CHALLENGE QUESTION 5

The vertices of a triangle are (1,0,0),(0,$\cfrac{1}{\sqrt{2}}$, 0), (0,0,$\cfrac{1}{\sqrt{3}}$). If its orthocentre is ($\alpha,\beta,\gamma$), then $\large{(\alpha\beta\gamma)^{\huge{\frac{-2}{5}}}}$ is

the answer ranges from 0 to 9...

Question

CHALLENGE QUESTION 5

The vertices of a triangle are (1,0,0),(0,$\cfrac{1}{\sqrt{2}}$, 0), (0,0,$\cfrac{1}{\sqrt{3}}$). If its orthocentre is ($\alpha,\beta,\gamma$), then $\large{(\alpha\beta\gamma)^{\huge{\frac{-2}{5}}}}$ is

the answer ranges from 0 to 9...

vishweshshrimali5 · Answer

@waterineyes

vishweshshrimali5 · Answer

The vertices of a triangle are (1,0,0),(0,$\cfrac{1}{\sqrt{2}}$, 0), (0,0,$\cfrac{1}{\sqrt{3}}$). If its orthocentre is ($\alpha,\beta,\gamma$), then $\large{(\alpha\beta\gamma)^{\huge{\frac{-2}{5}}}}$ is

the answer ranges from 0 to 9... 

This is the question

anonymous · Answer

It should be divided by 3..

anonymous · Answer

I think I am getting:

$$\color{green}{3 \times \sqrt[5]{18}}$$

vishweshshrimali5 · Answer

How did u get that @completeidiot

vishweshshrimali5 · Answer

$\huge\color{green}{3 \times\sqrt[5]{18}}$

vishweshshrimali5 · Answer

This is more clearly visible !!

vishweshshrimali5 · Answer

Random guess or any trick...... ?

anonymous · Answer

Orthocenter is given by Average of the respective coordinates.

vishweshshrimali5 · Answer

no........... @waterineyes

anonymous · Answer

my method was incorrect, ignore it

vishweshshrimali5 · Answer

that is what we call centroid @waterineyes

vishweshshrimali5 · Answer

All right @completeidiot

anonymous · Answer

Oh Then messed up sorry..

vishweshshrimali5 · Answer

np

anonymous · Answer

Here we have to do more..

vishweshshrimali5 · Answer

that's why it is a challenge problem :)

anonymous · Answer

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