Olympiad problems

2021 USA TSTST, 8

Let $ABC$ be a scalene triangle. Points $A_1,B_1$ and $C_1$ are chosen on segments $BC,CA$ and $AB$, respectively, such that $\triangle A_1B_1C_1$ and $\triangle ABC$ are similar. Let $A_2$ be the unique point on line $B_1C_1$ such that $AA_2=A_1A_2$. Points $B_2$ and $C_2$ are defined similarly. Prove that $\triangle A_2B_2C_2$ and $\triangle ABC$ are similar. [i]Fedir Yudin [/i]

2016 Iranian Geometry Olympiad, 2

Tags: circumcircle , equal segments , angle chasing , geometry

Let $\omega$ be the circumcircle of triangle $ABC$ with $AC > AB$. Let $X$ be a point on $AC$ and $Y$ be a point on the circle $\omega$, such that $CX = CY = AB$. (The points $A$ and $Y$ lie on different sides of the line $BC$). The line $XY$ intersects $\omega$ for the second time in point $P$. Show that $PB = PC$. by Iman Maghsoudi

2012 Indonesia TST, 3

Tags: geometry , angle chasing , complex number , cyclic quadrilateral

The incircle of a triangle $ABC$ is tangent to the sides $AB,AC$ at $M,N$ respectively. Suppose $P$ is the intersection between $MN$ and the bisector of $\angle ABC$. Prove that $BP$ and $CP$ are perpendicular.

2020 Romania EGMO TST, P1

Tags: radical axis , geometry , angle chasing

An acute triangle $ABC$ in which $AB<AC$ is given. The bisector of $\angle BAC$ crosses $BC$ at $D$. Point $M$ is the midpoint of $BC$. Prove that the line though centers of circles escribed on triangles $ABC$ and $ADM$ is parallel to $AD$.

2012 Dutch IMO TST, 4

Tags: geometry , right angle , angle chasing , equal angles

Let $\vartriangle ABC$ be a triangle. The angle bisector of $\angle CAB$ intersects$ BC$ at $L$. On the interior of line segments $AC$ and $AB$, two points, $M$ and $N$, respectively, are chosen in such a way that the lines $AL, BM$ and $CN$ are concurrent, and such that $\angle AMN = \angle ALB$. Prove that $\angle NML = 90^o$.

2020 Yasinsky Geometry Olympiad, 4

Tags: altitude , angle chasing , angle , geometry

Let $BB_1$ and $CC_1$ be the altitudes of the acute-angled triangle $ABC$. From the point $B_1$ the perpendiculars $B_1E$ and $B_1F$ are drawn on the sides $AB$ and $BC$ of the triangle, respectively, and from the point $C_1$ the perpendiculars $C_1 K$ and $C_1L$ on the sides $AC$ and $BC$, respectively. It turned out that the lines $EF$ and $KL$ are perpendicular. Find the measure of the angle $A$ of the triangle $ABC$. (Alexander Dunyak)

2018 Dutch BxMO TST, 4

Tags: geometry , circumcircle , angle , angle chasing

In a non-isosceles triangle $\vartriangle ABC$ we have $\angle BAC = 60^o$. Let $D$ be the intersection of the angular bisector of $\angle BAC$ with side $BC, O$ the centre of the circumcircle of $\vartriangle ABC$ and $E$ the intersection of $AO$ and $BC$. Prove that $\angle AED + \angle ADO = 90^o$.

2013 IMAC Arhimede, 5

Tags: angle , circumcircle , incircle , angle bisector , geometry , angle chasing

Let $\Gamma$ be the circumcircle of a triangle $ABC$ and let $E$ and $F$ be the intersections of the bisectors of $\angle ABC$ and $\angle ACB$ with $\Gamma$. If $EF$ is tangent to the incircle $\gamma$ of $\triangle ABC$, then find the value of $\angle BAC$.

