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 G Rai
4 years, 1 month ago

This discussion board is a place to discuss our Daily Challenges and the math and science related to those challenges. Explanations are more than just a solution — they should explain the steps and thinking strategies that you used to obtain the solution. Comments should further the discussion of math and science.

When posting on Brilliant:

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Problem 2 , 3 marks

Prove that $ab=gcd×lcm$ of $a$ and $b$

- 4 years, 1 month ago

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

- 4 years, 1 month ago

- 4 years, 1 month ago

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

- 4 years, 1 month ago

Attention:The actual problem 16 is not this.

- 4 years, 1 month ago

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 $AX.$

- 4 years, 1 month ago   - 4 years, 1 month ago

You too used a formula for the circumradius of the triangle

- 3 years, 11 months ago

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.

- 4 years, 1 month ago  - 4 years, 1 month ago

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

- 4 years, 1 month ago

here in the daigram it ab need not be the tangent

- 4 years, 1 month ago

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

- 4 years, 1 month ago

yes u are right

- 4 years, 1 month ago

[Problem 2, 3 marks]

Prove that $ab=gcd (a,b)×lcm (a,b)$

- 4 years, 1 month 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 numbers

- 4 years, 1 month ago

Wrong mathematical proof

- 4 years, 1 month ago

i can represent it using venn daigrams or sets.

- 4 years, 1 month 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 itself

- 4 years, 1 month 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 points

- 4 years, 1 month ago

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.

- 4 years, 1 month ago

i guess this is the mathematical proof

- 4 years, 1 month ago

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

- 4 years, 1 month ago

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

- 4 years, 1 month ago

- 4 years, 1 month ago

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

- 4 years, 1 month ago

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

- 4 years, 1 month ago

Correct solution! Post next problem

- 4 years, 1 month ago

c is hypotenuse=x^2+y^2

- 4 years, 1 month ago

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 $35/9>10/3.$

- 4 years, 1 month ago

Post next

- 4 years, 1 month ago

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

- 4 years, 1 month ago

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.

- 4 years, 1 month ago

correct solution.Post next problem.

- 4 years, 1 month ago

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 $a^2+b^2>c^2.$

- 4 years, 1 month ago

Wrong reasoning

- 4 years, 1 month ago

good try

- 4 years, 1 month ago

See 10th rule gphe cant try now haha 😂😂

- 4 years, 1 month ago

He*

- 4 years, 1 month 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 .

- 4 years, 1 month ago

Right you explained both sides.

- 4 years, 1 month ago

you post i dont have any problem

- 4 years, 1 month ago

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

- 4 years, 1 month 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 inequality

- 4 years, 1 month ago

ya you are right

- 4 years, 1 month ago

Post the next problem

- 4 years, 1 month ago

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

- 4 years, 1 month ago

the questio is 3 marks

- 4 years, 1 month ago

sorry guyes the question is wrong

- 4 years, 1 month ago

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

- 4 years, 1 month ago

the marks for the question is 3 marks

- 4 years, 1 month ago

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.

- 4 years, 1 month ago

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

- 4 years, 1 month ago

which question

- 4 years, 1 month ago

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 $\triangle ABC.$

- 4 years, 1 month ago - 4 years, 1 month ago

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

- 4 years, 1 month ago

- 4 years, 1 month ago

what did u do

- 4 years, 1 month ago

ab=bc=ca=8 root 3

- 4 years, 1 month ago

- 4 years, 1 month ago

Points?

- 4 years, 1 month ago

Wrong

- 4 years, 1 month ago

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 $x^{2}-x+1$

- 4 years, 1 month ago

1 hr and 15 minutes have passed

- 4 years, 1 month ago

It will be mid night after 6 hours :p

- 4 years, 1 month ago

See carefully the note again ;p

- 4 years, 1 month ago

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

- 4 years, 1 month ago

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

- 4 years, 1 month ago

@Prince Loomba the answer is 288

- 4 years, 1 month ago

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

- 4 years, 1 month ago

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

- 4 years, 1 month ago - 4 years, 1 month ago

which book is this?

- 4 years ago

Its FIITJEE 2012 RMO practice book

- 4 years ago

u go to fiit jee

- 4 years ago

No allen. My friend gave it to me.

- 4 years ago

oh is it good?

- 4 years ago

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

- 4 years ago

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.

- 4 years, 1 month ago

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 $p(\omega ^2)$

- 4 years, 1 month ago

Bro 6 hrs passed i suppose

- 4 years, 1 month ago

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)

- 4 years, 1 month ago

Where's the figure??

- 4 years, 1 month ago

Well sorry about that.Can u manage without that?

- 4 years, 1 month ago

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 $-xyz?$

- 4 years, 1 month ago

5892? Might be a calculating error Am I right?

- 4 years, 1 month ago

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

- 4 years, 1 month 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 answer

- 4 years, 1 month ago

Good one! Now post the next problem.

