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# IOQM 2022 Problems, Solutions and Discussion

Answer Key (This is a work in progress, please proceed with caution. We are reviewing some of these answers).

• 1st Problem - 45
• 2nd Problem - 68
• 3rd Problem - 8
• 4th Problem
• 5th Problem - 40
• 6th Problem - 30
• 7th Problem - 35
• 8th Problem - 24
• 9th Problem - 8
• 10th Problem - 40
• 11th Problem - 12
• 12th Problem
###### Problem 1

Three parallel lines $L_{1}, L_{2}, L_{3}$ are drawn in the plane such that the perpendicular distance between $L_{1}$ and $L_{2}$ is 3 and the perpendicular distance between $L_{2}$ and $L_{3}$ is also $3 .$ A square $A B C D$ is constructed such that $A$ lies on $L_{1}, B$ lies on $L_{3}$ and $C$ lies on $L_{2}$. Find the area of the square.

###### Solution

\begin{aligned} &\Rightarrow \quad A D=6 \\ &l \sin \theta=6 \\ &l \cos \theta=3 \\ &l^{2}=6^{2}+3^{2}=45 . \end{aligned}

###### Problem 2

Ria writes down the numbers $1,2, \ldots, 101$ in red and blue pens. The largest blue number is equal to the number of numbers written in blue and the smallest red number is equal to half the number of numbers written in red. How many numbers did Ria write with red pen?

###### Solution & Discussion

Claim: Numbers written in blue needs to be consecutive from $$1,2, .[$$ where $$B$$ is the largest blue no] The smallest required $$\operatorname{no}(\mathrm{r}) \rightarrow B+1$$ Total red number $$=2(B+1)$$ $\begin{gathered} 2(B+1)+B=101 \\ 2 B+B+2=101 \\ 3 B=99 \\ B=33 \end{gathered}$ $r=34$ total req no. $$\rightarrow 2 r=68$$

###### Problem 3

Consider the set $\mathcal{T}$ of all triangles whose sides are distinct prime numbers which are also in arithmetic progression. Let $\Delta \in \mathcal{T}$ be the triangle with the least perimeter. If $a^{\circ}$ is the largest angle of $\Delta$ and if $L$ is its perimeter, determine the value of $\frac{a}{L}$.

###### Solution

The triangles in $\tau=\{(a, b, c) | a+b>c, a+d=b, b+d=c, a, b, c \qquad \textrm{are prime}\}$ such that $a$ is less than $b$ is less than $c$. Least such prime tuple is $(3,5,7)$. Therefore, $7^{2}=3^{2}+5^{2}-2.3 .5 . \cos \alpha=>\alpha=120^{\circ}$ Then $\alpha / \mathrm{L}=120 / 15(\mathrm{~L}=$ perimeter $=15)$. Thus $\alpha / L=8$

###### Problem 4

Consider the set of all 6-digit numbers consisting of only 3 digits, $a, b, c$, where $a, b, c$ are distinct. Suppose the sum of all these numbers is 593999406 . What is the largest remainder when the three digit number $a b c$ is divided by $100 ?$

###### Solution

It is found that there are $$3^{6} = 729$$ numbers where each digit will repeat 243 times. Therefore ,sum of those numbers $$=111111 \times 243(a+b+c).$$ So now , $$\frac{593999406}{111111 \times 243}=\mathrm{a}+\mathrm{b}+\mathrm{c}=22.$$ To get the maximum remainder, number will be 598 , so $$598 \equiv 98 \text{(mod 100)}.$$

###### Problem 5

In parallelogram ABCD the longer side is twice the shorter side. Let XYZW be the quadrilateral formed by the internal bisectors of the angles of $A B C D$. If the area of $X Y Z W$ is 10 . find the area of $A B C D$.

###### Solution

\begin{aligned} \text { Connect } \mathrm{XZ} & \Rightarrow \operatorname(\Delta \mathrm{XWZ})=5=\operatorname(\Delta \mathrm{xyz}) \\ & \Rightarrow \operatorname(\Delta \mathrm{AWD})=5 \end{aligned}

Since $$\mathrm{A} \mathrm{X} \| \mathrm{DZ} \Rightarrow \angle \mathrm{AXD}=\angle \mathrm{ZDX}$$ \begin{aligned} &\angle \mathrm{ADX}=\angle \mathrm{ZDX} \\ &\angle \mathrm{AXD}=\angle \mathrm{ADX} \\ &\mathrm{AD}=\mathrm{AX} \end{aligned} $$A X Z D$$ will be rhombus

\begin{aligned} &\text ( \mathrm{AXZD})=20 \\ &\operatorname( \mathrm{ABCD})=40 \end{aligned}

###### Problem 6

Let $x, y, z$ be positive real numbers such that $x^{2}+y^{2}=49 \cdot y^{2}+y z+z^{2}=36$ and $x^{2}+\sqrt{3} x z+z^{2}=25$. If the value of $2 x y+\sqrt{3} y z+z x$ can be written as $p \sqrt{q}$ where $p, q$ are integers and $q$ is not divisible by square of any prime number, find $p+q$.

\begin{aligned} &x^{2}+y^{2}=z^{2} \\ &y^{2}+yz+z^{2}=36=6^{2} \\ &x^{2}+\sqrt{3} x z+z^{2}=25=5^{2} \end{aligned}

\begin{aligned} &\operatorname{Cos} 150^{\circ}=-\frac{\sqrt{3}}{2} \\ &\operatorname{Cos} 150^{\circ}=-1 / 2 \\ &\operatorname{Cos} 90^{\circ}=1 . \end{aligned}

Area of triangle (using sine Formula) $=[A O B]+[B O C]+[C O A]$

=$\frac{1}{2} y z \sin 150+\frac{1}{2} y x \sin 90$ $+\frac{1}{2} x z \sin 120$

$\Rightarrow \frac{1}{2} y z \frac{\sqrt{3}}{2}+\frac{1}{2} x y x+\frac{1}{2} \times 2 \times \frac{1}{2}$

$\Rightarrow \frac{\sqrt{3}}{4} y^{2}+\frac{1}{2} x+\frac{1}{4} x z$

$=\sqrt{5(1-A)(1-26)(5-6)} =\sqrt{5 \times \frac{5+6+7}{7}+9} =\sqrt{9(9-5)(9-6)(9-7)} =\sqrt{9 \times 84 * 3 * 1}$

$=\sqrt{216} =6 \frac{\sqrt{5}}{4} y z+\frac{1}{2} x+y+\frac{1}{4} \times z=6 \sqrt{c} \sqrt{3} y_{2}+17 z+x z=24 \sqrt{6}$