The professor tells Peter the product of two positive integers and Sam their sum. At first, nobody of them knows the number of the other. One of them says: You can't possibly guess my number. Then the other says: You are wrong, the number is 136. Which number did the professor tell them respectively? Give reasons for your claim.

Click for solution Hi, maskman, I think you have forgotten the statement, that the guessing person is not only using his number to deduce the number of the other, he also uses the knowledge, that the other knows that he can't deduce the other number by only his own number. I Hope that my solution is correct Obviously, if the product is 136, Sam cannot deduce the product by the knowledge, that Peter knows that he can't deduce the product with his sum, so the sum is 136. A possible solution would be e.g. if Peter had 135. When Sam said that Peter couldn't deduce the sum, he was right because the only possible case when Peter was able to was if the sum was 1+prime, which is wrong in this case (135 is not prime). But now Peter knows that Sam knows that Peter cannot deduce the sum by the mere knowledge of the product. Let Peter's number be $P$. If we consider the set of all $\{x+\frac{P}{x}\mid x|P\}$, then every number in this set except for one must be 1+prime. This is true for $P=135$, because in this case, Peter would know that Sam has one of the numbers 136, 48, 24, 32. Everyone of these numbers except for 136 is 1+prime. But if Sam had 48, 24 or 32 then Sam could not know that Peter cannot deduce the sum because Sam cannot tell for sure that Peter has no prime number. Hence Peter knows that Sam must have 136. Now we still have to prove that 135 is the only solution for $P$. But that's actually trivial because if there was another solution $P'$, then the set defined above would contain at least two numbers which do not have the form 1+prime (136 and $P'+1$) Hence Sam's number must be 136 and Peter's must be 135.

Let $k$ and $k'$ be concentric circles with center $O$ and radius $R$ and $R'$ where $R<R'$ holds. A line passing through $O$ intersects $k$ at $A$ and $k'$ at $B$ where $O$ is between $A$ and $B$. Another line passing through $O$ and distict from $AB$ intersects $k$ at $E$ and $k'$ at $F$ where $E$ is between $O$ and $F$. Prove that the circumcircles of the triangles $OAE$ and $OBF$, the circle with diameter $EF$ and the circle with diameter $AB$ are concurrent.

Click for solution ashegh wrote: i think inversion will help... i will think about it and will post the solution... but im not sure with it by inversion... Well, I have no idea how inversion (on a circle - as you probably meant) works (yes I know, I'm an uneducated geometrician), however I solved it by a mere angle-chasing (unspectacular but easy and stupit ) Let $T$ be the point of intersection of the circumcircles of $OAE$ and $OBF$. Let $\angle AOE = \alpha$ and $\angle BOF = \beta$. We have $\alpha+\beta = 180$. $\triangle OAE$ and $\triangle OBF$ are isosceles, thus $\angle OEA = \angle OAE = 90-\frac{\alpha}{2}$ and $\angle OFB = \angle OBF = 90-\frac{\beta}{2}$. $BFTO$ is cyclic, hence $\angle OTF = 180 - \angle OBF = 90+\frac{\beta}{2}$. Also, $OATE$ is cyclic, hence $\angle OTE = \angle OAE = 90-\frac{\alpha}{2}$. Thus, $\angle FTE = \angle OTF - \angle OTE = 90+\frac{\beta}{2} - 90 + \frac{\alpha}{2} = 90$, hence T lies on the circle with diameter $EF$. Similary, we'll show that $\angle ATB = 90$: $BOTF$ is cyclic, hence $\angle OTB = \angle OFB = 90-\frac{\beta}{2}$. $AOET$ is cyclic, hence $\angle OTA = \angle OEA = 90-\frac{\alpha}{2}$. Thus $\angle ATB = \angle OTA + \angle OTB = 90-\frac{\alpha}{2} + 90 - \frac{\beta}{2} = 90$, hence T lies on the circle with diameter $AB$. $\Longrightarrow q.e.d.$

Let $A_1,A_2,\ldots , A_n$ $(n\geq 3)$ be finite sets of positive integers. Prove, that \[ \displaystyle \frac{1}{n} \left( \sum_{i=1}^n |A_i|\right) + \frac{1}{{{n}\choose{3}}}\sum_{1\leq i < j < k \leq n} |A_i \cap A_j \cap A_k| \geq \frac{2}{{{n}\choose{2}}}\sum_{1\leq i < j \leq n}|A_i \cap A_j| \] holds, where $|E|$ is the cardinality of the set $E$

Click for solution let there be an element $a$ that's in exactly $k$ sets. this element will add to the first sum $k$, to the second sum ${k}\choose{3}$ and to the last sum ${k}\choose{2}$. if we prove te inequality by doing it one element at a time, then at the end it'll be true (since with start with empty set, which gives $0 \ge 0$). so we want to prove that: $\frac{k}{n} + \frac{{{k}\choose{3}}}{{{n}\choose{3}}} \ge \frac{2{{k}\choose{2}}}{{{n}\choose{2}}}$ $\frac{k}{n} + \frac{k(k-1)(k-2)}{n(n-1)(n-2)} \ge \frac{2k(k-1)}{n(n-1)}$ $1 + \frac{(k-1)(k-2)}{(n-1)(n-2)} \ge \frac{2(k-1)}{n-1}$ $(n-1)(n-2) + (k-1)(k-2) \ge 2(k-1)(n-2)$ let $n = 2 + s$ with $s \ge 1$ and $k = 1 + t$ with $t \ge 0$ $(s+1)s + t(t-1) \ge 2ts$ $(t-s)^2 \ge t-s$ which is true, since $t-s$ is an integer

Let $A$ be the set of all polynomials $f(x)$ of order $3$ with integer coefficients and cubic coefficient $1$, so that for every $f(x)$ there exists a prime number $p$ which does not divide $2004$ and a number $q$ which is coprime to $p$ and $2004$, so that $f(p)=2004$ and $f(q)=0$. Prove that there exists a infinite subset $B\subset A$, so that the function graphs of the members of $B$ are identical except of translations