Let $ABC$ be a triangle, and let $T$ be a point on the extension of $AB$ beyond $B$, and $U$ a point on the extension of $AC$ beyond $C$, such that $BT = CU$. Moreover, let $R$ and $S$ be points on the extensions of $AB$ and $AC$ beyond $A$ such that $AS = AT$ and $AR = AU$. Prove that $R$, $S$, $T$, $U$ lie on a circle whose centre lies on the circumcircle of $ABC$.

Let $ABCD$ be a convex quadrilateral. The circumcenter and the incenter of triangle $ABC$ coincide with the incenter and the circumcenter of triangle $ADC$ respectively. It is known that $AB = 1$. Find the remaining sidelengths and the angles of $ABCD$.

A point P in the interior of a cyclic quadrilateral ABCD satisfies ∠BPC = ∠BAP + ∠PDC. Denote by E, F and G the feet of the perpendiculars from P to the lines AB, AD and DC, respectively. Show that the triangles FEG and PBC are similar.

The point $D$ at the side $AB$ of triangle $ABC$ is given. Construct points $E,F$ at sides $BC, AC$ respectively such that the midpoints of $DE$ and $DF$ are collinear with $B$ and the midpoints of $DE$ and $EF$ are collinear with $C.$

The points $A$, $B$, $C$, $D$, and $E$ lie in one plane and have the following properties: $AB = 12, BC = 50, CD = 38, AD = 100, BE = 30, CE = 40$. Find the length of the segment $ED$.

Let $a, b, c$ be the lengths of the sides of a triangle, and let $S$ be its area. We know that $S = \frac14 (c^2 - a^2 - b^2)$. Prove that $\angle C = 135^o$.

Given a cyclic quadrilateral $ABCD$ with the circumcenter $O$, with $BC$ and $AD$ not parallel. Let $P$ be the intersection of $AC$ and $BD$. Let $E$ be the intersection of the rays $AB$ and $DC$. Let $I$ be the incenter of $EBC$ and the incircle of $EBC$ touches $BC$ at $T_1$. Let $J$ be the excenter of $EAD$ that touches $AD$ and the excircle of $EAD$ that touches $AD$ touches $AD$ at $T_2$. Let $Q$ be the intersection between $IT_1$ and $JT_2$. Prove that $O,P,Q$ are collinear.

Let $ABCD$ be a unit square. Consider an equilateral triangle $XYZ$ with $X,Y$ as (inner or boundary) points of the square. Determine the locus $M$ of vertices $Z$ of all these triangles $XYZ$ and compute the area of $M.$

Show that there is no polyhedron whose projection on the plane is a nondegenerate triangle.

In the trapezium $ABCD$ ($AD // BC$), the point $M$ lies on the side of $CD$, with $CM:MD=2:3$, $AB=AD$, $BC:AD=1:3$. Prove that $BD \perp AM$.

Square $ ABCD$ has side length $ s$, a circle centered at $ E$ has radius $ r$, and $ r$ and $ s$ are both rational. The circle passes through $ D$, and $ D$ lies on $ \overline{BE}$. Point $ F$ lies on the circle, on the same side of $ \overline{BE}$ as $ A$. Segment $ AF$ is tangent to the circle, and $ AF \equal{} \sqrt {9 \plus{} 5\sqrt {2}}$. What is $ r/s$? [asy]unitsize(6mm); defaultpen(linewidth(.8pt)+fontsize(10pt)); dotfactor=3; pair B=(0,0), C=(3,0), D=(3,3), A=(0,3); pair Ep=(3+5*sqrt(2)/6,3+5*sqrt(2)/6); pair F=intersectionpoints(Circle(A,sqrt(9+5*sqrt(2))),Circle(Ep,5/3))[0]; pair[] dots={A,B,C,D,Ep,F}; draw(A--F); draw(Circle(Ep,5/3)); draw(A--B--C--D--cycle); dot(dots); label("$A$",A,NW); label("$B$",B,SW); label("$C$",C,SE); label("$D$",D,SW); label("$E$",Ep,E); label("$F$",F,NW);[/asy]$ \textbf{(A) } \frac {1}{2}\qquad \textbf{(B) } \frac {5}{9}\qquad \textbf{(C) } \frac {3}{5}\qquad \textbf{(D) } \frac {5}{3}\qquad \textbf{(E) } \frac {9}{5}$

Let $\vartriangle ABC$ be a right-angled triangle with $\angle A = 90^o$ and circumcircle $\Gamma$. The inscribed circle is tangent to $BC$ in point $D$. Let $E$ be the midpoint of the arc $AB$ of $\Gamma$ not containing $C$ and let $F$ be the midpoint of the arc $AC$ of $\Gamma$ not containing $B$. (a) Prove that $\vartriangle ABC \sim \vartriangle DEF$. (b) Prove that $EF$ goes through the points of tangency of the incircle to $AB$ and $AC$.

Let $ABC$ be a scalene acute-angled triangle and $D$ be the point on its circumcircle such that $AD$ is a symmedian of triangle $ABC$. Let $E$ be the reflection of $D$ about $BC$, $C_0$ the reflection of $E$ about $AB$ and $B_0$ the reflection of $E$ about $AC$. Prove that the lines $AD$, $BB_0$ and $CC_0$ are concurrent if and only if $\angle BAC = 60^{\circ}.$

Let $H$ be the orthocenter of an acute-angled triangle $ABC$. Point $X_A$ lying on the tangent at $H$ to the circumcircle of triangle $BHC$ is such that $AH=AX_A$ and $X_A \not= H$. Points $X_B,X_C$ are defined similarly. Prove that the triangle $X_AX_BX_C$ and the orthotriangle of $ABC$ are similar.

