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:

- Suppose problem \(i\) is posted. The person who solves problem \(i\) and posts the solution can publish problem \(i+1\).
- 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.
- I request that problem-posters contribute problems that are solvable and do not post problems that will demotivate others.
- The contest will end once 20 problems have been posed and answered (problems not solved within 6 hours don't count toward this).
- Problem posters should know the solution of their posted problem in advance.
- If the new problem is not posted in 15 minutes of answering question \(i\), ANYBODY can post question \(i+1\).
- 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.
- I request all entries to be honest and fair. Do not check for the solution on the internet or copy-paste the solution.
- I would also request that you re-share this note so that other members can learn about and join this contest.
- 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.
- 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

## Comments

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TopNewestProblem 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 · 9 months ago

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– Vishwash Kumar · 7 months, 3 weeks ago

You too used a formula for the circumradius of the triangleLog in to reply

Problem 2 , 3 marksProve that \(ab=gcd×lcm\) of \(a\) and \(b\) – Prince Loomba · 9 months, 1 week ago

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Hello – Meera Somani · 5 months, 3 weeks ago

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– Ayush Rai · 5 months, 3 weeks ago

do u want something?Log in to reply

@Ayush Rai You posted the same question there ... Told told me it is from junio r ramanujan contest. – Rohit Camfar · 7 months, 3 weeks ago

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– Vishwash Kumar · 7 months, 3 weeks ago

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.Log in to reply

– Ayush Rai · 7 months, 3 weeks ago

Thanks for the suggestionLog in to reply

Where are you in slack @Ayush Rai – Vishwathiga Jayasankar · 8 months, 2 weeks ago

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– Ayush Rai · 8 months, 2 weeks ago

iam banned from slackLog in to reply

– Vishwathiga Jayasankar · 7 months, 4 weeks ago

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

@vishwathiga jayasankar – Anik Mandal · 7 months, 4 weeks ago

Why are you leaving?Log in to reply

@Anik Mandal I just dont fit in there. Please do reply. – Vishwathiga Jayasankar · 7 months, 4 weeks ago

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 thereLog in to reply

– Anik Mandal · 7 months, 4 weeks ago

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!Log in to reply

@Ayush Rai. do not fail to reply . Just say a goodbye atleast – Vishwathiga Jayasankar · 7 months, 4 weeks ago

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– Rohit Camfar · 7 months, 3 weeks ago

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.Log in to reply

@Rohit Camfar – Vishwathiga Jayasankar · 7 months, 3 weeks ago

hmm. How did you know that I have left slack?Log in to reply

– Rohit Camfar · 7 months, 3 weeks ago

hmm. How did you forget me. You had asked me to join your teamLog in to reply

– Vishwathiga Jayasankar · 7 months, 3 weeks ago

I know that . I remember you very well. did anyone say to u that I have left?Log in to reply

– Rohit Camfar · 7 months, 3 weeks ago

Nope I saw your account deactivatedLog in to reply

@Rohit Camfar good – Vishwathiga Jayasankar · 7 months, 3 weeks ago

okLog in to reply

– Rohit Camfar · 7 months, 3 weeks ago

What's your ageLog in to reply

– Vishwathiga Jayasankar · 7 months, 3 weeks ago

14 will b 15 on JAN 7 Why did u askLog in to reply

– Rohit Camfar · 7 months, 3 weeks ago

when I saw Vishwathiga Biotechnologist I thought you too be a professional whose age should be between 30 and 40Log in to reply

– Vishwathiga Jayasankar · 7 months, 3 weeks ago

hahaha no never. It is my dream.(biotech)Log in to reply

– Rohit Camfar · 7 months, 3 weeks ago

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.Log in to reply

– Rohit Camfar · 7 months, 3 weeks ago

b'bye I am making this discussion slack. Could be scolded by Calvin.Log in to reply

– Vishwathiga Jayasankar · 7 months, 3 weeks ago

byeLog in to reply

– Rohit Camfar · 7 months, 3 weeks ago

ok are you leaving brilliant or slackLog in to reply

– Vishwathiga Jayasankar · 7 months, 3 weeks ago

slack only why?Log in to reply

– Rohit Camfar · 7 months, 3 weeks ago

Why good?Log in to reply

– Vishwathiga Jayasankar · 7 months, 3 weeks ago

just like that Anyways goodbye (from slack)Log in to reply

– Ayush Rai · 7 months, 4 weeks ago

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.Log in to reply

@Ayush Rai – Vishwathiga Jayasankar · 7 months, 4 weeks ago

Its not that. You see there r very few who speak bio there. At times I feel mistreated tooLog in to reply

