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Math Olympiad Contest!-1

Hello everyone!,
I am happy to announce the Math Olympiad Contest which only allows High School Math problems. I am organizing this contest so as to get exposure to various topics and tricks to solve hard problems. The rules are as follows:

  1. Suppose problem \(i\) is posted. The person who solves problem \(i\) and posts the solution can publish problem \(i+1\).
  2. This will continue until the problem posted is not answered within 6 hours. If a solution is not posted, the problem maker themself will post a solution to the problem, and then pose a new problem.If the solution of a problem is not posted by the problem-poster within 10hrs.,then his/her marks will be deducted by 2.
  3. I request that problem-posters contribute problems that are solvable and do not post problems that will demotivate others.
  4. The contest will end once 20 problems have been posed and answered (problems not solved within 6 hours don't count toward this).
  5. Problem posters should know the solution of their posted problem in advance.
  6. If the new problem is not posted in 15 minutes of answering question \(i\), ANYBODY can post question \(i+1\).
  7. Each problem will be designated a number of points between 1 and 5, decided by the problem poster. Please include this value in the statement of the problem.
  8. I request all entries to be honest and fair. Do not check for the solution on the internet or copy-paste the solution.
  9. I would also request that you re-share this note so that other members can learn about and join this contest.
  10. If someone has posted a solution to a problem and if the problem-poster is not online,even others can post their solution till the problem-poster decides whose is the correct solution.The one who gives the fastest and correct solution gets the points.
  11. The solutions posted after 6hrs.[i.e the time limit of a problem] will be not be considered.

Points:

1 . Kaustubh Miglani : 9 marks

2 . Ayush Rai : 7 marks

3 . Abhishek Alva,Svatejas Shivakumar,Archit Agrawal,Govind Ramesh : 3 marks

4 . Prakhar Bindal : 1 mark

Note by Ayush Rai
1 year ago

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Problem 19 : [4 marks]
\(\triangle ABC\) has sides \(AB=168,BC=156,\) and \(CA=180.\)It is inscribed in a circle,which has center \(O.\)Let \(M\) be the midpoint of \(AB,\)let \(B'\) be the point on the circle diametrically opposite \(B,\)and let \(X\) be the intersection of \(AO\) and \(MB'.\)Find the length of \(AX.\)

Ayush Rai - 1 year ago

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Ayush Rai - 1 year ago

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You too used a formula for the circumradius of the triangle

Vishwash Kumar - 10 months, 3 weeks ago

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Problem 2 , 3 marks

Prove that \(ab=gcd×lcm\) of \(a\) and \(b\)

Prince Loomba - 1 year ago

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Hello

Meera Somani - 8 months, 3 weeks ago

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do u want something?

Ayush Rai - 8 months, 3 weeks ago

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@Ayush Rai You posted the same question there ... Told told me it is from junio r ramanujan contest.

Rohit Camfar - 10 months, 3 weeks ago

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If you found the circumradius of the triangle then use the property that the centroid of a triangle divides the median ib the ratio two is to one .No need to do such a long calculation.

Vishwash Kumar - 10 months, 3 weeks ago

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Thanks for the suggestion

Ayush Rai - 10 months, 3 weeks ago

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Where are you in slack @Ayush Rai

Vishwathiga Jayasankar - 11 months, 3 weeks ago

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iam banned from slack

Ayush Rai - 11 months, 3 weeks ago

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GOOD BYE . Please forgive me and please do reply to this comment .SORRY. @Ayush Rai.

Vishwathiga Jayasankar - 10 months, 4 weeks ago

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@Vishwathiga Jayasankar Why are you leaving?@vishwathiga jayasankar

Anik Mandal - 10 months, 4 weeks ago

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@Anik Mandal I dunno anik ji. I feel inferior and ignored there. No matter haw hard I try to console myself. I just cant take all anymore. I am not as good in math and physics like you guys. This makes me feel like I have got nothing to do there @Anik Mandal I just dont fit in there. Please do reply.

Vishwathiga Jayasankar - 10 months, 4 weeks ago

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@Vishwathiga Jayasankar Always learn to take the positives.Someone is good in some thing others are good at some other thing...But don't feel demotivated...Start from small steps...Soon you will make giant strides.Believe in yourself!

Anik Mandal - 10 months, 4 weeks ago

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@Vishwathiga Jayasankar @Ayush Rai. do not fail to reply . Just say a goodbye atleast

Vishwathiga Jayasankar - 10 months, 4 weeks ago

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@Vishwathiga Jayasankar Just see me I am good at nothing but I am present there . Hahahah I think you can contribute in biology there. No needto deactivate your account.

