In a wagon, every $m \geq 3$ people have exactly one common friend. (When $A$ is $B$'s friend, $B$ is also $A$'s friend. No one was considered as his own friend.) Find the number of friends of the person who has the most friends.

Click for solution I don't get it. If three people have exactly one friend $X$, then won't four people have the exact same friend X? Because if four people have a common friend $Y \neq X$, then the three people in that set of four that have a common friend $X$ also have a common friend $Y$, contradicting that they have exactly one common friend. But this is obviously too easy for chinese TST.. am I missing something? --- (edit) oops: m is fixed, i get it.

Finitely many polygons are placed in the plane. If for any two polygons of them, there exists a line through origin $O$ that cuts them both, then these polygons are called "properly placed". Find the least $m \in \mathbb{N}$, such that for any group of properly placed polygons, $m$ lines can drawn through $O$ and every polygon is cut by at least one of these $m$ lines.

Click for solution This looks very familiar. I think something similar appeared on the forum recently. First of all, notice that $m=1$ doesn't work: We may assume WLOG that none of the polygons contain $O$. Consider a line $AB$ through $O$ with $A,B$ on opposite sides of $O$, and rays $(OC,(OD$ on the same side of $AB$, with $(OC$ between $(OB$ and $(OD$. We can now take our "polygons" to be $BC,CD,DA$ (they can, of course, be "thickened" into proper polygons and not just segments). Next, we can show that $m=2$ works: For each polygon in the collection consider the minimal region contained between two lines through $O$ which contains the polygon. Now pick a polygon $\mathcal P$ in the collection whose region is minimal wrt inclusion, in the sense that there's no other polygon whose region is properly included in the $\mathcal P$'s region. The two lines through $O$ which bound the region of $\mathcal P$ have the required properties, as can easily be seen. I hope it's correct.

In set $S$, there is an operation $'' \circ ''$ such that $\forall a,b \in S$, a unique $a \circ b \in S$ exists. And (i) $\forall a,b,c \in S$, $(a \circ b) \circ c = a \circ (b \circ c)$. (ii) $a \circ b \neq b \circ a$ when $a \neq b$. Prove that: a.) $\forall a,b,c \in S$, $(a \circ b) \circ c = a \circ c$. b.) If $S = \{1,2, \ldots, 1990\}$, try to define an operation $'' \circ ''$ in $S$ with the above properties.

Click for solution Observe first that $(a\circ a)\circ a=a\circ (a\circ a)$ implies $a\circ a=a$, for all $a$. Now, we have \[ a\circ\left( a\circ\left( b\circ a\right) \right) =\left( a\circ a\right) \circ\left( b\circ a\right) =a\circ\left( b\circ a\right) \] and% \[ \left( a\circ\left( b\circ a\right) \right) \circ a=\left( a\circ b\right) \circ\left( a\circ a\right) =\left( a\circ b\right) \circ a=a\circ\left( b\circ a\right) \] Again, we deduce that $a\circ\left( b\circ a\right) =a,$ for all $a,b.$ Finally,% \[ \left( a\circ c\right) \circ\left( a\circ\left( b\circ c\right) \right) =\left( a\circ\left( c\circ a\right) \right) \circ\left( b\circ c\right) =a\circ\left( b\circ c\right) \] and% \[ \left( a\circ\left( b\circ c\right) \right) \circ\left( a\circ c\right) =\left( a\circ b\right) \circ\left( c\circ\left( a\circ c\right) \right) =\left( a\circ b\right) \circ c, \] therefore% \[ a\circ\left( b\circ c\right) =a\circ c \] For the second part, a trivial example is $a\circ b=a,$ for all $a,b$ in $S.$ A less trivial example can be obtained by considering a set $S^{\prime }=\left\{ \left( x,y\right) |1\leq x\leq n,1\leq y\leq m\right\} ,$ with $nm=1990,$ defining a one-to-one function $f: S\rightarrow S^{\prime}$ and transporting through $f$ the structure defined on $S^{\prime}$ as follows: $\left( x,y\right) \circ\left( x^{\prime},y^{\prime}\right) =\left( x,y^{\prime}\right) ,$ which clearly satisfy the requested conditions. The previous trivial example is obtained for $n=1990$ and $m=1$

