Olympiad problems

2021 Indonesia TST, G

Tags: geometry , trapezoid

Given points $A$, $B$, $C$, and $D$ on circle $\omega$ such that lines $AB$ and $CD$ intersect on point $T$ where $A$ is between $B$ and $T$, moreover $D$ is between $C$ and $T$. It is known that the line passing through $D$ which is parallel to $AB$ intersects $\omega$ again on point $E$ and line $ET$ intersects $\omega$ again on point $F$. Let $CF$ and $AB$ intersect on point $G$, $X$ be the midpoint of segment $AB$, and $Y$ be the reflection of point $T$ to $G$. Prove that $X$, $Y$, $C$, and $D$ are concyclic.

2021 Iranian Geometry Olympiad, 4

Tags: geometry , reflection , trapezoid , concentric circles

In isosceles trapezoid $ABCD$ ($AB \parallel CD$) points $E$ and $F$ lie on the segment $CD$ in such a way that $D, E, F$ and $C$ are in that order and $DE = CF$. Let $X$ and $Y$ be the reflection of $E$ and $C$ with respect to $AD$ and $AF$. Prove that circumcircles of triangles $ADF$ and $BXY$ are concentric. [i]Proposed by Iman Maghsoudi - Iran[/i]

VI Soros Olympiad 1999 - 2000 (Russia), 9.5

Tags: geometry , trapezoid , concurrency , concyclic

A straight line is drawn through an arbitrary internal point $K$ of the trapezoid $ABCD$, intersecting the bases of $BC$ and $AD$ at points $P$ and $Q$, respectively. The circles circumscribed around the triangles $BPK$ and $DQK$ intersect, besides the point $K$, also at the point $L$. Prove that the point $L$ lies on the diagonal $BD$.

2001 Czech And Slovak Olympiad IIIA, 5

Tags: geometry , trapezoid , isosceles trapezoid , area

A sheet of paper has the shape of an isosceles trapezoid $C_1AB_2C_2$ with the shorter base $B_2C_2$. The foot of the perpendicular from the midpoint $D$ of $C_1C_2$ to $AC_1$ is denoted by $B_1$. Suppose that upon folding the paper along $DB_1, AD$ and $AC_1$ points $C_1,C_2$ become a single point $C$ and points $B_1,B_2$ become a point $B$. The area of the tetrahedron $ABCD$ is $64$ cm$^3$ . Find the sides of the initial trapezoid.

2019 Saudi Arabia Pre-TST + Training Tests, 1.3

Tags: geometry , trapezoid , right angle , incircle , equal segments

Let $ABCD$ be a trapezoid with $\angle A = \angle B = 90^o$ and a point $E$ lies on the segment $CD$. Denote $(\omega)$ as incircle of triangle $ABE$ and it is tangent to $AB,AE,BE$ respectively at $P, F,K$. Suppose that $KF$ cuts $BC,AD$ at $M,N$ and $PM,PN$ cut $(\omega)$ at $H, T$. Prove that $PH = PT$.

1982 Dutch Mathematical Olympiad, 2

Tags: trapezoid , geometry

In a triangle $ ABC$, $ M$ is the midpoint of $ AB$ and $ P$ an arbitrary point on side $ AC$. Using only a straight edge, construct point $ Q$ on $ BC$ such that $ P$ and $ Q$ are at equal distance from $ CM$.

2014 Vietnam National Olympiad, 4

Tags: circumcircle , parallelogram , geometric transformation , reflection , trapezoid , geometry

Let $ABC$ be an acute triangle, $(O)$ be the circumcircle, and $AB<AC.$ Let $I$ be the midpoint of arc $BC$ (not containing $A$). $K$ lies on $AC,$ $K\ne C$ such that $IK=IC.$ $BK$ intersects $(O)$ at the second point $D,$ $D\ne B$ and intersects $AI$ at $E.$ $DI$ intersects $AC$ at $F.$ a) Prove that $EF=\frac{BC}{2}.$ b) $M$ lies on $DI$ such that $CM$ is parallel to $AD.$ $KM$ intersects $BC$ at $N.$ The circumcircle of triangle $BKN$ intersects $(O)$ at the second point $P.$ Prove that $PK$ passes through the midpoint of segment $AD.$

