This is my message board. I am moderator. My job is to keep the community clean and engaged. Any types of questions can be asked here. Any irrelevant comment , note or problem(report the problem and leave the link here) can be conveyed through here or slack.

\[\huge \color{red}{\text{Happy}} \quad \huge \color{blue}{\text{Brillianting !}}\]

No vote yet

1 vote

×

Problem Loading...

Note Loading...

Set Loading...

Easy Math Editor

`*italics*`

or`_italics_`

italics`**bold**`

or`__bold__`

boldNote: you must add a full line of space before and after lists for them to show up correctlyparagraph 1

paragraph 2

`[example link](https://brilliant.org)`

`> This is a quote`

Remember to wrap math in \( ... \) or \[ ... \] to ensure proper formatting.`2 \times 3`

`2^{34}`

`a_{i-1}`

`\frac{2}{3}`

`\sqrt{2}`

`\sum_{i=1}^3`

`\sin \theta`

`\boxed{123}`

## Comments

Sort by:

TopNewestRajdeep, it has come to my attention that you are aggressively @mentioning several members numerous times a day. Please refrain from doing so, especially if you notice that they have not responded to any of your past mentions.

In general, unless you have a strong reason, please refrain from @ mentioning any person more than once a day. If you continually bombard people with such requests, they will (similar to the boy who cried wolf) most likely start to ignore you.

Log in to reply

sorry

Log in to reply

@Vighnesh Raut The solution to your question \(\int{\frac{dx}{1 + \tan^{4}{x}}}\) is given below.

Solution : \[\int{\frac{dx}{1+ tan^4 {x}}} \\ \text{Substitute tan x = t . Then the integral will become } \\ \int{\frac{dt}{(1 + t^2)(1 + t^4 )}} \\ \text{ Separating using partial fractions we get} \\ \int{(\frac{1}{2(1 +t^2)} + \frac{2t - \sqrt{2}}{4\sqrt{2}(-t^2 + \sqrt{2}t - 1)} + \frac{2t + \sqrt{2}}{4\sqrt{2}(t^2 + \sqrt{2}t + 1)})} \\ \text{Integrating we get} \\ \frac{1}{2}\arctan{t} - \frac{1}{4\sqrt{2}}\log{(t^2 - \sqrt{2}t + 1)} + \frac{1}{4\sqrt{2}}\log{(t^2 + \sqrt{2}t + 1}) \\ \text{We can easily substitute back in t = tan x and then simplify.}\]

Log in to reply

Comment deleted Mar 17, 2015

Log in to reply

Fully agreed! @Azhaghu Roopesh M . Please try my 100 followers problem!

Log in to reply

Sorry for misprint here \(\phi = \varphi\)

Now If body is in equilibrium. Resultant = 0

So that means \(mg\sin{\theta} = T\cos{\phi}\)

Plugging in values and solving we get

\[T = \frac{10 \times 9.8 \times \sin{45^{\circ}}}{\cos{25^{\circ}}}\] If Doubt comment otherwise reply that you understood.

Log in to reply

ohh.. thank you.. I got it.. :D :D

Log in to reply

Your welcome!

Log in to reply

@Rajdeep Dhingra Could you suggest some resource for calculus?

Log in to reply

@Rajdeep Dhingra could you please help me and give some tips about nsejs? Mainly things like what to refer for electricity-is trignometry imp-which book is best for chemistry and such. I have the books for nsejs given by allen institute in their workshop. Are those helpfull?

Log in to reply

Complete those allen books , they cover the whole syllabus. For electricity if you know calculus refer University Physics Sears and Zemansky or HC Verma (although calculus is not at all used in NSEJS) , otherwise refer College Physics . Do trigonometry (Class 10th RD Sharma or Any good Textbook for NTSE) completely. For Chemistry mostly do it from Chemistry for the IB Diploma Coursebook or Chemistry by Raymond Chang (I did this book) and Chemistry - 1 , Advanced Chemistry (I really recommend these courses (They are free) , I learnt first time from them). For Biology check this out.

