2024 Indonesia Regional MO Short Answer Section There are 8 problems, time allowed is 60 minutes. Answers are always in integer form. 1. It is known that $\overline{ab}$ and $\overline{cd}$ are both two-digit numbers whose product is $777$. If $\overline{ab}<\overline{cd}$, find the value of $a+b$. 2. Let $f$ and $g$ be linear functions that satisfy the equation \[f(x+g(y)) = 7x+2y+11 \text{ for every real number } x,y \]If $g(7)=3$, find the value of $ g(-11+f(4)) $. Note: A linear function is a function of the form $h(x)=ax+b$ with real constants $a,b$. 3. Given a triangle $ABC$ with side lengths $AB=15, AC=13, BC=4$. There exists an equilateral triangle $PQR$ with $P,Q,\text{ and } R$ lying on sides $BC,CA, \text{ and } AB$ respectively such that $PQ$ is parallel to $AB$. The value $\dfrac{PQ}{AB} $ can be expressed in the form $\dfrac{a }{b+c\sqrt{d} }$ with $a,b,c,d$ such that $a$ is a positive integer, $d$ is squarefree, and $\text{GCD}(a,b,c)=1 $. Find value of $a+b+c+d$. 4. Each tile on a board of size $2023 \times 3$ will be colored either black or white, such that each $2\times 2$ sub-board has an odd number of black tiles and an odd number of white tiles. Suppose the number of possible ways of such coloring is $A$. Find the remainder of $A$ when divided by $1000$. 5. Find the number of positive integers $a<209$ such that $\text{GCD}(a,209)=1 $ and $a^2-1$ is not a multiple of $209$. 6. In a square $ABCD$ with side length $\sqrt{2}+\sqrt{6}$, $X$ lies on the diagonal $AC$ such that $AX>XC$. The internal bisector of angle $AXB$ intersects side $AB$ at $U$. The internal bisector of angle $CXD$ intersects side $CD$ at $V$. If $\angle UXV = 150^{\circ} $, find the value of $\lfloor 3 \times UV^2 \rfloor $. Note: the notation $\lfloor x \rfloor $ represents the largest integer that is less than or equal to $x$. 7. Given the set $S = \{1,2,\ldots,18\} $. Let $N$ be the number of ordered pairs $(A,B)$ of subsets $A,B\subseteq S$ such that $| A \cap B | \le 2 $. Find the value of $\dfrac{N}{3^{16} }$. Note: $|X|$ is defined as the number of elements in the set $X$. 8. Let $a,b,c$ be real numbers that satisfy the inequality: \[ |ax^2+bx+c|\le (18x-5)^2 \text{ for all real numbers } x \]Find the smallest possible value of $a+2b+5c$ .
Problem
Source:
Tags: algebra, geometry, combinatorics, number theory, Indonesia
06.05.2024 11:58
P1
08.05.2024 09:12
08.05.2024 10:35
Kscv wrote:
The case $| A \cap B| =2$ is slightly incorrect, as the number of way to choose two elements should be $\binom{18}{2}$, not $18 \times 17$
08.05.2024 10:40
08.05.2024 18:00
09.05.2024 19:08
no 8.)Set $x = 0$ to conclude that $\left|c\right| \le 25 \iff -25 \le c \le 25$ and set $x = \frac{1}{2}$ to conclude that $-16 \le \frac{a}{4}+ \frac{b}{2}+c \le 16$ or $-64 \le a+2b+4c \le 64$ which makes $-89 \le a+2b+5c = a+2b+4c+c \le 89$.
09.05.2024 20:25
Tweaked number 6 : https://artofproblemsolving.com/community/u37742h3315501p30640563 AXB + 90° = 150° ===> Angle AXB = 60° Given that AB = √2 + √6 Then AO = BO = (√2 + √6)/√2 (O the center of square) = 1 + √3 If we draw 30/60/90 RT named BOX, then OX = (1 + √3) × 1/√3 = 1/√3 + 1 Therefore AX = 2 + 4/√3 while CX = 2/√3 So ratio of AX – CX : AC = 2 + 2/√3 : √2 × (√2 + √6) = 2 × (1 + 1/√3) : 2 × (1 + √3) = 1 : √3 UV^2 = (√2 + √6)^2 × (1^2 + (1/√3)^2) = (8 + 4√3) × (1 + 1/3) = 16(2 + √3)/3 Answer = floor(3 × UV^2) = floor[ 16 × (2 + √3) ] = 59
09.05.2024 20:39
10.05.2024 05:14
Ronald Widjojo wrote: Kscv wrote:
The case $| A \cap B| =2$ is slightly incorrect, as the number of way to choose two elements should be $\binom{18}{2}$, not $18 \times 17$ Corrected.
12.05.2024 23:59
13.05.2024 00:52
aaravdodhia wrote:
Hi, aarav, don’t say that you’re not getting anywhere because you are just one step behind the finish line Plugging in $1/x=3,1$ gives $-9 \leq a+3b+9c \leq 9 , -169\leq a +b + c \leq 169$. Add them together to get –178 ≤ 2a + 4b + 10c ≤ 178 ===> –89 ≤ a + 2b + 5c ≤ 89
13.05.2024 01:03
Ho, why had I not seen that? I thought of adding those two inequalities and subtracting the second, but not of halving it directly.
13.05.2024 22:19
Alternately, once we fill in the topmost row and the leftmost column, the entire grid is determined so there are $2023 + 3 - 1 = 2025$ choices.