Rohit Camfar - 10 months, 3 weeks ago

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@Rohit Camfar hmm. How did you know that I have left slack? @Rohit Camfar

Vishwathiga Jayasankar - 10 months, 3 weeks ago

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@Vishwathiga Jayasankar hmm. How did you forget me. You had asked me to join your team

Rohit Camfar - 10 months, 3 weeks ago

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@Rohit Camfar I know that . I remember you very well. did anyone say to u that I have left?

Vishwathiga Jayasankar - 10 months, 3 weeks ago

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@Vishwathiga Jayasankar Nope I saw your account deactivated

Rohit Camfar - 10 months, 3 weeks ago

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@Rohit Camfar ok @Rohit Camfar good

Vishwathiga Jayasankar - 10 months, 3 weeks ago

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@Vishwathiga Jayasankar What's your age

Rohit Camfar - 10 months, 3 weeks ago

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@Rohit Camfar 14 will b 15 on JAN 7 Why did u ask

Vishwathiga Jayasankar - 10 months, 3 weeks ago

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@Vishwathiga Jayasankar when I saw Vishwathiga Biotechnologist I thought you too be a professional whose age should be between 30 and 40

Rohit Camfar - 10 months, 3 weeks ago

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@Rohit Camfar hahaha no never. It is my dream.(biotech)

Vishwathiga Jayasankar - 10 months, 3 weeks ago

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@Vishwathiga Jayasankar we tried our best rest upon u. But rememberif you leave slack hen not because ur not capabe to keep with physics maths chemistry but because you don't want to. Think from within your heart.

Rohit Camfar - 10 months, 3 weeks ago

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@Rohit Camfar b'bye I am making this discussion slack. Could be scolded by Calvin.

Rohit Camfar - 10 months, 3 weeks ago

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@Rohit Camfar bye

Vishwathiga Jayasankar - 10 months, 3 weeks ago

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@Vishwathiga Jayasankar ok are you leaving brilliant or slack

Rohit Camfar - 10 months, 3 weeks ago

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@Rohit Camfar slack only why?

Vishwathiga Jayasankar - 10 months, 3 weeks ago

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@Vishwathiga Jayasankar Why good?

Rohit Camfar - 10 months, 3 weeks ago

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@Rohit Camfar just like that Anyways goodbye (from slack)

Vishwathiga Jayasankar - 10 months, 3 weeks ago

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@Vishwathiga Jayasankar what? after soooooo much motivation still u are leaving. U are good in biology better than all of us. Why feel inferior? Fight for urself.Hope u change ur mind.

Ayush Rai - 10 months, 4 weeks ago

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@Ayush Rai Its not that. You see there r very few who speak bio there. At times I feel mistreated too @Ayush Rai

Vishwathiga Jayasankar - 10 months, 4 weeks ago

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@Vishwathiga Jayasankar ok then may god bless u and hope u have a bright future.

Ayush Rai - 10 months, 4 weeks ago

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@Ayush Rai What do u mean. Anyways thanks for blessing. I think I need some more time to think over it @Ayush Rai and @Anik Mandal

Vishwathiga Jayasankar - 10 months, 4 weeks ago

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@Vishwathiga Jayasankar are u here @Ayush Rai ???

Vishwathiga Jayasankar - 10 months, 4 weeks ago

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@Ayush Rai Good bye @Ayush Rai and @Anik Mandal till I am ready for slack. THANK U for being good to me.

Vishwathiga Jayasankar - 10 months, 4 weeks ago

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how?

Neel Khare - 11 months, 3 weeks ago

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@Neel Khare its a very long story.cant explain it to u now.preparing for NTSE

Ayush Rai - 11 months, 3 weeks ago

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@Ayush Rai ok so when are u going to be back

Neel Khare - 11 months, 3 weeks ago

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@Neel Khare back from where? iam banned from lounge

Ayush Rai - 11 months, 3 weeks ago

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@Ayush Rai OMG this is not good

will miss you

Neel Khare - 11 months, 3 weeks ago

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@Neel Khare can u give ur email

Neel Khare - 11 months, 3 weeks ago

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@Neel Khare raiayush234@gmail.com

Ayush Rai - 11 months, 3 weeks ago

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why? Do u want ntse qp

Vishwathiga Jayasankar - 11 months, 3 weeks ago

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@Vishwathiga Jayasankar yup.give me.

