Lattice points on a circle - No. of solution of x^2+y^2 = N

Author: Kazi Abu Rousan

There are some problems in number theory which are very important not only because they came in exams but also they hide much richer intuition inside them. Today, we will be seeing one of such problems.

Sources:
Problem:

The basic structure of the problem we are discussing is:

Find the number of solution of the equation $x^2+y^2 = N$ where $N \in \mathbb{N}$.

So, How to solve problems like this?, Here I will give you a formula to exactly find the number of solution for any general integer $N$.

Solution

The main idea behind these kind of problem is to use Fermat's Two-Square Theorem. So, what does the theorem says?

It says:

If any prime number $N$ is of the form $4k+1$, then $N$ can be written as $x^2+y^2$ for some $x,y \in \mathbb{I} $. This means if $N$ is of the form $4k+3$, then $x^2+y^2=N$ doesn't have any solution.

This gives us our answer if $N$ is a prime number. So, if the problem is to find the number of solution of $x^2+y^2 = 2003$ or maybe $3, 7, \cdots , 23, \cdots $, i.e., any prime with the remainder of $3$ when divided by 4, then we directly know that the number of solution is zero.

So, we have a method to solve these type of problems for prime $N$. But what about non-prime numbers?

For the case of non-prime numbers, I will just give you the formula and will show you how to use that. If you want to know how we got the formula or the details of the formula, you can just watch my lecture.

The formula to calculate the number of solution for $N = p_0^{a_0} p_1^{a_1} \cdots p_{n-1}^{a_{n-1}}$

$$\text{ No of soln } = 4\cdot \Pi_{i=0}^{n-1}\Big(\sum_{j=0}^{a_i} \chi(p_i ^{j}) \Big)$$

or in simple terms,

$$\text{ No of soln } = 4\cdot (\chi(p_0^0)+\chi(p_0^1)+\cdots + \chi(p_0^{a_0}))\cdot ( \chi(p_1^0)+\chi(p_1^1)+\cdots + \chi(p_1^{a_1}) ) \cdots $$

Where $\chi$ is a number theoretical function defined as,

$\chi(x) = -1$ if $x = 4k+3$

$\chi(x) = 0$ if $x = 2^k$

$\chi(x) = 1$ if $x = 4k+1$.

and also $\chi(ab) = \chi(a) \times \chi(b)$

Let's see an example,

Suppose $N = 2250$. Now, $2250 = 2\times 3^2 \times 5^3$.

So, the number of solutions is, $ n = 4\times (\chi(2^0)+\chi(2^1))\times (\chi(3^0)+\chi(3^1) + \chi(3^2)) \times (\chi(5^0)+\chi(5^1)+\chi(5^2)+\chi(5^3))$.

Now, from definition, $\chi(1) = 1$, $\chi(2) = 0$, $\chi(3) = -1$ and $\chi(5) = 1$. And also $\chi(x^n) = \chi(x)^n$, hence

$$n = 4\times (1+0) \times (1-1+1) \times (1+1+1+1) = 16$$.

So, I hope it is now clear on how to use that particular formula.

This is all for today. I hope you have learnt something new.

Collatz Conjecture and a simple program

Author: Kazi Abu Rousan

Mathematics is not yet ready for such problems.

Paul Erdos

Introduction

A problem in maths which is too tempting and seems very easy but is actually a hidden demon is this Collatz Conjecture. This problems seems so easy that it you will be tempted, but remember it is infamous for eating up your time without giving you anything. This problem is so hard that most mathematicians doesn't want to even try it, but to our surprise it is actually very easy to understand. In this article our main goal is to understand the scheme of this conjecture and how can we write a code to apply this algorithm to any number again and again.

Note: If you are curious, the featured image is actually a visual representation of this collatz conjecture \ $3n+1$ conjecture. You can find the code here.

Statement and Some Examples

The statement of this conjecture can be given like this:

Suppose, we have any random positive integer $n$, we now use two operations:

Mathematically, we can write it like this:

$f(n) = \frac{n}{2}$ if $n \equiv 0$ (mod 2)

$ f(n) = (3n+1) $ if $n \equiv 1$ (mod 2)

Now, this conjecture says that no-matter what value of n you choose, you will always get 1 at the end of this operation if you perform it again and again.

