Found problems: 1049
Given points $ A $, $ B $, $ C $ and $ D $ that do not lie in the same plane. Draw a plane through the point $ A $ such that the orthogonal projection of the quadrilateral $ ABCD $ on this plane is a parallelogram.
Let $ABC$ be a triangle whose angles $\alpha=\angle CAB$ and $\beta=\angle CBA$ are greater than $45^{\circ}$. Above the side $AB$ a right isosceles triangle $ABR$ is constructed with $AB$ as the hypotenuse, such that $R$ is inside the triangle $ABC$. Analogously we construct above the sides $BC$ and $AC$ the right isosceles triangles $CBP$ and $ACQ$, right at $P$ and in $Q$, but with these outside the triangle $ABC$. Prove that $CQRP$ is a parallelogram.
$ABC$ is a triangle and $R,Q,P$ are midpoints of $AB,AC,BC$. Line $AP$ intersects $RQ$ in $E$ and circumcircle of $ABC$ in $F$. $T,S$ are on $RP,PQ$ such that $ES\perp PQ,ET\perp RP$. $F'$ is on circumcircle of $ABC$ that $FF'$ is diameter. The point of intersection of $AF'$ and $BC$ is $E'$. $S',T'$ are on $AB,AC$ that $E'S'\perp AB,E'T'\perp AC$. Prove that $TS$ and $T'S'$ are perpendicular.
Let $ABCD$ be a parallelogram. Let $M$ be the midpoint of the arc $AC$ containing $B$ of the circumcircle of $ABC$ . Let $E$ be a point on segment $AD$ and $F$ a point on segment $CD$ such that $ME=MD=MF$. Show that $BMEF$ is cyclic.
[i]Proposed by A. Tereshin[/i]
$ABCD$ is a convex quadrilateral. $K$, $L$, $M$, $N$ are the midpoints of the sides $AB$, $BC$, $CD$, $DA$. $BD$ bisects $KM$ at $Q$. $QA = QB = QC = QD$ , and$\frac{LK}{LM} = \frac{CD}{CB}$. Prove that $ABCD$ is a square
In a triangle $ABC$, $I$ is the incenter. $D$ is the reflection of $A$ to $I$. the incircle is tangent to $BC$ at point $E$. $DE$ cuts $IG$ at $P$ ($G$ is centroid). $M$ is the midpoint of $BC$. prove that
a) $AP||DM$.(15 points)
b) $AP=2DM$. (10 points)
We're given two congruent, equilateral triangles $ABC$ and $PQR$ with parallel sides, but one has one vertex pointing up and the other one has the vertex pointing down. One is placed above the other so that the area of intersection is a hexagon $A_1A_2A_3A_4A_5A_6$ (labelled counterclockwise). Prove that $A_1A_4$, $A_2A_5$ and $A_3A_6$ are concurrent.
Given a convex pentagon $ABCDE$, with $\angle BAC = \angle ABE = \angle DEA - 90^o$, $\angle BCA = \angle ADE$ and also $BC = ED$. Prove that $BCDE$ is parallelogram.
Let $A$ be one of the two points where the circles whose centers are the points $M$ and $N$ intersect. The tangents in $A$ to such circles intersect them again in $B$ and $C$, respectively. Let $P$ be a point such that the quadrilateral $AMPN$ is a parallelogram. Show that $P$ is the circumcenter of triangle $ABC$.
In a convex pentagon every diagonal is parallel to one side. Show that the ratios between the lengths of diagonals and the sides parallel to them are equal and find their value.
Let $H$ be the orthocenter of a triangle $ABC$, and let $D$ be the midpoint of the segment $AH$.
The altitude $BH$ of triangle $ABC$ intersects the perpendicular to the line $AB$ through the point $A$ at the point $M$.
The altitude $CH$ of triangle $ABC$ intersects the perpendicular to the line $CA$ through the point $A$ at the point $N$.
The perpendicular bisector of the segment $AB$ intersects the perpendicular to the line $BC$ through the point $B$ at the point $U$.