Ayush Rai - 11 months, 3 weeks ago

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@Ayush Rai how?

Vishwathiga Jayasankar - 11 months, 3 weeks ago

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@Vishwathiga Jayasankar dont leave space. between the brackets

Ayush Rai - 11 months, 3 weeks ago

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@Vishwathiga Jayasankar ![] (link)

Ayush Rai - 11 months, 3 weeks ago

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@Ayush Rai link? Sorry I dont understand

Vishwathiga Jayasankar - 11 months, 3 weeks ago

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@Vishwathiga Jayasankar @Ayush Rai check generaldiscussions.

Vishwathiga Jayasankar - 11 months, 1 week ago

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@Vishwathiga Jayasankar @Ayush Rai check generaldiscussions.

Vishwathiga Jayasankar - 11 months, 1 week ago

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So our winner of this contest-1 is @Kaustubh Miglani.Congratulations.I would also congratulate @Prince Loomba as he was in the lead but eventually had drop out due to some misunderstandings.

Ayush Rai - 1 year ago

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Attention:
Problem 20 is the last problem of this contest.I will soon be conducting contest 2.Thank u for all those who have participated in this contest.

Ayush Rai - 1 year ago

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Problem 20 : [3 marks]
If \(x=\dfrac{p}{q}\) where \(p,q\) are integers having no common divisors other than \(1\),satisfies
\(\sqrt{x+\sqrt{x}}-\sqrt{x-\sqrt{x}}=\dfrac{3}{2}\sqrt{\dfrac{x}{x+\sqrt{x}}}\)

Ayush Rai - 1 year ago

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Multiplying both sides by √(x+√x). Then after transposing and squaring, the result follows

Govind Ramesh - 1 year ago

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good one!

Ayush Rai - 1 year ago

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5/4??

Govind Ramesh - 1 year ago

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I meant 25/16

Govind Ramesh - 1 year ago

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@Govind Ramesh Yes you are correct! Post your solution!

Ayush Rai - 1 year ago

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@Ayush Rai I'm trying to the image is not uploading

Govind Ramesh - 1 year ago

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Comment deleted Oct 20, 2016

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@Ayush Rai Multiplying both sides by √(x+√x). Then after transposing and squaring 2x= √x + 2√(x(x-1)) Squaring, 4x√(x-1) – 3x = 0 x= 1+ 9/25

Govind Ramesh - 1 year ago

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@Govind Ramesh Use imgur for uploading.Check this link to understand better.https://brilliant.org/math-formatting-guide/#

Ayush Rai - 1 year ago

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Attention: Since I dont have questions, and I requested Ayush to post them for me as he was angry as I posted from a book, and he said I was doing all this for points. So I am not participating anymore in this contest

I request Ayush to delete my points

Prince Loomba - 1 year ago

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Problem 18, 3 marks

Find all functions f such that:

\(f (x-y)^{2}=f^{2}(x)-2xf (y)+y^{2}\)

Prince Loomba - 1 year ago

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First put x=y=0 then we have f(0)=((f(0)))^2 Thus f(0)=1 or f(0)=0 Now put x=y but they are not equal to zero then we have f(0)=(f(x))^2 -2x(f(x))+x^2 write rhs as (f(x)-x))^2 So if f(0)=0 Then f(x)=x Or else f(x)-x=1 f(x)-x=-1 Thus possible values of f(x) are x,x+1,x-1 Now put values of f(x) To see which satisfy and u will get the answer

Kaustubh Miglani - 1 year ago

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Right, exact solution!

Prince Loomba - 1 year ago

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f(X)=X I dont think any other Is possible Am I right?

Kaustubh Miglani - 1 year ago

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Nope wrong

Prince Loomba - 1 year ago

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Clarification: its \(f ((x-y)^{2})\)

Prince Loomba - 1 year ago

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Problem 17 : [ 4 marks] Prove that if integer a is not divisible by \(5\) then \(x^5-x-a\) cannot be factorised as product of two nonconstant polynomials with integer coefficients

Kaustubh Miglani - 1 year ago

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Firstly if it has roots, one has to be integral as conjugate complex root pairs are formed and , now to prove no integral root is possible, consider y as a root, take it to be 5k,5k+1..,5k+4

Break given equation...

y (y-1)(y+1)(y^2+1)=a, a is not multiple of 5

Then y=5k is eliminated as first term (y)

y=5k+1 and 5k+4 are eliminated by 2nd and 3rd terms respectively.

y=5k+2 and 5k+3 are eliminated by the last term, forming type 5m. Square them and add 1 to get the result.