Let's take an example, suppose we take $n=12$. As $12$ is an even number, we divide it by $2$. $\frac{12}{2} = 6$, which is again even. So, we again divide it by $2$. This time we get, $\frac{6}{2} = 3$, which is odd, hence, we multiply it by 3 and add $1$, i.e., $3\times 3 + 1 = 10$, which is again even. Hence, repeating this process we get the series $5 \to 16 \to 8 \to 4 \to 2 \to 1$. Seems easy right?,

Let's take again, another number $n=19$. This one gives us $19 \to 58 \to 29 \to 88 \to 44 \to 22 \to 11 \to 34 \to 17 \to 52 \to 26 \to 13 \to 40 \to 20 \to 10 \to 5 \to 16 \to 8 \to 4 \to 2 \to 1$. Again at the end we get $1$.

This time, we get much bigger numbers. There is a nice trick to visualize the high values, the plot which we use to do this is called Hailstone Plot.

Let's take the number $n=27$. For this number, the list is $27 \to 82 \to 41 \to 124 \to 62 \to 31 \to 94 \to 47 \cdots \to 9232 \cdots$. As you can see, for this number it goes almost as high as 9232 but again it falls down to $1$. The hailstone plot for this is given below.

Hailstone for 27. The highest number and with it's iteration number is marked

So, what do you think is the conjecture holds for all number? , no one actually knows. Although, using powerful computers, we have verified this upto $2^{68}$. So, if you are looking for a counterexample, start from 300 quintillion!!

Generating the series for Collatz Conjecture for any number n using python

Let's start with coding part. We will be using numpy library.

import numpy as np
def collatz(n):
	result = np.array([n])
	while n >1:
		# print(n)
		if n%2 == 0:
			n = n/2
		else:
			n = 3*n + 1
		result = np.append(result,n)
	return result

The function above takes any number $n$ as it's argument. Then, it makes an array (think it as a box of numbers). It creates a box and then put $n$ inside it.

  1. Then for $n>1$, it check if $n$ is even or not.
  2. After this, if it is even then $n$ is divided by 2 and the previous value of n is replaced by $\frac{n}{2}$.
  3. And if $n$ is odd, the it's value is replaced by $3n+1$.
  4. For each case, we add the value of new $n$ inside our box of number. This whole process run until we have the value of $n=1$.
  5. Finally, we have the box of all number / series of all number inside the array result.

Let's see the result for $n = 12$.

n = 12
print(collatz(n))
#The output is: [12.  6.  3. 10.  5. 16.  8.  4.  2.  1.]

Similarly, for $n = 19$, the code gives us,

n = 19
print(collatz(n))
#The output is: [19. 58. 29. 88. 44. 22. 11. 34. 17. 52. 26. 13. 40. 20. 10.  5. 16.  8.  4.  2.  1.]

You can try out the code, it will give you the result for any number $n$.

Test this function.

Hailstone Plot

Now let's see how to plot Hailstone plot. Suppose, we have $n = 12$. In the image below, I have shown the series we get from $12$ using the algorithm.

Series and Step for $n=12$.

Now, as shown in the above image, we can assign natural numbers on each one of them as shown. So, Now we have two arraies (boxes), one with the series of numbers generated from $n$ and other one from the step number of the applied algorithm, i.e., series of natural numbers. Now, we can take elements from each series one by one and can generate pairs, i.e., two dimensional coordinates.

As shown in the above image, we can generate the coordinates using the steps as x coordinate and series a y coordinate. Now, if we plot the points, then that will give me Hailstone plot. For $n = 12$, we will get something like the image given below. Here I have simply added each dots with a red line.

The code to generate it is quite easy. We will be using matplotlib. We will just simply plot it and will mark the highest value along with it's corresponding step.

import numpy as np
import matplotlib.pyplot as plt
def collatz(n):
    result = np.array([n])
	while n >1:
		# print(n)
		if n%2 == 0:
			n = n/2
		else:
			n = 3*n + 1
		result = np.append(result,n)
	return result
n = 63728127
y_vals = collatz(n); x_vals = np.arange(1,len(y_vals)+1)
plt.plot(x_vals,y_vals)
x_val_max = np.where(y_vals == max(y_vals))[0][0]+1
plt.text(x_val_max, max(y_vals), '({}, {})'.format(x_val_max, int(max(y_vals))))
plt.grid(color='purple',linestyle='--')
plt.ylabel("Sequence Generated")
plt.xlabel("Number of Iteration")
plt.show()

The output image is given below.

Crazy right?, you can try it out. It's almost feels like some sort of physics thing right!, for some it seems like stock market.

This is all for today. In it's part-2, we will go into much more detail. I hope you have learn something new.

Maybe in the part-2, we will see how to create pattern similar to this using collatz conjecture.

