As shown below, there is a $40\times30$ paper with a filled $10\times5$ rectangle inside of it. We want to cut out the filled rectangle from the paper using four straight cuts. Each straight cut is a straight line that divides the paper into two pieces, and we keep the piece containing the filled rectangle. The goal is to minimize the total length of the straight cuts. How to achieve this goal, and what is that minimized length? Show the correct cuts and write the final answer. There is no need to prove the answer. Proposed by Morteza Saghafian

Convex hexagon $A_1A_2A_3A_4A_5A_6$ lies in the interior of convex hexagon $B_1B_2B_3B_4B_5B_6$ such that $A_1A_2 \parallel B_1B_2$, $A_2A_3 \parallel B_2B_3$,..., $A_6A_1 \parallel B_6B_1$. Prove that the areas of simple hexagons $A_1B_2A_3B_4A_5B_6$ and $B_1A_2B_3A_4B_5A_6$ are equal. (A simple hexagon is a hexagon which does not intersect itself.) Proposed by Hirad Aalipanah - Mahdi Etesamifard

In the given figure, $ABCD$ is a parallelogram. We know that $\angle D = 60^\circ$, $AD = 2$ and $AB = \sqrt3 + 1$. Point $M$ is the midpoint of $AD$. Segment $CK$ is the angle bisector of $C$. Find the angle $CKB$. Proposed by Mahdi Etesamifard

There are two circles with centers $O_1,O_2$ lie inside of circle $\omega$ and are tangent to it. Chord $AB$ of $\omega$ is tangent to these two circles such that they lie on opposite sides of this chord. Prove that $\angle O_1AO_2 + \angle O_1BO_2 > 90^\circ$. Proposed by Iman Maghsoudi

