Is the AREA imaginary or real?

Algebra Level 4

If the area of the polygon whose vertices are the solutions (in the complex plane) to the equation ${ x }^{ 7 }+{ x }^{ 6 }+{ x }^{ 5 }+{ x }^{ 4 }+{ x }^{ 3 }+{ x }^{ 2 }+x+1=0$ , can be expressed in the simplest form as $\cfrac { \xi \sqrt { \vartheta } +\Im }{ \forall }$ . Find the value of $\xi +\vartheta +\Im +\forall$ .

$Details$

$\xi ,\vartheta ,\Im ,\forall \quad \quad \epsilon \quad Positive Integers$

$\vartheta$ is a square free positive integer.

The answer is 8.

1 solution

Michael Ng
Dec 23, 2014

Great problem! Notice that the equation can be written like this: $\frac{x^8-1}{x-1} = 0 \implies x^8-1=0$ where $x \neq 1$ .

Now we know that the roots are arranged in an octagon with radius $1$ , but we must not include the root where $x=1$ . Finding the area: $6 \times \frac{1}{2}\times 1^2\times \sin 45^{\circ} + \frac{1}{2} = \frac{3\sqrt{2} +1}{2}$

This gives the answer of $\boxed{8}$ .

Same solution! I spend a while trying to chase up the side length but eventually I got it!

A Former Brilliant Member - 6 years, 5 months ago

How do you get 6? $\dfrac{1}{2}$ ?

Niranjan Khanderia - 6 years, 5 months ago

Here you are: diagram diagram

There are six shaded triangles so we multiply by $6$ , then we add the extra triangle of area $\frac{1}{2}$ . Hope this helps!

Michael Ng - 6 years, 5 months ago

Thanks. Six for imaginary solution and two for Real . That is how I understand. Thanks once more.

Niranjan Khanderia - 6 years, 5 months ago

So is the Area real or imaginary? just kidding BTW nice explanation.

Gautam Sharma - 6 years, 5 months ago

Is the AREA imaginary or real?

The answer is 8.

1 solution

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