Log in to reply

@Rajdeep Dhingra Thanks. Is trignometry asked directly or is it used to solve questions? And i know calculus but only the basics of differentiation (meaning i can differentitate) should i learn differentiation whole ? or should i leave it? i have heard that it helps in solving problems more quickly. I got 110 in 2011's paper when i tried it and 117 in 2014's paper. This was when i didn't attempt bio whole section and electricity problems (as i havent prepared them yet) and attempted only 10 chemistry questions(i am a little weak in chemistry). Do you think i can qualify if I complete atleast electricity and chemistry? I dont think i will be able to complete whole syllabus due to my school mid term exams. I have them in september. I will be studying during them of course for nsejs as my that month is free of any other competitions. I am from chhattisgarh.

Log in to reply

Log in to reply

Log in to reply

Log in to reply

@Harsh Shrivastava Please post your JOMC Questions here .

Log in to reply

In the above solution mass 1 = 1 kg , mass 2 = 2 kg , mass 3 = 3 kg. This will make calculation easier. You can try and generalize it for variables . If any doubt persists then do ask.

Log in to reply

How do we get that block m1 moves upwards? If we did m1 downwards we get a negative value? @Rajdeep Dhingra

Log in to reply

Yes. @Anik Mandal

Log in to reply

Comment deleted Dec 24, 2015

Log in to reply

Yes thanks for your co-operation

Log in to reply

Comment deleted Dec 24, 2015

Log in to reply

Well I would be happy if you post problems on Newton's sums.

Log in to reply

Try my new questions Find the Truth 7 and Dirty Algebra.

Log in to reply

Rajdeep try my question "I think factorization may help"

Log in to reply

I have already done it.

Log in to reply

Do "Dirty Maxima" then

Log in to reply

Here's the link to the staff list , in case you missed out on my comment on Calvin sir's Message Board .

Log in to reply

This is not the staff list. Also I read your comment on Sir's message board.

Just help me out in this: -

ABandCDare 2 straight lines meeting atOandXYis another straight line. Show that in general 2 points can be found inXYwhich are equidistant fromABandCD. When is there only one such point ?See this

Log in to reply

Did you check out the whole page ?

Also I see that Shashwat has explained it to you already .

Log in to reply

Guys I have a problem in questions like this one.

If \(\displaystyle f \left( \dfrac{x+y}{2} \right) = \dfrac{f(x) + f(y)}{2}. \)

\(\text{Find } f(2) \text{ given } f(0) = 1 \text{ and } f'(0) = -1. \)

Also give me links to many such problems so I can practice them. Any book or wiki or PDF also will work.

Log in to reply

You might want to see this .

Log in to reply

Also why do you want to become a mod ? Just curious .

Log in to reply

@Azhaghu Roopesh M , You can Solve the problem now... I have reposted it

Log in to reply

It was easy but good. When I tried the same problem with the wrong answer I wondered where had I went wrong.I thought of disputing it but didn't do it as the changed problem had come.

Log in to reply

Log in to reply

Well I wanted to become one since I came to know about it. I thought it was cool. Well know I have changed my mind(Looking at people like you and megh choksi).

P.S - If I get a request to become one well I won't deny it. \(\ddot\smile\)

Log in to reply

Frankly speaking , Megh has a better chance of becoming a mod . Look at the number of his followers !

Log in to reply

Log in to reply

Log in to reply

Could you help the guy out who posted a question on the RMO/INMO board. The \(\omega\) one.

Log in to reply

@Prasun Biswas @Rajdeep Dhingra help me in starting that question . Please help.

Log in to reply

@Prasun Biswas Help him. I am exhausted. Not because of you but other things. I typed almost 50 comments each no less than 200 - 300 words.

Log in to reply

rajdeep, I got my answer as 2013/2008 why did I go wrong?