Gaussian Prime Spiral and Its beautiful Patterns

Author: Kazi Abu Rousan

Mathematics is the science of patterns, and nature exploits just about every pattern that there is.

Ian Stewart

Introduction

If you are a math enthusiastic, then you must have seen many mysterious patterns of Prime numbers. They are really great but today, we will explore beautiful patterns of a special type of 2-dimensional primes, called Gaussian Primes. We will focus on a very special pattern of these primes, called the Gaussian Prime Spiral. First, we have understood what those primes are. The main purpose of this article is to show you guys how to utilize programming to analyze beautiful patterns of mathematics. So, I have not given any proof, rather we will only focus on the visualization part.

Gaussian Integers and Primes

We all know about complex numbers right? They are the number of form $z=a+bi$, were $i$ = $\sqrt{-1}$. They simply represent points in our good old coordinate plane. Like $z=a+bi$ represent the point $(a,b)$. This means every point on a 2d graph paper can be represented by a Complex number.

In python, we write $i$ = $\sqrt{-1}$ as $1j$. So, we can write any complex number $z$ = $2 + 3i$ as $z$ = $2+1j*3$. As an example, see the piece of code below.

z = 2 + 1j*3
print(z)

#output is (2+3j)
print(type(z))
#output is <class 'complex'>
#To verify that it indeed gives us complex number, we can see it by product law.
z1 = 2 + 1j*3
z2 = 5 + 1j*2
print(z1*z2)
#output is (4+19j)

To access the real and imaginary components individually, we use the real and imag command. Hence, z.real will give us 2 and z.imag will give us 3. We can use abs command to find the absolute value of any complex number, i.e., abs(z) = $\sqrt{a^2+b^2} = \sqrt{2^2+1^2}= \sqrt{5}$. Here is the example code.

print(z.real)
#Output is 2.0
print(z.imag)
#output is 3.0
print(abs(z))
#Output is 3.605551275463989

If lattice points, i.e., points with integer coordinates (eg, (2,3), (6,13),... etc) on the coordinate plane (like the graph paper you use in your class) are represented as complex numbers, then they are called Gaussian Integers. Like we can write (2,3) as $2+3i$. So, $2+3i$ is a Gaussian Integer.

Gaussian Primes are almost same in $\mathbb{Z}[i]$ as ordinary primes are in $\mathbb{Z_{+}}$, i.e., Gaussian Primes are complex numbers that cannot be further factorized in the field of complex numbers. As an example, $5+3i$ can be factorised as $(1+i)(3-2i)$. So, It is not a gaussian prime. But $3-2i$ is a Gaussian prime as it cannot be further factorized. Likewise, 5 can be factorised as $(2+i)(2-i)$. So it is not a gaussian prime. But we cannot factorize 3. Hence, $3+0i$ is a gaussian prime.

Note: Like, 1 is not a prime in the field of positive Integers, $i$ and also $1$ are not gaussian prime in the field of complex numbers. Also, you can define what a Gaussian Prime is, using the 2 condition given below (which we have used to check if any $a+ib$ is a gaussian prime or not).

Checking if a number is Gaussian prime or not

Now, the question is, How can you check if any given Gaussian Integer is prime or not? Well, you can use try and error. But it's not that great. So, to find if a complex number is gaussian prime or not, we will take the help of Fermat's two-square theorem.

A complex number $z = a + bi$ is a Gaussian prime, if:. As an example, $5+0i$ is not a gaussian prime as although its imaginary component is zero, its real component is not of the form $4n+3$. But $3i$ is a Gaussian prime, as its real component is zero but the imaginary component, 3 is a prime of the form $4n+3$.

  1. One of $|a|$ or $|b|$ is zero and the other one is a prime number of form $4n+3$ for some integer $n\geq 0$. As an example, $5+0i$ is a gaussian prime as although it's imaginary component is zero, it's real component is not of the form $4n+3$. But $3i$ is a gaussian prime, as it's real component is zero and imaginary component, 3 is a prime of form $4n+3$.
  2. If both a and b are non-zero and $(abs(z))^2=a^2+b^2$ is a prime. This prime will be of the form $4n+1$.

Using this simple idea, we can write a python code to check if any given Gaussian integer is Gaussian prime or not. But before that, I should mention that we are going to use 2 libraries of python:

  1. matplotlib $\to$ This one helps us to plot different plots to visualize data. We will use one of it's subpackage (pyplot).
  2. sympy $\to$ This will help us to find if any number is prime or not. You can actually define that yourself too. But here we are interested in gaussian primes, so we will take the function to check prime as granted.