2016 Iranian Geometry Olympiad, 1

Tags: geometry , trapezoid , perimeter

In trapezoid $ABCD$ with $AB || CD$, $\omega_1$ and $\omega_2$ are two circles with diameters $AD$ and $BC$, respectively. Let $X$ and $Y$ be two arbitrary points on $\omega_1$ and $\omega_2$, respectively. Show that the length of segment $XY$ is not more than half the perimeter of $ABCD$. [i]Proposed by Mahdi Etesami Fard[/i]

Kyiv City MO Juniors 2003+ geometry, 2018.8.41

Tags: angle bisector , trapezoid , parallel , geometry

In a trapezoid $ABCD$ with bases $AD$ and $BC$, the bisector of the angle $\angle DAB$ intersects the bisectors of the angles $\angle ABC$ and $\angle CDA$ at the points $P$ and $S$, respectively, and the bisector of the angle $\angle BCD$ intersects the bisectors of the angles $\angle ABC$ and $\angle CDA$ at the points $Q$ and $R$, respectively. Prove that if $PS\parallel RQ$, then $AB = CD$.

2023 HMNT, 10

Tags: geometry , trapezoid

Let $ABCD$ be a convex trapezoid such that $\angle ABC = \angle BCD = 90^o$, $AB = 3$, $BC = 6$, and $CD = 12$. Among all points $X$ inside the trapezoid satisfying $\angle XBC = \angle XDA$, compute the minimum possible value of $CX$.

2014 Taiwan TST Round 2, 1

Tags: geometry , circumcircle , trapezoid , symmetry

Let $\omega$ be the circumcircle of a triangle $ABC$. Denote by $M$ and $N$ the midpoints of the sides $AB$ and $AC$, respectively, and denote by $T$ the midpoint of the arc $BC$ of $\omega$ not containing $A$. The circumcircles of the triangles $AMT$ and $ANT$ intersect the perpendicular bisectors of $AC$ and $AB$ at points $X$ and $Y$, respectively; assume that $X$ and $Y$ lie inside the triangle $ABC$. The lines $MN$ and $XY$ intersect at $K$. Prove that $KA=KT$.

1978 Romania Team Selection Test, 1

Tags: geometry , perimeter , trapezoid , pure geometry

In a convex quadrilateral $ ABCD, $ let $ A’,B’ $ be the orthogonal projections to $ CD $ of $ A, $ respectively, $ B. $ [b]a)[/b] Assuming that $ BB’\le AA’ $ and that the perimeter of $ ABCD $ is $ (AB+CD)\cdot BB’, $ is $ ABCD $ necessarily a trapezoid? [b]b)[/b] The same question with the addition that $ \angle BAD $ is obtuse.

2006 Moldova National Olympiad, 10.3

Tags: trapezoid , geometry

A convex quadrilateral $ ABCD$ is inscribed in a circle. The tangents to the circle through $ A$ and $ C$ intersect at a point $ P$, such that this point $ P$ does not lie on $ BD$, and such that $ PA^{2}=PB\cdot PD$. Prove that the line $ BD$ passes through the midpoint of $ AC$.

2001 Brazil Team Selection Test, Problem 3

Tags: trapezoid , geometry

In a triangle $ABC,$ the internal and external bisectors of the angle $A$ intersect the line $BC$ at $D$ and $E$ respectively. The line $AC$ meets the circle with diameter $DE$ again at $F.$ The tangent line to the circle $ABF$ at $A$ meets the circle with diameter $DE$ again at $G.$ Show that $AF = AG.$

2000 Czech and Slovak Match, 5

Tags: isosceles , trapezoid , incircle , perpendicular , geometry

Let $ABCD$ be an isosceles trapezoid with bases $AB$ and $CD$. The incircle of the triangle $BCD$ touches $CD$ at $E$. Point $F$ is chosen on the bisector of the angle $DAC$ such that the lines $EF$ and $CD$ are perpendicular. The circumcircle of the triangle $ACF$ intersects the line $CD$ again at $G$. Prove that the triangle $AFG$ is isosceles.

2013 Moldova Team Selection Test, 3

Tags: trapezoid , geometry

The diagonals of a trapezoid $ABCD$ with $AD \parallel BC$ intersect at point $P$. Point $Q$ lies between the parallel lines $AD$ and $BC$ such that the line $CD$ separates points $P$ and $Q$, and $\angle AQD=\angle CQB$. Prove that $\angle BQP = \angle DAQ$.

