Found problems: 131
2002 IMO Shortlist, 8
Let two circles $S_{1}$ and $S_{2}$ meet at the points $A$ and $B$. A line through $A$ meets $S_{1}$ again at $C$ and $S_{2}$ again at $D$. Let $M$, $N$, $K$ be three points on the line segments $CD$, $BC$, $BD$ respectively, with $MN$ parallel to $BD$ and $MK$ parallel to $BC$. Let $E$ and $F$ be points on those arcs $BC$ of $S_{1}$ and $BD$ of $S_{2}$ respectively that do not contain $A$. Given that $EN$ is perpendicular to $BC$ and $FK$ is perpendicular to $BD$ prove that $\angle EMF=90^{\circ}$.
2013 Tournament of Towns, 2
Let $C$ be a right angle in triangle $ABC$. On legs $AC$ and$BC$ the squares $ACKL, BCMN$ are constructed outside of triangle. If $CE$ is an altitude of the triangle prove that $LEM$ is a right angle.
2024 Bulgaria National Olympiad, 2
Given is a triangle $ABC$ and the points $M, P$ lie on the segments $AB, BC$, respectively, such that $AM=BC$ and $CP=BM$. If $AP$ and $CM$ meet at $O$ and $2\angle AOM=\angle ABC$, find the measure of $\angle ABC$.
1979 Austrian-Polish Competition, 1
On sides $AB$ and $BC$ of a square $ABCD$ the respective points $E$ and $F$ have been chosen so that $BE = BF$. Let $BN$ be the altitude in triangle $BCE$. Prove that $\angle DNF = 90$.
2014 Danube Mathematical Competition, 3
Let $ABC$ be a triangle with $\angle A<90^o, AB \ne AC$. Denote $H$ the orthocenter of triangle $ABC$, $N$ the midpoint of segment $[AH]$, $M$ the midpoint of segment $[BC]$ and $D$ the intersection point of the angle bisector of $\angle BAC$ with the segment $[MN]$. Prove that $<ADH=90^o$
2007 Austria Beginners' Competition, 4
Consider a parallelogram $ABCD$ such that the midpoint $M$ of the side $CD$ lies on the angle bisector of $\angle BAD$. Show that $\angle AMB$ is a right angle.
Kharkiv City MO Seniors - geometry, 2015.11.3
In the rectangle $ABCD$, point $M$ is the midpoint of the side $BC$. The points $P$ and $Q$ lie on the diagonal $AC$ such that $\angle DPC = \angle DQM = 90^o$. Prove that $Q$ is the midpoint of the segment $AP$.
2012 Dutch IMO TST, 4
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 $\angle NML = 90^o$.
2016 Thailand TSTST, 3
Let $H$ be the orthocenter of acute-angled $\vartriangle ABC$, and $X, Y$ points on the ray $AB, AC$. ($B$ lies between $X, A$, and $C$ lies between $Y, A$.) Lines $HX, HY$ intersect $BC$ at $D, E$ respectively. Let the line through $D$ parallel to $AC$ intersect $XY$ at $Z$. Prove that $\angle XHY = 90^o$ if and only if $ZE \parallel AB$.
2023 Romanian Master of Mathematics Shortlist, C2
For positive integers $m,n \geq 2$, let $S_{m,n} = \{(i,j): i \in \{1,2,\ldots,m\}, j\in \{1,2,\ldots,n\}\}$ be a grid of $mn$ lattice points on the coordinate plane. Determine all pairs $(m,n)$ for which there exists a simple polygon $P$ with vertices in $S_{m,n}$ such that all points in $S_{m,n}$ are on the boundary of $P$, all interior angles of $P$ are either $90^{\circ}$ or $270^{\circ}$ and all side lengths of $P$ are $1$ or $3$.
2009 Bosnia And Herzegovina - Regional Olympiad, 1
In triangle $ABC$ such that $\angle ACB=90^{\circ}$, let point $H$ be foot of perpendicular from point $C$ to side $AB$. Show that sum of radiuses of incircles of $ABC$, $BCH$ and $ACH$ is $CH$
2009 China Northern MO, 2
In an acute triangle $ABC$ , $AB>AC$ , $ \cos B+ \cos C=1$ , $E,F$ are on the extend line of $AB,AC$ such that $\angle ABF = \angle ACE = 90$ .
(1) Prove :$BE+CF=EF$ ;
(2) Assume the bisector of $\angle EBC$ meet $EF$ at $P$ , prove that $CP$ is the bisector of $\angle BCF$.
[img]https://cdn.artofproblemsolving.com/attachments/a/2/c554c2bc0b4e044c45f88138568f5234d544a8.png[/img]
2018 Iranian Geometry Olympiad, 2
In convex quadrilateral $ABCD$, the diagonals $AC$ and $BD$ meet at the point $P$. We know that $\angle DAC = 90^o$ and $2 \angle ADB = \angle ACB$. If we have $ \angle DBC + 2 \angle ADC = 180^o$ prove that $2AP = BP$.