So, the function to check if any number is gaussian prime or not is,

import numpy as np
import matplotlib.pyplot as plt
from sympy import isprime
def gprime(z):#check if z is gaussian prime or not, return true or false
	re = int(abs(z.real)); im = int(abs(z.imag))
	if re == 0 and im == 0:
		return False
	d = (re**2+im**2) 
	if re != 0 and im != 0:
		return isprime(d)
	if re == 0 or im == 0:
		abs_val = int(abs(z))
		if abs_val % 4 == 3:
			return isprime(abs_val)
		else:
			return False

Let's test this function.

print(gprime(1j*3))
#output is True
print(gprime(5))
#output is False
print(gprime(4+1j*5))
#output is True

Let's now plot all the Gaussian primes within a radius of 100. There are many ways to do this, we can first define a function that returns a list containing all Gaussian primes within the radius n, i.e., Gaussian primes, whose absolute value is less than or equals to n. The code can be written like this (exercise: Try to increase the efficiency).

def gaussian_prime_inrange(n):
    gauss_pri = []
    for a in range(-n,n+1):
        for b in range(-n,n+1):
            gaussian_int = a + 1j*b
            if gprime(gaussian_int):
                gauss_pri.append(gaussian_int)
    return gauss_pri

Now, we can create a list containing all Gaussian primes in the range of 100.

gaussain_pri_array = gaussian_prime_inrange(100)
print(gaussain_pri_array)
#Output is [(-100-99j), (-100-89j), (-100-87j), (-100-83j),....., (100+83j), (100+87j), (100+89j), (100+99j)]

To plot all these points, we need to separate the real and imaginary parts. This can be done using the following code.

gaussian_p_real = [x.real for x in gaussain_pri_array]
gaussian_p_imag = [x.imag for x in gaussain_pri_array]

The code to plot this is here (exercise: Play around with parameters to generate a beautiful plot).

plt.axhline(0,color='Black');plt.axvline(0,color='Black') # Draw re and im axis
plt.scatter(gaussian_p_real,gaussian_p_imag,color='red',s=0.4)
plt.xlabel("Re(z)")
plt.ylabel("Im(z)")
plt.title("Gaussian Primes")
plt.savefig("gauss_prime.png", bbox_inches = 'tight',dpi = 300)
plt.show()
Gaussian Prime
Look closely, you can find a pattern... maybe try plotting some more primes

Gaussian Prime Spiral

Now, we are ready to plot the Gaussian prime spiral. I have come across these spirals while reading the book Learning Scientific Programming with Python, 2nd edition, written by Christian Hill. There is a particular problem, which is:

Gaussian Integer and Primes Problem

Although, history is far richer than just solving a simple problem. If you are interested try this article: Stepping to Infinity Along Gaussian Primes by Po-Ru Loh (The American Mathematical Monthly, vol-114, No. 2, Feb 2007, pp. 142-151). But here, we will just focus on the problem. Maybe in the future, I will explain this article in some video.

The plot of the path will be like this:

Gaussisan Primes and Gaussian Spirals

Beautiful!! Isn't it? Before seeing the code, let's try to understand how to draw this by hand. Let's take the initial point as $c_0=3+2i$ and $\Delta c = 1+0i=1$.

  1. For the first step, we don't care if $c_0$ is gaussian prime or not. We just add the step with it, i.e., we add $\Delta c$ with $c_0$. For our case it will give us $c_1=(3+2i)+1=4+2i$.
  2. Then, we check if $c_1$ is a gaussian prime or not. In our case, $c_1=4+2i$ is not a gaussian prime. So, we repeat step-1(i.e., add $\Delta c$ with it). This gives us $c_2=5+2i$. Again we check if $c_2$ is gaussian prime or not. In this case, $c_2$ is a gaussian prime. So, now we have to rotate the direction $90^{\circ}$ towards the left,i.e., anti-clockwise. In complex plane, it is very easy. Just multiply the $\Delta c$ by $i = \sqrt{-1}$ and that will be our new $\Delta c$. For our example, $c_3$ = $c_2+\Delta c$ = $5+2i+(1+0i)\cdot i$ = $5+3i$.
  3. From here, again we follow step-2, until we get the point from where we started with the same $\Delta c$ or you can do it for your required step.