There are some segments on the plane such that no two of them intersect each other (even at the ending points). We say segment $AB$ breaks segment $CD$ if the extension of $AB$ cuts $CD$ at some point between $C$ and $D$. [asy][asy] /* Geogebra to Asymptote conversion, documentation at artofproblemsolving.com/Wiki go to User:Azjps/geogebra */ import graph; size(4cm); real labelscalefactor = 0.5; /* changes label-to-point distance */ pen dps = linewidth(0.7) + fontsize(10); defaultpen(dps); /* default pen style */ pen dotstyle = black; /* point style */ real xmin = -5.267474904743955, xmax = 11.572179069738377, ymin = -10.642621257034536, ymax = 4.543526642434019; /* image dimensions */ /* draw figures */ draw((-4,-2)--(1.08,-2.03), linewidth(2)); draw(shift((-2.1866176795507295,-2.0107089507113147))*scale(0.21166666666666667)*(expi(pi/4)--expi(5*pi/4)^^expi(3*pi/4)--expi(7*pi/4))); /* special point */ draw((-0.16981767035094117,3.225314210196242)--(-2.1866176795507295,-2.0107089507113147), linewidth(2) + linetype("4 4")); draw((-0.16981767035094117,3.225314210196242)--(-0.8194002739586808,1.538865607509914), linewidth(2)); label("$A$",(-1.2684397405642523,3.860690076971137),SE*labelscalefactor,fontsize(16)); label("$B$",(-1.9211395070170559,2.002590777612728),SE*labelscalefactor,fontsize(16)); label("$C$",(-4.971261820527631,-1.6571211388676117),SE*labelscalefactor,fontsize(16)); label("$D$",(1.08925640451367566,-1.6571211388676117),SE*labelscalefactor,fontsize(16)); /* dots and labels */ dot((-4,-2),dotstyle); dot((1.08,-2.03),dotstyle); dot((-0.16981767035094117,3.225314210196242),dotstyle); dot((-0.8194002739586808,1.538865607509914),dotstyle); clip((xmin,ymin)--(xmin,ymax)--(xmax,ymax)--(xmax,ymin)--cycle); /* end of picture */ [/asy][/asy] $a)$ Is it possible that each segment when extended from both ends, breaks exactly one other segment from each way? [asy][asy] /* Geogebra to Asymptote conversion, documentation at artofproblemsolving.com/Wiki go to User:Azjps/geogebra */ import graph; size(4cm); real labelscalefactor = 0.5; /* changes label-to-point distance */ pen dps = linewidth(0.7) + fontsize(10); defaultpen(dps); /* default pen style */ pen dotstyle = black; /* point style */ real xmin = -6.8, xmax = 8.68, ymin = -10.32, ymax = 3.64; /* image dimensions */ /* draw figures */ draw((-2.56,1.24)--(-0.36,1.4), linewidth(2)); draw((-3.32,-2.68)--(-1.24,-3.08), linewidth(2)); draw(shift((-2.551651190956802,-2.8277593863544612))*scale(0.17638888888888887)*(expi(pi/4)--expi(5*pi/4)^^expi(3*pi/4)--expi(7*pi/4))); /* special point */ draw(shift((-0.8889576602618603,1.3615303519809556))*scale(0.17638888888888887)*(expi(pi/4)--expi(5*pi/4)^^expi(3*pi/4)--expi(7*pi/4))); /* special point */ draw((-2.551651190956802,-2.8277593863544612)--(-0.8889576602618603,1.3615303519809556), linewidth(2) + linetype("4 4")); draw((-1.4097008194020806,0.049476186483185636)--(-1.8514772275312024,-1.0636149148218605), linewidth(2)); /* dots and labels */ dot((-2.56,1.24),dotstyle); dot((-0.36,1.4),dotstyle); dot((-3.32,-2.68),dotstyle); dot((-1.24,-3.08),dotstyle); dot((-1.4097008194020806,0.049476186483185636),dotstyle); dot((-1.8514772275312024,-1.0636149148218605),dotstyle); clip((xmin,ymin)--(xmin,ymax)--(xmax,ymax)--(xmax,ymin)--cycle); /* end of picture */ [/asy][/asy] $b)$ A segment is called surrounded if from both sides of it, there is exactly one segment that breaks it. (e.g. segment $AB$ in the figure.) Is it possible to have all segments to be surrounded? [asy][asy] /* Geogebra to Asymptote conversion, documentation at artofproblemsolving.com/Wiki go to User:Azjps/geogebra */ import graph; size(7cm); real labelscalefactor = 0.5; /* changes label-to-point distance */ pen dps = linewidth(0.7) + fontsize(10); defaultpen(dps); /* default pen style */ pen dotstyle = black; /* point style */ real xmin = -10.70976151557872, xmax = 18.64292748469251, ymin = -16.354300717041443, ymax = 9.136192362141452; /* image dimensions */ /* draw figures */ draw((1.0313140845297686,0.748205038977829)--(-1.3,-4), linewidth(2.8)); draw((-5.780195085389632,-2.13088646583346)--(-2.549994860479401,-2.13088646583346), linewidth(2.8)); draw((4.121070821400425,-3.816208322308361)--(1.78,-1.88), linewidth(2.8)); draw(shift((-0.38228674372374466,-2.13088646583346))*scale(0.21166666666666667)*(expi(pi/4)--expi(5*pi/4)^^expi(3*pi/4)--expi(7*pi/4))); /* special point */ draw((-2.549994860479401,-2.13088646583346)--(-0.38228674372374466,-2.13088646583346), linewidth(2.8) + linetype("4 4")); draw(shift((0.32979226045261084,-0.6805897691262632))*scale(0.21166666666666667)*(expi(pi/4)--expi(5*pi/4)^^expi(3*pi/4)--expi(7*pi/4))); /* special point */ draw((4.121070821400425,-3.816208322308361)--(0.32979226045261084,-0.6805897691262632), linewidth(2.8) + linetype("4 4")); draw((-3.6313140845297687,-8.74820503897783)--(3.600422205681574,5.980726991931396), linewidth(2.8) + linetype("2 2")); label("$A$",(-0.397698406272906,1.754593418658662),SE*labelscalefactor,fontsize(16)); label("$B$",(-2.6377720405041316,-3.266261278756151),SE*labelscalefactor,fontsize(16)); /* dots and labels */ dot((1.0313140845297686,0.748205038977829),linewidth(6pt) + dotstyle); dot((-1.3,-4),linewidth(6pt) + dotstyle); dot((-5.780195085389632,-2.13088646583346),linewidth(6pt) + dotstyle); dot((-2.549994860479401,-2.13088646583346),linewidth(6pt) + dotstyle); dot((4.121070821400425,-3.816208322308361),linewidth(6pt) + dotstyle); dot((1.78,-1.88),linewidth(6pt) + dotstyle); clip((xmin,ymin)--(xmin,ymax)--(xmax,ymax)--(xmax,ymin)--cycle); /* end of picture */ [/asy][/asy] Proposed by Morteza Saghafian

