1. Let $ABC$ be an acute-angled triangle with $AB\neq AC$. The circle with diameter $BC$ intersects the sides $AB$ and $AC$ at $M$ and $N$ respectively. Denote by $O$ the midpoint of the side $BC$. The bisectors of the angles $\angle BAC$ and $\angle MON$ intersect at $R$. Prove that the circumcircles of the triangles $BMR$ and $CNR$ have a common point lying on the side $BC$.

Click for solution I needed 10 minutes to sort the trash out of the configuration. After that everything was trivial. If T is the point of intersection of the circumcircles of triangles BMR and CNR, then we must show that T lies on BC, i. e. that we have < BTR + < CTR = $180^{\circ}$. Since BMRT is a cyclic quadrilateral, we have < BTR = $180^{\circ}$ - < BMR, so < BTR = < AMR. Similarly < CTR = < ANR. So we have to show that < AMR + < ANR = $180^{\circ}$, or, in other words, we must prove that the quadrilateral AMRN is cyclic. Well, the points M and N lie on the circle with diameter BC, and the center of this circle is obviously the midpoint O of BC. Thus, MO = NO, so that the triangle MON is isosceles, and hence the line OR, being the angle bisector of the angle MON, coincides with the perpendicular bisector of the segment MN. Hence, the point R lies on the perpendicular bisector of the segment MN. In other words, the point R is the point of intersection of the perpendicular bisector of the segment MN with the angle bisector of the angle MAN. But a well-known fact states that the perpendicular bisector of a side of a triangle meets the angle bisector of the opposite angle at a point on the circumcircle of the triangle. Hence, the point R lies on the circumcircle of triangle MAN, and therefore, the quadrilateral AMRN is cyclic. Darij

Find all polynomials $f$ with real coefficients such that for all reals $a,b,c$ such that $ab+bc+ca = 0$ we have the following relations \[ f(a-b) + f(b-c) + f(c-a) = 2f(a+b+c). \]

Click for solution A reasonably clean solution: We want to find all polynomials f such that $f(a-b)+f(b-c)+f(c-a)=2f(a+b+c)$ whenever $ab+bc+ca=0$. Equivalently, $ab+bc+ca$ divides $f(a-b)+f(b-c)+f(c-a)-2f(a+b+c)$. Let $f(x)=r_nx^n+r_{n-1}x^{n-1}+...+r_0$ and $a'=ta,b'=tb,c'=tc$ Then $a'b'+b'c'+c'a'=0$ if $ab+bc+ca=0$ so $t^nr_n((a-b)^n+(b-c)^n+(c-a)^n-2(a+b+c)^n)+...+r_0(1+1+1-2)=0$ The functions of t $t^n,t^{n-1},...,1$ are linearly independent, so each coefficient $r_k((a-b)^k+(b-c)^k+(c-a)^k-2(a+b+c)^k)$ must be zero. This means that either $r_k=0$ or $ab+bc+ca$ divides $(a-b)^k+(b-c)^k+(c-a)^k-2(a+b+c)^k$ For $k=2$, $(a-b)^2+(b-c)^2+(c-a)^2-2(a+b+c)^2=-4(ab+bc+ca)$ For $k=4$, $(a-b)^4+(b-c)^4+(c-a)^4-2(a+b+c)^4= -6(ab+bc+ca)(2a^2+2b^2+2c^2+ab+bc+ca)$ Consider the case $a=2,b=2,c=-1$. Then $(-3)^k+(-3)^k=2*3^k$ This forces k to be even. Now take $a=6,b=3,c=-2$. This gives $3^k+5^k+(-8)^k=2*7^k$. Since $8^6>2*7^6$, $3^k+5^k+(-8)^k>2*7^k$ for all even $k>4$. By previous arguments, we must have $f(x)=dx^4+ex^2$

Define a "hook" to be a figure made up of six unit squares as shown below in the picture, or any of the figures obtained by applying rotations and reflections to this figure. [asy][asy] unitsize(0.5 cm); draw((0,0)--(1,0)); draw((0,1)--(1,1)); draw((2,1)--(3,1)); draw((0,2)--(3,2)); draw((0,3)--(3,3)); draw((0,0)--(0,3)); draw((1,0)--(1,3)); draw((2,1)--(2,3)); draw((3,1)--(3,3)); [/asy][/asy] Determine all $ m\times n$ rectangles that can be covered without gaps and without overlaps with hooks such that - the rectangle is covered without gaps and without overlaps - no part of a hook covers area outside the rectangle.