The list of all the complex number we will get for this particular example is:

IndexComplex No.Step Index Complex No. Step
0$3+2i$+17 $2+4i$ -1
1 $4+2i$ +18 $1+4i$ -i
2 $5+2i$ +i9$1+3i$-i
3 $5+3i$ +i10$1+2i$+1
4 $5+4i$ -111$2+2i$+1
5 $4+4i$ -112$3+2i$+i
6 $3+4i$ -113$3+3i$+i
Complex numbers and $\Delta c$'s along Gaussian prime spiral

Note that although $c_{12}$ is the same as $c_0$, as it is a Gaussian prime, the next gaussian integer will be different from $c_1$. This is the case because $\Delta c$ will be different.

To plot this, just take each $c_i$ as coordinate points, and add 2 consecutive points with a line. The path created by the lines is called a Gaussian prime spiral. Here is a hand-drawn plot.

Gaussian Prime Spiral hand-drawn plot

I hope now it is clear how to plot this type of spiral. You can use the same concept for Eisenstein primes, which are also a type of 2D primes to get beautiful patterns (Excercise: Try these out for Eisenstein primes, it will be a little tricky).

We can define our function to find points of the gaussian prime spiral such that it only contains as many numbers as we want. Using that, let's plot the spiral for $c_0 = 3+2i$, which only contains 30 steps.

Gaussian primes, integers and spirals

Here is the code to generate it. Try to analyze it.

def gaussian_spiral(seed, loop_num = 1, del_c = 1, initial_con = True):#Initial condition is actually the fact
#that represnet if you want to get back to the initial number(seed) at the end.
    d = seed; gaussian_primes_y = []; gaussian_primes_x = []
    points_x = [d.real]; points_y = [d.imag]
    if initial_con:
        while True:
            seed += del_c; real = seed.real; imagi = seed.imag
            points_x.append(real); points_y.append(imagi)
            if seed == d:
                break
            if gprime(seed):
                del_c *= 1j
                gaussian_primes_x.append(real); gaussian_primes_y.append(imagi)
    else:
        for i in range(loop_num):
            seed += del_c; real = seed.real; imagi = seed.imag
            points_x.append(real); points_y.append(imagi)
            if gprime(seed):
                del_c *= 1j ;
                gaussian_primes_x.append(real); gaussian_primes_y.append(imagi)
    gauss_p = [gaussian_primes_x,gaussian_primes_y]
    return points_x, points_y, gauss_p

Using this piece of code, we can literally generate any gaussian prime spiral. Like, for the problem of the book, here is the solution code:

seed1 = 5 + 23*1j
plot_x, plot_y, primes= gaussian_spiral(seed1)
loop_no = len(plot_x)-1
plt.ylim(21,96.5)
plt.xlim(-35,35)
plt.axhline(0,color='Black');plt.axvline(0,color='Black')
plt.plot(plot_x,plot_y,label='Gaussian spiral',color='mediumblue')
plt.scatter(primes[0][0],primes[1][0],c='Black',marker='X')#starting point
plt.scatter(primes[0][1::],primes[1][1::],c='Red',marker='*',label='Gaussian primes')
plt.grid(color='purple',linestyle='--')
plt.legend(loc='best',prop={'size':6})
plt.xlabel("Re(z) ; starting point = %s and loop number = %s "%(seed1,loop_no))
plt.ylabel("Im(z)")

plt.savefig("prob_sol.png", bbox_inches = 'tight',dpi = 300)
plt.show()

A few more of these patterns are:

Gaussian Prime and Spiral Pattern

One of the most beautiful patterns is generated for the seed: $277 + 232i$.

Gaussian Primes Spiral Patterns

๐Ÿ˜ณ Am I seeing a Bat doing back-flip?

All the codes for generating these can be found here:

Here is an interactive version. Play around with this: https://zurl.co/Hv3U

Also, you can use python using an android app - Pydroid 3 (which is free)

I have also written this in Julia. Julia is faster than Python. Here you can find Julia's version: https://zurl.co/wETi

Problem on Curve | AMC 10A, 2018 | Problem 21

Try this beautiful Problem on Algebra based on Problem on Curve from AMC 10 A, 2018. You may use sequential hints to solve the problem.

Curve- AMC 10A, 2018- Problem 21


Which of the following describes the set of values of $a$ for which the curves $x^{2}+y^{2}=a^{2}$ and $y=x^{2}-a$ in the real $x y$ -plane intersect at
exactly 3 points?