17.05.2024 12:38
Kscv wrote: Ronald Widjojo wrote: Kscv wrote:
The case $| A \cap B| =2$ is slightly incorrect, as the number of way to choose two elements should be $\binom{18}{2}$, not $18 \times 17$ Corrected. It should be 216
17.05.2024 12:39
miyukina wrote: Tweaked number 6 : https://artofproblemsolving.com/community/u37742h3315501p30640563 AXB + 90° = 150° ===> Angle AXB = 60° Given that AB = √2 + √6 Then AO = BO = (√2 + √6)/√2 (O the center of square) = 1 + √3 If we draw 30/60/90 RT named BOX, then OX = (1 + √3) × 1/√3 = 1/√3 + 1 Therefore AX = 2 + 4/√3 while CX = 2/√3 So ratio of AX – CX : AC = 2 + 2/√3 : √2 × (√2 + √6) = 2 × (1 + 1/√3) : 2 × (1 + √3) = 1 : √3 UV^2 = (√2 + √6)^2 × (1^2 + (1/√3)^2) = (8 + 4√3) × (1 + 1/3) = 16(2 + √3)/3 Answer = floor(3 × UV^2) = floor[ 16 × (2 + √3) ] = 59 i think it $3 UV^2$ should be $32+8\sqrt{3}$?
17.05.2024 18:17
GreenTea2593 wrote: miyukina wrote: Tweaked number 6 : https://artofproblemsolving.com/community/u37742h3315501p30640563 AXB + 90° = 150° ===> Angle AXB = 60° Given that AB = √2 + √6 Then AO = BO = (√2 + √6)/√2 (O the center of square) = 1 + √3 If we draw 30/60/90 RT named BOX, then OX = (1 + √3) × 1/√3 = 1/√3 + 1 Therefore AX = 2 + 4/√3 while CX = 2/√3 So ratio of AX – CX : AC = 2 + 2/√3 : √2 × (√2 + √6) = 2 × (1 + 1/√3) : 2 × (1 + √3) = 1 : √3 UV^2 = (√2 + √6)^2 × (1^2 + (1/√3)^2) = (8 + 4√3) × (1 + 1/3) = 16(2 + √3)/3 Answer = floor(3 × UV^2) = floor[ 16 × (2 + √3) ] = 59 i think it $3 UV^2$ should be $32+8\sqrt{3}$? Why?
29.09.2024 18:11
for P2. set the $f(x)$ and $g(y)$ as: $f(x)=ax+b$ and $g(y)=cy+d$ we substitute it in to $g(y)$ we get: $f(x+cy+d)=7x+2y+11$, we substitute it: $f(x+cy+d)=a(x+cy+d)+b=ax+acy+ad+b$, Nah set that for $ax+acy+ad+b=7x+2y+11$, compare it: $a=7$, $ac=2$ and $ad+b=11$, and we solve for $c$: $c=\frac{2}{7}$, after that because $a=7$ we get $7d+b=11$ and the other side from function $g(y)=cy+d$ we set the condition to: $a=y=7$ and $g(7)=3$, because $g(y)=cy+d$: $g(7)=\frac{2}{7}(7)+d=3$ and $d=1$, Now we subs back into $7d+b=11$. we get $b=4$. Nah we get the linear function as: $f(x)=ax+b=7x+4$ and $g(y)=cy+d=\frac{2}{7}y+1$. Now we have to calculate $g(-11+f(4))$. which is $g(-11+f(4))=g(21)$ thus. $g(21)=7$ Ya that is the solution. It tooks so loong to solve it and have so many tricky parts of there
29.09.2024 18:18
britishprobe17 wrote: for P2. set the $f(x)$ and $g(y)$ as: $f(x)=ax+b$ and $g(y)=cy+d$ we substitute it in to $g(y)$ we get: $f(x+cy+d)=7x+2y+11$, we substitute it: $f(x+cy+d)=a(x+cy+d)+b=ax+acy+ad+b$, Nah set that for $ax+acy+ad+b=7x+2y+11$, compare it: $a=7$, $ac=2$ and $ad+b=11$, and we solve for $c$: $c=\frac{2}{7}$, after that because $a=7$ we get $7d+b=11$ and the other side from function $g(y)=cy+d$ we set the condition to: $a=y=7$ and $g(7)=3$, because $g(y)=cy+d$: $g(7)=\frac{2}{7}(7)+d=3$ and $d=1$, Now we subs back into $7d+b=11$. we get $b=4$. Nah we get the linear function as: $f(x)=ax+b=7x+4$ and $g(y)=cy+d=\frac{2}{7}y+1$. Now we have to calculate $g(-11+f(4))$. which is $g(-11+f(4))=g(21)$ thus. $g(21)=7$ Ya that is the solution. It tooks so loong to solve it and have so many tricky parts of there just give me a some feedbacks to improve this solution
29.09.2024 18:54
For P8 i'm gonna approach for set $x$ for $0,\frac{1}{2},1$ we get: for $x=0$: $\left| c \right|\le 25$ for $x=\frac{1}{2}$: $\left| \frac{a}{4}+\frac{b}{2}+c \right|\le 16$ and for $x=1$:$\left|a+b+c\right|\le 169$ and i for $x=\frac{1}{2}$ we multiply it: $\left|a+2b+4c\right|\le 169$ and we get: $a+2b+4c+c\ge-89$
29.09.2024 19:22
For P1 i'm gonna factorize the $777$ which is $777=3\cdot 7\cdot 37$ which is 37 is prime thus. $21\cdot37$ and $a+b=3$