Number $a$ is such that $\forall a_1, a_2, a_3, a_4 \in \mathbb{R}$, there are integers $k_1, k_2, k_3, k_4$ such that $\sum_{1 \leq i < j \leq 4} ((a_i - k_i) - (a_j - k_j))^2 \leq a$. Find the minimum of $a$.

Given a triangle $ ABC$ with angle $ C \geq 60^{\circ}$. Prove that: $ \left(a + b\right) \cdot \left(\frac {1}{a} + \frac {1}{b} + \frac {1}{c} \right) \geq 4 + \frac {1}{\sin\left(\frac {C}{2}\right)}.$

Click for solution Attempt of a halfways nice solution. Problem. Let ABC be a triangle with $C\geq 60^{\circ}$. Prove the inequality $\left(a+b\right)\cdot\left(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\right)\geq 4+\frac{1}{\sin\frac{C}{2}}$. Solution. First, we equivalently transform the inequality in question: $\left(a+b\right)\cdot\left(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\right)\geq 4+\frac{1}{\sin\frac{C}{2}}$ $\Longleftrightarrow\ \ \ \ \ \left(a+b\right)\cdot\left(\frac{1}{a}+\frac{1}{b}\right)+\frac{a+b}{c}\geq 4+\frac{1}{\sin\frac{C}{2}}$ $\Longleftrightarrow\ \ \ \ \ \left(a+b\right)\cdot\left(\frac{1}{a}+\frac{1}{b}\right)-4\geq\frac{1}{\sin\frac{C}{2}}-\frac{a+b}{c}$ $\Longleftrightarrow\ \ \ \ \ \frac{\left(a-b\right)^2}{ab}\geq\frac{1}{\sin\frac{C}{2}}-\frac{a+b}{c}$. Now, by the Mollweide formulas, $\frac{a+b}{c}=\frac{\cos\frac{A-B}{2}}{\sin\frac{C}{2}}$ and $\frac{a-b}{c}=\frac{\sin\frac{A-B}{2}}{\cos\frac{C}{2}}$, so that $\frac{a-b}{a+b}=\frac{a-b}{c} : \frac{a+b}{c}=\frac{\sin\frac{A-B}{2}}{\cos\frac{C}{2}} : \frac{\cos\frac{A-B}{2}}{\sin\frac{C}{2}}=\frac{\sin\frac{A-B}{2}\sin\frac{C}{2}}{\cos\frac{A-B}{2}\cos\frac{C}{2}}$. Now, $\cos^2\frac{C}{2}\leq 1$ (as the square of every cosine is $\leq 1$). On the other hand, the AM-GM inequality yields $ab\leq\frac14\left(a+b\right)^2$. Hence, $\frac{\left(a-b\right)^2}{ab}\geq\frac{\left(a-b\right)^2}{\frac14\left(a+b\right)^2}$ (since $ab\leq\frac14\left(a+b\right)^2$) $=4\left(\frac{a-b}{a+b}\right)^2=4\left(\frac{\sin\frac{A-B}{2}\sin\frac{C}{2}}{\cos\frac{A-B}{2}\cos\frac{C}{2}}\right)^2=\frac{4\sin^2\frac{A-B}{2}\sin^2\frac{C}{2}}{\cos^2\frac{A-B}{2}\cos^2\frac{C}{2}}$ $\geq\frac{4\sin^2\frac{A-B}{2}\sin^2\frac{C}{2}}{\cos^2\frac{A-B}{2}}$ (since $\cos^2\frac{C}{2}\leq 1$). $=\frac{4\left(2\sin\frac{A-B}{4}\cos\frac{A-B}{4}\right)^2\sin^2\frac{C}{2}}{\cos^2\frac{A-B}{2}}=\frac{16\sin^2\frac{A-B}{4}\cos^2\frac{A-B}{4}\sin^2\frac{C}{2}}{\cos^2\frac{A-B}{2}}$. Thus, instead