  • $a=\frac{1}{4}$
  • $\frac{1}{4}<a<\frac{1}{2}$
  • $a>\frac{1}{4}$
  • $a=\frac{1}{2}$
  • $a>\frac{1}{2}$

Key Concepts


Algebra

greatest integer

Suggested Book | Source | Answer


Suggested Reading

Pre College Mathematics

Source of the problem

AMC-10A, 2018 Problem-14

Check the answer here, but try the problem first

$a>\frac{1}{2}$

Try with Hints


First Hint

We have to find out the value of \(a\)

Given that $y=x^{2}-a$ . now if we Substitute this value in $x^{2}+y^{2}=a^{2}$ we will get a quadratic equation of $x$ and \(a\). if you solve this equation you will get the value of \(a\)

Now can you finish the problem?

Second Hint

After substituting we will get $x^{2}+\left(x^{2}-a\right)^{2}$=$a^{2} \Longrightarrow x^{2}+x^{4}-2 a x^{2}=0 \Longrightarrow x^{2}\left(x^{2}-(2 a-1)\right)=0$

therefore we can say that either \(x^2=0\Rightarrow x=0\) or \(x^2-(2a-1)=0\)

\(\Rightarrow x=\pm \sqrt {2a-1}\). Therefore

Now Can you finish the Problem?

Third Hint

Therefore \(\sqrt {2a-1} > 0\)

\(\Rightarrow a>\frac{1}{2}\)

Subscribe to Cheenta at Youtube


ISI MStat PSB 2006 Problem 2 | Cauchy & Schwarz come to rescue

This is a very subtle sample problem from ISI MStat PSB 2006 Problem 2. After seeing this problem, one may think of using Lagrange Multipliers, but one can just find easier and beautiful way, if one is really keen to find one. Can you!

Problem- ISI MStat PSB 2006 Problem 2


Maximize \(x+y\) subject to the condition that \(2x^2+3y^2 \le 1\).

Prerequisites


Cauchy-Schwarz Inequality

Tangent-Normal

Conic section

Solution :

This is a beautiful problem, but only if one notices the trick, or else things gets ugly.

Now we need to find the maximum of \(x+y\) when it is given that \(2x^2+3y^2 \le 1\). Seeing the given condition we always think of using Lagrange Multipliers, but I find that thing very nasty, and always find ways to avoid it.

So let's recall the famous Cauchy-Schwarz Inequality, \((ab+cd)^2 \le (a^2+c^2)(b^2+d^2)\).

Now, lets take \(a=\sqrt{2}x ; b=\frac{1}{\sqrt{2}} ; c= \sqrt{3}y ; d= \frac{1}{\sqrt{3}} \), and observe our inequality reduces to,

\((x+y)^2 \le (2x^2+3y^2)(\frac{1}{2}+\frac{1}{3}) \le (\frac{1}{2}+\frac{1}{3})=\frac{5}{6} \Rightarrow x+y \le \sqrt{\frac{5}{6}}\). Hence the maximum of \(x+y\) with respect to the given condition \(2x^2+3y^2 \le 1\) is \(\frac{5}{6}\). Hence we got what we want without even doing any nasty calculations.

Another nice approach for doing this problem is looking through the pictures. Given the condition \(2x^2+3y^2 \le 1\) represents a disc whose shape is elliptical, and \(x+y=k\) is a family of straight parallel lines passing passing through that disc.

The disc and the line with maximum intercept.

Hence the line with the maximum intercept among all the lines passing through the given disc represents the maximized value of \(x+y\). So, basically if a line of form \(x+y=k_o\) (say), is a tangent to the disc, then it will basically represent the line with maximum intercept from the mentioned family of line. So, we just need to find the point on the boundary of the disc, where the line of form \(x+y=k_o\) touches as a tangent. Can you finish the rest and verify weather the maximum intercept .i.e. \(k_o= \sqrt{\frac{5}{6}}\) or not.


Food For Thought

Can you show another alternate solution to this problem ? No, Lagrange Multiplier Please !! How would you like to find out the point of tangency if the disc was circular ? Show us the solution we will post them in the comment.

Keep thinking !!


Similar Problems and Solutions



ISI MStat PSB 2008 Problem 10
Outstanding Statistics Program with Applications

Outstanding Statistics Program with Applications

Subscribe to Cheenta at Youtube


ISI MStat 2019 PSA Problem 14 | Reflection of a point

This is a problem from ISI MStat 2019 PSA Problem 14. First, try the problem yourself, then go through the sequential hints we provide.

Reflection of a point - ISI MStat Year 2019 PSA Question 14


The reflection of the point (1,2) with respect to the line \(x+2 y=15\) is

  • (3,6)
  • (6,3)
  • (10,5)
  • (5,10)

Key Concepts


Straight line

Check the Answer


Answer: is (5,10)

ISI MStat 2019 PSA Problem 14

Precollege Mathematics

Try with Hints


Find an algorithm to find the reflection,

Find the line perpendicular to \( x+2 y=15\) through (1,2).
Find the point of intersection.
Use Midpoint Segment Result.

