Junior Mathematics Olympiad questions

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

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

Please help me with your solutions..

Tell which one you couldn't solve .

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

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

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.

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

Where and when is this exam held?

is the answer of ques 8 34????

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