The expression \[ (x \plus{} y \plus{} z)^{2006} \plus{} (x \minus{} y \minus{} z)^{2006} \]is simplified by expanding it and combining like terms. How many terms are in the simplified expression? $ \textbf{(A) } 6018 \qquad \textbf{(B) } 671,676 \qquad \textbf{(C) } 1,007,514 \qquad \textbf{(D) } 1,008,016 \qquad \textbf{(E) } 2,015,028$

Each grid point of a cartesian plane is colored with one of three colors, whereby all three colors are used. Show that one can always find a right-angled triangle, whose three vertices have pairwise different colors.

Let $ABCD$ be a quadrilateral with an inscribed circle centered at $I$. Let $CI$ intersect $AB$ at $E$. If $\angle IDE=35^\circ$, $\angle ABC=70^\circ$, and $\angle BCD=60^\circ$, then what are all possible measures of $\angle CDA$?

Let $A$, $B$, $C$ be points on a circle whose centre is $O$ and whose radius is $1$, such that $\angle BAC = 45^\circ$. Lines $AC$ and $BO$ (possibly extended) intersect at $D$, and lines $AB$ and $CO$ (possibly extended) intersect at $E$. Prove that $BD \cdot CE = 2$.

In a right-angled triangle, $a$ and $b$ denote the lengths of the two catheti. A circle with radius $r$ has the center on the hypotenuse and touches both catheti. Show that $\frac{1}{a}+\frac{1}{b}=\frac{1}{r}$.

Triangle $GRT$ has $GR=5,$ $RT=12,$ and $GT=13.$ The perpendicular bisector of $GT$ intersects the extension of $GR$ at $O.$ Find $TO.$

A $n \times n \times n$ cube is constructed using $1 \times 1 \times 1$ cubes, some of them black and others white, such that in each $n \times 1 \times 1$, $1 \times n \times 1$, and $1 \times 1 \times n$ subprism there are exactly two black cubes, and they are separated by an even number of white cubes (possibly 0). Show it is possible to replace half of the black cubes with white cubes such that each $n \times 1 \times 1$, $1 \times n \times 1$ and $1 \times 1 \times n$ subprism contains exactly one black cube.

There is a chessboard with $m$ columns and $n$ rows. In each blanks, an integer is given. If a rectangle $R$ (in this chessboard) has an integer $h$ satisfying the following two conditions, we call $R$ as a 'shelf'. (i) All integers contained in $R$ are bigger than $h$. (ii) All integers in blanks, which are not contained in $R$ but meet with $R$ at a vertex or a side, are not bigger than $h$. Assume that all integers are given to make shelves as much as possible. Find the number of shelves.

A diameter $AK$ is drawn for the circumscribed circle $\omega$ of an acute-angled triangle $ABC$, an arbitrary point $M$ is chosen on the segment $BC$, the straight line $AM$ intersects $\omega$ at point $Q$. The foot of the perpendicular drawn from $M$ on $AK$ is $D$, the tangent drawn to the circle $\omega$ through the point $Q$, intersects the straight line $MD$ at $P$. A point $L$ (different from $Q$) is chosen on $\omega$ such that $PL$ is tangent to $\omega$. Prove that points $L$, $M$ and $K$ lie on the same line.

Let $ABCD$ be a quadrilateral inscribed in a circle $\omega$. The lines $AB$ and $CD$ meet at $P$, the lines $AD$ and $BC$ meet at $Q$, and the diagonals $AC$ and $BD$ meet at $R$. Let $M$ be the midpoint of the segment $PQ$, and let $K$ be the common point of the segment $MR$ and the circle $\omega$. Prove that the circumcircle of the triangle $KPQ$ and $\omega$ are tangent to one another.

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

Nondegenerate $ \triangle ABC$ has integer side lengths, $ BD$ is an angle bisector, $ AD \equal{} 3$, and $ DC \equal{} 8$. What is the smallest possible value of the perimeter? $ \textbf{(A)}\ 30 \qquad \textbf{(B)}\ 33 \qquad \textbf{(C)}\ 35 \qquad \textbf{(D)}\ 36 \qquad \textbf{(E)}\ 37$

(M.Volchkevich, 9--11) Given two circles and point $ P$ not lying on them. Draw a line through $ P$ which cuts chords of equal length from these circles.

