### ISI 2022 Objective Solutions

Hellooooooo people I am Rishad. I am a 12th grader. Passionate about pursuing a career in mathematics. I like doing and studying all types of math. Hobbies: Listening to music and doing absolutely nothing and of course reading articles about math(mainly) and related fields( physics, computer science, cryptography..... not chemistry, of course). Okay, now moving on to the objective of this blog. Here, I will be presenting to you the solutions for the Indian Statistical Institute( Objective Paper) 2022.

Q1.Any positive real number $x$ can be expanded as:

$x=a_{n}\cdot 2^{n} + a_{n-1}\cdot 2^{n-1}+ a_{n-2}\cdot 2^{n-2}+.........+a_{-1}\cdot 2^{-1}+a_{-2}\cdot 2^{-2}.....$

for some $n\geq0$ , where each $a_{i} \in \{0,1\}$. In the above-described expansion of 21.1875, the smallest positive integer k such that $a_{k}\neq0$ is:

Motivation/Overview: Okay, so this is a question based on a binary representation of numbers. For a number in base b the general representation is  $a_{n}\cdot b^{n} + a_{n-1}\cdot b^{n-1}+ a_{n-2}\cdot b^{n-2}+.........+a_{-1}\cdot b^{-1}+a_{-2}\cdot b^{-2}.....$. Here the $a_{i}$'s $\in{0,1,2,3,4.....,b-1}$. So for example, if you want to convert say a decimal number( i.e. a number in base 10) like $23$ in base $2$ then what you are actually doing is finding powers of 2 such that their sum is equal to 23. So for example $23=1.2^4+0.2^3+1.2^2+1.2^1+1.2^0$ So the binary representation of 23 is $10111$. These $1$'s and $0$'s are the coefficients of $2$.  Notice the powers of $2$ which have the coefficients $0$ are of no use and therefore according to the question we just need to find the smallest $k$ i.e. the largest power of 2(since as we increase the power in the denominator we decrease the value of the fraction) for which the coefficient of the decimal part's binary expansion is nonzero.

Solution:  We have to write the given number i.e., $21.1875$ in terms of sum of powers of $2$. Clearly $21.1875=21+0.1875.$

$21=2^4+2^2+2^0$ and to convert $0.1875$ in terms of powers of $2$ we can rewrite it as $1875/10000$

which is equal to $3/16$. Now to convert this in powers of $2$ we can rewrite $3$ as $2+1$ and then write $(2+1)/16$ as $2^{-3}+2^{-4}$ . Thus we can conclude that $21.1875=2^{4}+2^{2}+2^{0}+2^{-3}+2^{-4}$. Hence the answer that the smallest k such that $a_{k}\neq0=3$

Q2. Suppose for some $\theta \in [0,\pi/2]$, $\frac{cos3\theta}{cos\theta}$=1/3. Then $cot3\theta\cdot tan\theta$ equals:

Motivation/Overview: In such questions where we are asked to find some value of some trig expression it is always the case where we have to use trig identities. Also, we always want to remove cot, sec, cosec whenever and bring them in terms of tan, cos , sin respectively simply because they seem more convenient to work with. Of course, sometimes you might have to prove trig identities where you need cot or the other two. But still it's always nice to remove these 3 trig ratios. Also when we are asked to find the value of a complicated looking expression then we should try to simply that expression as well because many times that expression gets simplified a lot and the answer comes out straight. Here also its better for beginners to simply the expression first and then start working with the given equation.

Solution:

$\frac{cos3\theta}{cos\theta}$=$\frac{4cos^3\theta-3cos\theta}{cos\theta}$=$4cos^2\theta-3=1/3 \implies$ $cos^2\theta=5/6\implies sec^2\theta=6/5\implies tan^2\theta=1/5$

Now we reduce the expression for which we have to find the value:

$\frac{tan\theta}{tan3\theta}= \frac{ tan\theta\cdot (1-3tan^2\theta)}{3tan\theta-tan^3\theta}$=$\frac{1-3tan^2\theta}{3-tan^2\theta}$ which after plugging the value of $tan^2\theta$ gives$\frac{1}{7}$.

Q3. The locus of the point z in the complex plane satisfying $z^2+|z^2|=0$

Motivation/Overview: I personally find the locus question on complex numbers pretty interesting because some of them can have beautiful geometric solutions. Some can be done by just observing the equation. And well of course we have the algebraic method wherein we put z=x+iy. Pretty standard way of solving problems(very useful, sometimes). In equations involving complex numbers you should always know what the terms of the equation mean so for example the modulus of a complex number is a positive real number always because it indicates the distance of the complex number in the complex plane from the origin of the complex plane. Seeing the modulus and the 0 in the equation I thought that solving this question by observation would be a nice thing to do.