– Ayush Rai · 7 months, 3 weeks ago

ok then may god bless u and hope u have a bright future.Log in to reply

@Ayush Rai and @Anik Mandal – Vishwathiga Jayasankar · 7 months, 3 weeks ago

What do u mean. Anyways thanks for blessing. I think I need some more time to think over itLog in to reply

@Ayush Rai ??? – Vishwathiga Jayasankar · 7 months, 4 weeks ago

are u hereLog in to reply

@Ayush Rai and @Anik Mandal till I am ready for slack. THANK U for being good to me. – Vishwathiga Jayasankar · 7 months, 3 weeks ago

Good byeLog in to reply

– Neel Khare · 8 months, 2 weeks ago

how?Log in to reply

– Ayush Rai · 8 months, 2 weeks ago

its a very long story.cant explain it to u now.preparing for NTSELog in to reply

– Neel Khare · 8 months, 2 weeks ago

ok so when are u going to be backLog in to reply

– Ayush Rai · 8 months, 2 weeks ago

back from where? iam banned from loungeLog in to reply

will miss you – Neel Khare · 8 months, 2 weeks ago

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– Neel Khare · 8 months, 2 weeks ago

can u give ur emailLog in to reply

– Ayush Rai · 8 months, 2 weeks ago

raiayush234@gmail.comLog in to reply

– Vishwathiga Jayasankar · 8 months, 2 weeks ago

why? Do u want ntse qpLog in to reply

– Ayush Rai · 8 months, 2 weeks ago

yup.give me.Log in to reply

– Vishwathiga Jayasankar · 8 months, 2 weeks ago

how?Log in to reply

– Ayush Rai · 8 months, 2 weeks ago

dont leave space. between the bracketsLog in to reply

– Ayush Rai · 8 months, 2 weeks ago

![] (link)Log in to reply

– Vishwathiga Jayasankar · 8 months, 2 weeks ago

link? Sorry I dont understandLog in to reply

@Ayush Rai check generaldiscussions. – Vishwathiga Jayasankar · 8 months ago

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@Ayush Rai check generaldiscussions. – Vishwathiga Jayasankar · 8 months 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 · 9 months 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 · 9 months 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 · 9 months ago

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– Govind Ramesh · 9 months ago

Multiplying both sides by √(x+√x). Then after transposing and squaring, the result followsLog in to reply

– Ayush Rai · 9 months ago

good one!Log in to reply

– Govind Ramesh · 9 months ago

5/4??Log in to reply

– Govind Ramesh · 9 months ago

I meant 25/16Log in to reply

– Ayush Rai · 9 months ago

Yes you are correct! Post your solution!Log in to reply

– Govind Ramesh · 9 months ago

I'm trying to the image is not uploadingLog in to reply

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– Govind Ramesh · 9 months ago

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+ 9/25Log in to reply

– Ayush Rai · 9 months ago

Use imgur for uploading.Check this link to understand better.https://brilliant.org/math-formatting-guide/#Log in to reply

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 contestI request Ayush to

delete my points– Prince Loomba · 9 months agoLog in to reply

Problem 18, 3 marksFind all functions f such that:

\(f (x-y)^{2}=f^{2}(x)-2xf (y)+y^{2}\) – Prince Loomba · 9 months ago

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– Kaustubh Miglani · 9 months ago

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 answerLog in to reply

– Prince Loomba · 9 months ago

Right, exact solution!Log in to reply

– Kaustubh Miglani · 9 months ago

f(X)=X I dont think any other Is possible Am I right?Log in to reply

– Prince Loomba · 9 months ago

Nope wrongLog in to reply

– Prince Loomba · 9 months ago

Clarification: its \(f ((x-y)^{2})\)Log in to reply

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 · 9 months agoLog in to reply

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 · 9 months ago

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– Kaustubh Miglani · 9 months ago

wrong! it is possible that it has a quadratic factor and a cubic factor Pls give more details and explain more breiflyLog in to reply

