and $N$, respectively. Show that $BN$ and $CM$ meet on the bisector $AD$.

2020 Moldova EGMO TST, 4

Tags: incenter , geometry , angle bisector

The incircle of triangle $ABC$ touches $AC$ and $BC$ respectively $P$ and $Q$. Let $N$ and $M$ be the midpoints of the sides $AC$ and $BC$ respectively.$AM$ and $BP$,$BN$ and $AQ$ intersects at the points $X$ and $Y$ respectively. If the points $C,X$ and $Y$ are collinear , then prove that $CX$ is the angle bisector of $\angle ACB$.

2024 Israel TST, P1

Tags: angle bisector , tangent , geometry

Let $ABC$ be a triangle and let $D$ be a point on $BC$ so that $AD$ bisects the angle $\angle BAC$. The common tangents of the circles $(BAD)$, $(CAD)$ meet at the point $A'$. The points $B'$, $C'$ are defined similarly. Show that $A'$, $B'$, $C'$ are collinear.

2006 Bulgaria Team Selection Test, 1

Tags: parallelogram , symmetry , power of a point , geometry , angle bisector , circumcircle

[b]Problem 1.[/b] Points $D$ and $E$ are chosen on the sides $AB$ and $AC$, respectively, of a triangle $\triangle ABC$ such that $DE\parallel BC$. The circumcircle $k$ of triangle $\triangle ADE$ intersects the lines $BE$ and $CD$ at the points $M$ and $N$ (different from $E$ and $D$). The lines $AM$ and $AN$ intersect the side $BC$ at points $P$ and $Q$ such that $BC=2\cdot PQ$ and the point $P$ lies between $B$ and $Q$. Prove that the circle $k$ passes through the point of intersection of the side $BC$ and the angle bisector of $\angle BAC$. [i]Nikolai Nikolov[/i]

2024 Moldova EGMO TST, 5

Tags: geometric inequality , geometry , angle bisector , circumcircle , inequalities

$AD$ Is the angle bisector Of $\angle BAC$ Where $D$ lies on the The circumcircle of $\triangle ABC$. Show that $2AD>AB+AC$

2011 Belarus Team Selection Test, 2

Tags: geometry , angle bisector , altitude , ratio

Points $L$ and $H$ are marked on the sides $AB$ of an acute-angled triangle ABC so that $CL$ is a bisector and $CH$ is an altitude. Let $P,Q$ be the feet of the perpendiculars from $L$ to $AC$ and $BC$ respectively. Prove that $AP \cdot BH = BQ \cdot AH$. I. Gorodnin

2014 Bosnia Herzegovina Team Selection Test, 3

Tags: perpendicular bisector , angle bisector , circumcircle , geometry

Let $D$ and $E$ be foots of altitudes from $A$ and $B$ of triangle $ABC$, $F$ be intersection point of angle bisector from $C$ with side $AB$, and $O$, $I$ and $H$ be circumcenter, center of inscribed circle and orthocenter of triangle $ABC$, respectively. If $\frac{CF}{AD}+ \frac{CF}{BE}=2$, prove that $OI = IH$.

2015 Czech-Polish-Slovak Junior Match, 4

Tags: midpoint , right triangle , angle bisector , incenter , geometry

Let $ABC$ ne a right triangle with $\angle ACB=90^o$. Let $E, F$ be respecitvely the midpoints of the $BC, AC$ and $CD$ be it's altitude. Next, let $P$ be the intersection of the internal angle bisector from $A$ and the line $EF$. Prove that $P$ is the center of the circle inscribed in the triangle $CDE$ .

2013 Kyiv Mathematical Festival, 3

Tags: parallelogram , perpendicular , angle bisector , geometry

Let $ABCD$ be a parallelogram ($AB < BC$). The bisector of the angle $BAD$ intersects the side $BC$ at the point K; and the bisector of the angle $ADC$ intersects the diagonal $AC$ at the point $F$. Suppose that $KD \perp BC$. Prove that $KF \perp BD$.

