The test this year was held on Monday, 22 August 2022 on 09.10-11.40 (GMT+7) for the essay section and was held on 12.05-13.30 (GMT+7) for the short answers section, which was to be done in an hour using the Moodle Learning Management System. Each problem in this section has a weight of 2 points, with 0 points for incorrect or unanswered problems, whereas in the essay section each problem has a weight of 7 points. As with last year, calculators, protractors and set squares are prohibited. (Of course abacus is also prohibited, but who uses abacus?) Anyway here are the problems. Part 1: Speed Round (60 minutes) The problem was presented in no particular order, although here I will order it in a rough order of increasing difficulty. Problem 1. The number of positive integer solutions $(m,n)$ to the equation \[ m^n = 17^{324} \]is $\ldots.$ Problem 2. Consider the increasing sequence of all 7-digit numbers consisting of all of the following digits: $1,2,3,4,5,6,7$. The 2024th term of the sequence is $\ldots.$ Problem 3. Suppose $(a,b)$ is a positive integer solution of the equation \[ \sqrt{a + \frac{15}{b}} = a \sqrt{ \frac{15}{b}}. \]The sum of all possible values of $b$ is $\ldots.$ Problem 4. It is known that $ABCD$ is a trapezoid such that $AB$ is parallel to $CD$, with the length of $AB = 6$ and $CD = 7$. It is known that points $P$ and $Q$ are on $AD$ and $BC$ respectively such that $PQ$ is parallel to $AB$. If the perimeter of trapezoid $ABQP$ is the same as the perimeter of $PQCD$ and $AD + BC = 10$, then the length of $20PQ$ is $\ldots.$ Problem 5. Suppose $ABC$ is a triangle with side lengths $AB = 16$, $AC=23$, and $\angle{BAC} = 30^{\circ}$. The maximum possible area of a rectangle whose one of its sides lies on the line $BC$, and the two other vertices each lie on $AB$ and $AC$ is $\ldots.$ Problem 6. Suppose $a,b,c$ are natural numbers such that $a+2b+3c=73$. The minimum possible value of $a^2+b^2+c^2$ is $\ldots.$ Problem 7. An equilateral triangle with a side length of $21$ is partitioned into $21^2$ unit equilateral triangles, and the sides of the small equilateral triangles are all parallel to the original large triangle. The number of paralellograms which are made up of the unit equilateral triangles is $21k$. Then the value of $k = \ldots.$ Problem 8. Define the sequence $\{a_n\}$ with $a_1 > 3$, and for all $n \geq 1$, the following condition: \[ 2a_{n+1} = a_n(-1 + \sqrt{4a_n - 3}) \]is satisfied. If $\vert a_1 - a_{2022} \vert = 2023$, then the value of \[ \sum_{i=1}^{2021} \frac{a_{i+1}^3}{a_i^2 + a_i a_{i+1} + a_{i+1}^2} = \ldots. \] Problem 9. Suppose $P(x)$ is an integer polynomial such that $P(6)P(38)P(57) + 19$ is divisible by $114$. If $P(-13) = 479$, and $P(0) \geq 0$, then the minimum value of $P(0)$ is $\ldots.$ Problem 10. The number of nonempty subsets of $S = \{1,2, \ldots, 21\}$ such that the sum of its elements is divisble by 4 is $2^k - m$; where $k, m \in \mathbb{Z}$ and $0 \leq m < 2022$. The value of $10k + m$ is $\ldots.$
Problem
Source: Olimpiade Sains Nasional Tingkat Provinsi SMA/MA (sederajat)
Tags: Indonesia, RMO, 2022, regional, geometry, trapezoid, perimeter
24.08.2022 16:04
is it just me or the problem 9 i got was with different numbers?
24.08.2022 16:55
To clarify, yes there were slightly different versions to each problem, mostly with some of the numbers changed (so that they have integer answers as well). Oh yea, one important thing I forgot to say is that the answers to the problems are all integers (despite none of them actually having an answer of 0 or less).
30.05.2023 11:24
P6 by Cauchy Schwarz inequality we have $$(1^2+2^2+3^2)(a^2+b^2+c^2)\geq (a+2b+3c)^2 \iff a^2+b^2+c^2\geq \frac{5329}{14}$$Bacause $a,b,c$ are natural numbers so $$a^2+b^2+c^2\geq 381$$Next we want to check if this equality holds Notice that the equality happens if and only if $$\frac{a}{1}=\frac{b}{2}=\frac{c}{3}=k$$for real number $k$ we get that $$14k^2=381$$which is not a natural number, however we have $5^2+10^2+15^2=350$ that is close enough to the equality but if we set $c=16$ we have $$5^2+10^2+16^2=381$$so the equality happens if and only if $a=5,b=10,c=16$ the minimum value of $a^2+b^2+c^2=381$
30.05.2023 11:32
P3 squaring both sides and multiply them by $b$ we get $$ab+15=15a^2\iff b=15a-\frac{15}{a}$$this means that $a\mid 15 \Rightarrow a=1,3,5,15$ $\Rightarrow b=0,40,72,224$ we get that the sum of all $b=336$
30.09.2024 14:54
P8 is the probably the hardest from the 10 problems (for me)
30.09.2024 16:32
I'm not sure with this answer, whoever please reply this massage and correct it as the exact solution
02.11.2024 17:06
P2. 3672154