– Prince Loomba · 9 months ago

Thats what I said. A real root must be thereLog in to reply

– Kaustubh Miglani · 9 months ago

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 rootLog in to reply

– Prince Loomba · 9 months ago

Real root must be there, as your 2nd term has!Log in to reply

– Kaustubh Miglani · 9 months ago

Ok I got it now The soln is correct Though I had another soln in mindLog in to reply

– Prince Loomba · 9 months ago

Finally somebody understood hahaLog in to reply

– Kaustubh Miglani · 9 months ago

Now post prob 18 Be quick plsLog in to reply

– Prince Loomba · 9 months ago

If or if and only ifLog in to reply

– Kaustubh Miglani · 9 months ago

i think if is fine,Log in to reply

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 · 9 months ago

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– Kaustubh Miglani · 9 months ago

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

– Kaustubh Miglani · 9 months ago

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

– Kaustubh Miglani · 9 months ago

2? a=1 b=2Log in to reply

– Ayush Rai · 9 months ago

Post your solutionLog in to reply

the area on which axis the x or the y – Abhishek Alva · 9 months ago

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– Prince Loomba · 9 months ago

I think graph itself is bounded as rectanglesLog in to reply

Problem 16:[1 marks] Find area of the graph [|(3x+4y)/5|]+[|4y-3x|/5]=3. Where [ ] denote greatest integer. – Archit Agrawal · 9 months ago

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Attention:The actual problem 16 is not this. – Ayush Rai · 9 months agoLog in to reply

– Ayush Rai · 9 months ago

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

attentionsand such rules – Prince Loomba · 9 months agoLog in to reply

496 – Navneet Prabhat · 9 months, 1 week ago

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– Archit Agrawal · 9 months ago

NoLog in to reply

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 · 9 months, 1 week ago

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– Archit Agrawal · 9 months ago

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.Log in to reply

I hope from here on its quite simple! – Miraj Shah · 9 months, 1 week ago

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– Archit Agrawal · 9 months ago

There is a short soln also.Log in to reply

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– Archit Agrawal · 9 months, 1 week ago

NoLog in to reply

– Abhinav Jha · 9 months, 1 week ago

can we repeat the numbersLog in to reply

– Archit Agrawal · 9 months ago

YesLog in to reply

– Archit Agrawal · 9 months, 1 week ago

x,y,z are integers.Log in to reply

– Ayush Rai · 9 months, 1 week ago

Are x,y,z integers?Log in to reply

– Govind Ramesh · 9 months, 1 week ago

They should beLog in to reply

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 · 9 months, 1 week ago

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– Archit Agrawal · 9 months, 1 week ago

449Log in to reply

– Ayush Rai · 9 months, 1 week ago

You are absolutely correct.Give your solution and then post the next problem.Log in to reply

(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 · 9 months, 1 week agoLog in to reply

– Ayush Rai · 9 months, 1 week ago

correct solution..+1 post next problem with the points mentionedLog in to reply

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 · 9 months, 1 week ago

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– Kaustubh Miglani · 9 months, 1 week ago

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 answerLog in to reply

– Ayush Rai · 9 months, 1 week ago

Good one! Now post the next problem.Log in to reply

– Kaustubh Miglani · 9 months, 1 week ago

5892? Might be a calculating error Am I right?Log in to reply

– Ayush Rai · 9 months, 1 week ago

Yes you are correct.Now u must write the solution for it!Log in to reply

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 · 9 months, 1 week ago

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– Harsh Shrivastava · 9 months, 1 week ago

Where's the figure??Log in to reply

– Ayush Rai · 9 months, 1 week ago

Well sorry about that.Can u manage without that?Log in to reply

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 · 9 months, 1 week ago

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– Prince Loomba · 9 months, 1 week ago

Bro 6 hrs passed i supposeLog in to reply

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 · 9 months, 1 week ago

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Problem 11, 5 marksFind 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 · 9 months, 1 week ago

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Nobody got right. @Ayush Rai give my points to me. – Prince Loomba · 9 months, 1 week ago

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– Ayush Rai · 9 months, 1 week ago

