Found problems: 1049
Let $ABCD$ and $A^{\prime }B^{\prime}C^{\prime }D^{\prime }$ be two arbitrary parallelograms in the space, and let $M,$ $N,$ $P,$ $Q$ be points dividing the segments $AA^{\prime },$ $BB^{\prime },$ $CC^{\prime },$ $DD^{\prime }$ in equal ratios.
[b]a.)[/b] Prove that the quadrilateral $MNPQ$ is a parallelogram.
[b]b.)[/b] What is the locus of the center of the parallelogram $MNPQ,$ when the point $M$ moves on the segment $AA^{\prime }$ ?
(Consecutive vertices of the parallelograms are labelled in alphabetical order.
The solutions to the equations $z^2=4+4\sqrt{15}i$ and $z^2=2+2\sqrt 3i,$ where $i=\sqrt{-1},$ form the vertices of a parallelogram in the complex plane. The area of this parallelogram can be written in the form $p\sqrt q-r\sqrt s,$ where $p,$ $q,$ $r,$ and $s$ are positive integers and neither $q$ nor $s$ is divisible by the square of any prime number. What is $p+q+r+s?$
$\textbf{(A) } 20 \qquad
\textbf{(B) } 21 \qquad
\textbf{(C) } 22 \qquad
\textbf{(D) } 23 \qquad
\textbf{(E) } 24 $
A convex quadrilateral $ABCD$ is inscribed in a circle with center $O$. The diagonals $AC$, $BD$ of $ABCD$ meet at $P$. Circumcircles of $\triangle ABP$ and $\triangle CDP$ meet at $P$ and $Q$ ($O,P,Q$ are pairwise distinct). Show that $\angle OQP=90^{\circ}$.
Let $ABC$ be a triangle with incenter $I$, and let $D$ be a point on side $BC$. Points $X$ and $Y$ are chosen on lines $BI$ and $CI$ respectively such that $DXIY$ is a parallelogram. Points $E$ and $F$ are chosen on side $BC$ such that $AX$ and $AY$ are the angle bisectors of angles $\angle BAE$ and $\angle CAF$ respectively. Let $\omega$ be the circle tangent to segment $EF$, the extension of $AE$ past $E$, and the extension of $AF$ past $F$. Prove that $\omega$ is tangent to the circumcircle of triangle $ABC$.
In a quadrilateral $ABCD$ with an incircle, $AB = CD; BC < AD$ and $BC$ is parallel to $AD$. Prove that the bisector of $\angle C$ bisects the area of $ABCD$.
Let $ABC$ be a triangle inscribed in circle $\omega$, and let the medians from $B$ and $C$ intersect $\omega$ at $D$ and $E$ respectively. Let $O_1$ be the center of the circle through $D$ tangent to $AC$ at $C$, and let $O_2$ be the center of the circle through $E$ tangent to $AB$ at $B$. Prove that $O_1$, $O_2$, and the nine-point center of $ABC$ are collinear.
[i]Proposed by Michael Kural[/i]
Isosceles, similar triangles $QPA$ and $SPB$ are constructed (outwards) on the sides of parallelogram $PQRS$ (where $PQ = AQ$ and $PS = BS$). Prove that triangles $RAB$, $QPA$ and $SPB$ are similar.
From point $D$ of parallelogram $ABCD$ were drawn an arbitrary line $\ell_1$, intersecting the segment $AB$ and the line $BC$ at points $C_1$ and $A_1$, respectively, and an arbitrary line $\ell_2$ intersecting the segment $BC$ and the line $AB$ at the points $A_2$ and $C_2$, respectively. Find the locus of the intersection points of the circles $(A_1BC_2)$ and $(A_2BC_1)$ (other than point $B$).
Let the intersections of $\odot O_1$ and $\odot O_2$ be $A$ and $B$. Point $R$ is on arc $AB$ of $\odot O_1$ and $T$ is on arc $AB$ on $\odot O_2$. $AR$ and $BR$ meet $\odot O_2$ at $C$ and $D$; $AT$ and $BT$ meet $\odot O_1$ at $Q$ and $P$. If $PR$ and $TD$ meet at $E$ and $QR$ and $TC$ meet at $F$, then prove: $AE \cdot BT \cdot BR = BF \cdot AT \cdot AR$.
Inside parallelogram $ABCD$ is point $P$, such that $PC = BC$. Show that line $BP$ is perpendicular to line which connects middles of sides of line segments $AP$ and $CD$.
