problem: Let $p_{1}, p_{2}, p_{3}, \ldots$ be the prime numbers listed in increasing order, and let $0<x_{0}<1$ be a real number between 0 and 1 . For each positive integer $k$, define  $$ x_{k}= \begin{cases}0 & \text { if } x_{k-1}=0 \\ \left\{\frac{p_{k}}{x_{k-1}}\right\} & \text { if } x_{k-1} \neq 0\end{cases} $$  where $\{x\}$ denotes the fractional part of $x$. Find, with proof, all $x_{0}$ satisfying $0<x_{0}<1$ for which the sequence $x_{0}, x_{1}, x_{2}, \ldots$ eventually becomes 0 .

solution: ## The answer is $x_{0}$ rational.  If $x_{0}$ is irrational, then all $x_{i}$ are irrational by induction. So the sequence cannot become zero.  If $x_{0}$ is rational, then all are. Now one simply observes that the denominators of $x_{n}$ are strictly decreasing, until we reach $0=\frac{0}{1}$. This concludes the proof.  Remark. The sequence $p_{k}$ could have been any sequence of integers.


problem: Let $A B C$ be a triangle. Take noncollinear points $D, E, F$ on the perpendicular bisectors of $B C, C A, A B$ respectively. Show that the lines through $A, B, C$ perpendicular to $E F, F D, D E$ respectively are concurrent.

solution: The three lines are the radical axii of the three circles centered at $D, E, F$, so they concur.


problem: Prove that for any integer $n$, there exists a unique polynomial $Q$ with coefficients in $\{0,1, \ldots, 9\}$ such that $Q(-2)=Q(-5)=n$.

solution: If we let  $$ Q(x)=\sum_{k \geq 0} a_{k} x^{k} $$  then $a_{k}$ is uniquely determined by $n\left(\bmod 2^{k}\right)$ and $n\left(\bmod 5^{k}\right)$. Indeed, we can extract the coefficients of $Q$ exactly by the following algorithm:  - Define $b_{0}=c_{0}=n$. - For $i \geq 0$, let $a_{i}$ be the unique digit satisfying $a_{i} \equiv b_{i}(\bmod 2), a_{i} \equiv c_{i}(\bmod 5)$. Then, define  $$ b_{i+1}=\frac{b_{i}-a_{i}}{-2}, \quad c_{i+1}=\frac{c_{i}-a_{i}}{-5} . $$   In fact, we will prove the following claim: Claim - Suppose $b_{0}$ and $c_{0}$ are any integers such that  $$ b_{0} \equiv c_{0} \quad(\bmod 3) . $$  Then defining $b_{i}$ and $c_{i}$ as above, we have $b_{i} \equiv c_{i}(\bmod 3)$ for all $i$, and $b_{N}=c_{N}=0$ for large enough $N$.   $$ (b, c) \mapsto\left(\frac{b-a}{-2}, \frac{c-a}{-5}\right) $$  for some $0 \leq a \leq 9$ (a function in $b$ and $c$ ). The $b \equiv c(\bmod 3)$ is clearly preserved. Also, examining the size,  - If $|c|>2$, we have $\left|\frac{c-a}{-5}\right| \leq \frac{|c|+9}{5}<|c|$. Thus, we eventually reach a pair with $|c| \leq 2$. - Similarly, if $|b|>9$, we have $\left|\frac{b-a}{-2}\right| \leq \frac{|b|+9}{2}<|b|$, so we eventually reach a pair with $|b| \leq 9$. this leaves us with $5 \cdot 19=95$ ordered pairs to check (though only about one third have $b \equiv c(\bmod 3))$. This can be done by the following code:  ``` import functools @functools.lru_cache() def f(x0, yO):     if x0 == 0 and y0 == 0:         return 0     if x0 % 2 == (y0 % 5) % 2:         d = y0 % 5     else:         d = (y0 % 5) + 5     x1 = (x0 - d) // (-2)     y1 = (y0 - d) // (-5)     return 1 + f(x1, y1) for x in range(-9, 10): for y in range(-2, 3):     if (x % 3 == y % 3):         print(f"({x:2d}, {y:2d}) finished in {f(x,y)} moves") ```  As this gives the output ![](https://cdn.mathpix.com/cropped/2024_11_19_35408d86cf32c99e435fg-06.jpg?height=1317&width=664&top_left_y=1169&top_left_x=296) we are done.


problem: To clip a convex $n$-gon means to choose a pair of consecutive sides $A B, B C$ and to replace them by the three segments $A M, M N$, and $N C$, where $M$ is the midpoint of $A B$ and $N$ is the midpoint of $B C$. In other words, one cuts off the triangle $M B N$ to obtain a convex $(n+1)$-gon. A regular hexagon $\mathcal{P}_{6}$ of area 1 is clipped to obtain a heptagon $\mathcal{P}_{7}$. Then $\mathcal{P}_{7}$ is clipped (in one of the seven possible ways) to obtain an octagon $\mathcal{P}_{8}$, and so on. Prove that no matter how the clippings are done, the area of $\mathcal{P}_{n}$ is greater than $\frac{1}{3}$, for all $n \geq 6$.