The perpendicular bisector of the segment $CA$ intersects the perpendicular to the line $BC$ through the point $C$ at the point $V$.
Finally, let $E$ be the midpoint of the side $BC$ of triangle $ABC$.
Prove that the points $D$, $M$, $N$, $U$, $V$ all lie on one and the same perpendicular to the line $AE$.
[i]Extensions.[/i] In other words, we have to show that the points $M$, $N$, $U$, $V$ lie on the perpendicular to the line $AE$ through the point $D$. Additionally, one can find two more points on this perpendicular:
[b](a)[/b] The nine-point circle of triangle $ABC$ is known to pass through the midpoint $E$ of its side $BC$. Let $D^{\prime}$ be the point where this nine-point circle intersects the line $AE$ apart from $E$. Then, the point $D^{\prime}$ lies on the perpendicular to the line $AE$ through the point $D$.
[b](b)[/b] Let the tangent to the circumcircle of triangle $ABC$ at the point $A$ intersect the line $BC$ at a point $X$. Then, the point $X$ lies on the perpendicular to the line $AE$ through the point $D$.
[i]Comment.[/i] The actual problem was created by Victor Thébault around 1950 (cf. Hyacinthos messages #1102 and #1551). The extension [b](a)[/b] initially was a (pretty trivial) lemma in Thébault's solution of the problem. Extension [b](b)[/b] is rather new; in the form "prove that $X\in UV$", it was [url=http://www.mathlinks.ro/Forum/viewtopic.php?t=3659]proposed by Valentin Vornicu for the Balkan MO 2003[/url], however it circulated in the Hyacinthos newsgroup before (Hyacinthos messages #7240 and #7242), where different solutions of the problem were discussed as well. Hereby, "Hyacinthos" always refers to the triangle geometry newsgroup "Hyacinthos", which can be found at http://groups.yahoo.com/group/Hyacinthos .
I proposed the problem for the QEDMO math fight wishing to draw some attention to it. It has a rather short and elementary solution, by the way (without using radical axes or inversion like the standard solutions).
Darij
Let $ABCD$ be a trapezoid such that $|AC|=8$, $|BD|=6$, and $AD \parallel BC$. Let $P$ and $S$ be the midpoints of $[AD]$ and $[BC]$, respectively. If $|PS|=5$, find the area of the trapezoid $ABCD$.
Let $ABCD$ be a parallelogram with $AB>BC$ and $\angle DAB$ less than $\angle ABC$. The perpendicular bisectors of sides $AB$ and $BC$ intersect at the point $M$ lying on the extension of $AD$. If $\angle MCD=15^{\circ}$, find the measure of $\angle ABC$
A quadrilateral $ABCD$ is inscribed into a circle with center $O.$ Points $P$ and $Q$ are opposite to $C$ and $D$ respectively. Two tangents drawn to that circle at these points meet the line $AB$ in points $E$ and $F.$ ($A$ is between $E$ and $B$, $B$ is between $A$ and $F$). The line $EO$ meets $AC$ and $BC$ in points $X$ and $Y$ respectively, and the line $FO$ meets $AD$ and $BD$ in points $U$ and $V$ respectively. Prove that $XV=YU.$
Let $ABC$ be a triangle. The incircle of $ABC$ touches the sides $AB$ and $AC$ at the points $Z$ and $Y$, respectively. Let $G$ be the point where the lines $BY$ and $CZ$ meet, and let $R$ and $S$ be points such that the two quadrilaterals $BCYR$ and $BCSZ$ are parallelogram.
Prove that $GR=GS$.
[i]Proposed by Hossein Karke Abadi, Iran[/i]
The graph of $y^2+2xy+40|x|=400$ partitions the plane into several regions. What is the area of the bounded region?
