Let $x$, $y$ be two positive integers such that $\displaystyle 3x^2+x=4y^2+y$. Prove that $x-y$ is a perfect square.

Click for solution We can rewrite $ 3x^2 + x = 4y^2 + y$ as: $ (8x + 1)^2 = (4x)^2 + (8y + 1)^2$ but that is no more than a pythagorean triple. Exist $ u,v$ positive integers whit some conditions such that: $ u^2 + v^2 = 8x + 1$ $ 2uv = 4x$ $ u^2 - v^2 = 8y + 1$ because $ 4x$ is even, while $ 8y + 1$ is odd. From the first and the third ecuation: $ 2v^2 = 8(x - y)$, then ($ 2 \mid v$ and) $ (\frac {v}{2})^2 = x - y$

Two right angled triangles are given, such that the incircle of the first one is equal to the circumcircle of the second one. Let $S$ (respectively $S'$) be the area of the first triangle (respectively of the second triangle). Prove that $\frac{S}{S'}\geq 3+2\sqrt{2}$.

Click for solution Igor wrote: Two right angled triangles are given, such that the incircle of the first one is equal to the circumcircle of the second one. Let $S$ (respectively $S'$) be the area of the first triangle (respectively of the second triangle). Prove that $\displaystyle \frac{S}{S'}\geq 3+2\sqrt{2}$. It seems to be difficult to give a non-bash solution of this problem, but let me try: Let ABC be the first right-angled triangle, with the right angle at C, and with the sides a = BC, b = CA, c = AB and the angles A = < CAB, B = < ABC, C = < BCA. Let A'B'C' be the second right-angled triangle, with the right angle at C', and with the sides a' = B'C', b' = C'A', c' = A'B' and the angles A' = < C'A'B', B' = < A'B'C', C' = < B'C'A'. According to the conditions of the problem, the inradius r of triangle ABC equals to the circumradius of triangle A'B'C'. The circumradius of a right-angled triangle equals $\frac12$ of its hypotenuse. Since the triangle A'B'C' is right-angled with the hypotenuse c', its circumradius thus equals $\frac{c^{\prime}}{2}$; on the other hand, we know that its circumradius is r. Thus, $r=\frac{c^{\prime}}{2}$, so that 2r = c'. But by the Pythagorean theorem, applied to the right-angled triangle A'B'C', we have $c^{\prime}\ ^2=a^{\prime}\ ^2+b^{\prime}\ ^2$. On the other hand, the area S' of the right-angled triangle A'B'C' equals $\frac12$ of the product of its catets a' and b'; in other words, $S^{\prime}=\frac12 a^{\prime}b^{\prime}$. By the inequality $2xy\leq x^2+y^2$ for any two real numbers x and y, we thus have $4S^{\prime}=4\cdot\frac12 a^{\prime}b^{\prime}=2a^{\prime}b^{\prime}\leq a^{\prime}\ ^2+b^{\prime}\ ^2=c^{\prime}\ ^2=\left(2r\right)^2=4r^2$, so that $S^{\prime}\leq r^2$. On the other hand, by Lemma 1 from http://www.mathlinks.ro/Forum/viewtopic.php?t=4972 post #19, we have $S=r^2\cdot\cot\frac{A}{2}\cdot\cot\frac{B}{2}\cdot\cot\frac{C}{2}$, since S is the area and r is the inradius of triangle ABC. Since C is a right angle, we have $\frac{C}{2}=\frac{90^{\circ}}{2}=45^{\circ}$, so that $\cot\frac{C}{2}=\cot 45^{\circ}=1$, and this equation becomes $S=r^2\cdot\cot\frac{A}{2}\cdot\cot\frac{B}{2}$. The angles A and B are the two acute angles of the right-angled triangle ABC; thus, they satisfy 0° < A < 90° and 0° < B < 90°. Consequently, $0^{\circ}<\frac{A}{2}<45^{\circ}$ and $0^{\circ}<\frac{B}{2}<45^{\circ}$. Consequently, by the inequality from http://www.mathlinks.ro/Forum/viewtopic.php?t=21384 , we have $\sqrt{\tan\frac{A}{2}\tan\frac{B}{2}}\leq\tan\frac{\frac{A}{2}+\frac{B}{2}}{2}$. Since the triangle ABC is right-angled with the right angle at C, we have A + B = 90°, so that $\tan\frac{\frac{A}{2}+\frac{B}{2}}{2}=\tan\frac{A+B}{4}=\tan\frac{90^{\circ}}{4}=\tan 22,5^{\circ}=\sqrt2-1$, and thus this becomes $\sqrt{\tan\frac{A}{2}\tan\frac{B}{2}}\leq \sqrt2-1$. Since both sides of this inequality are nonnegative, we can square it: $\tan\frac{A}{2}\tan\frac{B}{2}\leq\left(\sqrt2-1\right)^2$. Hence, $S=r^2\cdot\cot\frac{A}{2}\cdot\cot\frac{B}{2}=\frac{r^2}{\tan\frac{A}{2}\tan\frac{B}{2}}\geq \frac{r^2}{\left(\sqrt2-1\right)^2}=r^2\cdot\left(3+2\sqrt2\right)$. Combining this with the inequality $S^{\prime}\leq r^2$ obtained before, we get $\frac{S}{S^{\prime}}\geq\frac{r^2\cdot\left(3+2\sqrt2\right)}{r^2}=3+2\sqrt2$, and the problem is solved. Darij

In an international meeting of $n \geq 3$ participants, 14 languages are spoken. We know that: - Any 3 participants speak a common language. - No language is spoken more that by the half of the participants. What is the least value of $n$?