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

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– Neel Khare · 8 months, 3 weeks ago

which book is this?Log in to reply

– Prince Loomba · 8 months, 3 weeks ago

Its FIITJEE 2012 RMO practice bookLog in to reply

– Neel Khare · 8 months, 3 weeks ago

u go to fiit jeeLog in to reply

– Prince Loomba · 8 months, 3 weeks ago

No allen. My friend gave it to me.Log in to reply

– Neel Khare · 8 months, 3 weeks ago

oh is it good?Log in to reply

– Prince Loomba · 8 months, 3 weeks ago

Very good. Talk in slack. Here its difficult to reply multiple timesLog in to reply

@Prince Loomba the answer is 288 – Abhishek Alva · 9 months, 1 week ago

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– Ayush Rai · 9 months, 1 week ago

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

– Prince Loomba · 9 months, 1 week ago

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

– Prince Loomba · 9 months, 1 week ago

1 hr and 15 minutes have passedLog in to reply

– Anik Mandal · 9 months, 1 week ago

It will be mid night after 6 hours :pLog in to reply

– Prince Loomba · 9 months, 1 week ago

See carefully the note again ;pLog in to reply

ab=bc=ca=8 root 3 – Abhishek Alva · 9 months, 1 week ago

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– Prince Loomba · 9 months, 1 week ago

WrongLog in to reply

– Ayush Rai · 9 months, 1 week ago

Wrong answer!Log in to reply

– Prince Loomba · 9 months, 1 week ago

Points?Log in to reply

Problem 10 : 2 marksIn 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 · 9 months, 1 week ago

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– Miraj Shah · 9 months, 1 week ago

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

– Ayush Rai · 9 months, 1 week ago

Correct answer! Post next problemLog in to reply

– Abhishek Alva · 9 months, 1 week ago

what did u doLog in to reply

which question – Abhishek Alva · 9 months, 1 week ago

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Problem 9 : 3 marksWhat 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 · 9 months, 1 week ago

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– Ayush Rai · 9 months, 1 week ago

You have to delete your solution as there are several hints in it.Log in to reply

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– Ayush Rai · 9 months, 1 week ago

Your circum-Radius is wrong.Its too high.Log in to reply

– Svatejas Shivakumar · 9 months, 1 week ago

Oh yes you are right. I entered it wrong into my calculator. The Circumradius is 8.125 hence the distance is approximately 1.008.Log in to reply

– Ayush Rai · 9 months, 1 week ago

Actually you are going against the rules.But still i give u the points.Post a new problem.No problem of the comments.Log in to reply

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@Svatejas Shivakumar – Ayush Rai · 9 months, 1 week ago

Ok.Log in to reply

– Ayush Rai · 9 months, 1 week ago

You have to delete your solution as there are several hints in it.Log in to reply

– Prince Loomba · 9 months, 1 week ago

I posted exact solution acc to rulesLog in to reply

– Prince Loomba · 9 months, 1 week ago

Give points to meLog in to reply

– Ayush Rai · 9 months, 1 week ago

You cannot give any more solutionsLog in to reply

– Ayush Rai · 9 months, 1 week ago

Sorry you're answer is wrong.Log in to reply

OI=SQRT (R (R-2r))=1.008 approximately – Prince Loomba · 9 months, 1 week ago

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the marks for the question is 3 marks – Abhishek Alva · 9 months, 1 week ago

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the questio is 3 marks – Abhishek Alva · 9 months, 1 week ago

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– Abhishek Alva · 9 months, 1 week ago

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

– Abhishek Alva · 9 months, 1 week ago

sorry guyes the question is wrongLog in to reply

Let a and b be positive real numbers. Prove that8a^4+b^4≥(a+b)^4 . – Abhishek Alva · 9 months, 1 week ago

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– Svatejas Shivakumar · 9 months, 1 week ago

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

– Svatejas Shivakumar · 9 months, 1 week ago

I don't have any problems currently. Anybody can post the next problem.Log in to reply

– Ayush Rai · 9 months, 1 week ago

Post the next problemLog in to reply

– Abhishek Alva · 9 months, 1 week ago

ya you are rightLog in to reply

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– Ayush Rai · 9 months, 1 week ago

delete other commentsLog in to reply

– Abhishek Alva · 9 months, 1 week ago

what is all this left and rightLog in to reply

– Ayush Rai · 9 months, 1 week ago

use latexLog in to reply

– Prince Loomba · 9 months, 1 week ago

Its \geqLog in to reply

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 · 9 months, 1 week ago

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– Abhishek Alva · 9 months, 1 week ago

