Prove that the inequality \[ \frac{ x+y}{x^2-xy+y^2 } \leq \frac{ 2\sqrt 2 }{\sqrt{ x^2 +y^2 } } \] holds for all real numbers $x$ and $y$, not both equal to 0.

Click for solution You can also prove the inequality by squaring it (in fact, the right hand side of the inequality is obviously $\geq 0$; if the left hand side is $\leq 0$, then the inequality is trivial, so it is enough to consider the case when it is $\geq 0$ as well, and then we can square the inequality); this leads to $\frac{\left( x+y\right) ^2}{\left( x^2-xy+y^2\right) ^2}\leq \frac{8}{x^2+y^2}$. This is obviously equivalent to $\left( x+y\right) ^2\left( x^2+y^2\right) \leq 8\left( x^2-xy+y^2\right) ^2$. But actually, an easy calculation shows that $8\left( x^2-xy+y^2\right) ^2-\left( x+y\right) ^2\left( x^2+y^2\right) =\left( x-y\right) ^2\left( 7x^2-4xy+7y^2\right)$ $=\left( x-y\right) ^2\left( 2\left( x-y\right) ^2+5x^2+5y^2\right) \geq 0$, so everything is proven. Darij

Let $ABC$ be an isosceles triangle with $AC=BC$, let $M$ be the midpoint of its side $AC$, and let $Z$ be the line through $C$ perpendicular to $AB$. The circle through the points $B$, $C$, and $M$ intersects the line $Z$ at the points $C$ and $Q$. Find the radius of the circumcircle of the triangle $ABC$ in terms of $m = CQ$.

Click for solution This is a solution of this problem posted by vedran6 in a different thread: It is obvious that the BCMQ is cyclic and because of that is and we have that angle <CQM=<CBM (1) let $CZ\cap BM=\left(T\right)$ ; and $CZ\cap AB=\left(D\right)$ it is obvious that is $CT=\frac23\cdot CD$ and because of AC=BC we have <ACQ=<BCT (2) , from (1) and (2) we have that trinagle CQM is similar to triangle CBT. From similarity we have that is $\frac{CQ}{CB}=\frac{CM}{CT}$ which is equal to $CQ=\frac{3\cdot CB\cdot CB}{4\cdot CD}$ (3). It is well known that in every triangle is $R=\frac{a\cdot b\cdot c}{4P}$ so we use it at this triangle ABC and got $R=\frac{BC\cdot BC}{2\cdot CD}$ (4) . After dividing (3) and (4) we get $CQ=1,5\cdot R$ and the problem is finished

If the positive integers $x$ and $y$ are such that $3x + 4y$ and $4x + 3y$ are both perfect squares, prove that both $x$ and $y$ are both divisible with $7$.

Click for solution My solution: Note that the quadratic residues modulo 7 are 0, 1, 2, 4, and thus we immediately see: Lemma. The only residue t modulo 7 such that t and -t are both quadratic residues modulo 7 is t = 0. Now, we have $3x+4y\equiv 3x-3y=3\left( x-y\right)\ \text{mod}\ 7$, and $4x+3y\equiv -3x+3y=-3\left( x-y\right)\ \text{mod}\ 7$. Therefore, both 3 (x - y) mod 7 and -3 (x - y) mod 7 must be quadratic residues modulo 7. Thus, after the Lemma, we must have $3\left( x-y\right) \equiv 0\ \text{mod}\ 7$; hence, $x-y\equiv 0\ \text{mod}\ 7$, and the number x - y is divisible by 7. From $x-y\equiv 0\ \text{mod}\ 7$, it follows that $3x+4y\equiv 3x-3y=3\left( x-y\right) \equiv 0\ \text{mod}\ 7$. Therefore, the number 3x + 4y is divisible by 7. Since this number is a perfect square, it must also be divisible by 49. Similarly, 4x + 3y is divisible by 49. Thus, (3x + 4y) + (4x + 3y) = 7x + 7y = 7 (x + y) is also divisible by 49. This means that x + y is divisible by 7. Since x - y is also divisible by 7, the number 2x = (x + y) + (x - y) must be divisible by 7, too. Since 2 and 7 are coprime, this entails that x is divisible by 7. Similarly, y is divisible by 7. Darij

Consider a convex polygon having $n$ vertices, $n\geq 4$. We arbitrarily decompose the polygon into triangles having all the vertices among the vertices of the polygon, such that no two of the triangles have interior points in common. We paint in black the triangles that have two sides that are also sides of the polygon, in red if only one side of the triangle is also a side of the polygon and in white those triangles that have no sides that are sides of the polygon. Prove that there are two more black triangles that white ones.