Solution:

Now there are different ways to approach this for example a purely geometric method or a purely algebraic method or the method that I used.

Observe that the modulus of a complex number is positive (as it is the distance form the origin therefore $z^2$ is negative real number(if we subtract $|z^2|$ on both sides of the equation). This is only possible when z is a purely imaginary number i.e. lying on the imaginary axis, therefore, the locus of z is a straight line i.e. the imaginary axis.

Q4.Amongst all polynomials $p(x) = c_{0} + c_{1}\cdot x + · · · + c_{10}\cdot x^{10}$  with real coefficients satisfying $|p(x)| \leq|x|$  $\forall x\in[-1, 1]$, what is the maximum possible value of $(2c_{0} + c_{1}) ^{10}$?

Motivation/Overview: Putting $0,-1,1$ in polynomials is a nice thing because they tell a lot about the coefficients of the polynomial and we can reduce a lot of our work. For that matter, we can eliminate $x$ by putting any number but these numbers make our job easy. Also since its valid to put $x=0,1$ so we put it because....well, why not !!?

Solution:

Plug in 0 in the polynomial $\implies |p(0)|\leq0$ . This is only possible is $|p(0)|\leq0 \implies |c_{0}|=0$. Similarly $|p(1)|\leq1 \implies |c_{0}+c_{1}+c_{3}+c_{4}+c_{5}+c_{6}+c_{7}+c_{8}+c_{9}+c_{10}|\leq1$

Clearly, to find the maximum value of the given expression we need to find the maximum value of ${c_{1}}$. So we should minimize the other $c_{i}$'s which means all are 0.

$\implies c_{1}=1$

So the maximum value of the given expression becomes 1.

Q5. Let $\mathbb{Z}$ denote the set of integers. Let f : $\mathbb{Z}\to\mathbb{Z}$ be such that $f(x)f(y) = f(x + y) + f (x -y)\forall x, y \in \mathbb{Z}$.If $f(1) = 3$, then $f(7)$ equals:

Motivation/Overview: We clearly cannot just work with $x$'s and $y$'s if we want to find some nice information about functions. So we just put in nice values to get some information about the function. Here in this question it doesn't ask us something like "determine all functions such that.......". No its just asking us to find the value of the function at a particular point. So we should put in some values smartly and try to get to 7. If $x+y=7$ then we can isolate this on one side and then calculate the value. Some of the things which work here are $x=4 ,y=3$, $x=6 ,y=1$,$x=5,y=2$. So just decide one pair with which you want to work and try to find out values which will be required in finding $f(7)$.

Solution:

Functional equations need some practice. its a different kind of a topic so for beginners this might be hard to understand of how one can think of such ideas. Okay, so the questions asks us to find the $f(7)$ which clearly we cannot find it directly by just inputting $7$ in the equation. so what is a nice way to find $f(7)$ . So what we do in functional equations is plug in nice values in the equation(of course those which are allowed in the domain) so that we get to the  answer. So plug in $x=1$ and $y=0$ in the functional equation. you get $f(1)f(0)=f(1)+f(1)=6 \implies f(0)=2$  . next plug in$x=1$and $y=1$

$\implies f(2)=7$. Observe if we plug $x=3$ and $y=4$ we can get to $f(7)$ therefore we need to find $f(4)$ and $f(3)$. Plug $x=2$and $y=1$ $\implies f(3)=18$. Now plug $x=3$ and $y=1$ $\implies f(4)=47$ $x=4$and $y=3$ $\implies f(4)f(3)=f(7)+f(1) \implies f(7)=843$

Q6.Let $A$ and $B$ be two 3 x 3 matrices such that $(A+ B) ^{2} = A^{2} + B^{2}$. Which of the following must be true?

(A) $A$ and $B$ are zero matrices.

(B) $AB$ is the zero matrix.

(C) $(A - B)^2 = A^2 - B^2$

(D) $(A - B)^2 = A^2 + B^2$

#### $(A+ B) ^{2} = A^{2} + B^{2} + AB +BA \implies AB+BA=0$. In case you don't know this formula you can just start multiplying the matrices like $(A+B)^2=(A+B)\cdot (A+B)$ and since multiplication of matrices is non-commutative(i.e.$AB\neq BA$) hence we will get the formula. Now we can find $(A-B)^2= A^2 +B^2 -AB-BA \implies (A-B)^2 =A^2+B^2$.

Q7.Let ($n_1, n_2, ......, n_{12}$) be a permutation of the numbers $1, 2, ..... , 12$. The number of arrangements with

$n_{1}>n_{2}>n_{3}>n_{4}>n_{5}>n_{6}$

$n_{6}<n_{7}<n_{8}<n_{9}<n_{10}<n_{11}<n_{12}$

equals:

Motivation/Overview : Observation. Observationnn. Observationnnn.......

