Let $P$ be a regular $2n$-sided polygon. A [b]rhombus-ulation[/b] of $P$ is dividing $P$ into rhombuses such that no two intersect and no vertex of any rhombus is on the edge of other rhombuses or $P$. (a) Prove that number of rhombuses is a function of $n$. Find the value of this function. Also find the number of vertices and edges of the rhombuses as a function of $n$. (b) Prove or disprove that there always exists an edge $e$ of $P$ such that by erasing all the segments parallel to $e$ the remaining rhombuses are connected. (c) Is it true that each two rhombus-ulations can turn into each other using the following algorithm multiple times? Algorithm: Take a hexagon -not necessarily regular- consisting of 3 rhombuses and re-rhombus-ulate the hexagon. (d) Let $f(n)$ be the number of ways to rhombus-ulate $P$. Prove that:\[\Pi_{k=1}^{n-1} ( \binom{k}{2} +1) \leq f(n) \leq \Pi_{k=1}^{n-1} k^{n-k} \]

Determine all positive integers $n$ with the following property: for each triple $(a, b, c)$ of positive real numbers there is a triple $(k, \ell, m)$ of non-negative integer numbers so that $an^k$, $bn^{\ell}$ and $cn^m$ are the lengths of the sides of a (non-degenerate) triangle shapes.

In the plane, four points are marked. It is known that these points are the centers of four circles, three of which are pairwise externally tangent, and all these three are internally tangent to the fourth one. It turns out, however, that it is impossible to determine which of the marked points is the center of the fourth (the largest) circle. Prove that these four points are the vertices of a rectangle.

On the sides of a convex hexagon $ ABCDEF$ , equilateral triangles are constructd in its exterior. Prove that the third vertices of these six triangles are vertices of a regular hexagon if and only if the initial hexagon is [i]affine regular[/i]. (A hexagon is called affine regular if it is centrally symmetric and any two opposite sides are parallel to the diagonal determine by the remaining two vertices.)

Using a ruler with a length of $20$ cm and a compass with a maximum deviation of $10$ cm to connect the segment given two points lying at a distance of $1$ m.

In convex quadrilateral $ABCD$, where $AB=AD$, on $BD$ point $K$ is chosen. On $KC$ point $L$ is such that $\bigtriangleup BAD \sim \bigtriangleup BKL$. Line parallel to $DL$ and passes through $K$, intersect $CD$ at $P$. Prove that $\angle APK = \angle LBC$. @below edited

The perimeter of a semicircular region, measured in centimeters, is numerically equal to its area, measured in square centimeters. The radius of the semicircle, measured in centimeters, is $\text{(A)} \pi \qquad \text{(B)} \frac{2}{\pi} \qquad \text{(C)} 1 \qquad \text{(D)} \frac{1}{2} \qquad \text{(E)} \frac{4}{\pi} + 2$

Let $ABC$ be a triangle with incenter $I$. The line $AI$ intersects $BC$ and the circumcircle of $ABC$ at the points $T$ and $S$, respectively. Let $K$ and $L$ be the incenters of $SBT$ and $SCT$, respectively, $M$ be the midpoint of $BC$ and $P$ be the reflection of $I$ with respect to $KL$. a) Prove that $M$, $T$, $K$ and $L$ are concyclic. b) Determine the measure of $\angle BPC$.

Logan is constructing a scaled model of his town. The city's water tower stands $ 40$ meters high, and the top portion is a sphere that holds $ 100,000$ liters of water. Logan's miniature water tower holds $ 0.1$ liters. How tall, in meters, should Logan make his tower? $ \textbf{(A)}\ 0.04\qquad \textbf{(B)}\ \frac{0.4}{\pi}\qquad \textbf{(C)}\ 0.4\qquad \textbf{(D)}\ \frac{4}{\pi}\qquad \textbf{(E)}\ 4$

In a triangle $ABC$ the incenter, the $B$-excenter and the $C$-excenter are $I, K$ and $L$, respectively. The perpendiculars at $B$ and $C$ to $BC$ intersect the lines $AC$ and $AB$ at $E$ and $F$, respectively. Prove that the circumcircles of $AEF, FIL, EIK$ concur.

In a cyclic quadrilateral $ABCD$ with circumcenter $O$, the diagonals $AC$ and $BD$ intersect at $X$. Let the circumcircles of triangles $AXD$ and $BXC$ intersect at $Y$. Let the circumcircles of triangles $AXB$ and $CXD$ intersect at $Z$. If $O$ lies inside $ABCD$ and if the points $O,X,Y,Z$ are all distinct, prove that $O,X,Y,Z$ lie on a circle.

Three rays with a common origin are drawn on the plane, dividing the plane into three angles. One point is marked inside each corner. Using one ruler, construct a triangle whose vertices lie on the given rays and whose sides contain the given points.

Let $\, ABCD \,$ be a convex quadrilateral such that diagonals $\, AC \,$ and $\, BD \,$ intersect at right angles, and let $\, E \,$ be their intersection. Prove that the reflections of $\, E \,$ across $\, AB, \, BC, \, CD, \, DA \,$ are concyclic.

