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

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** (Canada 1993) Show that the number $x$ is rational if and only if three distinct terms that form a geometric progression can be chosen from the sequence

**Solution:** One direction of the problem is much easier than the other. If there exist three distinct terms $x+a,x+b,x+c$ such that $(x+a)(x+c) = (x+b)^2$ then $x$ can be shown to be equal to $\frac{b^2-ac}{a+c-2b}$. Therefore $x$ is rational.

Let $x = \frac{p}{q}$ where $p,q$ are integers and $q \neq 0$. For the converse, consider the following triple:

It is easy to verify that this triple works. The triple can be found by considering a general triple $(x,x+a,x+b)$ and then working out the divisibility constraints.

However, we are not done yet! We need to ensure that each term of the triple is $\geq \frac{p}{q}$. Thus our proof fails if $x < 0$. In fact, it fails for $x = 0$ as well, as the terms of the triple won't be distinct. For $x = 0$ we can just pick $(1,2,4)$. For negative $x$ the trick is to see that we can just consider a large positive integer $m$ such that $x+m > 0$. Then letting $x+m = \frac{p}{q}$ and picking our tuple as before works.

**Problem 2** (239 2014 S6) Given positive real numbers $a_1,a_2,\dots,a_n$ such that $a_1^2+2a_2^3+\dots+na_n^{n+1} < 1$. Prove that $2a_1+3a_2^2+\dots+(n+1)a_{n}^n < 3$.

**Solution:** The difficult observation is that the infinite sum:

is equal to $1$. This suggests that all $a_i$ equal to $\frac{1}{2}$ is a pseudo equality case. Therefore, we are motivated to apply AM-GM Inequality to $a_k^{k+1}$ and $\frac{1}{2^{k+1}}$:

(Had we taken $a_k^k$ and $\frac{1}{2^{k}}$, the exponents would not be nice after AM-GM). This implies that $(k+1) a_k^k \le 2ka_k^{k+1} + \frac{1}{2^k}$. Summing this inequality over $1 \leq k \leq n$ gives:

**Problem 3** (HMMT 2023 Feb Teams) Let $A B C D$ be a convex quadrilateral such that $\angle A B C=\angle B C D=\theta$ for some angle $\theta < 90^{\circ}$. Point $X$ lies inside the quadrilateral such that $\angle X A D=\angle X D A=90^{\circ}-\theta$. Prove that $B X=X C$.

**Solution:** The official source provides 4 amazing solutions to the problem here. My solution was the same as the first one.

**Problem 4** (All Russian Olympiad 2000) Let $-1 < x_1 < x_2 , \cdots < x_n < 1$ and $x_1^{13} + x_2^{13} + \cdots + x_n^{13} = x_1 + x_2 + \cdots + x_n$. Prove that if $y_1 < y_2 < \cdots < y_n$, then

**Solution:** This is a non standard inequality as all regular methods fail. The key observation is that $S = \sum_{i =1}^n y_i(x^{13}_i-x_i)$is invariant when we shift the $\{y_i\}_{i=1}^n$ sequence by any constant. Further observe that $x^{12} - 1$ is always $< 0$ if $-1 < x < 1$. It would be easy to show $S < 0$ if each of the individual summands are negative. For that we need that $y_ix_i$ is positive. But we can ensure that by shifting the sequence $y_i$ such that $y_i$ and $x_i$ have the same sign for all $1 \leq i \leq n$. Thus we are done!

**Problem 5** (HMMT 2023 Feb Teams) Let $A B C$ be a triangle. Point $D$ lies on segment $B C$ such that $\angle B A D=\angle D A C$. Point $X$ lies on the opposite side of line $B C$ as $A$ and satisfies $X B=X D$ and $\angle B X D=\angle A C B$. Analogously, point $Y$ lies on the opposite side of line $B C$ as $A$ and satisfies $Y C=Y D$ and $\angle C Y D=\angle A B C$. Prove that lines $X Y$ and $A D$ are perpendicular.

**Solution:** The official source provides $3$ beautiful proofs to this superb problem here

**Problem 6** (239 2017 J1) Let $\{a_1,a_2, \cdots, a_n\}$ be a permutation of $\{1,2,3 \cdots,n \}$. Prove that the sum

taken over all permutations equals $\frac{1}{n!}$.

**Solution:** We show the result by induction on $n$. The base case $n=1$ is clear. We would, in fact, show the general result for a permutation of any set of real numbers (The reason for this would be soon clear). For any $1 \leq i \leq n+1$, consider all the permutations in which $a_i$ appears last. By using the induction hypothesis, the corresponding sum would be equal to $\frac{1}{(a_1 + a_2 + \cdots + a_{n+1})\prod\limits_{j \neq i}^{n+1}a_j}$. Note that we used the general form of the result here. Hence the total sum is:

which completes the induction hypothesis.

Side note: I had not anticipated the huge workload that is associated with the college admissions process. So I was not able to put out any mocks in the past two months. Nonetheless, I plan to post 1-2 more mocks in the coming days. Have a look out for those!