Log in to reply

\(P(1)=P(2)=P(3)=P(4)=P(5)=1\)

\(\Rightarrow P(x) = A(x-1) (x-2) (x-3) (x-4) (x-5)+1\)

Therefore,

\(\dfrac {P(2015)-1}{P(2014)-1} \)

\(= \dfrac {A(2015-1) (2015-2) (2015-3) (2015-4) (2015-5)+1-1}{A(2014-1) (2014-2) (2014-3) (2014-4) (2014-5)+1-1} \)

\(= \dfrac {A(2014) (2013) (2012) (2011) (2010)}{A(2013) (2012) (2011) (2010) (2009)} = \dfrac {2014}{2009} \)

\(\Rightarrow a = 2014\), \(b = 2009\) and \(a+b = 2014+2009 = \boxed{4023}\)

Log in to reply

@Chew-Seong Cheong Sir's solution... XD

Hahaha.. I truly understood how exhausted u are... This isLog in to reply

Log in to reply

Log in to reply

P.S - I will post 3 problems tomorrow on this. Then you can check your newly learnt skill. \(\ddot\smile\)

Log in to reply

Log in to reply

See , A polynomial is anything in the form \(a_0 x^n + a_1 x^{n-1} + ......... + a_{n-1} x + a_n\).Provided that only \(a_0 \neq 0\). Rest anything is possible.There no particular constant.

Log in to reply

Log in to reply

Log in to reply

Log in to reply

link.

Sorry class 10th. Here's theoffline

Log in to reply

@Rajdeep Dhingra :*

Hell man.... I got that as 4022 !!! Though I finally understood all this.... Thank you so much for your helpLog in to reply

@Rajdeep Dhingra Can You please help me with this ?

Log in to reply

Hint: Use the Remainder-Factor Theorem for polynomials.

Extra hint: Try forming a new polynomial whose roots are the given \(x\)'s.

Log in to reply

\[\lim _{x \rightarrow \infty}{\left( \frac{x!}{x^x} \right) ^{1/x}}\]

Log in to reply

\[\ln L=\lim_{x\to\infty}\frac{1}{x}\cdot \sum_{i=1}^x f\left(\frac{i}{x}\right)\]

This is a Riemann sum and can be evaluated by using limit sum definition of definite integral with \(f(x)=\ln x\). We proceed as follows:

\[\ln L=\int\limits_0^1\ln x\,dx\]

Using IBP, we have,

\[\ln L=\left[x\ln x-x\right]_0^1\implies \ln L=(1\times \ln 1-1)-\lim_{x\to 0^+}(x\ln x-x)\]

The limit \(\displaystyle\lim_{x\to 0^+}(x\ln x)\) can be evaluated to be \(0\) using L'Hopital's rule. Thus,

\[\ln L=(0-1)-(0-0)=(-1)\implies \boxed{L=e^{-1}}\]

Log in to reply

You are watching four people play bridge, where a hand begins by dealing each player 13 cards.

a. After a hand is dealt, you ask a player, “Do you have at least one Ace?” She says, “Yes.” What is the probability that she’s holding more than one Ace?

b. On a later hand, you ask another player, “Do you have the Ace of Spades?” She says, “Yes.” What is the probability that she’s holding more than one Ace?

See when we apply bayes theorem I can't understand how to evaluate P(A) and P(B).The probabilities I asked you.So, instead just solve the whole question please.

Log in to reply

Hey bro . Did you solve that probability problem I had asked ?

Log in to reply

Log in to reply

Have you tried solving it ?

Log in to reply

I have a really low level of knowledge in Functions and all so can you please help me out? with this and the basics and all....

Log in to reply

Its gonna take time to type but if you have patience I will help you.

Log in to reply

Log in to reply

Log in to reply

Log in to reply

So for values up to 5 the function is just \(\large f(n) = n + 1\)

Now we can set up are polynomial

\[\large f(x) = C(x - 1)(x - 2)(x - 3)(x - 4)(x - 5) + x + 1\]

Now try the question. I am sure you'll be able to do it. If not comment.(I am going offline will be online in half an hour)

Log in to reply

Log in to reply

Log in to reply

Log in to reply

Now we have to find \(C \text{ from } f(x)\).