The line perpendicular to \( x+2 y=15\) is of the form \(-2x+y=k \) .Now it passes through (1,2) . So, \( -2+2=k \Rightarrow k=0 \)

Hence the line perpendicular to \( x+2 y=15\) through (1,2) is y=2x.

Now we will find point of intersection (Foot of Perpendicular )

(3,6) is the point of intersection i.e the foot of perpendicular.

Use Mid-Point Formula (special case of Section formula) to get required point (Foot of perpendicular is mid-point of reflection and original point)

\( (3,6)=( \frac{x+1}{2} , \frac{y+2}{2} ) \) \( \Rightarrow x=5 , y=10 \)

Therefore the reflection point is (5,10) .

Similar Problems and Solutions



ISI MStat 2019 PSA Problem 14
Outstanding Statistics Program with Applications

Outstanding Statistics Program with Applications

Subscribe to Cheenta at Youtube


ISI MStat 2016 PSA Problem 9 | Equation of a circle

This is a problem from ISI MStat 2016 PSA Problem 9 based on equation of a circle. First, try the problem yourself, then go through the sequential hints we provide.

Equation of a circle- ISI MStat Year 2016 PSA Question 9


Given \( \theta \) in the range \( 0 \leq \theta<\pi,\) the equation \( 2 x^{2}+2 y^{2}+4 x \cos \theta+8 y \sin \theta+5=0\) represents a circle for all \( \theta\) in the interval

  • \( 0 < \theta <\frac{\pi}{3} \)
  • \( \frac{\pi}{4} < \theta <\frac{3\pi}{4} \)
  • \( 0 < \theta <\frac{\pi}{2} \)
  • \( 0 \le \theta <\frac{\pi}{2} \)

Key Concepts


Equation of a circle

Trigonometry

Basic Inequality

Check the Answer


Answer: is \( \frac{\pi}{4} < \theta <\frac{3\pi}{4} \)

ISI MStat 2016 PSA Problem 9

Precollege Mathematics

Try with Hints


Complete the Square.

We get ,

\(2{(x+\cos \theta)}^2 + 2{(y+ 2\sin \theta)}^2 = (6{\sin \theta}^2-3)) \)
\(6{\sin \theta}^2-3 > 0 \Rightarrow {\sin^2 \theta} \geq \frac{1}{2} \)

We are given that \( 0 \leq \theta<\pi,\) . So, \( {\sin^2 \theta} \geq \frac{1}{2} \) \( \Rightarrow \frac{\pi}{4} < \theta <\frac{3\pi}{4} \).

ISI MStat 2016 PSA Problem 9
Outstanding Statistics Program with Applications

Outstanding Statistics Program with Applications

Subscribe to Cheenta at Youtube


Sides of Quadrilateral | PRMO 2017 | Question 20

Try this beautiful problem from the Pre-RMO, 2017 based on Sides of Quadrilateral.

Sides of Quadrilateral - PRMO 2017


What is the number of triples (a,b,c) of positive integers such that (i) a<b<c<10 and (ii) a,b,c,10 form the sides of a quadrilateral?

  • is 107
  • is 73
  • is 840
  • cannot be determined from the given information

Key Concepts


Largest number of triples

Quadrilateral

Distance

Check the Answer


Answer: is 73.

PRMO, 2017, Question 20

Geometry Vol I to IV by Hall and Stevens

Try with Hints


a+b+c>10

(a+b+c) can be

a b c

1 2 8,9

1 3 7,8,9

1 4 6 ,7,8,9

1 5 6,7,8,9

1 6 7,8,9

1 7 8,9

1 8 9

2 3 6,7,8,9

2 4 5,6,7,8,9

2 5 6,7,8,9

2 6 7,8,9

2 7 8,9

2 8 9

3 4 5,6,7,8,9

3 5 6,7,8,9

3 6 7,8,9

3 7 8,9

3 8 9

4 5 6,7,8,9

4 6 7,8,9

4 7 8,9

4 8 9

5 6 7,8,9

5 7 8,9

5 8 9

6 7 8,9

6 8 9

7 8 9

Total 73 cases.

Subscribe to Cheenta at Youtube


Ordered pair Problem | AIME I, 1987 | Question 1

Try this beautiful problem from the American Invitational Mathematics Examination I, AIME I, 1987 based on Ordered pair.