Let the three sides of a triangle be $\ell, m, n$, respectively, satisfying $\ell>m>n$ and $\left\{\frac{3^\ell}{10^4}\right\}=\left\{\frac{3^m}{10^4}\right\}=\left\{\frac{3^n}{10^4}\right\}$, where $\{x\}=x-\lfloor{x}\rfloor$ and $\lfloor{x}\rfloor$ denotes the integral part of the number $x$. Find the minimum perimeter of such a triangle.

A parallelogram $ABCD$ is split by the diagonal $BD$ into two equal triangles. A regular hexagon is inscribed into the triangle $ABD$ so that two of its consecutive sides lie on $AB$ and $AD$ and one of its vertices lies on $BD$. Another regular hexagon is inscribed into the triangle $CBD{}$ so that two of its consecutive vertices lie on $CB$ and $CD$ and one of its sides lies on $BD$. Which of the hexagons is bigger? [i]Konstantin Knop[/i]

$(FRA 4)$ A right-angled triangle $OAB$ has its right angle at the point $B.$ An arbitrary circle with center on the line $OB$ is tangent to the line $OA.$ Let $AT$ be the tangent to the circle different from $OA$ ($T$ is the point of tangency). Prove that the median from $B$ of the triangle $OAB$ intersects $AT$ at a point $M$ such that $MB = MT.$

Let $a,b,c$ be the side lengths of any triangle. Prove that $$\frac{a}{\sqrt{2b^2+2c^2-a^2}}+\frac{b}{\sqrt{2c^2+2a^2-b^2 }}+\frac{c}{\sqrt{2a^2+2b^2-c^2}}\ge \sqrt{3}.$$ [i](Zhuge Liang)[/i]

Let $ABC$ be an isosceles triangle with $AC = BC$. On the arc $CA$ of its circumcircle, which does not contain $ B$, there is a point $ P$. The projection of $C$ on the line $AP$ is denoted by $E$, the projection of $C$ on the line $BP$ is denoted by $F$. Prove that the lines $AE$ and $BF$ have equal lengths. (W. Janous, WRG Ursulincn, Innsbruck)

In triangle $ABC$, we have $\angle ABC=\angle ACB=44^o$. Point $M$ is in its interior such that $\angle MBC=16^o$ and $\angle MCB=30^o$. Prove that $\angle MAC=\angle MBC$. [i] (Andra Elena Mircea)[/i]

Describe all closed bounded figures $\Phi$ in the plane any two points of which are connectable by a semicircle lying in $\Phi$.

A pentagon $ABCDE$ is circumscribed about a circle. The angles at the vertices $A{}$, $C{}$ and $E{}$ of the pentagon are equal to $100^\circ$. Find the measure of the angle $\angle ACE$.

A $30$-gon $A_1A_2\cdots A_{30}$ is inscribed in a circle of radius $2$. Prove that one can choose a point $B_k$ on the arc $A_kA_{k+1}$ for $1 \leq k \leq 29$ and a point $B_{30}$ on the arc $A_{30}A_1$, such that the numerical value of the area of the $60$-gon $A_1B_1A_2B_2 \dots A_{30}B_{30}$ is equal to the numerical value of the perimeter of the original $30$-gon.

[b]p1.[/b] Anne purchased yesterday at WalMart in Puerto Rico $6$ identical notebooks, $8$ identical pens and $7$ identical erasers. Anne remembers that each eraser costs $73$ cents. She did not buy anything else. Anne told her mother that she spent $12$ dollars and $76$ cents at Walmart. Can she be right? Note that in Puerto Rico there is no sales tax. [b]p2.[/b] Two men ski one after the other first in a flat field and then uphill. In the field the men run with the same velocity $12$ kilometers/hour. Uphill their velocity drops to $8$ kilometers/hour. When both skiers enter the uphill trail segment the distance between them is $300$ meters less than the initial distance in the field. What was the initial distance between skiers? (There are $1000$ meters in 1 kilometer.) [b]p3.[/b] In the equality $** + **** = ****$ all the digits are replaced by $*$. Restore the equality if it is known that any numbers in the equality does not change if we write all its digits in the opposite order. [b]p4.[/b] If a polyleg has even number of legs he always tells truth. If he has an odd number of legs he always lies. Once a green polyleg told a dark-blue polyleg ”- I have $8$ legs. And you have only $6$ legs!” The offended dark-blue polyleg replied ”-It is me who has $8$ legs, and you have only $7$ legs!” A violet polyleg added ”-The dark-blue polyleg indeed has $8$ legs. But I have $9$ legs!” Then a stripped polyleg started: ”-None of you has $8$ legs. Only I have 8 legs!” Which polyleg has exactly $8$ legs? [b]p5.[/b] Cut the figure shown below in two equal pieces. (Both the area and the form of the pieces must be the same.) [img]https://cdn.artofproblemsolving.com/attachments/e/4/778678c1e8748e213ffc94ba71b1f3cc26c028.png[/img] PS. You should use hide for answers. Collected [url=https://artofproblemsolving.com/community/c5h2760506p24143309]here[/url].

In the acute-angled triangle $ABC$, with $AB \ne AC$, $D$ is the foot of the angle bisector of angle $A$, and $E, F$ are the feet of the altitudes from $B$ and $C$, respectively. The circumcircles of triangles $DBF$ and $DCE$ intersect for the second time at $M$. Prove that $ME = MF$. Leonard Giugiuc