1985 Tournament Of Towns, (080) T1

Tags: angle bisector , concurrency , equilateral , median , altitude , geometry

A median , a bisector and an altitude of a certain triangle intersect at an inner point $O$ . The segment of the bisector from the vertex to $O$ is equal to the segment of the altitude from the vertex to $O$ . Prove that the triangle is equilateral .

2004 Junior Balkan MO, 2

Tags: symmetry , perpendicular bisector , circumcircle , geometry , angle bisector

Let $ABC$ be an isosceles triangle with $AC=BC$, let $M$ be the midpoint of its side $AC$, and let $Z$ be the line through $C$ perpendicular to $AB$. The circle through the points $B$, $C$, and $M$ intersects the line $Z$ at the points $C$ and $Q$. Find the radius of the circumcircle of the triangle $ABC$ in terms of $m = CQ$.

2010 China Team Selection Test, 1

Tags: symmetry , angle bisector , geometric transformation , geometry

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 $\frac{AE}{AB}-\frac{ED}{AD}=1$.

2006 Argentina National Olympiad, 2

Tags: perpendicular , angle bisector , ratio , geometry

In triangle $ABC, M$ is the midpoint of $AB$ and $D$ the foot of the bisector of angle $\angle ABC$. If $MD$ and $BD$ are known to be perpendicular, calculate $\frac{AB}{BC}$.

2008 IMAC Arhimede, 2

Tags: circumcircle , geometric inequality , angle bisector , geometry , inequalities

In the $ ABC$ triangle, the bisector of $A $ intersects the $ [BC] $ at the point $ A_ {1} $ , and the circle circumscribed to the triangle $ ABC $ at the point $ A_ {2} $. Similarly are defined $ B_ {1} $ and $ B_ {2} $ , as well as $ C_ {1} $ and $ C_ {2} $. Prove that $$ \frac {A_{1}A_{2}}{BA_{2} + A_{2}C} + \frac {B_{1}B_{2}}{CB_{2} + B_{2}A} + \frac {C_{1}C_{2}}{AC_{2} + C_{2}B} \geq \frac {3}{4}$$

2019 India IMO Training Camp, P1

Tags: geometry , rectangle , radical axis , angle bisector , circumcircle

Let the points $O$ and $H$ be the circumcenter and orthocenter of an acute angled triangle $ABC.$ Let $D$ be the midpoint of $BC.$ Let $E$ be the point on the angle bisector of $\angle BAC$ such that $AE\perp HE.$ Let $F$ be the point such that $AEHF$ is a rectangle. Prove that $D,E,F$ are collinear.

2013 CentroAmerican, 2

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

Let $ABC$ be an acute triangle and let $\Gamma$ be its circumcircle. The bisector of $\angle{A}$ intersects $BC$ at $D$, $\Gamma$ at $K$ (different from $A$), and the line through $B$ tangent to $\Gamma$ at $X$. Show that $K$ is the midpoint of $AX$ if and only if $\frac{AD}{DC}=\sqrt{2}$.

2006 Regional Competition For Advanced Students, 3

Tags: circumcircle , angle bisector , exterior angle , geometry

In a non isosceles triangle $ ABC$ let $ w$ be the angle bisector of the exterior angle at $ C$. Let $ D$ be the point of intersection of $ w$ with the extension of $ AB$. Let $ k_A$ be the circumcircle of the triangle $ ADC$ and analogy $ k_B$ the circumcircle of the triangle $ BDC$. Let $ t_A$ be the tangent line to $ k_A$ in A and $ t_B$ the tangent line to $ k_B$ in B. Let $ P$ be the point of intersection of $ t_A$ and $ t_B$. Given are the points $ A$ and $ B$. Determine the set of points $ P\equal{}P(C )$ over all points $ C$, so that $ ABC$ is a non isosceles, acute-angled triangle.