Click for solution And here is my solution: At first we will show a rather simple lemma: Lemma. Let $n\geq 4$ be an integer. Every triangulation of a convex $n$-gon contains at least $2$ black triangles. Proof. The proof will use induction after $n$: Induction base: The case $n = 4$ is not what most of you would call difficult... Induction step: Now assume that we have proved our lemma for all $n\leq k$, where $k$ is a certain number with $k\geq 4$, and now we are going to prove it for $n = k + 1$. In fact, consider a triangulated convex $\left(k + 1\right)$-gon $P$. Assume that it doesn't contain $2$ black triangles. Every side of $P$ belongs either to a black triangle or to a red triangle. Since $k\geq 4$, we have $k+1\geq 5$, and our $\left(k + 1\right)$-gon $P$ has, at least, $5$ sides. Only $2$ of these sides can belong to black triangles (else, we would have at least $2$ black triangles, contradicting our assumption!). Hence, we will always find, at least, one side of $P$ which belongs to a red triangle $T$. If we remove this red triangle, we are left with two triangulated convex polygons $Q$ and $Q^{\prime}$ having one common vertex (namely the third vertex of the removed red triangle). Each of these two polygons $Q$ and $Q^{\prime}$ has $\leq k$ vertices. Now, by our induction assumption, each of these two polygons must have at least two black triangles. Hence we have four black triangles altogether. Now, if we put in the removed red triangle $T$ again, then at most two of these black triangles may possibly become red triangles (in fact, every of our two polygons $Q$ and $Q^{\prime}$ had a side bordering with the removed red triangle $T$; now, if such a side belongs to a black triangle in $Q$ or in $Q^{\prime}$, then, after putting in the red triangle $T$ again, this black triangle becomes a red triangle). But at least $4 - 2 = 2$ black triangles remain. Hence, we have found $2$ black triangles in our $\left(k + 1\right)$-gon $P$, contradicting our assumption. Hence, our $\left(k + 1\right)$-gon $P$ must have $2$ black triangles (at least). This completes the induction step, and the Lemma is proven. Now to the solution of the problem: By the niveau of a triangle in our triangulation, we will mean the number defined according to the following rule: If the triangle is white, then its niveau is -1; if the triangle is red, then its niveau is 0; if the triangle is black, then its niveau is 1. The summary niveau of a triangulated polygon will denote the sum of the niveaus of all triangles involved in the triangulation. Equivalently, it is the number of the black triangles minus the number of the white triangles. Now we will prove that the summary niveau is always equal to $2$ (when $n\geq 4$). This will be obviously enough to solve the problem. I will prove my assertion by induction. For $n = 4$, it is more than trivial. Now we assume that it holds for a certain $n\geq 4$ (i. e. it holds for all $n$-gons), and we have to show that it holds for $n + 1$ (i. e. for all $\left(n + 1\right)$-gons). Consider a triangulated $\left(n + 1\right)$-gon $P$. Clearly, $n\geq 4$ yields $n+1\geq 5$. On the other hand, after the Lemma, the triangulated $\left(n + 1\right)$-gon $P$ contains at least two black triangles. Let $D$ be one of them. Now remove this black triangle $D$. What you then get is a triangulated $n$-gon $P^{\prime}$. All black, white, red triangles in the triangulation of $P$ preserve their color in $P^{\prime}$, except for the initial black triangle $D$, which disappears, and the triangle $D^{\prime}$ directly bordering with $D$. In fact, if $D^{\prime}$ was a red triangle for $P$, then it becomes a black triangle for $P^{\prime}$, and if $D^{\prime}$ was a white triangle for $P$, then it becomes a red triangle for $P^{\prime}$. (You can easily figure out that $D^{\prime}$ could not have been a black triangle for $P$; since otherwise, all three sides of $D^{\prime}$ would be sides of $P^{\prime}$, and thus, the polygon $P^{\prime}$ would be a triangle, and $P$ would be a quadrilateral, contradicting $n+1\geq 5$.) Now, we see that the triangulated polygons $P$ and $P^{\prime}$ have the same summary niveau. This is because when we remove the black triangle $D$, then on one hand, we lose a niveau of $1$ (since $D$ had a niveau of $1$), but on the other hand, we gain a niveau of $1$ again (in fact, if $D^{\prime}$ was a red triangle for $P$, then it becomes a black triangle for $P^{\prime}$, so that the niveau changes from $0$ to $1$, and if $D^{\prime}$ was a white triangle for $P$, then it becomes a red triangle for $P^{\prime}$, so that the niveau changes from $-1$ to $0$). Hence, the sum of the niveaus of all triangles remains constant, i. e., the summary niveaus of the triangulated polygons $P$ and $P^{\prime}$ are equal. But since $P^{\prime}$ has $n$ edges, then by our induction assumption, its summary niveau is $2$, and hence the summary niveau of $P$ is also $2$. Proof complete. Darij