Ordered pair Problem - AIME I, 1987


An ordered pair (m,n) of non-negative integers is called simple if the additive m+n in base 10 requires no carrying. Find the number of simple ordered pairs of non-negative integers that sum to 1492.

  • is 107
  • is 300
  • is 840
  • cannot be determined from the given information

Key Concepts


Integers

Ordered pair

Algebra

Check the Answer


Answer: is 300.

AIME I, 1987, Question 1

Elementary Algebra by Hall and Knight

Try with Hints


for no carrying required

the range of possible values of any digit m is from 0 to 1492 where the value of n is fixed

Number of ordered pair (1+1)(4+1)(9+1)(2+1)

=(2)(5)(10)(3)

=300.

Subscribe to Cheenta at Youtube


Hyperbola & Tangent | ISI MStat 2016 Problem 1 | PSB Sample

This is a beautiful sample problem from ISI MStat 2016 PSB Problem 1. This is based on finding the minimum value of a function subjected to the restriction.

ISI MStat 2016 Problem 1

Let \( x, y\) be real numbers such that \( x y=10 \) . Find the minimum value of \(|x+y|\) and all also find all the points \((x, y)\) where this minimum value is achieved.
Justify your answer.

Prerequisites

Solution

(a) Using graph

The equation \( xy=10={(\sqrt{10})}^2 \) represents the equation of rectangular hyperbola with foci are \( (- \sqrt{10},- \sqrt{10}) \) and \( (\sqrt{10}, \sqrt{10} ) \) .

ISI MStat 2016 Problem 1 graph
Fig-1

Now , \( |x+y|=c \Rightarrow \) \( x+y=\pm c \) , which looks somewhat like this ,

ISI MStat 2016 Problem 1 Figure 2
Fig-2

we have to find the the minimum value of \(|x + y|\) subject to the restriction that \( xy=10 \) . If we move \( |x+y|=c \) along \( xy =10 \) by varying c , then local minimum can occur at the points where the level curve \( |x+y|=c \) touch \( xy =10 \) . Now as both the rectangular hyperbola and |x+y|=c are symmetric about \( x+y=0 \) for \( c \ne 0 \) and \(x=y\) , the level curve will touch \( xy =10 \) when x+y=c and x+y=-c both are tangent to the curve \( xy =10 \) . And tangents occurs at the foci of \( xy =10 \) i.e at \( (- \sqrt{10},- \sqrt{10}) \) and \( (\sqrt{10}, \sqrt{10} ) \) .

ISI MStat 2016 Problem 1 Figure 3
Fig-3

Hence , the minimum value of \(|x + y|\) are \(|\sqrt{10} +\sqrt {10} |\) and \(|-\sqrt{10}- \sqrt{10}|\) both gives the same value \( 2 \sqrt{10} \) .

Therefore , the minimum value of \(|x + y|\) is \( 2 \sqrt{10}\) and it attains it's minimum at \( (\sqrt{10} , \sqrt{10} ) \) and \( ( - \sqrt{10} , -\sqrt{10}) \) .

(b) Using Derivative test

\( |x+y|= |x+ \frac{10}{x} |\) as we are given that \( xy=10 \)

Let, \( f(x) =|x+ \frac{10}{x}|\) then we have to find the minimum value of f(x)

Now ,as the function is not defined at x=0 and also x=0 can't give the minimum value of |x+y| due to the condition that xy=10. So, we will study f(x) for two cases when x>0 and when x<0 .

\( f(x) = \begin{cases} x+ \frac{10}{x} & ,x > 0 \\ -(x+ \frac{10}{x} ) & ,x < 0 \end{cases} \)

\( f'(x)=\begin{cases} 1- \frac{10}{x^2} & ,x > 0 \\ -(1- \frac{10}{x^2} ) & ,x < 0 \end{cases} \)

\( f'(x)=0 \Rightarrow \) \( x= \pm \sqrt{10} \)

\( f''(x)= \begin{cases} \frac{20}{x^3} & ,x > 0 \\ -\frac{20}{x^3} & ,x < 0 \end{cases} \)

So,\( f''(x) >0\) for \( x= \pm \sqrt{10} \)

Hence f(x) attains it's minimum value at \( (\sqrt{10} , \sqrt{10} ) \) and \( - \sqrt{10} , -\sqrt{10}) \) and minimum value is \( 2 \sqrt{10} \).

Challenge Problem

Let \( x_1, x_2 ,..., x_n \) be be positive real numbers such that \( \prod_{i=1}^{n} x_{i} = 10 \) . Find the minimum value of \( \sum_{i=1}^{n} x_{i} \).