Found problems: 131
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
1999 Swedish Mathematical Competition, 2
Circle $C$ center $O$ touches externally circle $C'$ center $O'$. A line touches $C$ at $A$ and $C'$ at $B$. $P$ is the midpoint of $AB$. Show that $\angle OPO' = 90^o$.
Geometry Mathley 2011-12, 10.1
Let $ABC$ be a triangle with two angles $B,C$ not having the same measure, $I$ be its incircle, $(O)$ its circumcircle. Circle $(O_b)$ touches $BA,BC$ and is internally tangent to $(O)$ at $B_1$. Circle $(O_c)$ touches $CA,CB$ and is internally tangent to $(O)$ at $C_1$. Let $S$ be the intersection of $BC$ and $B_1C_1$. Prove that $\angle AIS = 90^o$.
Nguyễn Minh Hà
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 $\angle ABC$.
2009 All-Russian Olympiad Regional Round, 9.3
In an acute triangle $ABC$ the altitudes $AA_1$, $BB_1$, $CC_1$ are drawn. A line perpendicular to side $AC$ and passing through a point $A_1$, intersects the line $B_1C_1$ at point $D$. Prove that angle $ADC$ is right.
2019 Adygea Teachers' Geometry Olympiad, 4
From which two statements about the trapezoid follows the third:
1) the trapezoid is tangential,
2) the trapezoid is right,
3) its area is equal to the product of the bases?
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 $<ADH=90^o$
2006 Oral Moscow Geometry Olympiad, 3
On the sides $AB, BC$ and $AC$ of the triangle $ABC$, points $C', A'$ and $B'$ are selected, respectively, so that the angle $A'C'B'$ is right. Prove that the segment $A'B'$ is longer than the diameter of the inscribed circle of the triangle $ABC$.
(M. Volchkevich)
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 $\frac{1}{3}\sqrt6 h \le d \le h$.
2016 Sharygin Geometry Olympiad, P15
Let $O, M, N$ be the circumcenter, the centroid and the Nagel point of a triangle. Prove that angle $MON$ is right if and only if one of the triangle’s angles is equal to $60^o$.
2011 Sharygin Geometry Olympiad, 7
Let a point $M$ not lying on coordinates axes be given. Points $Q$ and $P$ move along $Y$ - and $X$-axis respectively so that angle $P M Q$ is always right. Find the locus of points symmetric to $M$ wrt $P Q$.
2017 Saudi Arabia Pre-TST + Training Tests, 9
Let $ABC$ be a triangle inscribed in circle $(O)$, with its altitudes $BH_b, CH_c$ intersect at orthocenter $H$ ($H_b \in AC$, $H_c \in AB$). $H_bH_c$ meets $BC$ at $P$. Let $N$ be the midpoint of $AH, L$ be the orthogonal projection of $O$ on the symmedian with respect to angle $A$ of triangle $ABC$. Prove that $\angle NLP = 90^o$.
2003 IMO Shortlist, 3
Let $n \geq 5$ be a given integer. Determine the greatest integer $k$ for which there exists a polygon with $n$ vertices (convex or not, with non-selfintersecting boundary) having $k$ internal right angles.
[i]Proposed by Juozas Juvencijus Macys, Lithuania[/i]
Brazil L2 Finals (OBM) - geometry, 2004.5
Let $D$ be the midpoint of the hypotenuse $AB$ of a right triangle $ABC$. Let $O_1$ and $O_2$ be the circumcenters of the $ADC$ and $DBC$ triangles, respectively.
a) Prove that $\angle O_1DO_2$ is right.
b) Prove that $AB$ is tangent to the circle of diameter $O_1O_2$ .
2021 Polish Junior MO Second Round, 2
Given is the square $ABCD$. Point $E$ lies on the diagonal $AC$, where $AE> EC$. On the side $AB$, a different point from $B$ has been selected for which $EF = DE$. Prove that $\angle DEF = 90^o$.
2013 Sharygin Geometry Olympiad, 1
Let $ABCDE$ be a pentagon with right angles at vertices $B$ and $E$ and such that $AB = AE$ and $BC = CD = DE$. The diagonals $BD$ and $CE$ meet at point $F$. Prove that $FA = AB$.
2019 Grand Duchy of Lithuania, 3
Let $ABC$ be an acute triangle with orthocenter $H$ and circumcenter $O$. The perpendicular bisector of segment $CH$ intersects the sides $AC$ and $BC$ in points $X$ and $Y$ , respectively. The lines $XO$ and $YO$ intersect the side $AB$ in points $P$ and $Q$, respectively. Prove that if $XP + Y Q = AB + XY$ then $\angle OHC = 90^o$.
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 $3$.
2012 Peru MO (ONEM), 4
In a circle $S$, a chord $AB$ is drawn and let $M$ be the midpoint of the arc $AB$. Let $P$ be a point in segment $AB$ other than its midpoint. The extension of the segment $MP$ cuts $S$ in $Q$. Let $S_1$ be the circle that is tangent to the AP segments and $MP$, and also is tangent to $S$, and let $S_2$ be the circle that is tangent to the segments $BP$ and $MP$, and also tangent to $S$. The common outer tangent lines to the circles $S_1$ and $S_2$ are cut at $C$. Prove that $\angle MQC = 90^o$.
2004 Germany Team Selection Test, 2
Let $n \geq 5$ be a given integer. Determine the greatest integer $k$ for which there exists a polygon with $n$ vertices (convex or not, with non-selfintersecting boundary) having $k$ internal right angles.
[i]Proposed by Juozas Juvencijus Macys, Lithuania[/i]
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 $PH = PT$.
2020 China Team Selection Test, 2
Given an isosceles triangle $\triangle ABC$, $AB=AC$. A line passes through $M$, the midpoint of $BC$, and intersects segment $AB$ and ray $CA$ at $D$ and $E$, respectively. Let $F$ be a point of $ME$ such that $EF=DM$, and $K$ be a point on $MD$. Let $\Gamma_1$ be the circle passes through $B,D,K$ and $\Gamma_2$ be the circle passes through $C,E,K$. $\Gamma_1$ and $\Gamma_2$ intersect again at $L \neq K$. Let $\omega_1$ and $\omega_2$ be the circumcircle of $\triangle LDE$ and $\triangle LKM$. Prove that, if $\omega_1$ and $\omega_2$ are symmetric wrt $L$, then $BF$ is perpendicular to $BC$.
Kyiv City MO Juniors Round2 2010+ geometry, 2021.9.2
In an acute triangle $AB$ the heights $ BE$ and $CF$ intersect at the orthocenter $H$, and $M$ is the midpoint of $BC$. The line $EF$ intersects the lines $MH$ and $BC$ at the points $P$ and $T$ , respectively. $AP$ intersects the cirumcscribed circle of $\vartriangle ABC$ for second time at the point $Q$ . Prove that $\angle AQT= 90^o$.
(Fedir Yudin)
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 $AI=ND$
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 $|TA|^2 =|TB| \cdot |TC|$.