RMO 2023 Orange Mock Two

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! :)