They have given us the value of the function at \(f(8)\) which is \(f(8) = 8-1 = x-1\).

So now we have \[\large f(x) = C(x - 1)(x - 2)(x - 3)(x - 4)(x - 5) + x + 1 \\ \large f(8) = C(8-1)(8-2)(8-3)(8-4)(8-5) + 8 + 1 \\ \text{Plugging in value of the function at 8 , we get} \\ 7 = C\dfrac{7!}{2!} + 9 \\ \text{Finding value of C we get} \\ C = (-2)\dfrac{2!}{7!} \\ \text{Now our polynomial changes to n} \\ \large f(x) = (-2)(\dfrac{2!}{7!})(x - 1)(x - 2)(x - 3)(x - 4)(x - 5) + x +1 \\ \text{Plugging 9 we get} \\ \large f(9) = \dfrac{14}{3} \]

Comment if doubt, otherwise reply that you understood. Here is a problem same as this . Try this to clear concepts

Log in to reply

Log in to reply

Log in to reply

Log in to reply

ndegree polynomial can be expressed in the form \[P(x) = C(x - a_1)(x - a_2)(x - a_3).........(x - a_n) + k\].Try the problem now. If unable to solve I will help you.

Log in to reply

Will you please tell me how did you get thiscorrect?

Log in to reply

But the person who has posted it has written himself that this question is incorrect .Is the question correct?

Log in to reply

Yes it is but you need to assume it in complex plane.

here is the solution:

Log in to reply

Log in to reply

Comment deleted Mar 01, 2016

Log in to reply

A square always ends in 21 41 61 01 or 81 ok Any no is in form of 10k+1....................................10k+9 Hope u can square them now

Log in to reply

Nishi Sofat Arushi Sofat

Please ask Rahul Sir to solve this.\[\lim _{x \rightarrow \infty}{\left( \frac{x!}{x^x} \right) ^{1/x}}\]

Log in to reply

Please Find \[\int{\frac{1+ x^4}{( 1 -x^4)^{3/2}}dx}\] Please help @Ronak Agarwal @Prasun Biswas @megh choksi

Log in to reply

That's a simple one : Put \( \displaystyle x=\sqrt{sin\theta} \) to get :

\( \displaystyle I=\frac { 1 }{ 2 } \int { (2\sec ^{ 2 }{ \theta } -1)\sqrt { \csc { \theta } } d\theta } \)

\(\displaystyle I = \frac { 1 }{ 2 } (\int { (2\sec ^{ 2 }{ \theta } \sqrt { \csc { \theta } } d\theta } -\int { \sqrt { \csc { \theta } } d\theta } ) \)

Applying product rule to first integral we have :

\( \displaystyle u = \sqrt { \csc { \theta } } , v'=\sec ^{ 2 }{ \theta } \)

\( \displaystyle I = \frac { 1 }{ 2 } (2\tan { \theta } \sqrt { \csc { \theta } } -\int { 2\tan { \theta } \frac { -\csc { \theta } \cot { \theta } }{ 2\sqrt { \csc { \theta } } } } -\int { \sqrt { \csc { \theta } } d\theta } )+C \)

\( \displaystyle I = \frac { 1 }{ 2 } (2\tan { \theta } \sqrt { \csc { \theta } } +\int { \sqrt { \csc { \theta } } d\theta } -\int { \sqrt { \csc { \theta } } d\theta } )+C \)

\( \displaystyle \Rightarrow I = \tan { \theta } \sqrt { \csc { \theta } } +C \)

Reverting back to our substitution we have :

\( I = \dfrac { x }{ \sqrt { 1-{ x }^{ 4 } } } +C \)

Log in to reply

Today, I also found a new way. \[ \int{\frac{1+ x^4}{( 1 -x^4)^{3/2}}dx} \\ \text{Dividing the inner things by (x^2) we get} \\ \int{\frac{\frac{1 + x^4}{x^3}}{(\frac{ 1 - x^4}{x^2})^{3/2}}dx} \\ \int{\frac{\frac{1}{x^3} + x}{(\frac{1}{x^2} + x^2)^{3/2}}dx} \\ \text{Now we have a derivative, So we substitute the down expression. Integrating and substituting back we get} \\ \frac{1}{\sqrt{\frac{1}{x^2} - x^2}} + C \\ \frac{x}{\sqrt{1 - x^4}} + C \]

Log in to reply

Hi Guys, I read your conversation and was fascinated to see your knowledge?