2017 Flanders Math Olympiad, 2

Tags: geometry , circumcircle , angle , angle chasing

In triangle $\vartriangle ABC$, $\angle A = 50^o, \angle B = 60^o$ and $\angle C = 70^o$. The point $P$ is on the side $[AB]$ (with $P \ne A$ and $P \ne B$). The inscribed circle of $\vartriangle ABC$ intersects the inscribed circle of $\vartriangle ACP$ at points $U$ and $V$ and intersects the inscribed circle of $\vartriangle BCP$ at points $X$ and $Y$. The rights $UV$ and $XY$ intersect in $K$. Calculate the $\angle UKX$.

2019 USEMO, 1

Tags: geometry , cyclic quadrilateral , angle chasing

Let $ABCD$ be a cyclic quadrilateral. A circle centered at $O$ passes through $B$ and $D$ and meets lines $BA$ and $BC$ again at points $E$ and $F$ (distinct from $A,B,C$). Let $H$ denote the orthocenter of triangle $DEF.$ Prove that if lines $AC,$ $DO,$ $EF$ are concurrent, then triangle $ABC$ and $EHF$ are similar. [i]Robin Son[/i]

2019 Belarus Team Selection Test, 8.1

Tags: symmetry , angle chasing , miquel point , geometry

Let $ABC$ be a triangle with $AB=AC$, and let $M$ be the midpoint of $BC$. Let $P$ be a point such that $PB<PC$ and $PA$ is parallel to $BC$. Let $X$ and $Y$ be points on the lines $PB$ and $PC$, respectively, so that $B$ lies on the segment $PX$, $C$ lies on the segment $PY$, and $\angle PXM=\angle PYM$. Prove that the quadrilateral $APXY$ is cyclic.

2018 Estonia Team Selection Test, 5

Tags: similar triangles , geometry , angle chasing , projective geometry

Let $O$ be the circumcenter of an acute triangle $ABC$. Line $OA$ intersects the altitudes of $ABC$ through $B$ and $C$ at $P$ and $Q$, respectively. The altitudes meet at $H$. Prove that the circumcenter of triangle $PQH$ lies on a median of triangle $ABC$.

2012 India PRMO, 14

Tags: geometry , incenter , angle chasing

$O$ and $I$ are the circumcentre and incentre of $\vartriangle ABC$ respectively. Suppose $O$ lies in the interior of $\vartriangle ABC$ and $I$ lies on the circle passing through $B, O$, and $C$. What is the magnitude of $\angle B AC$ in degrees?

2018 Yasinsky Geometry Olympiad, 5

Tags: geometry , angle chasing , rhombus

The point $M$ lies inside the rhombus $ABCD$. It is known that $\angle DAB=110^o$, $\angle AMD=80^o$, $\angle BMC= 100^o$. What can the angle $\angle AMB$ be equal?

2022 Taiwan TST Round 3, 1

Tags: concurrency , cyclic quadrilateral , angle chasing , geometry

Let $ABCD$ be a quadrilateral inscribed in a circle $\Omega.$ Let the tangent to $\Omega$ at $D$ meet rays $BA$ and $BC$ at $E$ and $F,$ respectively. A point $T$ is chosen inside $\triangle ABC$ so that $\overline{TE}\parallel\overline{CD}$ and $\overline{TF}\parallel\overline{AD}.$ Let $K\ne D$ be a point on segment $DF$ satisfying $TD=TK.$ Prove that lines $AC,DT,$ and $BK$ are concurrent.

1996 APMO, 3

Tags: rectangle , incenter , cyclic quadrilateral , geometry , angle chasing

If $ABCD$ is a cyclic quadrilateral, then prove that the incenters of the triangles $ABC$, $BCD$, $CDA$, $DAB$ are the vertices of a rectangle.