of proving the inequality $\frac{\left(a-b\right)^2}{ab}\geq\frac{1}{\sin\frac{C}{2}}-\frac{a+b}{c}$, it will be enough to show the stronger inequality $\frac{16\sin^2\frac{A-B}{4}\cos^2\frac{A-B}{4}\sin^2\frac{C}{2}}{\cos^2\frac{A-B}{2}}\geq\frac{1}{\sin\frac{C}{2}}-\frac{a+b}{c}$. Noting that $\frac{1}{\sin\frac{C}{2}}-\frac{a+b}{c}=\frac{1}{\sin\frac{C}{2}}-\frac{\cos\frac{A-B}{2}}{\sin\frac{C}{2}}=\frac{1-\cos\frac{A-B}{2}}{\sin\frac{C}{2}}=\frac{2\sin^2\frac{A-B}{4}}{\sin\frac{C}{2}}$, we transform this inequality into $\frac{16\sin^2\frac{A-B}{4}\cos^2\frac{A-B}{4}\sin^2\frac{C}{2}}{\cos^2\frac{A-B}{2}}\geq\frac{2\sin^2\frac{A-B}{4}}{\sin\frac{C}{2}}$, what, upon multiplication by $\frac{\cos^2\frac{A-B}{2}\sin\frac{C}{2}}{16\sin^2\frac{A-B}{4}}$ and rearrangement of terms, becomes $\sin^3\frac{C}{2}\cos^2\frac{A-B}{4}\geq\frac18\cos^2\frac{A-B}{2}$. But this trivially follows by multiplying the two inequalities $\sin^3\frac{C}{2}\geq\frac18$ (equivalent to $\sin\frac{C}{2}\geq\frac12$, what is true because $60^{\circ}\leq C\leq 180^{\circ}$ yields $30^{\circ}\leq\frac{C}{2}\leq 90^{\circ}$) and $\cos^2\frac{A-B}{4}\geq\cos^2\frac{A-B}{2}$ (follows from the obvious fact that $\left|\frac{A-B}{4}\right|\leq\left|\frac{A-B}{2}\right|$ since $\left|\frac{A-B}{2}\right|<90^{\circ}$, what is true because $\left|A-B\right|<180^{\circ}$, as the angles A and B, being angles of a triangle, lie between 0° and 180°). Hence, the problem is solved. Darij

Find all functions $f,g,h: \mathbb{R} \mapsto \mathbb{R}$ such that $f(x) - g(y) = (x-y) \cdot h(x+y)$ for $x,y \in \mathbb{R}.$

Click for solution Take $x=y$ and you find $f=g$. Next, take $y=0$ and you find $ xh(x)=f(x)-f(0)$. Put $ t(x)=f(x)-f(0)$. Then $ (x-y)t(x+y)=(x+y)(t(x)-t(y))$. Now, take $ y=1$ and you find $ t(x+1)=\frac{x+1}{x-1} (t(x)-t(1))$. Substitute here $x+1$ instead of $x$ and obtain $ t(x+2)=\frac{x+2}{x} (\frac{x+1}{X-1}(t(x)-t(1))-t(1))$. But $ t(x+2)=\frac{x+2}{x-2} (t(x)-t(2))$. From the two relations we find that $t$ is of the form $ ax+b$ and since $ t(0)=0$ we have $t(x)=ax$ and then $ f(x)=ax+b$.

Prove that for every integer power of 2, there exists a multiple of it with all digits (in decimal expression) not zero.

There are arbitrary 7 points in the plane. Circles are drawn through every 4 possible concyclic points. Find the maximum number of circles that can be drawn.