solution: Call the original hexagon $A B C D E F$. We show the area common to triangles $A C E$ and $B D F$ is in every $\mathcal{P}_{n}$; this solves the problem since the area is $1 / 3$.  For every side of a clipped polygon, we define its foundation recursively as follows:  - $A B, B C, C D, D E, E F, F A$ are each their own foundation (we also call these original sides). - When a new clipped edge is added, its foundation is the union of the foundations of the two edges it touches.  Hence, any foundations are nonempty subsets of original sides. Claim - All foundations are in fact at most two-element sets of adjacent original sides.   Now, if a side has foundation contained in $\{A B, B C\}$, say, then the side should be contained within triangle $A B C$. Hence the side does not touch $A C$. This proves the problem.


solution: From $a^{3}+b^{3} \geq a b(a+b)$, the left-hand side becomes  $$ \sum_{\mathrm{cyc}} \frac{1}{a^{3}+b^{3}+a b c} \leq \sum_{\mathrm{cyc}} \frac{1}{a b(a+b+c)}=\frac{1}{a b c} \sum_{\mathrm{cyc}} \frac{c}{a+b+c}=\frac{1}{a b c} $$


problem: Suppose the sequence of nonnegative integers $a_{1}, a_{2}, \ldots, a_{1997}$ satisfies  $$ a_{i}+a_{j} \leq a_{i+j} \leq a_{i}+a_{j}+1 $$  for all $i, j \geq 1$ with $i+j \leq 1997$. Show that there exists a real number $x$ such that $a_{n}=\lfloor n x\rfloor$ for all $1 \leq n \leq 1997$.

solution: We are trying to show there exists an $x \in \mathbb{R}$ such that  $$ \frac{a_{n}}{n} \leq x<\frac{a_{n}+1}{n} \quad \forall n $$  This means we need to show  $$ \max _{i} \frac{a_{i}}{i}<\min _{j} \frac{a_{j}+1}{j} . $$  Replace 1997 by $N$. We will prove this by induction, but we will need some extra hypotheses on the indices $i, j$ which are used above.  ## Claim - Suppose that  - Integers $a_{1}, a_{2}, \ldots, a_{N}$ satisfy the given conditions. - Let $i=\operatorname{argmax}_{n} \frac{a_{n}}{n}$; if there are ties, pick the smallest $i$. - Let $j=\operatorname{argmin}_{n} \frac{a_{n}+1}{n}$; if there are ties, pick the smallest $j$.  Then  $$ \frac{a_{i}}{i}<\frac{a_{j}+1}{j} $$  Moreover, these two fractions are in lowest terms, and are adjacent in the Farey sequence of order $\max (i, j)$.   Now, consider the new number $a_{N}$. We have two cases:  - Suppose $i+j>N$. Then, no fraction with denominator $N$ can lie strictly inside the interval; so we may write for some integer $b$  $$ \frac{b}{N} \leq \frac{a_{i}}{i}<\frac{a_{j}+1}{j} \leq \frac{b+1}{N} $$  For purely algebraic reasons we have  $$ \frac{b-a_{i}}{N-i} \leq \frac{b}{N} \leq \frac{a_{i}}{i}<\frac{a_{j}+1}{j} \leq \frac{b+1}{N} \leq \frac{b-a_{j}}{N-j} . $$  Now,  $$ \begin{aligned} a_{N} & \geq a_{i}+a_{N-i} \geq a_{i}+(N-i) \cdot \frac{a_{i}}{i} \\ & \geq a_{i}+\left(b-a_{i}\right)=b \\ a_{N} & \leq a_{j}+a_{N-j}+1 \leq\left(a_{j}+1\right)+(N-j) \cdot \frac{a_{j}+1}{j} \\ & =\left(a_{j}+1\right)+\left(b-a_{j}\right)=b+1 \end{aligned} $$  Thus $a_{N} \in\{b, b+1\}$. This proves that $\frac{a_{N}}{N} \leq \frac{a_{i}}{i}$ while $\frac{a_{N}+1}{N} \geq \frac{a_{j}+1}{j}$. Moreover, the pair $(i, j)$ does not change, so all inductive hypotheses carry over.  - On the other hand, suppose $i+j=N$. Then we have  $$ \frac{a_{i}}{i}<\frac{a_{i}+a_{j}+1}{N}<\frac{a_{j}+1}{j} . $$  Now, we know $a_{N}$ could be either $a_{i}+a_{j}$ or $a_{i}+a_{j}+1$. If it's the former, then $(i, j)$ becomes $(i, N)$. If it's the latter, then $(i, j)$ becomes $(N, j)$. The properties of Farey sequences ensure that the $\frac{a_{i}+a_{j}+1}{N}$ is reduced, either way.