Thus no y is possible.

So no real root. So cant be broken into factors, as one has to be linear if it can be broken ( the other has imaginary roots)

Prince Loomba - 1 year ago

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wrong! it is possible that it has a quadratic factor and a cubic factor Pls give more details and explain more breifly

Kaustubh Miglani - 1 year ago

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@Kaustubh Miglani Thats what I said. A real root must be there

Prince Loomba - 1 year ago

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@Prince Loomba Your soln is wrong Suppose it can be factorised as (x2-3x+4)(x3+2) None of which has integer roots It isnt necessary to have an integer root

Kaustubh Miglani - 1 year ago

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@Kaustubh Miglani Real root must be there, as your 2nd term has!

Prince Loomba - 1 year ago

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@Prince Loomba Ok I got it now The soln is correct Though I had another soln in mind

Kaustubh Miglani - 1 year ago

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@Kaustubh Miglani Finally somebody understood haha

Prince Loomba - 1 year ago

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@Prince Loomba Now post prob 18 Be quick pls

Kaustubh Miglani - 1 year ago

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If or if and only if

Prince Loomba - 1 year ago

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i think if is fine,

Kaustubh Miglani - 1 year ago

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Problem 16 : [3 marks]
Given the sum :
\(\dfrac{1}{1^4+1^2+1}+\dfrac{2}{2^4+2^2+1}+\dfrac{3}{3^4+3^2+1}+\dots\)(upto infinity)\(=\dfrac{a}{b}.\)
What is the product \(ab?\)

Ayush Rai - 1 year ago

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n^4+n^2+1=(n^2+n+1)(n^2-n+1) now convert it into telescoping sum to get answer

Kaustubh Miglani - 1 year ago

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LaTeX: \(\begin{equation} \begin{split} S & = \displaystyle \sum^{infty}_{n=1}\frac{n}{n^4+n^2+1} \\ & = \displaystyle \sum^{infty}_{n=1} \frac{n}{(n^2+1+n)(n^2+1-n)} \\ & = \frac{1}{2} \displaystyle \sum^{infty}_{n=1} \left(\frac{1}{n^2+1-n} - \frac{1}{n^2+1+n}\right) \\ & = \frac{1}{2} \left(1-\frac{1}{3}+\frac{1}{3} - \frac{1}{7} + \frac{1}{7} -\frac{1}{13} + \dots =1/2\)

Kaustubh Miglani - 1 year ago

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2? a=1 b=2

Kaustubh Miglani - 1 year ago

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Post your solution

Ayush Rai - 1 year ago

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the area on which axis the x or the y

Abhishek Alva - 1 year ago

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I think graph itself is bounded as rectangles

Prince Loomba - 1 year ago

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Problem 16:[1 marks] Find area of the graph [|(3x+4y)/5|]+[|4y-3x|/5]=3. Where [ ] denote greatest integer.

Archit Agrawal - 1 year ago

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Attention:The actual problem 16 is not this.

Ayush Rai - 1 year ago

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Even this does not come under Olypmpiad problems.Please delete this question and post another OLYMPIAD problem.

Ayush Rai - 1 year ago

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Archit wont be more in the contest I suppose because of your foolish attentions and such rules

Prince Loomba - 1 year ago

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496

Navneet Prabhat - 1 year ago

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No

Archit Agrawal - 1 year ago

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Problem 15:[1 marks]

Find the number of solutions to \[\begin{align} x+y+z&\geq150 \\ 0\leq x,y,z &\leq 60 \end{align}\]

Archit Agrawal - 1 year ago

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From above condition we can get (60-x)+(60-y)+(60-z)<=30 Then introduce a dummy variable w and change the equation to (60-x)+(60-y)+(60-z)+w=30. Counting these cases is equal to counting the ways of distributing 30 apples in 4 people such that anyone can get any number of apples which are equal to 33C3.

Archit Agrawal - 1 year ago

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I am feeling too lazy to write the whole solution and even finding the correct answer. I will only share the idea that I had in mind. This question actually boils down to finding coefficient of \(m^k\) (\(k=x+y+z\)) in the expansion of : \((1+m+m^2+m^3+\dots m^60)^3=(1-m^{61})^3(1-m)^{-3}\)

I hope from here on its quite simple!

Miraj Shah - 1 year ago

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There is a short soln also.

Archit Agrawal - 1 year ago

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Comment deleted Oct 19, 2016

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No

Archit Agrawal - 1 year ago

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@Archit Agrawal can we repeat the numbers

Abhinav Jha - 1 year ago

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@Abhinav Jha Yes

Archit Agrawal - 1 year ago

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x,y,z are integers.