Click for solution It is easy to construct the required sets using a 7 point projective plane + an extra point. Languages are spoken by lines + the extra point or by complements of lines.

Let $X$ be a non empty subset of $\mathbb{N} = \{1,2,\ldots \}$. Suppose that for all $x \in X$, $4x \in X$ and $\lfloor \sqrt{x} \rfloor \in X$. Prove that $X=\mathbb{N}$.

Click for solution Obviously for any integer k we have 4^k \in X. Thus \lfloor \sqrt{4^k} \rfloor = 2 k \in X, so for any integer m, we also have 4 m \in X. Thus \lfloor \sqrt{4 m} \rfloor = \lfloor 2 \sqrt{m} \rfloor \in X, for any integer m. Let's take m = \lceil \frac{a^2}{4} \rceil for a given integer a \geqslant 2 then \lfloor 2 \sqrt{m} \rfloor = a, the conclusion follows...

Let $ABC$ be a triangle such that $BC=AC+\frac{1}{2}AB$. Let $P$ be a point of $AB$ such that $AP=3PB$. Show that $\widehat{PAC} = 2 \widehat{CPA}.$

Click for solution Ok, so here is my solution: On $AB$, take $I$ such that $IC=AC$. It sufficient to prove that $PI=IC$. But $PI=PA-AI=\frac{3}{4}AB-AI$ Hence : $IC=AC \Longleftrightarrow \frac{3}{4}AB-AC=AI\Longleftrightarrow \frac{3AB}{4AC}-1=\frac{AI}{AC}$. $IC=AC \Longleftrightarrow \frac{3AB}{4AC}-1=2.\textrm{cos}(\widehat{PAC})$ by the sine law. $IC=AC \Longleftrightarrow \frac{3AB}{4AC}-1=\frac{AB^2+AC^2-BC^2}{AB.AC}$ by the cosine law. $IC=AC \Longleftrightarrow 3AB^2-4AC.AB=4AB^2+4AC^2-4BC^2$. $IC=AC \Longleftrightarrow BC^2=(AC+\frac{1}{2}AB)^2$, and we are done.

Let $P$ be a polynom of degree $n \geq 5$ with integer coefficients given by $P(x)=a_{n}x^n+a_{n-1}x^{n-1}+\cdots+a_0 \quad$ with $a_i \in \mathbb{Z}$, $a_n \neq 0$. Suppose that $P$ has $n$ different integer roots (elements of $\mathbb{Z}$) : $0,\alpha_2,\ldots,\alpha_n$. Find all integers $k \in \mathbb{Z}$ such that $P(P(k))=0$.

Click for solution Nice problem! Let me post my solution of this: Let $ P(x) = a_nx(x - \alpha_2)\cdots (x - \alpha_n)$. We shall prove that the equation $ P(x) = \alpha_2$ has no integer solution, so WLOG, none of the equations $ P(x) = \alpha_k$ has integer solutions. Thus all the integer solutions of $ P(P(x)) = 0$ are exactly the roots of $ P(x)$. Since all the $ \alpha_k$-s are distinct, and $ n > 4$ it follows that if $ |P(k)| = |a_nk(k - \alpha_2)\cdots (k - \alpha_n)| \geq 4 = 1 \cdot 1 \cdot 2 \cdot | - 2| \cdot | - 1|$ for all integers $ 0\neq k \neq \alpha_i$, $ i = \overline{2,n}$. Therefore if $ k$ is an integer solution of $ P(x) = \alpha_2$ then $ |\alpha_2|\geq 4$. Now we have $ (|k|\cdot |k - \alpha_2|) \mid \alpha_2$, and as such $ |k| \neq |\alpha_2|$ and also $ |k - \alpha_2| \neq |\alpha_2|$ (for obvious reasons), therefore we have $ 2 |k| \leq |\alpha_2|$ and also $ 2 |k - \alpha_2 | \leq |\alpha_2|$. Adding these last two relations up side by side we obtain \[ |\alpha_2| \leq |k| + |k - \alpha_2| \leq |\alpha_2| \] by the triangle inequality, so we must have $ |\alpha_2 | = 2|k|$ and also $ |\alpha_2| = 2|k - \alpha_2|$ but then $ |k|\cdot |k - \alpha_2| = \frac {|\alpha_2|^2}4 \geq |\alpha_2|$ which means that $ 2\leq |a_n (k - \alpha_3)\cdots (k - \alpha_n)| \leq 1$ which is a contradiction. The problem is thus finished