Solution:

$n_6=1$ because of the ordered list of numbers we have been given we can see that $n_{6}$ obtains the least value out of all the other $n_{i}$'s.

Now if we choose any 5 out of the other 11 numbers they will be automatically ordered in a particular order therefore we need not worry about making any cases  or anything. The answer just comes out to be ${11\choose 5}={11\choose 6}$.

Q8.The sides of a regular hexagon $ABCDEF$ is extended by doubling them to form a bigger hexagon $A' B' C' D' E' F'$ as in the figure below:

Then the ratio of the areas of the bigger to the smaller hexagon is:

Motivation/Overview: Since we are talking of areas in the question why not find the areas of the figures inside the bigger hexagon since all those add up to the area of the bigger hexagon. And also triangles and rectangles are all very nice figures because finding their area is not a tough job. Since in all areas we have the side of figure has an important role to play therefore we have to assume the side of the hexagon to be $x$. And then we use some trig ( useful a lot of the times when the standard angles are given) and find areas in terms of only one variable i.e. $x$.

Solution:

$\angle BC'D'=\angle CD'E'=\angle DE'F'=\angle EF'A'=\angle FA'B'=\angle AB'C' =90$. Think why these should be 90.

Let the side of the smaller hexagon be x units. By using trigonometry we can find that the side of the bigger hexagon is $\sqrt{3}x$.

Area of triangle E'DF'=${\sqrt{3}x^2}/2$

By symmetry we get that area of all triangles of the size of triangle E'DF' inside the bigger hexagon is ${\sqrt{3}x^2}/2$. See that the smaller hexagon can be divided into 2 triangles of the same size each of area$\sqrt{3}x^2/4$ and a rectangle of area$\sqrt{3}x^2$. So the total area of the smaller hexagon is $3\sqrt{3}x^2/2$. To find the area of the bigger hexagon multiply 6 to the Area of triangle E'DF'(of course we'll have to multiply 6 because there are 6 such triangles) and add the area of the smaller hexagon. Now, all we have to do is to calculate the ratio of areas and this turns out to be $3$.

Q9.In how many ways can we choose$a_1 < a_2 < a_3 < a_4$from the set ${1, 2, ... , 30}$ such that $a_1, a_2, a_3, a_4$ are in arithmetic progression?

Motivation/Overview: To form an AP we need an initial term and a common difference. Those are the key ingredients for forming an AP (besides, a pen and a paper). We have four terms of an AP and now the question asks us to form an AP from these four. So the common difference is missing. Now, all we have to do is to see the maximum and the minimum that the common difference can be, based on the set that has been provided and the number of terms given. Casework will be done for different values of the common difference.

Solution:

We have our initial term according to the question. Now we need to make cases about the different possibilities of the common difference(call it "d") which in our case $\in \{1,2,3,4,5,6,7,8,9\}$ If we have $d=1$ then there will be 27 ways in which we can form an AP If we $d=2$ then there will be 24 ways in which we can form an AP. If $d=3$, there will be 21 ways in which we can form an AP. No you can start seeing the pattern the number of AP we can form as d increases by 1 is decreasing by 3. So when $d=9$ we can form 3 AP's. $\implies$ the total number of AP's will be$27+24+21+18+15+.....+3$ which turns out to be $135$.

Q10. Suppose the numbers 71,104 and 159 leave the same remainder r when divided by ·a certain number N > 1. Then, the value of 3N + 4r must equal:

Motivation/Overview: We can always use the ideas of Euclid Divison Lemma divisibility and congruence( of numbers) in questions regarding remainders. We can do a lot of back and forth with these things like jumping from one to the other is an easy task.

Solution:

By using Euclid's Division Lemma we have $71=Na+r$, $104=Nb+r$, $159=Nc+r$. Now eliminate $r$ by subtracting the equations. $Nb+r-Na-r=33$ and $Nc+r-Na-r=88$ and $Nc+r-Nb-r = 55$. Observe that $N \mid 88$. $N \mid 55$, $N \mid 33$. The only possible number which divide all three of them is $11$ therefore $N=11$ and $r=5 \implies 3N+4r=53$

Q11.If x , y are positive real numbers such that $3x + 4y < 72$, then the maximum possible value of $12xy(72 - 3x - 4y)$ is:

Motivation/Overview: Some weird expressions and the question is about talking about the maximum of that weird expression. Aha!! AM-GM inequality should be our best guess. In questions where we have to find the maximum and minimum and which deals with an expression which are multiplied or added then AM-GM might be a nice first step towards finding a solution.

Solution:

$3x+4y<72\implies 72-3x-4y>0$.