Could you solve one of my questions? Ques1. Prove that \(n^{4}\) + \(4^{n}\) is composite for all integer value of \(n\) \(\gt\) \(1\)

Please help.@Prasun Biswas @Ronak Agarwal @Rajdeep Dhingra @Satvik Golechha

Log in to reply

Here's the solution to your problem.

Log in to reply

Prasun Have you checked it out !

Will you participate ? Check the edit in it.

Log in to reply

@PRIYANSHU MISHRA For all n \(\in\) Even . It can be easily proved that it composite(As it is a multiple of \(2\) ). For the odd I Have an approach but is not complete. Still working on that .

Here is my approach for odd.

now the cyclicity is \(\boxed{10}\) . So we find remainders for n = 1 to 10.

Here I am getting problems.Solving this would take time . Could you simplify the mod expression.@Calvin Lin @Satvik Golechha @Peter Taylor @Prasun Biswas @Ronak Agarwal Please help.

Log in to reply

Interesting Question. On it.

Log in to reply

@Prasun Biswas @Ronak Agarwal @Calvin Lin @Satvik Golechha

I have a doubt in the wiki on Chinese Remainder Theorem

I didn't understand how did this theorem is applied. In the example please tell this step.Which values we multipied in this?

Image

Log in to reply

@Prasun Biswas @Ronak Agarwal @Satvik Golechha @satvik pandey @Calvin Lin @Sreejato Bhattacharya Please help

Log in to reply

The sum is, \(x\equiv \displaystyle \sum_i a_iy_iz_i\) where \(z_i\equiv y_i^{-1}\pmod{n_i}\) for the \(i^{\textrm{th}}\) congruence of the system and the other symbols have their usual meanings as given in that wiki. Read it carefully and try the example problems yourself to get a better grasp on how to utilize CRT to solve a system of congruencies. A level 5 like you would've no problem in doing that. Cheers! :)

Log in to reply

Just read the theorem and went on to try the example.

Log in to reply

@Prasun Biswas @Ronak Agarwal @Calvin Lin Help. I solved it using basic prime factorization. But I bet there has got to be a shorter solution.

Log in to reply

If a,b,c and d are any 4 positive real numbers, then prove that \[\frac{a}{b} + \frac{b}{c} + \frac{c}{d} + \frac{d}{a} \ge 4\]

@Prasun Biswas @Ronak Agarwal @Satvik Golechha @Calvin Lin

Log in to reply

You can also use the Rearrangement Inequality. That trivializes the problem.

Consider the following sequences: \(\{a,b,c,d\}\) and \(\left \{\dfrac{1}{d},\dfrac{1}{c},\dfrac{1}{b},\dfrac{1}{a}\right \}\) of positive reals and without loss of generality, assume \(a\geq b\geq c\geq d\). Then, we also have \(\dfrac{1}{d}\geq \dfrac{1}{c}\geq \dfrac{1}{b}\geq \dfrac{1}{a}\)

Using the Rearrangement Inequality, we have,

\[\text{Directed Sum} \ge \text{Random Sum} \ge \text{Reverse Sum}\\ \implies \frac{a}{d}+\frac{b}{c}+\frac{c}{b}+\frac{d}{a}\geq \frac{a}{b}+\frac{b}{c}+\frac{c}{d}+\frac{d}{a}\geq \frac{a}{a}+\frac{b}{b}+\frac{c}{c}+\frac{d}{d}=1+1+1+1=4\]

\[\therefore \frac{a}{b}+\frac{b}{c}+\frac{c}{d}+\frac{d}{a}\geq 4\]

and equality holds iff \(a=b=c=d\)

Log in to reply

@Prasun Biswas Nice solution.

