This is the second RMO Mock test from MOMC Season 2. All the credit and authorship of the problems belongs to their respective sources.

Sticking to my usual, I will be discussing the solutions of the problems along with how I motivated them. So if you are planning to attempt the mock, beware of the spoilers ahead!

**Problem 1** (Simon Marais 2017 A2) Let $a_1,a_2,a_3,\ldots$ be the sequence of real numbers deﬁned by $a_1=1$ and

Determine whether there exists a positive integer $N$ such that

**Solution:** Suppose for the sake of contradiction that the sum of the sequence was bounded. Thus there doesn't exist a constant $c$ such that $a_i \geq c$ for all $i$. Hence $a_i \rightarrow 0$ as $i \rightarrow \infty$. Now we could relate $\sum a_i$ with $\sum a_i^2$ if $0 < a_m \leq 1$ or $a_m \geq 1$ for all $m$. The former luckily holds as $a_1^2 + a_2^2 + \dots + a_{m-1}^2 \geq a_1^2 = 1 \implies a_m \leq 1$. Hence:

A contradiction!

**Problem 2** (China Northern Mathematical Olympiad 2020 P4) In $\triangle ABC$, $\angle BAC = 60^{\circ}$, point $D$ lies on side $BC$, $O_1$ and $O_2$ are the centers of the circumcircles of $\triangle ABD$ and $\triangle ACD$, respectively. Lines $BO_1$ and $CO_2$ intersect at point $P$. If $I$ is the incenter of $\triangle ABC$ and $H$ is the orthocenter of $\triangle PBC$, then prove that the four points $B,C,I,H$ are on the same circle.

**Solution:** This is a straightforward angle chasing problem. It suffices to show that $\angle BHC = \angle BIC$. It is well-known that $\angle BIC = 90 + \frac{1}{2} \angle BAC$ which is $120$ in our case. It is also well-known that $\angle BHC = 180 - \angle BPC$. So $\angle BIC = 120 \iff \angle BPC = 60 \iff \angle O_1BC + \angle O_2CB = 120$. Now $\angle O_1BD = \frac{180-\angle BO_1D}{2} = \frac{180 - 2\angle BAD}{2} = 90 - \angle BAD$. Likewise $\angle O_2CB = 90 - \angle CAD$. Adding we get the desired result. For more insightful solutions you can check out the AoPS thread.

**Problem 3:** (China Northern Math Olympiad 2017 P4) Let $Q$ be a set of permutations of $1,2,...,100$ such that for all $1\leq a,b \leq 100$, $a$ can be found to the left of $b$ and adjacent to $b$ in at most one permutation in $Q$. Find the largest possible number of elements in $Q$.

**Solution:** Let $M$ be the maximum number of elements in $Q$. For each permutation, there exist exactly $99$ consecutive pairs of numbers. So, there are $99M$ total pairs. Each consecutive pair of numbers appears exactly once. Hence $99M \leq 99\cdot 100$ as there are $99 \cdot 100$ total possible pairs of consecutive integers in a permutation. Therefore $M \leq 100$. It remains to show that a construction exists.

Consider a $100 \times 100$ matrix where each row represents a permutation of $1,2, \dots 100$. Initially, all elements of the first row are $1$, all elements of the second row are $2$ and so on. We cycle upwards, the $k^{\text{th}}$ column other than the first one, by $t_k$ units relative to the $k-1^{\text{th}}$ column. It suffices to find **distinct** positive integers $t_k$ for $2 \leq k \leq 100$ such that the sum of any contiguous sum of $t_k$ isn't divisible by $100$. The following sequence works:

**Problem 4** (Austrian Math Olympiad 2006 P4) The function $f: \mathbb{R} \rightarrow \mathbb{R}$ is defined as $f(x) = \lfloor x^2 \rfloor + \{ x \}$. Call a real number *regional* if it doesn't lie in the range of $f$. Show that there exists an infinite arithmetic progression of distinct regional positive rational numbers, which all have denominator $3$ when written in lowest form.

**Solution:** If $3$ is the denominator of a rational number, then it can be written as $m+\frac{1}{3}$ or $m + \frac{2}{3}$. For simplicity, consider the former one. Then $f(x) = m + \frac{1}{3} \implies \{x\} = \frac{1}{3}$. Hence $x = \ell + \frac{1}{3}$ for some integer $\ell$. By using the definition of $f(x)$ and taking cases on $\ell$ modulo $3$, we deduce that $9m+1, 9m+4$ and $9m+7$ are perfect squares. Picking $m \equiv 6 \pmod 8 \implies 9m+1 \equiv 7, 9m+4 \equiv 2, 9m+7 \equiv 5 \pmod 8$. None of $2,5,7$ are quadratic residues modulo $8$, as required. Hence the arithmetic sequence $6 + \frac{1}{3}, 14 + \frac{1}{3}, 22 + \frac{1}{3}, \dots$ works.

**Problem 5** (China Northern Math Olympiad) Find all integers $n$ such that there exists a concave pentagon which can be dissected into $n$ congruent triangles.

**Solution:** The solution to this difficult problem can be found here.

**Problem 6** (Hong Kong National Olympiad 2013 P1) Let $a,b,c$ be positive real numbers such that $ab+bc+ca=1$. Prove that

**Solution:** Check out the solution of this inequality here.

The final mock would be posted on **26th October**. Stay in the loop for that! :)