2017 All-Russian Olympiad, 8

Tags: geometry , incenter , angle chasing , incircle

Given a convex quadrilateral $ABCD$. We denote $I_A,I_B, I_C$ and $I_D$ centers of $\omega_A, \omega_B,\omega_C $and $\omega_D$,inscribed In the triangles $DAB, ABC, BCD$ and $CDA$, respectively.It turned out that $\angle BI_AA + \angle I_CI_AI_D = 180^\circ$. Prove that $\angle BI_BA + \angle I_CI_BI_D = 180^{\circ}$. (A. Kuznetsov)

2019 Yasinsky Geometry Olympiad, p4

Tags: cyclic quadrilateral , angle , angle chasing , geometry

Find the angles of the cyclic quadrilateral if you know that each of its diagonals is a bisector of one angle and a trisector of the opposite one (the trisector of the angle is one of the two rays that lie in the interior of the angle and divide it into three equal parts). (Vyacheslav Yasinsky)

STEMS 2021 Math Cat A, Q3

Tags: geometry , angle chasing

An acute scalene triangle $\triangle{ABC}$ with altitudes $\overline{AD}, \overline{BE},$ and $\overline{CF}$ is inscribed in circle $\Gamma$. Medians from $B$ and $C$ meet $\Gamma$ again at $K$ and $L$ respectively. Prove that the circumcircles of $\triangle{BFK}, \triangle{CEL}$ and $\triangle{DEF}$ concur.

2013 India Regional Mathematical Olympiad, 4

Tags: geometry , angle chasing , similar triangles

Let $ABC$ be a triangle with $\angle A=90^{\circ}$ and $AB=AC$. Let $D$ and $E$ be points on the segment $BC$ such that $BD:DE:EC = 1:2:\sqrt{3}$. Prove that $\angle DAE= 45^{\circ}$

2019 Yasinsky Geometry Olympiad, p1

Tags: geometry , inradius , angle chasing , angle

It is known that in the triangle $ABC$ the distance from the intersection point of the angle bisector to each of the vertices of the triangle does not exceed the diameter of the circle inscribed in this triangle. Find the angles of the triangle $ABC$. (Grigory Filippovsky)

2009 Balkan MO Shortlist, G1

Tags: angle bisector , geometry , angle chasing , equal segments , angle

In the triangle $ABC, \angle BAC$ is acute, the angle bisector of $\angle BAC$ meets $BC$ at $D, K$ is the foot of the perpendicular from $B$ to $AC$, and $\angle ADB = 45^o$. Point $P$ lies between $K$ and $C$ such that $\angle KDP = 30^o$. Point $Q$ lies on the ray $DP$ such that $DQ = DK$. The perpendicular at $P$ to $AC$ meets $KD$ at $L$. Prove that $PL^2 = DQ \cdot PQ$.

2017 IMO Shortlist, G3

Tags: projective geometry , angle chasing , similar triangles , geometry

Let $O$ be the circumcenter of an acute triangle $ABC$. Line $OA$ intersects the altitudes of $ABC$ through $B$ and $C$ at $P$ and $Q$, respectively. The altitudes meet at $H$. Prove that the circumcenter of triangle $PQH$ lies on a median of triangle $ABC$.

2018 Romania Team Selection Tests, 1

Tags: projective geometry , geometry , angle chasing , similar triangles

Let $O$ be the circumcenter of an acute triangle $ABC$. Line $OA$ intersects the altitudes of $ABC$ through $B$ and $C$ at $P$ and $Q$, respectively. The altitudes meet at $H$. Prove that the circumcenter of triangle $PQH$ lies on a median of triangle $ABC$.

2006 Singapore Junior Math Olympiad, 4

Tags: angle bisector , geometry , incenter , angle chasing

In $\vartriangle ABC$, the bisector of $\angle B$ meets $AC$ at $D$ and the bisector of $\angle C$ meets $AB$ at $E$. These bisectors intersect at $O$ and $OD = OE$. If $AD \ne AE$, prove that $\angle A = 60^o$.