In this Rearrangement Inequality can we change the random sum any way we like ? Should it be the Product of the squence in any order?

Like in this case \(Directed Sum \ge \frac{a}{c} + \frac{b}{a} + \frac{d}{b} + \frac{c}{d} \ge Reverse Sum \)

Help @Ronak Agarwal @Calvin Lin

Log in to reply

Log in to reply

Simply apply \(A.M \ge G.M. \)

Log in to reply

How did you expand (a+b+c+d)^4

Log in to reply

Log in to reply

@Ronak Agarwal Please give a solution to the problem Logarithms, Integrals,Trigonometry

First give a hint.Like what to set as a variable to use Leibniz's integral rule.

Log in to reply

Log in to reply

Log in to reply

Log in to reply

Log in to reply

@Ronak Agarwal How is this related to your question ?

Log in to reply

Log in to reply

Log in to reply

Log in to reply

\(F(n)\) is the nth Fibonacci number.Here are some values :

\(F(0) = 0 , F(1) = 1 , F(2) = 1 , F(3) = 2 , \dots \dots , F(n) = F(n-1) + F(n-2)\)

Log in to reply

@Rajdeep Dhingra , Its not a good idea to ask problems for which no close form exist. This is quite famous, and called "Reciprocal Fibonacci Constant" . There is NO CLOSE form to this summation. You earlier asked , \( \zeta(3) \), which also has no close form, inspite of knowing that. Trying to waste anyone time who is helping you is not a good thing.

Log in to reply

I don't remember the problems name (as i did it a month ago) but I remember the problem . Also there was no solution to it but had quite a many solvers.

The problem was \(2.2.2^{\frac{1}{2}}.2^{\frac{1}{3}}.2^{\frac{1}{5}}\dots \dots\) here the power is fibonacci numbers. hence to evaluate this we needed "Reciprocal Fibonacci Constant".

Sorry @Ronak Agarwal I didn't know it.

P.S- \(\zeta{(3)}\) was just a joke. Please don't take it seriously and please keep helping me Ronak agarwal. I won't reapeat the \(\zeta(3)\) thing. Thank you for helping me now in the future.

I have a doubt that I will ask you later Ronak Agarwal. I have proceeded with the solution till some extent but could you help me out with it. Hopefully will post in 3-4 hours.

Log in to reply

Log in to reply

Two boxes contain between them 65 balls of several different sizes. Each ball is white, black, red or yellow. If you take any five balls of the same colour, at least 2 of them will always be of the same size. Prove that there are at least 3 balls which lie in the same box, have the same colour and the same size.

Log in to reply

Log in to reply

@Rajdeep Dhingra

I have added solution to my problem. You can check it outLog in to reply

Log in to reply

Log in to reply

Log in to reply

@Ronak Agarwal Its not working.

Please help @Calvin Lin @Satvik Golechha @Sreejato Bhattacharya

Log in to reply

Log in to reply

@Ronak Agarwal Have you tried my problem

Log in to reply

Easy Integral

You should try this@Rajdeep Dhingra

Log in to reply

@Ronak Agarwal Try this. \(\text{ Harmonic }\) at its best

Please Contribute a Solution.@Prasun Biswas @Calvin Lin

Log in to reply

@Ronak Agarwal I solved it using hyperbolic tangent integral. How did you solve it?

Log in to reply

Log in to reply

Simple use of Rearrangemet inequality.

Log in to reply

Could you give a algebraic solution to this \(\text{ Harmonic }\) at its best

Log in to reply

Which question ?

Log in to reply

@Ronak Agarwal @Satvik Golechha @Prasun Biswas @megh choksi @Calvin Lin

Do you know from where I can learn about 'Mod' ,the remainder operator, and it's properties.

Is there a wiki.

Log in to reply

Yes, there are lots of wikis on the Modular Arithmetic. These are all listed under Number Theory - Modular Arithmetic.

You can also do a search and filter by wiki, to see the relevant pages that appear. Often, the suggested page will be the most relevant to you, or it will contain the most information that will be useful to you.

