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
6 years, 1 month ago

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[6] Let cosα+isinα=eiα\cos \alpha + i\cdot \sin \alpha = e^{i\alpha}

cosβ+isinβ=eiβ\cos \beta + i\cdot \sin \beta = e^{i\beta}

cosγ+isinγ=eiγ\cos \gamma + i\cdot \sin \gamma = e^{i\gamma}

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

and sinα+sinβ+sinγ=0.......................................(2)×i\sin \alpha + \sin \beta + \sin \gamma = 0.......................................(2) \times i

Add These two equations::

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

Now Square both Side

e2iα+e2iβ+e2iγ+2(ei(α+β)+ei(β+γ)+ei(γ+α))=0e^{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 α+β+γ=πα+β=πγ\alpha+\beta+\gamma = \pi\Leftrightarrow \alpha +\beta = \pi-\gamma

So e2iα+e2iβ+e2iγ+ei(πγ)+ei(πβ)+ei(πα)=0e^{2i\alpha}+e^{2i\beta}+e^{2i\gamma} +e^{i(\pi-\gamma)}+e^{i(\pi-\beta)}+e^{i(\pi-\alpha)} = 0

e2iα+e2iβ+e2iγ+ei(πγ)2(eiα+eiβ+eiγ)=0e^{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)(3), Taking Congugate on both side, we get

eiα+eiβ+eiγ=0e^{-i\alpha}+e^{-i\beta}+e^{-i\gamma} = 0

So e2iα+e2iβ+e2iγ=0e^{2i\alpha}+e^{2i\beta}+e^{2i\gamma} = 0

So cos2α+cos2β+cos2γ+i(sin2α+sin2β+sin2γ)=0+i0\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 cos2α+cos2β+cos2γ=0\cos 2\alpha + \cos 2\beta+\cos 2 \gamma = 0

and sin2α+sin2β+sin2γ=0\sin 2\alpha + \sin 2\beta+\sin 2 \gamma = 0

jagdish singh - 6 years ago

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Question 10.

Firstly r1=18b(a2+4b2)r_1 = \tfrac{1}{8b}(a^2 + 4b^2) and r2=18a(4a2+b2)r_2 = \tfrac{1}{8a}(4a^2 + b^2). To see this, for example, suppose that PP has coordinates (0,12a)(0,\tfrac12a), QQ coordinates (0,12a)(0,-\tfrac12a). Then the centre of the first circle will have coordinates (x,0)(x,0) for some xx, and the circle will touch RSRS at its midpoint (b,0)(b,0). Then x2+14a2=(bx)2x^2 + \tfrac14a^2 = (b-x)^2, which gives us xx. Then r1=bxr_1=b-x. While xx could be negative, r1r_1 cannot.

Then r1+r2=a(a2+4b2)+b(4a2+b2)8ab  =  a3+b3+4a2b+4ab28ab=(a+b)(a2ab+b2+4ab)8ab  =  (a+b)(a2+3ab+b2)8ab=18(a+b)[3+ab+ba]    18(a+b)[3+2]  =  58(a+b) \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 - 6 years, 1 month ago

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Please help me with your solutions..

Kumar Ashutosh - 6 years, 1 month ago

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

Tim Vermeulen - 6 years, 1 month 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 - 6 years, 1 month 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 - 6 years, 1 month ago

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Tell which one you couldn't solve .

kushagraa aggarwal - 6 years, 1 month 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 - 6 years, 1 month ago

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

2(a) 153846\rightarrow 153846

4(b) \rightarrow 7!4!(62)7!4!\binom{6}{2}

5 \rightarrow 90

7 πx2\rightarrow \pi x^2

8(a) 184\rightarrow 184

8(b) xy32=0\rightarrow x - y\sqrt{3} - 2 = 0 , (2+32,12)(2 + \sqrt{\frac{3}{2}} , \frac{1}{\sqrt{2}})

kushagraa aggarwal - 6 years, 1 month 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, tan60=y0x2 \tan{60} = \frac {y - 0}{x - 2} Thus equation is y=3x23 y= \sqrt{3}x - 2 \sqrt{3} which is my answer, and the coordinate also has some error. Am I right?

Kumar Ashutosh - 6 years, 1 month 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 4×3 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 77 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 (62)2!{6 \choose 2}2! ways.

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

Total no. of ways = 4×3×7×(62)2!×6!=7!4!(62)4\times3\times7\times{6 \choose 2}2!\times6! = 7!4!{6 \choose 2}

kushagraa aggarwal - 6 years, 1 month 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 - 6 years, 1 month 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 = 4×3 4\times 3.

8(b) . (2+12,32)( 2 + \frac{1}{\sqrt{2}} , \sqrt{\frac{3}{2}})

kushagraa aggarwal - 6 years, 1 month 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! 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 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! 6! ways

Therefore total possibilities= 4!.5.6! 4!.5.6!

Kumar Ashutosh - 6 years, 1 month 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 - 6 years, 1 month ago

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

Kumar Ashutosh - 6 years, 1 month 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 - 6 years, 1 month ago

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

Siddharth Kumar - 6 years, 1 month ago

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how to do it? if we can find the number of cubes less than 108 10^8 and number of fifth power numbers less than 108 10^8 then only we can solve it. Is there another way?

Kumar Ashutosh - 6 years, 1 month ago

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Where and when is this exam held?

Priyatam Roy - 6 years, 1 month ago

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is the answer of ques 8 34????

varnika chaturvedi - 6 years, 1 month ago

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

Kumar Ashutosh - 6 years, 1 month ago

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(A)

varnika chaturvedi - 6 years, 1 month ago

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@Varnika Chaturvedi its 184

Kumar Ashutosh - 6 years, 1 month ago

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2[b]:: Here a,bRa,b\in \mathbb{R} and 3a44a3b+b403a^4-4a^3b+b^4\geq 0

Now this is a Homogeneous equation, so Let a=bxa = bx and Let y=3b4x44b3x3b+b4y = 3b^4x^4-4b^3x^3\cdot b+b^4

So y=b4(3x44x3+1)=b4(x1)2(3x2+2x+1)0y = b^4\cdot \left(3x^4-4x^3+1\right) = b^4\cdot (x-1)^2\cdot (3x^2+2x+1)\geq 0

bcz 3x2+2x+1=3(x+13)2+83>0  xR3x^2+2x+1 = 3\left(x+\frac{1}{3}\right)^2+\frac{8}{3}>0\;\forall x\in \mathbb{R}

jagdish singh - 6 years ago

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

We have, 3a44a3b+b43a^4-4a^3b+b^4 =a42a2b2+b4+2a2b2+2a44a3b=a^4-2a^2b^2+b^4+2a^2b^2+2a^4-4a^3b =(a2b2)2+2a2(b2+a22ab)=(a^2-b^2)^2+2a^2(b^2+a^2-2ab) =(a2b2)2+2a2(ab)20= (a^2-b^2)^2+2a^2(a-b)^2 \geq 0

Kumar Ashutosh - 6 years ago

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