There are three rectangles in the following figure. The lengths of some segments are shown. Find the length of the segment $XY$ . Proposed by Hirad Aalipanah

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

Let $\omega_1,\omega_2$ be two circles with centers $O_1$ and $O_2$, respectively. These two circles intersect each other at points $A$ and $B$. Line $O_1B$ intersects $\omega_2$ for the second time at point $C$, and line $O_2A$ intersects $\omega_1$ for the second time at point $D$ . Let $X$ be the second intersection of $AC$ and $\omega_1$. Also $Y$ is the second intersection point of $BD$ and $\omega_2$. Prove that $CX = DY$ . Proposed by Alireza Dadgarnia

We have a polyhedron all faces of which are triangle. Let $P$ be an arbitrary point on one of the edges of this polyhedron such that $P$ is not the midpoint or endpoint of this edge. Assume that $P_0 = P$. In each step, connect $P_i$ to the centroid of one of the faces containing it. This line meets the perimeter of this face again at point $P_{i+1}$. Continue this process with $P_{i+1}$ and the other face containing $P_{i+1}$. Prove that by continuing this process, we cannot pass through all the faces. (The centroid of a triangle is the point of intersection of its medians.) Proposed by Mahdi Etesamifard - Morteza Saghafian

Suppose that $ABCD$ is a parallelogram such that $\angle DAC = 90^o$. Let $H$ be the foot of perpendicular from $A$ to $DC$, also let $P$ be a point along the line $AC$ such that the line $PD$ is tangent to the circumcircle of the triangle $ABD$. Prove that $\angle PBA = \angle DBH$. Proposed by Iman Maghsoudi

Two circles $\omega_1,\omega_2$ intersect each other at points $A,B$. Let $PQ$ be a common tangent line of these two circles with $P \in \omega_1$ and $Q \in \omega_2$. An arbitrary point $X$ lies on $\omega_1$. Line $AX$ intersects $ \omega_2$ for the second time at $Y$ . Point $Y'\ne Y$ lies on $\omega_2$ such that $QY = QY'$. Line $Y'B$ intersects $ \omega_1$ for the second time at $X'$. Prove that $PX = PX'$. Proposed by Morteza Saghafian

In acute triangle $ABC, \angle A = 45^o$. Points $O,H$ are the circumcenter and the orthocenter of $ABC$, respectively. $D$ is the foot of altitude from $B$. Point $X$ is the midpoint of arc $AH$ of the circumcircle of triangle $ADH$ that contains $D$. Prove that $DX = DO$. Proposed by Fatemeh Sajadi

Find all possible values of integer $n > 3$ such that there is a convex $n$-gon in which, each diagonal is the perpendicular bisector of at least one other diagonal. Proposed by Mahdi Etesamifard

Quadrilateral $ABCD$ is circumscribed around a circle. Diagonals $AC,BD$ are not perpendicular to each other. The angle bisectors of angles between these diagonals, intersect the segments $AB,BC,CD$ and $DA$ at points $K,L,M$ and $N$. Given that $KLMN$ is cyclic, prove that so is $ABCD$. Proposed by Nikolai Beluhov (Bulgaria)

$ABCD$ is a cyclic quadrilateral. A circle passing through $A,B$ is tangent to segment $CD$ at point $E$. Another circle passing through $C,D$ is tangent to $AB$ at point $F$. Point $G$ is the intersection point of $AE,DF$, and point $H$ is the intersection point of $BE$, $CF$. Prove that the incenters of triangles $AGF$, $BHF$, $CHE$, $DGE$ lie on a circle. Proposed by Le Viet An (Vietnam)