Archit Agrawal - 1 year ago

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Are x,y,z integers?

Ayush Rai - 1 year ago

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They should be

Govind Ramesh - 1 year ago

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Problem 14 : [3 marks]
\(\dfrac{3}{1^2.2^2}+\dfrac{5}{2^2.3^2}+\dots+\dfrac{29}{14^2.15^2}=\dfrac{m}{n}.\)Find the value of \(m+n.\)

Ayush Rai - 1 year ago

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449

Archit Agrawal - 1 year ago

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You are absolutely correct.Give your solution and then post the next problem.

Ayush Rai - 1 year ago

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@Ayush Rai It's general term is (2n+1)/n^2(n+1)^2 =((n+1)^2-n^2)/n^2(n+1)^2 =1/n^2-1/(n+1)^2 =1-1/15^2 =224/225

Archit Agrawal - 1 year ago

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@Archit Agrawal correct solution..+1 post next problem with the points mentioned

Ayush Rai - 1 year ago

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Problem 13 : [3 marks]
If \(a=\dfrac{b+c}{x-2},b=\dfrac{c+a}{y-2},c=\dfrac{a+b}{z-2},xy+yz+xz=67\) and \(x+y+z=2010.\)What is the value of \(-xyz?\)

Ayush Rai - 1 year ago

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First get that x-2=b+c/a -->x-1=(a+b+c)/a --->1/(X-1)=a/(a+b+c) Now get eqns for 1/(y-1) ,1/(z-1) Their sum is 1 Thus(x-1)(y-1)+(y-1)(z-1)+(x-1)(z-1)=(x-1)(y-1)(z-1 Now expand and put values to get answer

Kaustubh Miglani - 1 year ago

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Good one! Now post the next problem.

Ayush Rai - 1 year ago

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5892? Might be a calculating error Am I right?

Kaustubh Miglani - 1 year ago

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Yes you are correct.Now u must write the solution for it!

Ayush Rai - 1 year ago

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PROBLEM 12 In an isosceles triangle ABC (AB = BC), N is the midpoint of the median BM and MD is perpendicular to CN (see the figure below). Prove that the angles BAD and ACN are equal.(2 MARKS)

Abhishek Alva - 1 year ago

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Where's the figure??

Harsh Shrivastava - 1 year ago

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Well sorry about that.Can u manage without that?

Ayush Rai - 1 year ago

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Let the polynomial \(p(x)=a_5x^5+a_4x^4+a_3x^3+a_2x^2+a_1x+a_0. \)
Then \((x^2-x+1)|p(x)\Rightarrow p(-w)=0.\)
So,\(a_5(1+w)+a_4(w)-a_3-a_2(1+w)-a_1w+a_0=0. \)
\(a_5-a_3-a_2+a_0=0\Rightarrow a_5+a_0=a_2+a_3 \)
\(a_5+a_4=a_2+a_1.\)Now we have to brute force according the condition.Well i just gave an idea.I am not so interested in combi as such.So i dont know the answer.

Ayush Rai - 1 year ago

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Bro 6 hrs passed i suppose

Prince Loomba - 1 year ago

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Along the same lines I had a different idea: \(p(\omega) = -a_{5}\omega ^2 + a_4 \omega - a_3 +a_2 \omega ^2 - a_1\omega + a_0 = 0\)

Therefore,

\((a_2 - a_5)\omega ^2 + (a_4 - a_1)\omega + (a_0 - a_3) = 0\)

Hence we get,

\(a_2 - a_5 = a_4-a_1=a_0-a_3\)

And similarly for \(p(\omega ^2)\)

Miraj Shah - 1 year ago

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Problem 11, 5 marks

Find the number of polynomials of degree 5 with distinct coeffecients from the set {1,2,3,4,...,9} that are divisible by \(x^{2}-x+1\)

Prince Loomba - 1 year ago

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ANSWER IS 636

Nobody got right. @Ayush Rai give my points to me.

Prince Loomba - 1 year ago

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Nobody gets the points as i have not mentioned it in the rules.You must also give the solution.

Ayush Rai - 1 year ago

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@Ayush Rai

Prince Loomba - 1 year ago

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@Prince Loomba which book is this?

Neel Khare - 11 months, 3 weeks ago

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@Neel Khare Its FIITJEE 2012 RMO practice book

Prince Loomba - 11 months, 3 weeks ago

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@Prince Loomba u go to fiit jee

Neel Khare - 11 months, 3 weeks ago

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@Neel Khare No allen. My friend gave it to me.

