Find all real numbers $ a,b,c,d$ such that \[ \left\{\begin{array}{cc}a + b + c + d = 20, \\ ab + ac + ad + bc + bd + cd = 150. \end{array} \right.\]

Click for solution That is really triviality $ (a-b)^2 + (b-c)^2 + (c-d)^2 + (b-d)^2 + (a-c)^2 \ge 0$ $ \implies \frac{3}{8}(a+b+c+d)^2 \ge ab+bc+cd+ad+bd+ac$ with equality iff $ a=b=c=d$. In this case we have equality, so $ a=b=c=d=5$.

The vertices $ A$ and $ B$ of an equilateral triangle $ ABC$ lie on a circle $k$ of radius $1$, and the vertex $ C$ is in the interior of the circle $ k$. A point $ D$, different from $ B$, lies on $ k$ so that $ AD=AB$. The line $ DC$ intersects $ k$ for the second time at point $ E$. Find the length of the line segment $ CE$.

Click for solution Let $ S$ be center of circle $ k$. Let $ \mathcal{C}$ be circle with center at $ A$ and $ r=AB$. Then $ C,D\in \mathcal{C}$. Let $ \angle CBD = \alpha$. Then we have: $ \angle CAD = 2\alpha$ and $ \angle ABD = 60-\alpha$. Beacuse triangles $ ABD$ and $ ACD$ are isoscale we have $ \angle ADB = 60-\alpha$ and $ \angle ACD =\angle ADC = 90-\alpha$. Then $ \angle BSE = 2\cdot \angle BDC = 60$. So $ \overline{BE} =1$. Now $ \angle BCE = 180- 60-(90-\alpha) = 30+\alpha$ and $ \angle BEC = 120-2\alpha$ since $ ABED$ is cyclic. Finaly $ \angle CBE =30+\alpha$ wich means $ \overline{CE} = \overline{BE} =1$.

Find all prime numbers $ p,q,r$, such that $ \frac{p}{q}-\frac{4}{r+1}=1$

Click for solution As suggested by Taidoda-X, note that $ \dfrac{p}{q}=\dfrac{r+5}{r+1}$. Euclid's algorithm immediately yields that $ \lambda(r)=(r+5,r+1)=(4,r+1)$. Hence, $ \lambda(r)\in\{1;2;4\}$. Note also that $ p>q$. The rest is little more than casework: $ \lambda(r)=1$: This readily implies $ r+1=q$, whence the solution $ (p,q,r)=(7,3,2)$ $ \lambda(r)=2$: We get $ 2p=r+5$ and $ 2q=r+1$. Thus, additionning and subtracting the two equalities, $ p+q=r+3$ and $ p-q=2$. Suppose $ p,q,r\geq5$. Hence, $ p+q\equiv 0,2\mbox{ mod }6$ and $ r+3\equiv 2,4\mbox{ mod }6$, implying that $ p\equiv q\equiv 1\mbox{ mod }6$, contradicting the fact that $ p$ and $ q$ are twin primes. Checking the remaining cases, we find the solution $ (p,q,r)=(5,3,5)$. $ \lambda(r)=4$: Subtracting the equalities $ 4p=r+5$ and $ 4q=r+1$, we find that $ p-q=1$, which readility implies the solution $ (p,q,r)=(3,2,7)$.

A $ 4\times 4$ table is divided into $ 16$ white unit square cells. Two cells are called neighbors if they share a common side. A move consists in choosing a cell and the colors of neighbors from white to black or from black to white. After exactly $ n$ moves all the $ 16$ cells were black. Find all possible values of $ n$.

Click for solution By considering parity, $ n$ must be even. If $ a$ is a possible value of $ n$, then $ a+2$ is also a possible value of $ n$, because we can choose the same square for the $ a+1$-th and the $ a+2$-th move. If $ a$ is the least possible value of $ n$, then the answer is $ a,a+2,a+4,\ldots$. Now, we have to find the least possible value of $ n$. I have yet to find this least value. It is quite hard for me. By the way, I played this kind of game in my computer, and I finished it after about 80 moves.