Extensions of opposite sides of a convex quadrilateral $ABCD$ intersect at points $P$ and $Q$. Points are marked on the sides of $ABCD$ (one per side), which are the vertices of a parallelogram with a side parallel to $PQ$. Prove that the intersection point of the diagonals of this parallelogram lies on one of the diagonals of quadrilateral $ABCD$.
(E. Bakaev)
Let $ABC$ be an acute triangle with $AB<AC$, and let $D$ and $E$ be points on side $BC$ such that $BD=CE$ and $D$ lies between $B$ and $E$. Suppose there exists a point $P$ inside $ABC$ such that $PD\parallel AE$ and $\angle PAB=\angle EAC$. Prove that $\angle PBA=\angle PCA$.
[i]Calvin Deng.[/i]
Let $\omega$ be the circumcircle of $\triangle{ABC}$. $P$ is an interior point of $\triangle{ABC}$. $A_{1}, B_{1}, C_{1}$ are the intersections of $AP, BP, CP$ respectively and $A_{2}, B_{2}, C_{2}$ are the symmetrical points of $A_{1}, B_{1}, C_{1}$ with respect to the midpoints of side $BC, CA, AB$.
Show that the circumcircle of $\triangle{A_{2}B_{2}C_{2}}$ passes through the orthocentre of $\triangle{ABC}$.
$\textbf{Bake Sale}$
Four friends, Art, Roger, Paul and Trisha, bake cookies, and all cookies have the same thickness. The shapes of the cookies di ffer, as shown.
$\circ$ Art's cookies are trapezoids:
[asy]size(80);defaultpen(linewidth(0.8));defaultpen(fontsize(8));
draw(origin--(5,0)--(5,3)--(2,3)--cycle);
draw(rightanglemark((5,3), (5,0), origin));
label("5 in", (2.5,0), S);
label("3 in", (5,1.5), E);
label("3 in", (3.5,3), N);[/asy]
$\circ$ Roger's cookies are rectangles:
[asy]size(80);defaultpen(linewidth(0.8));defaultpen(fontsize(8));
draw(origin--(4,0)--(4,2)--(0,2)--cycle);
draw(rightanglemark((4,2), (4,0), origin));
draw(rightanglemark((0,2), origin, (4,0)));
label("4 in", (2,0), S);
label("2 in", (4,1), E);[/asy]
$\circ$ Paul's cookies are parallelograms:
[asy]size(80);defaultpen(linewidth(0.8));defaultpen(fontsize(8));
draw(origin--(3,0)--(2.5,2)--(-0.5,2)--cycle);
draw((2.5,2)--(2.5,0), dashed);
draw(rightanglemark((2.5,2),(2.5,0), origin));
label("3 in", (1.5,0), S);
label("2 in", (2.5,1), W);[/asy]
$\circ$ Trisha's cookies are triangles:
[asy]size(80);defaultpen(linewidth(0.8));defaultpen(fontsize(8));
draw(origin--(3,0)--(3,4)--cycle);
draw(rightanglemark((3,4),(3,0), origin));
label("3 in", (1.5,0), S);
label("4 in", (3,2), E);[/asy]
Each friend uses the same amount of dough, and Art makes exactly 12 cookies. How many cookies will be in one batch of Trisha's cookies?
$ \textbf{(A)}\ 10\qquad\textbf{(B)}\ 12\qquad\textbf{(C)}\ 16\qquad\textbf{(D)}\ 18\qquad\textbf{(E)}\ 24$
Triangle $ ABC$ has $ AC \equal{} 450$ and $ BC \equal{} 300$. Points $ K$ and $ L$ are located on $ \overline{AC}$ and $ \overline{AB}$ respectively so that $ AK \equal{} CK$, and $ \overline{CL}$ is the angle bisector of angle $ C$. Let $ P$ be the point of intersection of $ \overline{BK}$ and $ \overline{CL}$, and let $ M$ be the point on line $ BK$ for which $ K$ is the midpoint of $ \overline{PM}$. If $ AM \equal{} 180$, find $ LP$.
Let $ABCD$ be a convex quadrilateral. Assume line $AB$ and $CD$ intersect at $E$, and $B$ lies between $A$ and $E$. Assume line $AD$ and $BC$ intersect at $F$, and $D$ lies between $A$ and $F$. Assume the circumcircles of $\triangle BEC$ and $\triangle CFD$ intersect at $C$ and $P$. Prove that $\angle BAP=\angle CAD$ if and only if $BD\parallel EF$.
