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# Annoying algebra in today's pre-RMO !

Hello !

If

$\sin^{-1} x + \sin^{-1} y + \sin^{-1} z = \dfrac{3 \pi}{2}$

Then find the value of :

$x^9 + y^9 + z^9 + \dfrac{1}{x^9y^9z^9}$

Well , I had got the following result from this : ( Might be wrong )

$x + y + z = -1$

Note by Priyansh Sangule
3 years, 10 months ago

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The range of $$\sin^{-1}$$ is $$[-\tfrac12\pi,\tfrac12\pi]$$. Since $$\sin^{-1}x+\sin^{-1}y+\sin^{-1}z = \tfrac32\pi$$, we know the values of $$\sin^{-1}x,\sin^{-1}y,\sin^{-1}z$$ straight away (what three numbers between $$-a$$ and $$a$$ add to $$3a$$?). Thus we know $$x,y,z$$. · 3 years, 10 months ago

Thank You ! · 3 years, 10 months ago

Very simple one! As, $$\sin^{-1} x \to [\frac{- \pi}{2}, \frac{\pi}{2}]$$

So, it leads us to the conclusion that $$x=1, y=1, z=1$$

Then, required value becomes $$\boxed{4}$$ · 3 years, 10 months ago

i thought same . quite easy. · 3 years, 9 months ago

max value of the following equation is 3pi/2 and hence each angle should be pi/2. if angle is pi/2 then x=y=z=1 then 1 + 1 + 1 + 1 = 4. That's it · 3 years, 10 months ago

Priansh, can I get the copy of your PRMO paper?? · 3 years, 10 months ago

Well they didn't allow us to take it back home :/ · 3 years, 9 months ago

What is Pre-RMO? Is the full paper available somewhere?

Thanks! · 3 years, 10 months ago

Promo paper · 1 week, 2 days ago

In Mumbai region there is Pre-RMO, as the no. of students taking part was much more. While the other regions in the country have RMO directly (including other-than-Mumbai Maharashtra part. We are going to give it on 1st December. Let's discuss some good problems thereafter...!!! · 3 years, 10 months ago

Well this time , the government has allowed every state to set their own papers for pre-RMO . That is the qualification stage for RMO level 1 . · 3 years, 10 months ago