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 .Log in to reply

– Prakhar Bindal · 9 months, 1 week ago

you post i dont have any problemLog in to reply

– Prince Loomba · 9 months, 1 week ago

Right you explained both sides.Log in to reply

– Abhishek Alva · 9 months, 1 week ago

good tryLog in to reply

– Prince Loomba · 9 months, 1 week ago

See 10th rule gphe cant try now haha 😂😂Log in to reply

– Prince Loomba · 9 months, 1 week ago

He*Log in to reply

– Prince Loomba · 9 months, 1 week ago

Wrong reasoningLog in to reply

Problem 6 : 2 marksIf \(\alpha,\beta,\gamma\) 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 · 9 months, 1 week ago

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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 · 9 months, 1 week ago

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– Ayush Rai · 9 months, 1 week ago

correct solution.Post next problem.Log in to reply

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 · 9 months, 1 week ago

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– Prince Loomba · 9 months, 1 week ago

Post nextLog in to reply

Problem 4: 3 marksProve that the inradius of a right angled triangle with integer sides is an integer. – Ayush Rai · 9 months, 1 week agoLog in to reply

– Prince Loomba · 9 months, 1 week ago

c is hypotenuse=x^2+y^2Log in to reply

– Prince Loomba · 9 months, 1 week ago

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

– Ayush Rai · 9 months, 1 week ago

Correct solution! Post next problemLog in to reply

Problem 3: 2 marksSolve 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 · 9 months, 1 week ago

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– Prince Loomba · 9 months, 1 week ago

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

– Ayush Rai · 9 months, 1 week ago

Correct answer!Post next problemLog in to reply

[Problem 2, 3 marks]

Prove that \(ab=gcd (a,b)×lcm (a,b)\) – Prince Loomba · 9 months, 1 week ago

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Case 1:When \(a\) and \(b\) are co-prime to each otherSince \(a\) 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 otherLet 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 · 9 months, 1 week ago

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– Abhishek Alva · 9 months, 1 week ago

i guess this is the mathematical proofLog in to reply

– Abhishek Alva · 9 months, 1 week ago

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 numbersLog in to reply

– Abhishek Alva · 9 months, 1 week ago

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 itselfLog in to reply

– Prince Loomba · 9 months, 1 week ago

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

– Prince Loomba · 9 months, 1 week ago

Wrong mathematical proofLog in to reply

– Abhishek Alva · 9 months, 1 week ago

i can represent it using venn daigrams or sets.Log in to reply

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 · 9 months, 1 week ago

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– Ayush Rai · 9 months, 1 week ago

Good solution..+1,you are correct.you can proceed with the next problemLog in to reply

– Abhishek Alva · 9 months, 1 week ago

here in the daigram it ab need not be the tangentLog in to reply

– Prince Loomba · 9 months, 1 week ago

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

– Ayush Rai · 9 months, 1 week ago

yes u are rightLog in to reply

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– Kaustubh Miglani · 9 months, 1 week ago

4 polynomials?Log in to reply

– Prince Loomba · 9 months, 1 week ago

Nope much greater than that hahaLog in to reply

– Anik Mandal · 9 months, 1 week ago

9^5 ?Log in to reply

– Prince Loomba · 9 months, 1 week ago

Obviously not...Log in to reply

– Anik Mandal · 9 months, 1 week ago

yah i was wrong!Log in to reply

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 · 9 months, 1 week ago

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– Kaustubh Miglani · 9 months, 1 week ago

i think noLog in to reply

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 · 9 months, 1 week ago

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– Ayush Rai · 9 months, 1 week ago

You are wrong.The points A and D do not coincide..Log in to reply

Problem 7 (1 mark)Prove that for any acute triangle, \(a^{2}+b^{2}>c^{2}\), where c is the longest side. – Prince Loomba · 9 months, 1 week ago

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– Prakhar Bindal · 9 months, 1 week ago

Directly use cosine rule and put cos>0 so simple! :)Log in to reply

– Prince Loomba · 9 months, 1 week ago

But you And abhishek posted simultaneouslyLog in to reply

– Prince Loomba · 9 months, 1 week ago

RightLog in to reply