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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 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 $\sin 2\alpha + \sin 2\beta+\sin 2 \gamma = 0$

- 6 years ago

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

- 6 years, 1 month ago

- 6 years, 1 month ago

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

- 6 years, 1 month ago

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

- 6 years, 1 month ago

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.

Staff - 6 years, 1 month ago

Tell which one you couldn't solve .

- 6 years, 1 month ago

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

- 6 years, 1 month ago

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

- 6 years, 1 month ago

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?

- 6 years, 1 month ago

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 = $4\times3\times7\times{6 \choose 2}2!\times6! = 7!4!{6 \choose 2}$

- 6 years, 1 month ago

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..

- 6 years, 1 month ago

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

- 6 years, 1 month ago

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= $4!.5.6!$

- 6 years, 1 month ago

- 6 years, 1 month ago

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

- 6 years, 1 month ago

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.

- 6 years, 1 month ago

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

- 6 years, 1 month ago

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?

- 6 years, 1 month ago

Where and when is this exam held?

- 6 years, 1 month ago

is the answer of ques 8 34????

- 6 years, 1 month ago

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

- 6 years, 1 month ago

(A)

- 6 years, 1 month ago

its 184

- 6 years, 1 month ago

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

- 6 years ago

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

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

- 6 years ago