Found problems: 131
Cono Sur Shortlist - geometry, 2003.G3
An interior $P$ point to a square $ABCD$ is such that $PA = a, PB = b$ and $PC = b + c$, where the numbers $a, b$ and $c$ satisfy the relationship $a^2 = b^2 + c^2$. Prove that the angle $BPC$ is right.
Indonesia MO Shortlist - geometry, g5
Two circles intersect at points $A$ and $B$. The line $\ell$ through A intersects the circles at $C$ and $D$, respectively. Let $M, N$ be the midpoints of arc $BC$ and arc $BD$. which does not contain $A$, and suppose that $K$ is the midpoint of the segment $CD$ . Prove that $\angle MKN=90^o$.
Kyiv City MO Juniors 2003+ geometry, 2018.7.4
Inside the triangle $ABC $, the point $P $ is selected so that $BC = AP $ and $\angle APC = 180 {} ^ \circ - \angle ABC $. On the side $AB $ there is a point $K $, for which $AK = KB + PC $. Prove that $\angle AKC = 90 {} ^ \circ $.
(Danilo Hilko)
2005 Cuba MO, 5
On the circumcircle of triangle $ABC$, point $P$ is taken in such a way that the perpendicular drawn by the point $P$ to the line $AC$ cuts the circle also at the point $Q$, the perpendicular drawn by the point $Q$ to the line $AB$ cuts the circle also at point R and the perpendicular drawn by point $R$ to the line BC cuts the circle also at the point $P$. Let $O$ be the center of this circle. Prove that $\angle POC = 90^o$ .
2021 Saudi Arabia IMO TST, 5
Let $ABC$ be a non isosceles triangle with incenter $I$ . The circumcircle of the triangle $ABC$ has radius $R$. Let $AL$ be the external angle bisector of $\angle BAC $with $L \in BC$. Let $K$ be the point on perpendicular bisector of $BC$ such that $IL \perp IK$.Prove that $OK=3R$.
Kyiv City MO 1984-93 - geometry, 1992.7.2
Inside a right angle is given a point $A$. Construct an equilateral triangle, one of the vertices of which is point $A$, and two others lie on the sides of the angle (one on each side).
2000 Junior Balkan Team Selection Tests - Moldova, 7
Let a triangle $ABC, A_1$ be the midpoint of the segment $[BC], B_1 \in (AC)$ ¸and $C_1 \in (AB)$ such that $[A_1B_1$ is the bisector of the angle $AA_1C$ and $A_1C_1$ is perpendicular to $AB$. Show that the lines $AA_1, BB_1$ and $CC_1$ are concurrent if and only if $ \angle BAC = 90^o$
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 $\frac{1}{3}\sqrt6 h \le d \le h$.
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 $AP$.
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 $3$.
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 $|MA| = |MC|$.
2015 India Regional MathematicaI Olympiad, 1
In a cyclic quadrilateral $ABCD$, let the diagonals $AC$ and $BD$ intersect at $X$. Let the circumcircles of triangles $AXD$ and $BXC$ intersect again at $Y$ . If $X$ is the incentre of triangle $ABY$ , show that $\angle CAD = 90^o$.
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 $\angle NML = 90^o$.
2007 Estonia Math Open Junior Contests, 7
The center of square $ABCD$ is $K$. The point $P$ is chosen such that $P \ne K$ and the angle $\angle APB$ is right . Prove that the line $PK$ bisects the angle between the lines $AP$ and $BP$.
2024 Bulgaria MO Regional Round, 12.1
Let $ABC$ be an acute triangle with midpoint $M$ of $AB$. The point $D$ lies on the segment $MB$ and $I_1, I_2$ denote the incenters of $\triangle ADC$ and $\triangle BDC$. Given that $\angle I_1MI_2=90^{\circ}$, show that $CA=CB$.
2017 Ukrainian Geometry Olympiad, 4
Let $ABCD$ be a parallelogram and $P$ be an arbitrary point of the circumcircle of $\Delta ABD$, different from the vertices. Line $PA$ intersects the line $CD$ at point $Q$. Let $O$ be the center of the circumcircle $\Delta PCQ$. Prove that $\angle ADO = 90^o$.
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 $AD$.
2006 MOP Homework, 2
Points $P$ and $Q$ lies inside triangle $ABC$ such that $\angle ACP =\angle BCQ$ and $\angle CAP = \angle BAQ$. Denote by $D,E$, and $F$ the feet of perpendiculars from $P$ to lines $BC,CA$, and $AB$, respectively. Prove that if $\angle DEF = 90^o$, then $Q$ is the orthocenter of triangle $BDF$.
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]
Kyiv City MO Juniors Round2 2010+ geometry, 2021.7.41
Point $C$ lies inside the right angle $AOB$. Prove that the perimeter of triangle $ABC$ is greater than $2 OC$.
Estonia Open Senior - geometry, 2004.2.4
On the circumcircle of triangle $ABC$, point $P$ is chosen, such that the perpendicular drawn from point $P$ to line $AC$ intersects the circle again at a point $Q$, the perpendicular drawn from point $Q$ to line $AB$ intersects the circle again at a point $R$ and the perpendicular drawn from point $R$ to line $BC$ intersects the circle again at the initial point $P$. Let $O$ be the centre of this circle. Prove that $\angle POC = 90^o$.
May Olympiad L2 - geometry, 2001.2
On the trapezoid $ABCD$ , side $DA$ is perpendicular to the bases $AB$ and $CD$ . The base $AB$ measures $45$, the base $CD$ measures $20$ and the $BC$ side measures $65$. Let $P$ on the $BC$ side such that $BP$ measures $45$ and $M$ is the midpoint of $DA$. Calculate the measure of the $PM$ segment.
2005 Oral Moscow Geometry Olympiad, 4
Given a hexagon $ABCDEF$, in which $AB = BC, CD = DE, EF = FA$, and angles $A$ and $C$ are right. Prove that lines $FD$ and $BE$ are perpendicular.
(B. Kukushkin)
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.
2019 Polish Junior MO First Round, 5
A parallelogram $ABCD$ is given. On the diagonal BD, a point $P$ is selected such that $AP = BD$ is satisfied. Point $Q$ is the midpoint of segment $CP$. Prove that $\angle BQD = 90^o$.
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