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Junior Mathematics Olympiad questions

Hello all, This is the question paper of JMO. Please help me with the solutions. Post solutions in the comment providing question number. Thanks.

Note by Kumar Ashutosh
4 years, 2 months ago

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[6] Let \(\cos \alpha + i\cdot \sin \alpha = e^{i\alpha}\)

\(\cos \beta + i\cdot \sin \beta = e^{i\beta}\)

\(\cos \gamma + i\cdot \sin \gamma = e^{i\gamma}\)

Now Given \(\cos \alpha + \cos \beta + \cos \gamma = 0 .........................(1)\)

and \(\sin \alpha + \sin \beta + \sin \gamma = 0.......................................(2) \times i\)

Add These two equations::

\(e^{i\alpha}+e^{i\beta}+e^{i\gamma} = 0...................................................(3)\)

Now Square both Side

\(e^{2i\alpha}+e^{2i\beta}+e^{2i\gamma} + 2 \left(e^{i(\alpha+\beta)}+e^{i(\beta+\gamma)}+e^{i(\gamma+\alpha)}\right) = 0\)

Now in a \(\triangle\) Sum of angle i.e \(\alpha+\beta+\gamma = \pi\Leftrightarrow \alpha +\beta = \pi-\gamma\)

So \(e^{2i\alpha}+e^{2i\beta}+e^{2i\gamma} +e^{i(\pi-\gamma)}+e^{i(\pi-\beta)}+e^{i(\pi-\alpha)} = 0\)

\(e^{2i\alpha}+e^{2i\beta}+e^{2i\gamma} +e^{i(\pi-\gamma)}-2\left(e^{-i\alpha}+e^{-i\beta}+e^{-i\gamma}\right) = 0\)

Now from \((3)\), Taking Congugate on both side, we get

\(e^{-i\alpha}+e^{-i\beta}+e^{-i\gamma} = 0\)

So \(e^{2i\alpha}+e^{2i\beta}+e^{2i\gamma} = 0\)

So \(\cos 2\alpha + \cos 2\beta+\cos 2 \gamma+i\left(\sin 2 \alpha+\sin 2 \beta+\sin 2 \gamma\right) = 0+i\cdot 0\)

we Get \(\cos 2\alpha + \cos 2\beta+\cos 2 \gamma = 0\)

and \(\sin 2\alpha + \sin 2\beta+\sin 2 \gamma = 0\)

Jagdish Singh - 4 years, 1 month ago

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Question 10.

Firstly \(r_1 = \tfrac{1}{8b}(a^2 + 4b^2)\) and \(r_2 = \tfrac{1}{8a}(4a^2 + b^2)\). To see this, for example, suppose that \(P\) has coordinates \((0,\tfrac12a)\), \(Q\) coordinates \((0,-\tfrac12a)\). Then the centre of the first circle will have coordinates \((x,0)\) for some \(x\), and the circle will touch \(RS\) at its midpoint \((b,0)\). Then \(x^2 + \tfrac14a^2 = (b-x)^2\), which gives us \(x\). Then \(r_1=b-x\). While \(x\) could be negative, \(r_1\) cannot.

Then \[ \begin{array}{rcl} r_1+r_2 & = & \frac{a(a^2+4b^2)+b(4a^2+b^2)}{8ab} \; = \; \frac{a^3+b^3 + 4a^2b + 4ab^2}{8ab} \\ & = & \frac{(a+b)(a^2-ab + b^2 + 4ab)}{8ab} \; = \; \frac{(a+b)(a^2 + 3ab + b^2)}{8ab} \\ & = & \tfrac{1}{8}(a+b)\big[3 + \tfrac{a}{b} + \tfrac{b}{a}\big] \; \ge \; \tfrac{1}{8}(a+b)[3+2] \; = \; \tfrac{5}{8}(a+b) \end{array} \]

Mark Hennings - 4 years, 2 months ago

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2[b]:: Here \(a,b\in \mathbb{R}\) and \(3a^4-4a^3b+b^4\geq 0\)

Now this is a Homogeneous equation, so Let \(a = bx\) and Let \(y = 3b^4x^4-4b^3x^3\cdot b+b^4\)

So \(y = b^4\cdot \left(3x^4-4x^3+1\right) = b^4\cdot (x-1)^2\cdot (3x^2+2x+1)\geq 0\)

bcz \(3x^2+2x+1 = 3\left(x+\frac{1}{3}\right)^2+\frac{8}{3}>0\;\forall x\in \mathbb{R}\)

Jagdish Singh - 4 years, 1 month ago

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There is also a simple answer to this. It goes like this:

We have, \(3a^4-4a^3b+b^4\) \(=a^4-2a^2b^2+b^4+2a^2b^2+2a^4-4a^3b\) \(=(a^2-b^2)^2+2a^2(b^2+a^2-2ab)\) \(= (a^2-b^2)^2+2a^2(a-b)^2 \geq 0 \)

Kumar Ashutosh - 4 years, 1 month ago

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is the answer of ques 8 34????

Varnika Chaturvedi - 4 years, 2 months ago

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which part, 8(a) or 8(b)

Kumar Ashutosh - 4 years, 2 months ago

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(A)

Varnika Chaturvedi - 4 years, 2 months ago

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@Varnika Chaturvedi its 184

Kumar Ashutosh - 4 years, 2 months ago

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Where and when is this exam held?