Log in to reply

Thank you very much

Log in to reply

@Calvin Lin Sir please change my age. It is 12 years.My DOB is 14th August 2002. Thanks! In advance

Log in to reply

@Calvin Lin There is a mistake in the typing of my date of birth. It is actually 14th august 2001. Please restore my streak, account and points. Or transfer those things in this account.

Log in to reply

Log in to reply

@Calvin Lin My DOB is 14th August 2002 not 2001

Sorry for the trouble

Log in to reply

@Calvin Lin Sir My parents have replied you back. Sir , Please restore my account.

Log in to reply

@Calvin Lin Sir i have sent you there Email IDs Please reach out to them Fast as i don't want to lose my streak

Log in to reply

note earlier. I wanted someone to verify whether my method is correct or not. I even tagged you there, but no response.

Can you please help me with this? I posted aLog in to reply

@Calvin Lin Please help

Log in to reply

Could You Guys help me?

Find the remainder when \(2^{1990}\) is divided by 1990.

@Calvin Lin @Satvik Golechha @Prasun Biswas @Ronak Agarwal @Rajdeep Dhingra @Sreejato Bhattacharya

Log in to reply

You can see the solution here

Log in to reply

A square sheet of paper ABCD is so folded that B falls on the mid-point M of CD.Prove that the crease will divide BC in the ratio 3:5.

@Calvin Lin @Sandeep Bhardwaj @jaikirat sandhu @Prasun Biswas @Ronak Agarwal @Satvik Golechha @megh choksi @Peter Taylor @Sreejato Bhattacharya @Suyeon Khim @Agnishom Chattopadhyay @Mursalin Habib

Please help.

Log in to reply

Well a neat diagram and simple geometry would do it.

Look at this diagram.

Yo Yo

Here \(AP\) is the crease along which paper is to be folded.Hence \(M\) becomes the reflection of \(B\)

Since \(M\) is the reflection of \(B\) hence \(BN=NM\).

Also let side of square be \(x\) then \(MC=\frac{x}{2}\)

Also \(tan(\theta)=\frac{1}{2},\Rightarrow sec(\theta)=\frac{\sqrt{5}}{2}\)

Now \(BM=\frac{\sqrt{5}x}{2}\) (By applying pythagorean theorom in \(BMC\))

Since \(N\) is the mid-point of \(BM\) hence \(BN=\frac{\sqrt{5}x}{4}\)

Finally in triangle \(BNP\) by simple trignometry we have :

\(BNsec(\theta)=BP=\frac{\sqrt{5}x}{4} \times \frac{\sqrt{5}}{2} = \frac{5x}{8}\)

\(PC=BC-BP=\frac{3x}{8}\)

Finally we are getting \(PC : PB :: 3:5\)

Hence proved.

Log in to reply

And by the way how did you make the figure.

Log in to reply

Log in to reply

Log in to reply

Log in to reply

Thanks!!

Log in to reply

I think this is a previous RMO (maybe 1991) problem, you can find a solution online.

Log in to reply

Its a RMO 1990 problem

Tried but couldn't find a solution.Please help @Satvik Golechha

Log in to reply

You are much more likely to get a response by posting that as a note instead of just as a comment on your messageboard.

Log in to reply

Thanks for the suggestion. Anyways, do you know how to solve it.

Log in to reply

Archit Boobna Calvin lin liked and reshared my integration cauchy

a

b

Log in to reply

do you like S.S.T?

Log in to reply

Comment deleted Jan 23, 2015

Log in to reply

well,same here!just like that!

Log in to reply

@Archit Boobna Please tell me your FTRE marks

Log in to reply

333

Log in to reply

@Archit Boobna Is class there tomorrow ? Tomorrow's date is 14/2/2015

Log in to reply

Archit Boobna In your new question Be Careful.

To evaluate C can the integers be same.

Like-\((-1)^{1000} = 1 \\ also \quad 1^{1000} = 1\).

Do we have take both of them as different cases or same.