Prince Loomba - 11 months, 3 weeks ago

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@Prince Loomba oh is it good?

Neel Khare - 11 months, 3 weeks ago

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@Neel Khare Very good. Talk in slack. Here its difficult to reply multiple times

Prince Loomba - 11 months, 3 weeks ago

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@Prince Loomba the answer is 288

Abhishek Alva - 1 year ago

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As the problem is too hard to solve and going against rule 3,we are continuing with a new problem.

Ayush Rai - 1 year ago

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It not that tough. I will upload solution in night. Its actually just simple forming of equation and then permutation-combination

Prince Loomba - 1 year ago

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1 hr and 15 minutes have passed

Prince Loomba - 1 year ago

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It will be mid night after 6 hours :p

Anik Mandal - 1 year ago

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@Anik Mandal See carefully the note again ;p

Prince Loomba - 1 year ago

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ab=bc=ca=8 root 3

Abhishek Alva - 1 year ago

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Wrong

Prince Loomba - 1 year ago

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Wrong answer!

Ayush Rai - 1 year ago

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Points?

Prince Loomba - 1 year ago

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Problem 10 : 2 marks
In a \(\triangle ABC,\)the incircle touches the sides \(BC,CA\) and \(AB\) respectively at \(D,E\) and \(F.\)If the radius of the incircle is \(4\) units and if \(BD,CE\) and \(AF\) are consecutive integers,find the sides of the \(\triangle ABC.\)

Ayush Rai - 1 year ago

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Prince Loomba - 1 year ago

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Another approach could have been by using the fact that in a triangle \(tan\frac{A}{2} tan\frac{B}{2} +tan\frac{B}{2}tan\frac{C}{2} +tan\frac{C}{2} tan\frac{A}{2}= 1\)

Miraj Shah - 1 year ago

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Correct answer! Post next problem

Ayush Rai - 1 year ago

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@Ayush Rai what did u do

Abhishek Alva - 1 year ago

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which question

Abhishek Alva - 1 year ago

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Problem 9 : 3 marks
What is the distance between the incenter and circumcenter of the triangle with sides \(13,14\) and \(15?\) Give your answer to three decimal places.

Ayush Rai - 1 year ago

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Comment deleted Oct 18, 2016

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@Brilliant Member You have to delete your solution as there are several hints in it.

Ayush Rai - 1 year ago

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Comment deleted Oct 18, 2016

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@Brilliant Member Your circum-Radius is wrong.Its too high.

Ayush Rai - 1 year ago

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@Ayush Rai Oh yes you are right. I entered it wrong into my calculator. The Circumradius is 8.125 hence the distance is approximately 1.008.

Brilliant Member - 1 year ago

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@Brilliant Member Actually you are going against the rules.But still i give u the points.Post a new problem.No problem of the comments.

Ayush Rai - 1 year ago

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Comment deleted Oct 18, 2016

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@Brilliant Member You have to delete your solution as there are several hints in it.

Ayush Rai - 1 year ago

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@Ayush Rai I posted exact solution acc to rules

Prince Loomba - 1 year ago

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@Ayush Rai Give points to me

Prince Loomba - 1 year ago

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@Brilliant Member You cannot give any more solutions

Ayush Rai - 1 year ago

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@Brilliant Member Sorry you're answer is wrong.

Ayush Rai - 1 year ago

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Circumradius=abc/4delta=65/8.

OI=SQRT (R (R-2r))=1.008 approximately

Prince Loomba - 1 year ago

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the marks for the question is 3 marks

Abhishek Alva - 1 year ago

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the questio is 3 marks

Abhishek Alva - 1 year ago

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8(a^4+b^4)greater than or equal to (a+b)^2 if a, b, are positive numbers

Abhishek Alva - 1 year ago

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sorry guyes the question is wrong

Abhishek Alva - 1 year ago

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Let a and b be positive real numbers. Prove that8a^4+b^4≥(a+b)^4 .

Abhishek Alva - 1 year ago

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\(8(a^4+b^4) \ge 4(a^2+b^2)^2 \ge 4 \left(\dfrac{(a+b)^2}{2} \right)^2=(a+b)^4\). Used Holder and QM-AM inequality

Brilliant Member - 1 year ago

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I don't have any problems currently. Anybody can post the next problem.