Priyatam Roy - 4 years, 2 months ago

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Question 9 is nothing but inclusion-exclusion principle...perhaps!

Siddharth Kumar - 4 years, 2 months ago

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how to do it? if we can find the number of cubes less than \( 10^8 \) and number of fifth power numbers less than \( 10^8 \) then only we can solve it. Is there another way?

Kumar Ashutosh - 4 years, 2 months ago

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The first question is very easy to solve. First of all, p^6 - 1 is divisible by 7, can be proved by fermat's little theorem. Next, p^6 -1 can be factorised to find whether it is divisible by 3^2, next the factorisation can also help to find whther it is divisble by 2^3. So 504 = 7* (3^2) * (2^3). This means p^6-1 is divisible by 504.

Siddharth Kumar - 4 years, 2 months ago

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Challenge masters please give some hints!!!!!!!!!!

Kumar Ashutosh - 4 years, 2 months ago

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Here are the anwers ( not for the proof - type questions.)

2(a) \(\rightarrow 153846\)

4(b) \(\rightarrow \)\(7!4!\binom{6}{2} \)

5 \(\rightarrow\) 90

7 \(\rightarrow \pi x^2\)

8(a) \(\rightarrow 184\)

8(b) \(\rightarrow x - y\sqrt{3} - 2 = 0 \) , \((2 + \sqrt{\frac{3}{2}} , \frac{1}{\sqrt{2}})\)

Kushagraa Aggarwal - 4 years, 2 months ago

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yes, thanks, question 2(a), 5, 7, 8(a) matches with my solution, but can you give some explanation in 4(b). and I think your 8(b) is wrong. the angle of AC with x-axis is 60 degree. So, \( \tan{60} = \frac {y - 0}{x - 2} \) Thus equation is \( y= \sqrt{3}x - 2 \sqrt{3} \) which is my answer, and the coordinate also has some error. Am I right?

Kumar Ashutosh - 4 years, 2 months ago

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ok , i went wrong in q.8(b) , it is rotated anti - clockwise , i rotated it clockwise.

4(b) Give end seats to grandchildren in \( 4\times3\) ways.

Now choose 1 seat out of 7 for grandfather( you can't select the seat which second from left or right.).in \(7\) ways.

Then choose 2 seats out of 6 for remaining two grandchildren ( as we can't choose from grandfather's left or right and 3 seats are already occupied) and permutate in \({6 \choose 2}2!\) ways.

Now , six seats for sons and daughters remain , \(6!\) ways for them.

Total no. of ways = \(4\times3\times7\times{6 \choose 2}2!\times6! = 7!4!{6 \choose 2}\)

Kushagraa Aggarwal - 4 years, 2 months ago

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@Kushagraa Aggarwal I am getting your logic. Please check: https://brilliant.org/discussions/thread/an-olympiad-level-question-of-combinomatrics/

Kumar Ashutosh - 4 years, 2 months ago

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@Kushagraa Aggarwal am I right in question 8(b), then what is the coordinate of C? In 4. how did you write- Give end seats to grandchildren in 4×3 ways? as far as I have understood the question,two seats each on the left and right ends of row is occupied by children. Am I wrong or You are mistaken..

Kumar Ashutosh - 4 years, 2 months ago

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@Kumar Ashutosh Left and Right end seats are occupied by grand children .

For left seat it has four options and right has three options ( as one is already taken by left one.). Hence by Fundamental principle of counting , no. of ways = \( 4\times 3\).

8(b) . \(( 2 + \frac{1}{\sqrt{2}} , \sqrt{\frac{3}{2}})\)

Kushagraa Aggarwal - 4 years, 2 months ago

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@Kushagraa Aggarwal okay, 8(b) is correct and also matches my solution.
In 4, I am still confused.Look and find error in my solution, it goes like this:

The extreme 2 seats on both ends are occupied by grand children. So 4 grand children occupy 4 seats in \( 4! \) ways. Numbering the seats 1,2,3.....11, we find that Grandfather canno sit on 3rd and 9th seat. So grandfather can sit on any one of 4th,5th,6th,7th,8th seats. This is done in \( 5 \) ways.

Now after this arrangement, we have 6 seats left on which 6 son/daughter is to sit. This can be done in \( 6! \) ways

Therefore total possibilities= \( 4!.5.6! \)

Kumar Ashutosh - 4 years, 2 months ago

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Tell which one you couldn't solve .

Kushagraa Aggarwal - 4 years, 2 months ago

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Please give solutions to all the questions you can, I will check the questions done by me and will understand what didn't solved.. Thanks

Kumar Ashutosh - 4 years, 2 months ago

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Please help me with your solutions..

Kumar Ashutosh - 4 years, 2 months ago

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Did you try them already? Which did you manage to solve?

Tim Vermeulen - 4 years, 2 months ago

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I did q.1,2,4(b), 5,7,8, but try to give solutions of every question, so that I could verify my answers

Kumar Ashutosh - 4 years, 2 months ago

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@Kumar Ashutosh It would be better for you to state your solutions / approaches, and have others comment on it.

There can be subtleties in proof based questions, which would result in a 'similar-looking' answer being marked wrong.

Calvin Lin Staff - 4 years, 2 months ago

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