Proposed by Iman Maghsoudi
2019 Saudi Arabia Pre-TST + Training Tests, 3.2
Let $ABC$ be a triangle, the circle having $BC$ as diameter cuts $AB,AC$ at $F,E$ respectively. Let $P$ a point on this circle. Let $C',B$' be the projections of $P$ upon the sides $AB,AC$ respectively. Let $H$ be the orthocenter of the triangle $AB'C'$. Show that $\angle EHF = 90^o$.
2013 Tournament of Towns, 5
In a quadrilateral $ABCD$, angle $B$ is equal to $150^o$, angle $C$ is right, and sides $AB$ and $CD$ are equal. Determine the angle between $BC$ and the line connecting the midpoints of sides $BC$ and $AD$.
1971 Czech and Slovak Olympiad III A, 2
Let $ABC$ be a triangle. Four distinct points $D,A,B,E$ lie on the line $AB$ in this order such that $DA=AB=BE.$ Find necessary and sufficient condition for lengths $a=BC,b=AC$ such that the angle $\angle DCE$ is right.
2017 Saudi Arabia IMO TST, 1
Let $ABC$ be a triangle inscribed in circle $(O),$ with its altitudes $BE, CF$ intersect at orthocenter $H$ ($E \in AC, F \in AB$). Let $M$ be the midpoint of $BC, K$ be the orthogonal projection of $H$ on $AM$. $EF$ intersects $BC$ at $P$. Let $Q$ be the intersection of tangent of $(O)$ which passes through $A$ with $BC, T$ be the reflection of $Q$ through $P$. Prove that $\angle OKT = 90^o$.
2017 Junior Regional Olympiad - FBH, 5
Points $K$ and $L$ are on side $AB$ of triangle $ABC$ such that $KL=BC$ and $AK=LB$. Let $M$ be a midpoint of $AC$. Prove that $\angle KML = 90^{\circ}$
2017 Bosnia and Herzegovina Junior BMO TST, 3
Let $ABC$ be a triangle such that $\angle ABC = 90 ^{\circ}$. Let $I$ be an incenter of $ABC$ and let $F$, $D$ and $E$ be points where incircle touches sides $AB$, $BC$ and $AC$, respectively. If lines $CI$ and $EF$ intersect at point $M$ and if $DM$ and $AB$ intersect in $N$, prove that $AI=ND$
2019 Saudi Arabia BMO TST, 2
Let $ABCD$ is a trapezoid with $\angle A = \angle B = 90^o$ and let $E$ is a point lying on side $CD$. Let the circle $\omega$ is inscribed to triangle $ABE$ and tangents sides $AB, AE$ and $BE$ at points $P, F$ and $K$ respectively. Let $KF$ intersects segments $BC$ and $AD$ at points $M$ and $N$ respectively, as well as $PM$ and $PN$ intersect $\omega$ at points $H$ and $T$ respectively. Prove that $PH = PT$.
1992 Austrian-Polish Competition, 7
Consider triangles $ABC$ in space.
(a) What condition must the angles $\angle A, \angle B , \angle C$ of $\triangle ABC$ fulfill in order that there is a point $P$ in space such that $\angle APB, \angle BPC, \angle CPA$ are right angles?
(b) Let $d$ be the longest of the edges $PA,PB,PC$ and let $h$ be the longest altitude of $\triangle ABC$. Show that $\frac{1}{3}\sqrt6 h \le d \le h$.
2012 Grand Duchy of Lithuania, 2
The base $AB$ of a trapezium $ABCD$ is longer than the base $CD$, and $\angle ADC$ is a right angle. The diagonals $AC$ and $BD$ are perpendicular. Let $E$ be the foot of the altitude from $D$ to the line $BC$. Prove that
$$\frac{AE}{BE} =\frac{ AC \cdot CD}{AC^2 - CD^2}$$
.
2015 Dutch IMO TST, 1
In a quadrilateral $ABCD$ we have $\angle A = \angle C = 90^o$. Let $E$ be a point in the interior of $ABCD$. Let $M$ be the midpoint of $BE$. Prove that $\angle ADB = \angle EDC$ if and only if $|MA| = |MC|$.
Indonesia MO Shortlist - geometry, g2
Given an acute triangle $ABC$. The inscribed circle of triangle $ABC$ is tangent to $AB$ and $AC$ at $X$ and $Y$ respectively. Let $CH$ be the altitude. The perpendicular bisector of the segment $CH$ intersects the line $XY$ at $Z$. Prove that $\angle BZC = 90^o.$
2018 Dutch IMO TST, 3
Let $ABC$ be an acute triangle, and let $D$ be the foot of the altitude through $A$. On $AD$, there are distinct points $E$ and $F$ such that $|AE| = |BE|$ and $|AF| =|CF|$. A point$ T \ne D$ satis es $\angle BTE = \angle CTF = 90^o$. Show that $|TA|^2 =|TB| \cdot |TC|$.