Show that for every convex polygon whose area is less than or equal to $1$, there exists a parallelogram with area $2$ containing the polygon.
Geodude wants to assign one of the integers $1,2,3,\ldots,11$ to each lattice point $(x,y,z)$ in a 3D Cartesian coordinate system. In how many ways can Geodude do this if for every lattice parallelogram $ABCD$, the positive difference between the sum of the numbers assigned to $A$ and $C$ and the sum of the numbers assigned to $B$ and $D$ must be a multiple of $11$? (A [i]lattice point[/i] is a point with all integer coordinates. A [i]lattice parallelogram[/i] is a parallelogram with all four vertices lying on lattice points. Here, we say four not necessarily distinct points $A,B,C,D$ form a [i]parallelogram[/i] $ABCD$ if and only if the midpoint of segment $AC$ coincides with the midpoint of segment $BD$.)
[hide="Clarifications"]
[list]
[*] The ``positive difference'' between two real numbers $x$ and $y$ is the quantity $|x-y|$. Note that this may be zero.
[*] The last sentence was added to remove confusion about ``degenerate parallelograms.''[/list][/hide]
[i]Victor Wang[/i]
A regular hexagon is cut up into $N$ parallelograms of equal area. Prove that $N$ is divisible by three.
(V. Prasolov, I. Sharygin, Moscow)
Let $\omega$ be the circumcircle of acute triangle $ABC$. Two tangents of $\omega$ from $B$ and $C$ intersect at $P$, $AP$ and $BC$ intersect at $D$. Point $E$, $F$ are on $AC$ and $AB$ such that $DE \parallel BA$ and $DF \parallel CA$.
(1) Prove that $F,B,C,E$ are concyclic.
(2) Denote $A_{1}$ the centre of the circle passing through $F,B,C,E$. $B_{1}$, $C_{1}$ are difined similarly. Prove that $AA_{1}$, $BB_{1}$, $CC_{1}$ are concurrent.
In a fixed plane, consider a convex quadrilateral $ABCD$. Choose a point $O$ in the plane and let $K, L, M$, and $N$ be the circumcentres of triangles $AOB, BOC, COD$, and $DOA$, respectively. Prove that there exists exactly one point $O$ in the plane such that $KLMN$ is a parallelogram.
The circles $\mathcal C_1$ and $\mathcal C_2$ touch each other externally at $D$, and touch a circle $\omega$ internally at $B$ and $C$, respectively. Let $A$ be an intersection point of $\omega$ and the common tangent to $\mathcal C_1$ and $\mathcal C_2$ at $D$. Lines $AB$ and $AC$ meet $\mathcal C_1$ and $\mathcal C_2$ again at $K$ and $L$, respectively, and the line $BC$ meets $\mathcal C_1$ again at $M$ and $\mathcal C_2$ again at $N$. Prove that the lines $AD$, $KM$, $LN$ are concurrent.
[i]Greece[/i]
Let $ABCD$ be a convex quadrilateral whose sides $AD$ and $BC$ are not parallel. Suppose that the circles with diameters $AB$ and $CD$ meet at points $E$ and $F$ inside the quadrilateral. Let $\omega_E$ be the circle through the feet of the perpendiculars from $E$ to the lines $AB,BC$ and $CD$. Let $\omega_F$ be the circle through the feet of the perpendiculars from $F$ to the lines $CD,DA$ and $AB$. Prove that the midpoint of the segment $EF$ lies on the line through the two intersections of $\omega_E$ and $\omega_F$.
[i]Proposed by Carlos Yuzo Shine, Brazil[/i]
The sides $AB$, $BC$, $CD$ and $DA$ of the quadrilateral $ABCD$ are respectively equal to the sides $A'B'$, $B'C'$, $C'D' $ and $D'A'$ of the quadrilateral $A'B'CD$' and it is known that $AB \parallel CD$ and $B'C' \parallel D'A'$. Prove that both quadrilaterals are parallelograms.
(V Proizvolov, Moscow)
Let $ABCD$ be a rhombus, let $APQC$ be a parallelogram such that the point $B$ lies inside it and the side $AP$ is equal to the side of the rhombus. Prove that $B$ is the orthocenter of the triangle $DPQ$.
Egor Bakaev