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

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## 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 \(AX.\)

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You too used a formula for the circumradius of the triangle

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

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Attention:The actual problem 16 is not this.Log in to reply

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

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Archit wont be more in the contest I suppose because of your foolish

attentionsand such rulesLog in to reply

Problem 2 , 3 marksProve that \(ab=gcd×lcm\) of \(a\) and \(b\)

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Hello

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do u want something?

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

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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.

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Thanks for the suggestion

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Where are you in slack @Ayush Rai

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iam banned from slack

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

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@vishwathiga jayasankar

Why are you leaving?Log in to reply

@Anik Mandal I just dont fit in there. Please do reply.

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

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

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@Rohit Camfar

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

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@Rohit Camfar good

okLog in to reply

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@Ayush Rai

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

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@Ayush Rai and @Anik Mandal

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

@Ayush Rai ???

are u hereLog in to reply

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

Good byeLog in to reply

how?

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will miss you

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why? Do u want ntse qp

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@Ayush Rai check generaldiscussions.

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@Ayush Rai check generaldiscussions.

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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.

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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.

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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

\(\sqrt{x+\sqrt{x}}-\sqrt{x-\sqrt{x}}=\dfrac{3}{2}\sqrt{\dfrac{x}{x+\sqrt{x}}}\)

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

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good one!

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5/4??

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I meant 25/16

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Comment deleted Oct 20, 2016

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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 contestI request Ayush to

delete my pointsLog in to reply

Problem 18, 3 marksFind all functions f such that:

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

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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

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Right, exact solution!

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

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Nope wrong

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

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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 coefficientsLog in to reply

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)

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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

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If or if and only if

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i think if is fine,

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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 \(ab?\)

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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

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

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2? a=1 b=2

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Post your solution

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the area on which axis the x or the y

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I think graph itself is bounded as rectangles

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496

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No

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Problem 15:[1 marks]

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

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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.

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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!

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There is a short soln also.

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Comment deleted Oct 19, 2016

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No

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x,y,z are integers.

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Are x,y,z integers?

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They should be

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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 \(m+n.\)

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449

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

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(n+1)^2 =((n+1)^2-n^2)/n^2(n+1)^2 =1/n^2-1/(n+1)^2 =1-1/15^2 =224/225Log in to reply

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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 \(-xyz?\)

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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

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Good one! Now post the next problem.

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5892? Might be a calculating error Am I right?

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

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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)

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Where's the figure??

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Well sorry about that.Can u manage without that?

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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.

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Bro 6 hrs passed i suppose

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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 \(p(\omega ^2)\)

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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 \(x^{2}-x+1\)

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ANSWER IS 636

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

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

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@Prince Loomba the answer is 288

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

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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

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1 hr and 15 minutes have passed

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It will be mid night after 6 hours :p

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ab=bc=ca=8 root 3

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Wrong

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Wrong answer!

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Points?

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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 \(\triangle ABC.\)

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

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Correct answer! Post next problem

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which question

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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.

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Comment deleted Oct 18, 2016

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Comment deleted Oct 18, 2016

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Comment deleted Oct 18, 2016

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@Svatejas Shivakumar

Ok.Log in to reply

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Circumradius=abc/4delta=65/8.

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

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the marks for the question is 3 marks

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the questio is 3 marks

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

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sorry guyes the question is wrong

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

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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

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

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Post the next problem

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ya you are right

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Comment deleted Oct 18, 2016

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delete other comments

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what is all this left and right

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use latex

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Its \geq

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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 \(a^2+b^2>c^2.\)

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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 .

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you post i dont have any problem

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Right you explained both sides.

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good try

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See 10th rule gphe cant try now haha 😂😂

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Wrong reasoning

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

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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

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correct solution.Post next problem.

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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 \(35/9>10/3.\)

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Post next

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Problem 4: 3 marksProve that the inradius of a right angled triangle with integer sides is an integer.Log in to reply

c is hypotenuse=x^2+y^2

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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.

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Correct solution! Post next problem

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Problem 3: 2 marksSolve the equations for \(x,y\) and \(z.\)

\(x+y+z=9\)

\(x^2+y^2+z^2=29\)

\(x^3+y^3+z^3=99\)

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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

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Correct answer!Post next problem

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[Problem 2, 3 marks]

Prove that \(ab=gcd (a,b)×lcm (a,b)\)

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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.

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i guess this is the mathematical proof

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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

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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

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Wrong mathematical proof

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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.

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

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here in the daigram it ab need not be the tangent

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

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Comment deleted Oct 18, 2016

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4 polynomials?

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Nope much greater than that haha

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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?

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i think no

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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.

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

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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.

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

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But you And abhishek posted simultaneously

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Right

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