Let $ABC$ be an acute triangle with its altitudes $BE,CF$. $M$ is the midpoint of $BC$. $N$ is the intersection of $AM$ and $EF. X$ is the projection of $N$ on $BC$. $Y,Z$ are respectively the projections of $X$ onto $AB,AC$. Prove that $N$ is the orthocenter of triangle $AYZ$. Nguyễn Minh Hà

There are four collinear spots: $ A,B,C,D $ $ \bar{AB}=\bar{BC}=\frac{\bar{CD}}{4}=\sqrt{5} $ There are two circles; One which has $ \bar{AC} $ as a diameter, and the other having $ \bar{BD} $ as a diameter. Let's put $ \odot (AC) \cap \odot (BD) = E,F $ Let's put the area of $ EAFD $ $ S $ Find $ S^2 $.

Find the side of the smallest regular triangle that can be inscribed in a right triangle with an acute angle of $30^o$ and a hypotenuse of $2$. (All vertices of the required regular triangle must be located on different sides of this right triangle.)

A square $ABCD$ is given. Two circles are inscribed into angles $A$ and $B$, and the sum of their diameters is equal to the sidelength of the square. Prove that one of their common tangents passes through the midpoint of $AB$.

Let $I$ be the incentre of a non-equilateral triangle $ABC$, $I_A$ be the $A$-excentre, $I'_A$ be the reflection of $I_A$ in $BC$, and $l_A$ be the reflection of line $AI'_A$ in $AI$. Define points $I_B$, $I'_B$ and line $l_B$ analogously. Let $P$ be the intersection point of $l_A$ and $l_B$. [list=a] [*] Prove that $P$ lies on line $OI$ where $O$ is the circumcentre of triangle $ABC$. [*] Let one of the tangents from $P$ to the incircle of triangle $ABC$ meet the circumcircle at points $X$ and $Y$. Show that $\angle XIY = 120^{\circ}$. [/list]

Let $ABC$ be a triangle, and let $M$ be a point on the side $(AC)$ .The line through $M$ and parallel to $BC$ crosses $AB$ at $N$. Segments $BM$ and $CN$ cross at $P$, and the circles $BNP$ and $CMP$ cross again at $Q$. Show that angles $BAP$ and $CAQ$ are equal.

Let there be given a regular hexagon of side length $1$. Six circles with the sides of the hexagon as diameters are drawn. Find the area of the part of the hexagon lying outside all the circles.

Line $\frac{x}{4}+\frac{y}{3}=1$ and ellipse $\frac{x^2}{16}+\frac{y^2}{9}=1$ intersect at $A$ and $B$. A point on the ellipse $P$ satisties that the area of $\triangle PAB$ is $3$. The number of such points is $\text{(A)}1\qquad\text{(B)}2\qquad\text{(C)}3\qquad\text{(D)}4$

Let $ABC$ be an acute-angled triangle with circumcenter $O$. The circumcircle of $\triangle{BOC}$ meets the lines $AB, AC$ at points $A_1, A_2$, respectively. Let $\omega_{A}$ be the circumcircle of triangle $AA_1A_2$. Define $\omega_B$ and $\omega_C$ analogously. Prove that the circles $\omega_A, \omega_B, \omega_C$ concur on $\odot(ABC)$.

$ABC$ is a triangle and $D$ is an internal point such that $\angle DAB=\angle DBC =\angle DCA$. $O_a$ is the circumcenter of $DBC$. $O_b$ is the circumcenter of $DAC$. $O_c$ is the circumcenter of $DAB$. Show that if the area of $ABC$ and $O_aO_bO_c$ are equal then $ABC$ is equilateral.

Pentagon $ABCDE$ consists of a square $ACDE$ and an equilateral triangle $ABC$ that share the side $\overline{AC}$. A circle centered at $C$ has area 24. The intersection of the circle and the pentagon has half the area of the pentagon. Find the area of the pentagon. [asy]/* File unicodetex not found. */ /* Geogebra to Asymptote conversion, documentation at artofproblemsolving.com/Wiki, go to User:Azjps/geogebra */ import graph; size(4.26cm); 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 = -1.52, xmax = 2.74, ymin = -2.18, ymax = 6.72; /* image dimensions */ draw((0,1)--(2,1)--(2,3)--(0,3)--cycle); draw((0,3)--(2,3)--(1,4.73)--cycle); /* draw figures */ draw((0,1)--(2,1)); draw((2,1)--(2,3)); draw((2,3)--(0,3)); draw((0,3)--(0,1)); draw((0,3)--(2,3)); draw((2,3)--(1,4.73)); draw((1,4.73)--(0,3)); draw(circle((0,3), 1.44)); label("$C$",(-0.4,3.14),SE*labelscalefactor); label("$A$",(2.1,3.1),SE*labelscalefactor); label("$B$",(0.86,5.18),SE*labelscalefactor); label("$D$",(-0.28,0.88),SE*labelscalefactor); label("$E$",(2.1,0.8),SE*labelscalefactor); /* dots and labels */ clip((xmin,ymin)--(xmin,ymax)--(xmax,ymax)--(xmax,ymin)--cycle); /* end of picture */[/asy]

Given is an obtuse isosceles triangle $ABC$ with $CA=CB$ and circumcenter $O$. The point $P$ on $AB$ is such that $AP<\frac{AB} {2}$ and $Q$ on $AB$ is such that $BQ=AP$. The circle with diameter $CQ$ meets $(ABC)$ at $E$ and the lines $CE, AB$ meet at $F$. If $N$ is the midpoint of $CP$ and $ON, AB$ meet at $D$, show that $ODCF$ is cyclic.