Observe the expression very carefully. $12xy=(3x)(4y)$. Consider $(72-3x-4y)+(3x)+(4y)/(3)$ what will be the geometric mean of these three numbers. YES!!!! It will be $\sqrt[3]{(12xy(72-3x-4y)}$. which implies the maximum value of the given expression will be given by this inequality $(72)/(3) \geq$ $\sqrt[3]{12xy(72-3x-4y)}$ $\implies 12xy(72-3x-4y)$ has a maximum value of $24^{3}=13824$

Q12.What is the minimum value of the function $|x-3|+|x+2|+|x+1|+|x|$ $\forall x\in\mathbb{R}$?

Motivation/Overview: Redefining the function over intervals is a nice idea because then we have to work with straight lines only and we can easily check the minimum or maximum value at just the endpoints of the intervals since at those points the value of the straight line will be minimum or maximum for that interval.

Solution:

Oh. Let's see. One way is to use Desmos (No, its a joke don't do that).  We first need to define the function piecewise i.e. we need to redefine the function for different intervals according to the critical points i.e. places where the definition of the function changes, critical points here are $-2,-1,0,3$. These are obtained by equating each term of the function to $0$.

Note that the function can be drawn as straight lines with different slopes in different intervals.

$$f(x)= \begin{cases} -4x & \text{if} x \leq{-2} \\ -2x+4 & \text{if} x\in[-2,-1] \\ 6 & \text{if} x \in [-1,0] \\ 2x+6 & \text{if} x \in [0,3] \\ 4x & \text{if} x \geq{3} \end{cases}$$

From here we can easily obtain that the minimum value for the function $f(x)$ is $6$.

Q13. Consider a differentiable function $u:[0,1] \rightarrow \mathbb{R}$. Assume the function u satisfies:

$u(a)= \frac{(\int_{a-r}^{a+r} u(x) \,dx)}{2r}$  $\forall a \in (0,1)$ and all $r < min(a,1-a)$. Which of the following statements is true?

(A) u attains its maximum but not its minimum on the set {0, l}.

(B) u attains its minimum but not maximum on the set {0, 1}.

(C) If attains either its maximum or its minimum on the set {0, 1}, then it must be constant.

(D) u attains both its maximum and its minimum on the set {0, l}.

Motivation/Overview: Observation is again the key here. Since the average value function is the same for all values in a particular that is only possible if the function is linear because that's when the average will be the same for all points in the interval.

Solution:

This integral is a very nice one. It is telling us the average of the function in the interval $a-r$ to $a+r$

Why?? Because it can be rewritten as:

$u(a)= \frac{(\int_{a-r}^{a+r} u(x) \,dx)}{(a+r)-(a-r)}$

This integral shows up while studying The Mean Value Theorem for Integrals. Since the integral is true $\forall a \in (0,1) \implies u(x)$ is a linear function. Now a linear function can be of decreasing or increasing slope which means it attains both its maximum and minimum value on the set ${0,1}$.

Q14. A straight road has walls on both sides of height 8 feet and 4 feet respectively. Two ladders are placed from the top of one wall to the foot of the other as in the figure below. What is the height (in feet) of the maximum clearance x below the ladders?

#### This is a simple question based on the concept of similarity in triangles.  From the ladder of height 8 to the point where there is  maximum clearance below the ladder, let that distance be $a$and the distance from the ladder of height 4 to the place where there is maximum clearance below the ladders, be $b$. By similarity we get $\frac{b}{a+b}=\frac{x}{4}$ and then by applying similarity to the other set of triangles we get that $\frac{a}{a+b}=\frac{x}{8}$. Add both equations to get $\frac{x}{4}+\frac{x}{8}=1$ $\implies x=(8\cdot 4)/(8+4)$ $\implies x=\frac{8}{3}$

Q15.Let $y = x + c_{1} , y = x + c_{2}$ be the two tangents to the ellipse $x^{2} + 4{y^{2}} = 1$ What is the value of $|c_{1}-c_{2}|$?

#### Solution: The equation of ellipse may be rewritten as $x^2/(1)^2 +y^2/(1/2)^2=1$> on comparing this with $(x^2/a^2)+(y^2/b^2)=1$ we get that $a=\pm 1$ and $b=\pm (1/2)$ . Now the equation of parallel tangents on an ellipse with slope m is given by the formula $y=mx \pm \sqrt{a^2\cdot m^2 + b^2}$.

According to formula the equation of the tangents are

#### $y=x \pm \sqrt[2]{1^2\cdot 1^2 + (1/2)^2}$. = $x \pm \sqrt{5}/2$

$\implies |c_1-c_2|=\sqrt{5}/2+\sqrt{5}/2=\sqrt{5}$

Q16.In the figure below, ABCD is a square and BCEF is a triangle with given sides inscribed as in the figure. Find the length BE.