Brilliant Member - 1 year ago

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Post the next problem

Ayush Rai - 1 year ago

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ya you are right

Abhishek Alva - 1 year ago

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Comment deleted Oct 18, 2016

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delete other comments

Ayush Rai - 1 year ago

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what is all this left and right

Abhishek Alva - 1 year ago

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use latex

Ayush Rai - 1 year ago

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Its \geq

Prince Loomba - 1 year ago

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For a right-angled triangle \(a^2+b^2=c^2\) where c is the largest side.But here they have asked for acute angled triangle by the angle opposite to \(c\) becomes smaller and smaller.So the \(a^2+b^2>c^2.\)

Ayush Rai - 1 year ago

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for a right angled triangle a^2+b^2=c^2 where c is the largest side. the equality holds even for the angles that is 90 =90 now for an acute angled triangle , when we decrease the 90 degree angle the other degrees will be added to the other two angles ,thus the equality breaks and would point towards a and b .

Abhishek Alva - 1 year ago

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you post i dont have any problem

Prakhar Bindal - 1 year ago

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Right you explained both sides.

Prince Loomba - 1 year ago

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good try

Abhishek Alva - 1 year ago

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See 10th rule gphe cant try now haha 😂😂

Prince Loomba - 1 year ago

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@Prince Loomba He*

Prince Loomba - 1 year ago

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Wrong reasoning

Prince Loomba - 1 year ago

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Problem 6 : 2 marks
If \(\alpha,\beta,\gamma\) be the roots of \(x^3+2x^2-3x-1=0.\)Find the value of \(\dfrac{1}{\alpha^3}+\dfrac{1}{\beta^3}+\dfrac{1}{\gamma^3}.\)

Ayush Rai - 1 year ago

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Make a polynomial with 1/alpha,1/beta,1/gamma as roots.

We have to find A^3+B^3+C^3=3ABC+(A+B+C)((A+B+C)^2-3(AB+BC+CA). from the new polynomial we know each of the following term.

Answer is -42

Prince Loomba - 1 year ago

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correct solution.Post next problem.

Ayush Rai - 1 year ago

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No, she can´t. Cinderella, on her turn, empties the two neighboring buckets with maximum sum of water. Assume the stepmother wins. She wins if at some point three or more buckets have more than one liter of water each. That happens if in her turn there are three buckets that have two liters of water between the three of them. Obviously this can`t happen in the first turn. Since Cinderella always empties the two neighboring buckets with maximum sum of water, Cinderella always empties at least \(2/5\) of the total amount of water. Therefore, before Cinderella emptied her buckets, there had to be at least \(2:3/5=10/3\) liters of water.Assume this was the first time there was at least \(10/3\) liters. Then on the previous turn, the stepmother added \(1\) liter, so before the stepmother played there were \(7/3\) liters. But the same calculation gives that the turn before there were at least \(7/3:3/5=35/9\) liters. But \(35/9>10/3.\)

Ayush Rai - 1 year ago

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Post next

Prince Loomba - 1 year ago

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Problem 4: 3 marks Prove that the inradius of a right angled triangle with integer sides is an integer.

Ayush Rai - 1 year ago

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c is hypotenuse=x^2+y^2

Prince Loomba - 1 year ago

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r=(a+b-c)/2. Take general pythagorean triplet, x^2+y^2,x^2-y^2,2xy to get the result.

Prince Loomba - 1 year ago

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Correct solution! Post next problem

Ayush Rai - 1 year ago

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Problem 3: 2 marks
Solve the equations for \(x,y\) and \(z.\)
\(x+y+z=9\)
\(x^2+y^2+z^2=29\)
\(x^3+y^3+z^3=99\)

Ayush Rai - 1 year ago

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2,3,4 in any order. Solved by using two new terms, xy+yz+zx and xyz and then using newton's identities

Prince Loomba - 1 year ago

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Correct answer!Post next problem

Ayush Rai - 1 year ago

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[Problem 2, 3 marks]

Prove that \(ab=gcd (a,b)×lcm (a,b)\)

Prince Loomba - 1 year ago

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Case 1:When \(a\) and \(b\) are co-prime to each other
Since \(a\) and \(b\) are co-prime,the G.C.D. is \(1.\)The \(L.C.M.\) will be \(a\times b\) because \(a\) and \(b\) should divide the \(L.C.M.\)So since \(a\) and \(b\) do not have any common factors,the \(L.C.M\) should be the product of \(a\) and \(b.\)Therefore \(ab=1\times(a\times b).ab=ab.\)Hence proved.
Case 2:When a and b are not co-prime to each other
Let the \(G.C.D\) be \(k\) of \(a\) and \(b.\)Well,\(a\times b\) will contain \(k\) since it is the common divisor of \(a\) and \(b.\)So let \(a\times b=k\times m.\)Since \(k\) is already contained in \(a\) and \(b,\)the L.C.M will have to remaining part[that is \(m\)] for it to divisible by both \(a\) and \(b.\)So,\(a\times b=gcd(a,b)\times lcm(a,b)=k\times m=a\times b.\)Hence proved.