Prove this proposition: Center the sphere circumscribed around a tetrahedron which coincides with the center of a sphere inscribed in that tetrahedron if and only if the skew edges of the tetrahedron are equal.

Let $AC$ and $BD$ be two chords of a circle with center $O$ such that they intersect at right angles inside the circle at the point $M$. Suppose $K$ and $L$ are midpoints of the chords $AB$ and $CD$ respectively. Prove that $OKML$ is a parallelogram.

In rectangle $ABCD$, $AB=6$ and $BC=3$. Point $E$ between $B$ and $C$, and point $F$ between $E$ and $C$ are such that $BE=EF=FC$. Segments $\overline{AE}$ and $\overline{AF}$ intersect $\overline{BD}$ at $P$ and $Q$, respectively. The ratio $BP:PQ:QD$ can be written as $r:s:t$, where the greatest common factor of $r,s$ and $t$ is $1$. What is $r+s+t$? $\textbf{(A) } 7 \qquad \textbf{(B) } 9 \qquad \textbf{(C) } 12 \qquad \textbf{(D) } 15 \qquad \textbf{(E) } 20$

Let $ABCD$ be a parallelogram and let $AX$ and $AY$ be the altitudes from $A$ to $CB, CD$, respectively. A line $\ell \perp XY$ bisects $AX$ and meets $AB, BC$ at $K, L$. Similarly, a line $d \perp XY$ bisects $AY$ and meets $DA, DC$ at $P, Q$. Show that the circumcircles of $\triangle BKL$ and $\triangle DPQ$ are tangent to each other.

Let $ABC$ be an isosceles triangle with $AC=BC$, whose incentre is $I$. Let $P$ be a point on the circumcircle of the triangle $AIB$ lying inside the triangle $ABC$. The lines through $P$ parallel to $CA$ and $CB$ meet $AB$ at $D$ and $E$, respectively. The line through $P$ parallel to $AB$ meets $CA$ and $CB$ at $F$ and $G$, respectively. Prove that the lines $DF$ and $EG$ intersect on the circumcircle of the triangle $ABC$. [i]Proposed by Hojoo Lee[/i]

Let $AX$, $AY$ be tangents to circle $\omega$ from point $A$. Le $B$, $C$ be points inside $AX$ and $AY$ respectively, such that perimeter of $\triangle ABC$ is equal to length of $AX$. $D$ is reflection of $A$ over $BC$. Prove that circumcircle $\triangle BDC$ and $\omega$ are tangent to each other.

[color=darkred]The altitude $[BH]$ dropped onto the hypotenuse of a triangle $ABC$ intersects the bisectors $[AD]$ and $[CE]$ at $Q$ and $P$ respectively. Prove that the line passing through the midpoints of the segments $[QD]$ and $[PE]$ is parallel to the line $AC$ .[/color]

One convex quadrilateral is inside another. Can it turn out that the sum of the lengths of the diagonals of the outer quadrilateral is less than the sum of the lengths of the diagonals of the inner?

A circle with radius $1$ is circumscribed by a rhombus. What is the minimum possible area of this rhombus?

Are there two such quadrangles that the sides of the first are less than the corresponding sides of the second, and the corresponding diagonals are larger? (Arseniy Akopyan)

Isosceles trapezoid $ ABCD$, with $ AB \parallel CD$, is such that there exists a circle $ \Gamma$ tangent to its four sides. Let $ T \equal{} \Gamma \cap BC$, and $ P \equal{} \Gamma \cap AT$ ($ P \neq T$). If $ \frac{AP}{AT} \equal{} \frac{2}{5}$, compute $ \frac{AB}{CD}$.

Let $ABC$ an acute triangle and $\Gamma$ its circumcircle. The bisector of $BAC$ intersects $\Gamma$ at $M\neq A$. A line $r$ parallel to $BC$ intersects $AC$ at $X$ and $AB$ at $Y$. Also, $MX$ and $MY$ intersect $\Gamma$ again at $S$ and $T$, respectively. If $XY$ and $ST$ intersect at $P$, prove that $PA$ is tangent to $\Gamma$.