Click for solution solution from Christian Sattler(Germany): first we define an involution on the hooks: send every hook to the one which includes the square where the hook goes around. it is clear that this is an involution, so the number of hooks is even. it immediately follows that 12 | mn. there are three cases(wlog 3 divides m): 3 divides m and 4 divides n(trivial since we can do 3x4-rectangles), 12 divides m, then n can be arbritarily, but not equal to 1,2 or 5(it is clear that n is not 1,2, and 5 is easy to exclude. in all other cases we may write n as linear combination of 3 and 4 with positive coefficients and we can use 3x4-rectangles again taking rows of 3x4-rectangles or 4x3-rectangles). so we are left with the case 6 | m and 2 | n. so look again at our involution: it shows that our hooks appear in pairs of the possible forms: either 3x4-rectangles or something similar to an S: .=== .=== ===. ===. now mark every second row. it is easy to see that there are always 6 out of the 12 squares of the S are marked, the same for horizontal 3x4-rectangles, but vertical 3x4-rectangles have either 4 or 8 squares marked, that shows that there's an even number of them. doing the same with columns, we get that the total number of 3x4-rectangles is even. now, look at the coloring of .==. =..= .==. =..= and periodic. it is easy to see that 6 out of 12 squares are marked at 3x4-rectangles, the S upright, the S turned by 90 degree and even distance to the left boundary of the rectangle, but 4 or 8 for an S turned by 90 degree with odd distance to the left boundary. this shows that there's an even number of them and by translating the coloring by 1 to the right and turning it all around, we get that the number of S is even as well. this shows that 24 is a divisor of mn, so we get one of the two other cases. i only found all solutions but couldn't prove that it are all. Peter

Let $n \geq 3$ be an integer. Let $t_1$, $t_2$, ..., $t_n$ be positive real numbers such that \[n^2 + 1 > \left( t_1 + t_2 + \cdots + t_n \right) \left( \frac{1}{t_1} + \frac{1}{t_2} + \cdots + \frac{1}{t_n} \right).\] Show that $t_i$, $t_j$, $t_k$ are side lengths of a triangle for all $i$, $j$, $k$ with $1 \leq i < j < k \leq n$.

Click for solution solution by induction: the hardest part is to show that it works for n=3. so assume $t_1\geq t_2\geq t_3$. it is easy to see that the right is strictly increasing in $t_3$. so if we show that the reverse inequality holds for $t_1=t_2+t_3$, we are done. a quick calculation is enough to show that(it is too boring to post). but if we know is true for n-1, then let's do the following computation: $((t_1+...+t_{n-1})+t_n)(\frac{1}{t_1}+...+\frac{1}{t_{n-1}}+\frac{1}{t_n})\geq (t_1+...+t_{n-1})(\frac{1}{t_1}+...+\frac{1}{t_{n-1}})+2n-1$, by AM-GM and AM-HM. so we get the inequality reduces to that one for n-1 variables. Peter

In a convex quadrilateral $ABCD$, the diagonal $BD$ bisects neither the angle $ABC$ nor the angle $CDA$. The point $P$ lies inside $ABCD$ and satisfies \[\angle PBC=\angle DBA\quad\text{and}\quad \angle PDC=\angle BDA.\] Prove that $ABCD$ is a cyclic quadrilateral if and only if $AP=CP$.