#### Motivation/Overview: $3,4,5$  is just such a lovely Pythagorean triple that comes up in a lot of places. Note that we have been provided with very less information about the sides of the figures and if we start applying the Pythagoras theorem in this question( it might work as well, but I don't know. You can try if you want, though). We will have to introduce more variables to solve the problem. So why not just use only one variable and solve all our problems. So what is that thing that connects the side and angle of a triangle? Of course, it's our old friend trigonometry.

Solution:

Instead of using too many variables in a problem we just use a single variable which is $\theta$

and make use of the trig ratios. So say $\angle FCD=\theta$. $\angle EFC= 90$ because $3,4,5$ is a pythagorean triplet. By using the linear pair axiom we get that $\angle EFA=\theta$. Now we can use trig ratios. $DC=4 cos\theta$, $FD=4sin\theta$ and similarly we get $AF=3cos\theta$ ,$AE=3 sin\theta$

since ABCD is a square so we get that $4cos\theta=4sin\theta+3cos\theta \implies tan\theta=\frac{1}{4}$. By the definition of tan we can get that $sin\theta= 1/ (\sqrt[2]{17})$ and $cos\theta=4/(\sqrt[2]{17})$.  So as $BE= 4cos\theta - 3sin\theta \implies BE=\frac{13}{\sqrt{17}}$

Q17.Let p and q be two non-zero polynomials such that the $deg(p)\leq deg(q)$  and $p( a )q( a) = 0$ for $a= 0, 1, 2, ... , 10.$ Which of the following must be true?

Options:

A) degree of $q\neq10$

b) degree of $p\neq10$

C) degree of $q\neq5$

D) degree of $p\neq5$

#### Motivation/Overview:  If a polynomial has a degree n then it has n roots(simple statement, not the most precise one). This is a very important theorem and is also known as the fundamental theorem of algebra. Might be useful when we have to do something with the degree and roots of a polynomial.

Solution:

Notice that $p(a)\cdot q(a)$ has a degree of at least 11 since there are at least 11 values of a which are $\{0,1,2,3,4,5,6,7,8,9,10\}$for which the polynomial vanishes . But now say that $q$ has a degree of 5 , then $p$ should have a degree that is equal to or greater than 6. This is not possible because the question clearly states that the degree of $p$ is less than the degree of $q$.

Q18. For $n \in \mathbb{N}$, let  $a{n}$ be defined as:

$a_{n}=\int_{0}^n \frac{dx}{1+nx^2}$

Then $\lim_{n\to\infty} a_{n}$

#### Motivation/Overview :We know that $\int_\frac{dx}{1+x^2}= tan^{-1}{x}+C$. So Observation again. I cannot stress enough how important Observation is in math olympiads and math overall.

Solution:

This should be a nice motivation to substitute ${\sqrt{n}\cdot x}$ as $t$.This has been done so as to get $t^2$ in the denominator of $a_{n}$. After converting the integral in terms of $t$ and also applying the new limits ( WARNING: DO NOT forget to change the limits!!!!!) we get $\frac {tan^{-1}{(n^{\frac{3}{2})}}}{\sqrt{n}}$. Clearly $a_{n} \rightarrow 0$ as $n \rightarrow \infty$.

Q19.A 3 x 3 magic square is a 3 x 3 rectangular array of positive integers such that the sum of the three numbers in any row, any column or any of the two major diagonals, is the same. For the following incomplete magic square

#### Motivation/Overview: We have been given a lot of information in the form that the sum of numbers in row and column and the diagonal sum is equal. Since the number of numbers in row 1 and column 1 is maximum, why not introduce a variable for the third number and then repeatedly form equations where we eliminate x repeatedly to find different values in the grid. In the end, we should be able to find the column sum.

Solution:

Reference index of the table:

1   2   3

4   5   6

7   8   9

Let the number at index 3 be x. Let the number at index 4 be y. Then by the information in the question $27 + 36+x=27+31+y \implies y=x+5$ to find the number at fifth index we can use the following equation $27+36+x=x+ a_{5}+ 31 \implies a_{5}=32$. we can find the number on the sixth index we can use the following equation $x+5+32+a_{6}=x+36+27 \implies a_{6}=26$. Another equation $27+36+x=x+26+a_{9} \implies a_{9}=37$. Sum of column is the same as sum of diagonal $\implies sum=27+32+37=96$

Q20.The number of positive integers $n\leq22$ such that 7 divides $n_5 + 4n_4 + 3n_3 + 2022$ is:

#### Solution: When we divide 2022 by 7 we see that the remainder comes out to be 6. therefore if we want to find the n such that the expression is divisible by 7 we should have that the $n^5+4{n^4}+3{n^3}$ leaves a remainder 1 when divided by 7. $n^5+4{n^4}+3{n^3}= n^3{n+1}{n+3}$. Now after checking for values between 1 and 7 there are only 3 possible values of n such that $n^3{n+1}{n+3}$ leaves a remainder 1 on dividing by 7( for $n=1,2,5$). Now one may ask why we checked for values between 1 and 7. That is because we are talking about the expression mod 7 therefore values will be repeated i.e. we will get 3 more values in the case where we look for n in between 8 and 14 and then 3 more when we look for n in between 15 and 21. Also, n=22 should satisfy as n=1 did. Think deeply about why we took cases in which there were only 7 numbers and convince yourself why we will get 3 values in every set of 7 numbers. The answer is hence $10$.

Q21.In a class of $45$ students, three students can write well using either hand. The number of students who can write well only with the right hand is $24$ more than the number of those who write well only with the left hand. Then, the number of students who can write well with the right hand is:

#### Motivation/Overview: Just write the question in the form of mathematical equations.

Solution: Let the number of students who write well with only their left hand be $x$. Then the number of students who write well with "only" their right hand be$x+24$. With all this information a [pretty simple equation can be formed which is $x+x+24+3=45 \implies 2x=18 \rightarrow x=9$. Therefore the number of people who can write well with their right hand is $9+24+3=36$. Think about why we added $3$ to $x+24$.

Q22. Let 1,$\omega$, $\omega ^2$ be the cube roots of unity. Then the product

(1-$\omega$+ $\omega^{2}$)(1-$\omega^{2}$+$\omega^{4}$)(1-$\omega^{4}$+$\omega^{8}$)........(1-$\omega^{512}$+$\omega^{1024}$)

#### Motivation/Overview: Properties of roots of unity are pretty important. We have a seemingly difficult product to deal with. And in such cases where we have a huge sum or huge product which has roots of unity in them. There are high chances that these properties might be of some use.

Solution: we have to repeatedly use the property that the sum of the root of unity is always 0. Also note that $\omega^{3n}=1$, $\omega^{3n+1}=\omega$ ,$\omega^{3n+2}=\omega^{2}$.

By using these properties repeatedly for every bracket we get to this  expression:

$2^{10}(\omega^{15})=2^{10}(1)=2^{10}$

Q23.The function $x^2 log_{e} x$ in the interval $(0, 2)$ has:

(A) exactly one point of local maximum and no points of local minimum.

(B) exactly one point of local minimum and no points of the local maximum.

(C) points of local maximum as well as local minimum.

(D) neither a point of local maximum nor a point of local minimum.

#### Motivation/Overview:  The question asks about the maxima and minima of some functions. Well then please welcome:) derivatives. These things are important when determining the maxima and minima of a function. Of course, since we are talking about maxima and minima we have to consider the second order derivative as well since we want to know whether we have local maxima or minima.

Solution: The derivative of this function is $x(1+2lnx)$. We can use the product rule of the derivatives to obtain this. To find the minima or maxima of a function we must equate the derivative of the function equal to 0. This is because at the point of minima or maxima the tangent at the function at that point is parallel to the x-axis i.e. the slope of the tangent becomes 0. We can visualize a derivative as a slope of a function at a point(precisely, in the neighbourhood of that point) as well.

$f'(x)=x(1+2lnx)=0 \implies x=e^{-1/2}$.  By taking the second derivative we realize that $e^{-1/2}$ is a point of minima (the second derivative is positive). Therefore we can conclude that there is exactly one point of local minima and no points of local maxima.

Q24.The number of triples (a, b, c) of positive integers satisfying the equation

$(1/a)+(1/b)+(1/c)=(1)+2/(abc)$

and such that $a<b<c$ equals:

#### Motivation/Overview: It just seems bad that we have things in the denominator. And since the denominator cannot be 0 therefore we can easily cancel the denominator. Also factoring equations is an important idea because factors of an equation are easy to deal with in the sense that we can make divide the problems into some cases with the help of factors. making factors and working with them is a general problem-solving tip. Also in questions where it is written like " find order triples/ pairs such that...." then what we should try to do is to restrict the number of cases that we have to check and one way to do that( if not enough information is given in the question for restricting our cases) is using inequalities i.e. putting bounds.

Solution:

$a,b,c$ are positive which means we can clear the denominator without worrying whether it's 0 or non zero.

