Each vertex of a cube is assigned an 1 or a -1, and each face is assigned the product of the numbers assigned to its vertices. Determine the possible values the sum of these 14 numbers can attain.

A square is divided in four parts by two perpendicular lines, in such a way that three of these parts have areas equal to 1. Show that the square has area equal to 4.

Let $f: \ [0,\ 1] \rightarrow \mathbb{R}$ be an increasing function satisfying the following conditions: a) $f(0)=0$; b) $f\left(\frac{x}{3}\right)=\frac{f(x)}{2}$; c) $f(1-x)=1-f(x)$. Determine $f\left(\frac{18}{1991}\right)$.

Click for solution Jutaro wrote: Let $ f: \ [0,\ 1] \rightarrow \mathbb{R}$ be an increasing function satisfying the following conditions: a) $ f(0) = 0$; b) $ f\left(\frac {x}{3}\right) = \frac {f(x)}{2}$; c) $ f(1 - x) = 1 - f(x)$. Determine $ f\left(\frac {18}{1991}\right)$. We have : $ f(1) = f(1 - 0) = 1 - f(0) = 1$ $ f(\frac {1}{3}) = \frac {f(1)}{2} = \frac {1}{2} = > f(\frac {2}{3}) = \frac {1}{2}$ $ = > f(\frac {1}{9}) = \frac {1}{4} = > f(\frac {8}{9}) = \frac {3}{4}$ Similar , we also have : $ f(\frac {1}{27}) = f(\frac {2}{27}) = \frac {1}{8} ; f(\frac {7}{27}) = f(\frac {8}{27}) = \frac {3}{8}$ $ f(\frac {19}{27}) = f(\frac {20}{27}) = \frac {5}{8} ; f(\frac {25}{27}) = f(\frac {26}{27}) = \frac {7}{8}$ Because f is the decreasing on $ [0;1]$ , we have : $ f(x) = \frac {1}{2}$ for all $ x \in [\frac {1}{3};\frac {2}{3}]$ $ f(x) = \frac {1}{4}$ for all $ x \in [\frac {1}{9};\frac {2}{9}]$ .... $ f(x) = \frac {7}{8}$ for all $ x \in [\frac {25}{27};\frac {26}{27}]$ We have : $ \frac {19}{3^7} < \frac {18}{1991} < \frac {20}{3^7}$ But , $ f(\frac {19}{27}) = \frac {5}{8} = > f(\frac {19}{3^4}) = \frac {5}{16} = > ... = > f(\frac {19}{3^7}) = \frac {5}{128}$ $ f(\frac {20}{27}) = \frac {5}{8} = > f(\frac {20}{3^4}) = \frac {5}{16} = > ... = > f(\frac {20}{3^7}) = \frac {5}{128}$ Because f is the decreasing on $ [0;1]$ , we have : $ f(\frac {18}{1991}) = \frac {5}{128}$

Find a positive integer $n$ with five non-zero different digits, which satisfies to be equal to the sum of all the three-digit numbers that can be formed using the digits of $n$.

Let $P(x,\, y)=2x^{2}-6xy+5y^{2}$. Let us say an integer number $a$ is a value of $P$ if there exist integer numbers $b$, $c$ such that $P(b,\, c)=a$. a) Find all values of $P$ lying between 1 and 100. b) Show that if $r$ and $s$ are values of $P$, then so is $rs$.

Let $M$, $N$ and $P$ be three non-collinear points. Construct using straight edge and compass a triangle for which $M$ and $N$ are the midpoints of two of its sides, and $P$ is its orthocenter.

Click for solution Jutaro wrote: Let $ ABC$ a triangle, $ M$, $ N$ the midpoints of $ AB,AC$ and $ H$ the orthocenter. Given the points $ M,N,H$ construct the triangle If the triangle $ MHN$ is right then $ \triangle ABC$ is also right, which means that $ H$ coincides with some vertice (if it is right at $ H$ then $ H\equiv A$, if it is right at $ M$ then $ H\equiv B$, if it is right at $ N$ then $ H\equiv C$) So suppose that $ \triangle MHN$ is not right Let $ w$ the circumcircle of $ \triangle ABC$. Let $ F$ be the anti-diametric point of $ B$. It is $ FA \perp BA \Rightarrow FA \parallel HC$ and $ FC \perp BC \Rightarrow FC\parallel AH$. (notice that noone of the above segments is zero, because $ \triangle ABC$ is not right) So $ AHCF$ is a parallelogram. This means that $ N$ is the midpoint of $ FH$ So the points $ H,F$ are symmetric wrt $ N$ Construction Let $ \varepsilon$ be the line peprendicular to $ MN$ passing through $ H$. Clearly $ A\in \varepsilon$ Let $ F$ be the symmetric point of $ H$ wrt $ N$. The circle with diameter $ MF$ intersects the line $ \varepsilon$ at two points $ A_{1},A_{2}$ For $ A\in\{A_{1},A_{2}\}$ let $ B,C$ be the symmetric points of $ A$ wrt $ M,N$ respectively Proof From the construction it is $ AF\perp AB$. But $ AHCF$ is a parallelogram (because $ N$ is the midpoint of $ AC$ and $ HF$) So $ CH\perp AB$. On the other hand it is $ AH\perp MN \Rightarrow AH\perp BC$, hence $ H$ is the orthocenter of $ \triangle ABC$