Click for solution I needed over two hours for the solution, even though it is not so hard. Let $P_1$, $P_2$, $P_3$, $P_4$ be the orthogonal projections of the point P on the lines AB, BC, CD, DA, and let $Q_1$, $Q_2$, $Q_3$, $Q_4$ be the reflections of P in these lines, or, equivalently, the reflections of P in the points $P_1$, $P_2$, $P_3$, $P_4$. From < PBC = < DBA, you see that the line BD is the reflection of the line BP in the angle bisector of the angle ABC. Now you use the following (quite easy) lemma: If X is a point in the plane of an angle VUW, and V' and W' are the reflections of X in the lines UV and UW, then the perpendicular bisector of the line V'W' is the reflection of the line UX in the angle bisector of the angle VUW. Now, apply this lemma to the angle ABC and the point P, whose reflections in the lines BA and BC are $Q_1$ and $Q_2$, respectively, and you see that the perpendicular bisector of the line $Q_1 Q_2$ is the line BD. Hence, the points $Q_1$ and $Q_2$ are symmetric to each other with respect to the line BD. Similarly, the points $Q_4$ and $Q_3$ are symmetric to each other with respect to the line BD, too. Hence, since symmetry is a congruence transformation , we have $Q_1 Q_4 = Q_2 Q_3$. Now, since $Q_1$ and $Q_4$ are the reflections of P in $P_1$ and $P_4$, the points $P_1$ and $P_4$ are the midpoints of the segments $PQ_1$ and $PQ_4$, and thus $P_1 P_4 = \frac12 Q_1 Q_4$. Similarly $P_2 P_3 = \frac12 Q_2 Q_3$. Now, from $Q_1 Q_4 = Q_2 Q_3$, it follows that $P_1 P_4 = P_2 P_3$. Since $\measuredangle AP_4 P = 90^{\circ}$ and $\measuredangle AP_1 P = 90^{\circ}$, the points $P_4$ and $P_1$ lie on the circle with diameter AP. In other words, the circumcircle of triangle $AP_4 P_1$ has the segment AP as diameter. Therefore, by the Extended Law of Sines, we have $P_1 P_4 = AP \cdot \sin \measuredangle P_1 A P_4 = AP \cdot \sin A$, where A = < DAB and C = < BCD are two opposite angles of our quadrilateral. Similarly, $P_2 P_3 = CP \cdot \sin C$. Since $P_1 P_4 = P_2 P_3$, we conclude that AP = CP holds if and only if sin A = sin C. Now, sin A = sin C means either A = C, or A + C = $180^{\circ}$. In the case of A + C = $180^{\circ}$, the quadrilateral ABCD is cyclic, and we are happy. Remains to show that the case A = C cannot occur. In fact, a simple angle chase shows that in this case, we have < BPD = $180^{\circ}$, so that the point P (if it exists) lies on the diagonal BD, and from < PBC = < DBA we conclude that the diagonal BD bisects the angle ABC. This contradicts the assumption of the problem. Hence, AP = CP holds if and only if the quadrilateral ABCD is cyclic. Peter has another solution, applying the sine law one thousand times (later he admitted it were just four times or something like this ). Darij

We call a positive integer alternating if every two consecutive digits in its decimal representation are of different parity. Find all positive integers $n$ such that $n$ has a multiple which is alternating.

Click for solution not really nice, but anyway not too hard, if one doesn't forget something at the end. ok, it are all numbers not divisible by 20. actually, it is clear that there is no alternating multiple of them since the last two digits are always even. so, let's first do powers of 2 and 5. in both cases start with the number 25252525...2525 for 5 with "enough" digits(or 101010...1010 for 2). in the k-th step, change the k-th digit such that the new number is divisible by $2^{k+1}$(respectively $5^{k+1}$) if the number is not divisible by that term before(where even digits stay even and odd digits stay odd). if inductively follows that this is actually possible, so we get a alternating number which is divisible by $2^n$(respectively $5^n$) and has an odd digit at the end. now, take a power of 2 times an odd number not divisible by 5. take the alternating number divisible by the power of 2 several times to get by fermat a number divisible by the number we want to. now, take (10 times a) power of 5 times an odd number not divisible by 5. take the number divisible by to the power of 5 several times to get a number divisible by the power of 5 times the odd number not divisible by 5. add a 0 at the end if the number is divisible by 10. so, we have everything since numbers divisible by 20 have no alternating multiple. i hope the main argument gets clear. Peter