Let $A'$ and $B'$ be feet of altitudes from $A$ and $B$, respectively, in acute-angled triangle $ABC$ ($AC\not = BC$). Circle $k$ contains points $A'$ and $B'$ and touches segment $AB$ in $D$. If triangles $ADA'$ and $BDB'$ have the same area, prove that \[\angle A'DB'= \angle ACB.\]
On the diagonals $AC$ and $BD$ of the inscribed quadrilateral A$BCD$, the points $X$ and $Y$ are marked, respectively, so that the quadrilateral $ABXY$ is a parallelogram. Prove that the circumscribed circles of triangles $BXD$ and $CYA$ have equal radii.
(Vyacheslav Yasinsky)
Let $ABC$ be an acute-angled triangle with $AB<AC,$ and let $D$ be the foot of its altitude from$A,$ points $B'$ and $C'$ lie on the rays $AB$ and $AC,$ respectively , so that points $B',$ $C'$ and $D$ are collinear and points $B,$ $C,$ $B'$ and $C'$ lie on one circle with center $O.$ Prove that if $M$ is the midpoint of $BC$ and $H$ is the orthocenter of $ABC,$ then $DHMO$ is a parallelogram.
Let $\omega_P$ and $\omega_Q$ be two circles of radius $1$, intersecting in points $A$ and $B$. Let $P$ and $Q$ be two regular $n$-gons (for some positive integer $n\ge4$) inscribed in $\omega_P$ and $\omega_Q$, respectively, such that $A$ and $B$ are vertices of both $P$ and $Q$. Suppose a third circle $\omega$ of radius $1$ intersects $P$ at two of its vertices $C$, $D$ and intersects $Q$ at two of its vertices $E$, $F$. Further assume that $A$, $B$, $C$, $D$, $E$, $F$ are all distinct points, that $A$ lies outside of $\omega$, and that $B$ lies inside $\omega$. Show that there exists a regular $2n$-gon that contains $C$, $D$, $E$, $F$ as four of its vertices.
Let $ABCD$ be a convex quadrilateral with $AB$ is not parallel to $CD$. Circle $\omega_1$ with center $O_1$ passes through $A$ and $B$, and touches segment $CD$ at $P$. Circle $\omega_2$ with center $O_2$ passes through $C$ and $D$, and touches segment $AB$ at $Q$. Let $E$ and $F$ be the intersection of circles $\omega_1$ and $\omega_2$. Prove that $EF$ bisects segment $PQ$ if and only if $BC$ is parallel to $AD$.
Let ABCD be a parallelogram, M the midpoint of AB and N the intersection of CD
and the angle bisector of ABC. Prove that CM and BN are perpendicular iff AN is the angle bisector of DAB.
Two circles in the plane, $\gamma_1$ and $\gamma_2$, meet at points $M$ and $N$. Let $A$ be a point on $\gamma_1$, and let $D$ be a point on $\gamma_2$. The lines $AM$ and $AN$ meet again $\gamma_2$ at points $B$ and $C$, respectively, and the lines $DM$ and $DN$ meet again $\gamma_1$ at points $E$ and $F$, respectively. Assume the order $M$, $N$, $F$, $A$, $E$ is circular around $\gamma_1$, and the segments $AB$ and $DE$ are congruent. Prove that the points $A$, $F$, $C$ and $D$ lie on a circle whose centre does not depend on the position of the points $A$ and $D$ on the respective circles, subject to the assumptions above.
[i]***[/i]
Let $ ABCD$ be an inscribed quadrilateral and let $ E$ be the midpoint of the diagonal $ BD$. Let $ \Gamma_1,\Gamma_2,\Gamma_3,\Gamma_4$ be the circumcircles of triangles $ AEB$, $ BEC$, $ CED$ and $ DEA$ respectively. Prove that if $ \Gamma_4$ is tangent to the line $ CD$, then $ \Gamma_1,\Gamma_2,\Gamma_3$ are tangent to the lines $ BC,AB,AD$ respectively.
Let $O$ be the centre of a parallelogram $ABCD$ and $P$ be any point in the plane. Let $M, N$ be the midpoints of $AP, BP$, respectively and $Q$ be the intersection of $MC$ and $ND$. Prove that $O, P$ and $Q$ are collinear.