We get:
$ab+bc+ac=abc +2 \implies (a-1)(b-1)(c-1)=(a-1)+(b-1)+(c-1)$ .Think of how this has happened. Just try to go backwards from the factorized equation to the original one and you'll understand how it has been factorized. This idea is a very important and a very interesting one as well. Notice that if we increase values of $a, b, c$ then the RHS would become huge for huge values of $a,b,c$ but the LHS will be comparatively small. Therefore only some small values will satisfy such an equation can you think of the smallest pair which satisfies this. YESSSSS. It happens when $a=2 b=3 c=4$. To check if any other pair satisfies this equation we need to put bounds on each of its terms.

So assume $X=a-1, Y=b-1, Z=c-1$

Also by the inequality given in the question, we know that Z>Y>X

We have to find bounds on $X,Y, Z$

$XYZ-X=Y+Z \rightarrow Y+Z=X(YZ-1)>YZ-1 \implies Y+Z-YZ>-1$. $Y(1-Z) +Z -1+1>-1 \implies (Y-1)(Z-1) \leq2 \implies Y=2 and Z=3$. So now we can see that X=1. This will give that $a=2, b=3, c=4$ is the only solution. Done!!

Q25. An urn contains 30 balls out of which one is special. If 6 of these balls are taken out at random, what is the probability that the special ball is chosen?

#### Motivation/Overview:  6 balls are chosen out at random. Hmmmmm... number of ways.....Hmmmmmmmm combinatorics (specifically, combinations).

Solution: Whenever we want to perform a task in which subtasks (that are dependent on each other) have to be performed we must multiply the number of ways. Why is this so??? Well, see because we want every subtask to be performed to complete the entire task so we would want to find all possible ways in which the entire task can be completed and hence we should multiply. If the occurrence of one subtask does not affect the other then we can simply add. Because now to complete a task we have subtasks that are mutually exclusive.

Now, back to our question. The number of ways to select the special ball is just $30 \choose 1$. Noooooooooooooooooooooooo!!!!!!!!! That is a common mistake. The number of ways to choose a special ball is just 1( of course, because there is just one special ball). And now since we have to choose 6 balls the number of ways in which we can choose them is ${29 \choose 5} \cdot {1 \choose 1}$. And the probability of choosing 6 balls form 30 balls such that one of them is a special ball $={\frac{{29 \choose 5}{1 \choose 1}}{30 \choose 6 }} = \frac{1}{5}$.

Q26. A triangle has sides of length $\sqrt{5},2 \sqrt{2}, \sqrt{3}$ units. Then, the radius of its inscribed circle is:

Motivation/Overview: Circle and tangent lines in euclidean geometry ( Two of the most important theorems regarding the circle and the tangent lines are that 1)tangent to circle is perpendicular to the radius and the 2)tangents from an external point are equal in length). Yeah, I knew the formula for this as well so I directly did it but do not just learn the formula if this is your first time doing such a question. Try to apply simple logic.

Solution:

Ooooooh!! Do you what's special about the length of the sides of this triangle. Yes, it's a right-angled triangle. Why?? Because the sides satisfy the Pythagoras theorem. The radius of an inscribed circle in right-angled triangle:

$r=(P+B-H)/2 \implies r= (\sqrt{5}+\sqrt{3}+2\sqrt{2})$ where $P$ and $B$ are the two legs of the right triangle and hypotenuse is $H$ .Try to derive this formula using the fact that the radius of a circle is perpendicular to the tangent at a point and the fact that length of tangents from the same external point to a circle is the same.

Q27. Two ships are approaching a port along straight routes at constant velocities. Initially, the two ships and the port formed an equilateral triangle. After the second ship travelled $80 km$ , the triangle became right-angled. When the first ship reaches the port, the second ship was still $120 km$  from the port. Find the initial distance of the ships from the port.

Motivation/Overview: Again we are given angles and we need the sides of the triangle to find the distance of the ship from port ,so yeah, trigonometry. Also since we have been given some information that velocities are constant so we have to use a formula which relates velocities(constant) and distances which is that $v=\frac{distance}{time}$.

Solution: All we have to do in this question is to convert a statement from English into a mathematical equation by adding some variables removing them later when not needed and yes some basic use of trigonometry.

Let  the time taken for ship 2 to travel 80 km and ship 1 to travel x km be  $t_1$  and let the speed of ship 1 and ship 2 be v_1 and v_2 respectively. So $v_1 \cdot t_1=x$ and $v_2 \cdot t_1=80$.

by using some trig we can find that $d-x=(d-80)cos 60 \implies x=(d+80)/2$. Now when the first ship reaches the port in time t_2  the second ship travels $d-200$ km in time t2.$\implies v_2 \cdot t_2 =(d-200$ and the first ship has travelled $(d-80)/2$. $\implies v_1 \cdot t_2 = (d-80)/2$. $v_1 \cdot t_1 \cdot v_2 \cdot t_2= {(d+80)/2} \cdot {(d-200)}$. Why we multplied here?? Simply because this is the easiest way to eliminate the unwanted variables. Note that we will get the same expression in terms of v_1, v_2 , t_1, t_2 when we multiply the other 2 equations. So all of these variables will be eliminated at once. After equating,

we get,

$d^2-200d-9600=0 \implies d=240 km$

Q28. If $x_{1}>x_{2}>x_{3}>x_{4}>.........x_{10}$ are real numbers, what is the least possible value of