- 4 years, 1 month ago

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 $m+n.$

- 4 years, 1 month ago

449

- 4 years, 1 month ago

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

- 4 years, 1 month ago

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

- 4 years, 1 month ago

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

- 4 years, 1 month ago

Problem 15:[1 marks]

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

- 4 years, 1 month ago

Are x,y,z integers?

- 4 years, 1 month ago

They should be

- 4 years, 1 month ago

x,y,z are integers.

- 4 years, 1 month ago

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!

- 4 years, 1 month ago

There is a short soln also.

- 4 years, 1 month 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.

- 4 years, 1 month ago

496

- 4 years, 1 month ago

No

- 4 years, 1 month ago

the area on which axis the x or the y

- 4 years, 1 month ago

I think graph itself is bounded as rectangles

- 4 years, 1 month ago

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 $ab?$

- 4 years, 1 month ago

2? a=1 b=2

- 4 years, 1 month ago

- 4 years, 1 month ago

LaTeX: \begin{aligned} 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

- 4 years, 1 month ago

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

- 4 years, 1 month ago

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

- 4 years, 1 month ago

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)

- 4 years, 1 month ago

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

- 4 years, 1 month ago

Thats what I said. A real root must be there

- 4 years, 1 month 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 root

- 4 years, 1 month ago

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

- 4 years, 1 month ago

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

- 4 years, 1 month ago

Finally somebody understood haha

- 4 years, 1 month ago

Now post prob 18 Be quick pls

- 4 years, 1 month ago

If or if and only if

- 4 years, 1 month ago

i think if is fine,

- 4 years, 1 month ago

Problem 18, 3 marks

Find all functions f such that:

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

- 4 years, 1 month ago

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

- 4 years, 1 month ago

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

- 4 years, 1 month ago

Nope wrong

- 4 years, 1 month 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 answer

- 4 years, 1 month ago

Right, exact solution!

- 4 years, 1 month ago

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

- 4 years, 1 month ago

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

- 4 years, 1 month ago

5/4??

- 4 years, 1 month ago

I meant 25/16

- 4 years, 1 month ago

Yes you are correct! Post your solution!

- 4 years, 1 month ago

- 4 years, 1 month ago

- 4 years, 1 month ago

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

- 4 years, 1 month ago

good one!

- 4 years, 1 month ago

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.

- 4 years, 1 month ago

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.

- 4 years, 1 month ago

Where are you in slack @Ayush Rai

iam banned from slack

- 4 years ago

why? Do u want ntse qp

yup.give me.

- 4 years ago

how?

- 4 years ago

@Ayush Rai check generaldiscussions.

@Ayush Rai check generaldiscussions.

dont leave space. between the brackets

- 4 years ago

how?

- 4 years ago

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

- 4 years ago

ok so when are u going to be back

- 4 years ago

back from where? iam banned from lounge

- 4 years ago

OMG this is not good

will miss you

- 4 years ago

can u give ur email

- 4 years ago

raiayush234@gmail.com

- 4 years ago

- 3 years, 12 months ago

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

- 3 years, 12 months 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.

- 3 years, 12 months ago

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

- 3 years, 12 months ago

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

- 3 years, 12 months ago

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

- 3 years, 12 months ago

are u here @Ayush Rai ???

- 3 years, 12 months ago

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

- 3 years, 12 months 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.

- 3 years, 12 months ago

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

- 3 years, 11 months ago

- 3 years, 11 months ago

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

- 3 years, 11 months ago

Nope I saw your account deactivated

- 3 years, 11 months ago

ok @Rohit Camfar good

- 3 years, 11 months ago

Why good?

- 3 years, 11 months ago

just like that Anyways goodbye (from slack)

- 3 years, 11 months ago

- 3 years, 11 months ago

14 will b 15 on JAN 7 Why did u ask

- 3 years, 11 months ago

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

- 3 years, 11 months ago

hahaha no never. It is my dream.(biotech)

- 3 years, 11 months ago

ok are you leaving brilliant or slack

- 3 years, 11 months ago

slack only why?

- 3 years, 11 months 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.

- 3 years, 11 months ago

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

- 3 years, 11 months ago

bye

- 3 years, 11 months ago

Why are you leaving?@vishwathiga jayasankar

- 3 years, 12 months 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 there @Anik Mandal I just dont fit in there. Please do reply.

- 3 years, 12 months 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!

- 3 years, 12 months ago

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

- 3 years, 11 months 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.

- 3 years, 11 months ago

Thanks for the suggestion

- 3 years, 11 months ago

Hello

- 3 years, 10 months ago

do u want something?

- 3 years, 10 months ago

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.

- 4 years, 1 month ago

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

- 4 years, 1 month ago

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?

- 4 years, 1 month ago

i think no

- 4 years, 1 month ago

Problem 7 (1 mark)

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

- 4 years, 1 month ago

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

- 4 years, 1 month ago

Right

- 4 years, 1 month ago

But you And abhishek posted simultaneously

- 4 years, 1 month ago