Ayush Rai - 1 year ago

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i guess this is the mathematical proof

Abhishek Alva - 1 year ago

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the answer lies in the defination of hcf and lcm.we can represent it as a venn daigram.the elements which are perfectly common are taken out which is the hcf and what remains behind are the elements which not perfectly common which is the lcm . when we do the union of the set hcf and lcm we get the original set which the union of the 2 numbers

Abhishek Alva - 1 year ago

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if u are still not convienced then lcm and hcf are like we are taking some elements from the product of the 2 set such that the all the elements are utilized.so when we unite both the lcm and the hcf we get the number back.the proof of this is the defination itself

Abhishek Alva - 1 year ago

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@Abhishek Alva Yes you were writing exact answer, but in a theoritical way. So I can award only 2 points to you. And Ayush gets 3 points

Prince Loomba - 1 year ago

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Wrong mathematical proof

Prince Loomba - 1 year ago

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@Prince Loomba i can represent it using venn daigrams or sets.

Abhishek Alva - 1 year ago

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Problem 1:[3 marks]


\(ABC\) and \(DBC\) are two equilateral triangles on the same base \(BC\).A point \(P\) is taken on the circle with center \(D\) and radius as \(BD\).Show that \(PA,PB\) and \(PC\) are the sides of a right triangle where \(P\) does not coincide with the points \(B\) and \(C\) on the circle.

Ayush Rai - 1 year ago

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Prince Loomba - 1 year ago

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Good solution..+1,you are correct.you can proceed with the next problem

Ayush Rai - 1 year ago

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here in the daigram it ab need not be the tangent

Abhishek Alva - 1 year ago

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Actually not need not, its CAN not as angle ABD is 120 degree and for tangent angle is 90 degree!

Prince Loomba - 1 year ago

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@Prince Loomba yes u are right

Ayush Rai - 1 year ago

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Comment deleted Oct 18, 2016

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4 polynomials?

Kaustubh Miglani - 1 year ago

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Nope much greater than that haha

Prince Loomba - 1 year ago

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@Prince Loomba 9^5 ?

Anik Mandal - 1 year ago

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@Anik Mandal Obviously not...

Prince Loomba - 1 year ago

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@Prince Loomba yah i was wrong!

Anik Mandal - 1 year ago

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Problem 5(4 marks)

Five identical empty buckets of 2-liter capacity stand at the vertices of a regular pentagon. Cinderella and her wicked Stepmother go through a sequence of rounds: At the beginning of every round, the Stepmother takes one liter of water from the nearby river and distributes it arbitrarily over the five buckets. Then Cinderella chooses a pair of neighboring buckets, empties them into the river, and puts them back. Then the next round begins. The Stepmother’s goal is to make one of these buckets overflow. Cinderella’s goal is to prevent this. Can the wicked Stepmother enforce a bucket overflow?

Prince Loomba - 1 year ago

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i think no

Kaustubh Miglani - 1 year ago

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solution since abc dbc are equilateral triangles with the same base there is only one possibility that is both the triangles super impose on each other .so A =D.we know that BD=CD because equilateral triangle and BD is the radius so CD is also the radius. choose a point p ,such that it is perpendicular to AC or rather it is the extention of AC radius. now even PA also be the radius as the point p is on the circumference and a is the centre. join PB. since they are radius AC= BA and PA=AB (radius)since we have choosen the point p as the extention of AC and they are perpendicular .in triangle PBD PD=BD, and PAD =90 . so it is right angled triangle. hence proved.

Abhishek Alva - 1 year ago

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You are wrong.The points A and D do not coincide..

Ayush Rai - 1 year ago

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Problem 7 (1 mark)

Prove that for any acute triangle, \(a^{2}+b^{2}>c^{2}\), where c is the longest side.

Prince Loomba - 1 year ago

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Directly use cosine rule and put cos>0 so simple! :)

Prakhar Bindal - 1 year ago

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But you And abhishek posted simultaneously

Prince Loomba - 1 year ago

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Right

Prince Loomba - 1 year ago

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