#### $\frac{(x_1-x_{10})}{(x_1-x_2)} \cdot$ $\frac{(x_1-x_{10})}{(x_2-x_3)} \cdot$  $\frac {(x_1-x_{10})}{(x_3-x_4)} \cdot$  $\frac{(x_1-x_{10})}{(x_4-x_5)} \cdot$  $\frac{(x_1-x_{10})}{(x_5-x_6)} \cdot$ $\frac {(x_1-x_{10})}{(x_6-x_7)} \cdot$  $\frac{(x_1-x_{10})}{(x_7-x_8)} \cdot$  $\frac{ (x_1-x_{10})}{(x_8-x_9)}\cdot$ $\frac {(x_1-x_{10})}{(x_9-x_{10})}$

Motivation/Overview : Weird expression, least value of scary product, AM-GM (WARNING: this is just a problem solving trick, never be rigid while solving problems i.e. never just stick to one method which has been taught or learnt otherwise, always try to find alternate ways). Its just that I am finding AM-GM suitable here so I am using it.

Solution:
So, $x_{1}-x_{10}$ is common in all terms of the product $\implies$ the product will have a power of 9 on $x_{1}-x_{10}$ so we try to break $x_{1}- x_{10}$ as a sum of $9$ terms( You can guess this by keeping in mind The AM-GM Inequality) . So we say,
$x_{1}- x_{10}= (x_{1}-x_{2}) + (x_{2}-x_{3})...............(x_{9}-x_{10})$. By applying AM-GM we get that the required product is greater than of equal to $9^9$.

Q29. The range of the function

$f(x)= \frac {x^2+ 2x+4}{2{x^2}+4x+9}$

Motivation/Overview :  We have been given information about x in the question but we don't have the information about y. So we can try to form an equation which can be written in terms of x. That will only be possible if we cross multiply the denominator mix $x$ and $y$ terms to get a quadratic(here) equation in $x$, which will turn out be useful.

#### Solution: $f(x)=y=\frac {x^2+ 2x+4}{2{x^2}+4x+9}$. Now cross multiply.

$\implies x^2(2y-1)+x(4y-2) + (9y-4)=0$ notice this equation is a quadratic in x anis satisfied by real x therefore the discriminant($D \geq0$). Hence we get $(4y-2)^2-4(2y-1)(9y-4) \geq0$. On solving this quadratic inequality we get that $y \in [3/7,1/2)$

Q30. In the following diagram, four triangles and their sides are given. Areas of three of them are also given. Find the area x of the remaining triangle:

Motivation/Overview :  Since all these figures have a very nice symmetry in them therefore we should try to use it for our purpose and well instead of dealing with them individually we can actually try to deal with all of them together so we decide to join them to form one figure which turns out be a rhombus. Using symmetry in a way which might help us to find an answer is a nice technique since symmetry reduces our efforts and sometimes help in arriving at beautiful solutions.

Solution: This question has a very nice symmetry. We can put the triangles in such a manner that the sides of same length coincide. We can hence form a rhombus. Why?????????????  Because all of the triangles have one common side of length 6 which can actually form the 4 sides of a rhombus. Now call the point of intersection of all 4 triangles $P$. It is an interior point in the rhombus. Now draw perpendiculars to all sides from this point( call them $h_1$,$h_2$,$h_3$,$h_4$)  and write the areas of the four triangles in terms of these variables which will be $\frac{1}{2} \cdot 6 \cdot h_1$ ,  $\frac{1}{2} \cdot 6 \cdot h_2$ , $\frac{1}{2} \cdot 6 \cdot h_3$,  $\frac{1}{2} \cdot 6 \cdot h_4$ . So what might be a nice way to remove these 4 extra variables. Well, notice that the sum of areas of opposite triangles will become same. Can you see why that should be true????Simply because the sum of heights to opposite sides of a rhombus should be the same from any interior point ( the sides are equal in length and parallel as well).

So we conclude that $4+x=13+5 \implies x=14$

So here we come to THE END of the solutions of ISI 2022 Objective Paper. Do tell in the comments if you find any mistakes in any of the solutions. Keep doing mathematics and enjoy!! Lets end by a nice quote by Maryam Mirzakhani ,"The beauty of mathematics only shows itself to more patient followers". Signing off with that. Thanks for reading!!!

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