2007 Iran MO (3rd Round), 5

Tags: geometry , rectangle , search , trapezoid , circumcircle , parallelogram , ratio

Let $ ABC$ be a triangle. Squares $ AB_{c}B_{a}C$, $ CA_{b}A_{c}B$ and $ BC_{a}C_{b}A$ are outside the triangle. Square $ B_{c}B_{c}'B_{a}'B_{a}$ with center $ P$ is outside square $ AB_{c}B_{a}C$. Prove that $ BP,C_{a}B_{a}$ and $ A_{c}B_{c}$ are concurrent.

1969 IMO Longlists, 21

Tags: geometry , trapezoid , incenter , circumcircle , triangle

$(FRA 4)$ A right-angled triangle $OAB$ has its right angle at the point $B.$ An arbitrary circle with center on the line $OB$ is tangent to the line $OA.$ Let $AT$ be the tangent to the circle different from $OA$ ($T$ is the point of tangency). Prove that the median from $B$ of the triangle $OAB$ intersects $AT$ at a point $M$ such that $MB = MT.$

1998 IMO Shortlist, 7

Tags: trigonometry , geometry , trapezoid , parallelogram

Let $ABC$ be a triangle such that $\angle ACB=2\angle ABC$. Let $D$ be the point on the side $BC$ such that $CD=2BD$. The segment $AD$ is extended to $E$ so that $AD=DE$. Prove that \[ \angle ECB+180^{\circ }=2\angle EBC. \]

2013 NIMO Problems, 13

Tags: geometry , trapezoid , trigonometry

In trapezoid $ABCD$, $AD \parallel BC$ and $\angle ABC + \angle CDA = 270^{\circ}$. Compute $AB^2$ given that $AB \cdot \tan(\angle BCD) = 20$ and $CD = 13$. [i]Proposed by Lewis Chen[/i]

2011 Spain Mathematical Olympiad, 1

Tags: ratio , trapezoid , trigonometry , perpendicular bisector , angle bisector , geometry

In triangle $ABC$, $\angle B=2\angle C$ and $\angle A>90^\circ$. Let $D$ be the point on the line $AB$ such that $CD$ is perpendicular to $AC$, and let $M$ be the midpoint of $BC$. Prove that $\angle AMB=\angle DMC$.

Ukraine Correspondence MO - geometry, 2004.8

Tags: geometry , ratio , trapezoid , area

The extensions of the sides $AB$ and $CD$ of the trapezoid $ABCD$ intersect at point $E$. Denote by $H$ and $G$ the midpoints of $BD$ and $AC$. Find the ratio of the area $AEGH$ to the area $ABCD$.

2017 Peru Iberoamerican Team Selection Test, P5

Tags: geometry , trapezoid

Let $ABCD$ be a trapezoid of bases $AD$ and $BC$ , with $AD> BC$, whose diagonals are cut at point $E$. Let $P$ and $Q$ be the feet of the perpendicular drawn from $E$ on the sides $AD$ and $BC$, respectively, with $P$ and $Q$ in segments $AD$ and $BC,$ respectively. Let $I$ be the center of the triangle $AED$ and let $K$ be the point of intersection of the lines $AI$ and $CD$. If $AP + AE = BQ + BE$, show that $AI = IK$.

2019 IberoAmerican, 4

Tags: trapezoid , projective geometry , miquel point , geometry

Let $ABCD$ be a trapezoid with $AB\parallel CD$ and inscribed in a circumference $\Gamma$. Let $P$ and $Q$ be two points on segment $AB$ ($A$, $P$, $Q$, $B$ appear in that order and are distinct) such that $AP=QB$. Let $E$ and $F$ be the second intersection points of lines $CP$ and $CQ$ with $\Gamma$, respectively. Lines $AB$ and $EF$ intersect at $G$. Prove that line $DG$ is tangent to $\Gamma$.

2007 India IMO Training Camp, 1

Tags: geometry , trapezoid , circumcircle , ratio , homothety

Let $ ABCD$ be a trapezoid with parallel sides $ AB > CD$. Points $ K$ and $ L$ lie on the line segments $ AB$ and $ CD$, respectively, so that $AK/KB=DL/LC$. Suppose that there are points $ P$ and $ Q$ on the line segment $ KL$ satisfying \[\angle{APB} \equal{} \angle{BCD}\qquad\text{and}\qquad \angle{CQD} \equal{} \angle{ABC}.\] Prove that the points $ P$, $ Q$, $ B$ and $ C$ are concyclic. [i]Proposed by Vyacheslev Yasinskiy, Ukraine[/i]