Log in to reply

Log in to reply

@Prasun Biswas So you mean the value of C is \(\boxed{1}\) I really do feel that the answer is wrong.

Also look at the other question . posted in this note

Log in to reply

Log in to reply

@Prasun Biswas

ThanksLog in to reply

Log in to reply

P is any point inside a triangle ABC. The perimeter of the triangle ABC is \(2s\). Prove that \(s\) < \(AP + BP + CP\) < \(2s\)

Also could you give a solution for tanveen dhingra Problem Above

Log in to reply

Lemma: Whenever we have a point inside a triangle, the distance of it from a vertex is always less than any one of the sides of the triangle originating from that vertex.Image

Given that \(AB+BC+AC=2s\). Using the property specified in the beginning, we have,

\[AP+BP\gt AB\quad,\quad AP+CP\gt AC\quad,\quad BP+CP\gt BC\]

Adding the three inequalities, we get,

\[2(AP+BP+CP)\gt (AB+BC+AC)=2s \implies AP+BP+CP\gt \dfrac{2s}{2}=s\]

\[\therefore AP+BP+CP\gt s\ldots (i)\]

When \(P\) is a point inside \(\Delta\)\(ABC\), we always have,

\[AP\lt AB~\lor~AP\lt AC \quad \land \quad BP\lt BC~\lor~BP\lt AB\quad \land \quad CP\lt AC~\lor~CP\lt BC\]

Adding the three inequalities casewise, we have,

\[AP+BP+CP\lt AB+BC+AC\quad \lor \quad AP+BP+CP\lt AC+AB+BC\]

Note that both the cases imply the same thing which is,

\[AP+BP+CP\lt AB+BC+AC=2s\]

\[\therefore AP+BP+CP\lt 2s\ldots (ii)\]

From \((i)\) and \((ii)\), we have,

\[s\lt AP+BP+CP\lt 2s\]

P.S - I still cannot think of a rigorous proof of the lemma I stated though other than analytically verifying the validity of that lemma.

Log in to reply

@Prasun Biswas The Lower bound is easy.

I believe that the upper bound can only be prove by drawing a mirror image

Please also solve @tanveen dhingra Question

Log in to reply

Log in to reply

Log in to reply

Log in to reply

@tanveen dhingra Question.

Could you please solveLog in to reply

Log in to reply

here

Prasun Biswas Try my new integration problemLog in to reply

Log in to reply

Log in to reply

Log in to reply

The question is not of proof it is of finding the remainder

Log in to reply

Log in to reply

Log in to reply

Log in to reply

(Here "a" is \(1024({ 1024 }^{ 198 }-1)\) mod 1990

Now we can easily factorize 2^1980-1 and we will see that it is a multiple of 995=5x199, so a=0.

So x-1024 is a multiple of 1990. So x mod 1990 is 1024

Log in to reply

@Archit Boobna How would factorization help?

\(2^{1980} - 1\) = \((2^{990}-1)(2^{990} + 1)\) = \((2^{495} - 1)(2^{495}+1)(2^{990} + 1)\)

How did this help? Even further cubic factorization isn't working. @Prasun Biswas @Calvin Lin

Log in to reply

Log in to reply

Log in to reply

I checked using the Chinese Remainder Theorem. The result comes out to be \(2^{1980}\equiv 996 \pmod{1990}\)

Log in to reply

@Prasun Biswas

http://www.wolframalpha.com/input/?i=2%5E1980-1+mod+995&dataset=Log in to reply

Log in to reply

@Archit Boobna's method is correct (as far as I can see).

By the way, \(2^{1980}\equiv 1\pmod{995}\), soLog in to reply

Log in to reply

P.S - Check again. In his solution, he said that \(2^{1980}\equiv 1\pmod{995}\). That doesn't necessarily mean that we'll also have \(2^{1980}\equiv 1\pmod{1990}\).

Log in to reply

Log in to reply

Log in to reply

Log in to reply

Log in to reply

Log in to reply

Log in to reply

Log in to reply

Log in to reply