Find the area of the polygon whose vertices are the solutions in the complex plane of the equation $x^7+x^6+x^5+x^4+x^3+x^2+x+1=0$

Question

If the area of the polygon whose vertices are the solutions in the complex plane of the equation $x^7+x^6+x^5+x^4+x^3+x^2+x+1=0$ can be expressed as $\frac{a\sqrt b+c}{d}$.Find $a+b+c+d$

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The polynomial is $\frac{x^8-1}{x-1}$ has roots $\operatorname{cis}(2\pi k/8)$ for $k \in \{1, \ldots, 7\}$.

Thus the value is the area of the regular octagon minus the area of a triangle formed by two adjacent sides.

The area of an octagon (by splitting into triangles) with radius $1$ is $8 \cdot \frac{1}{2} \cdot \frac{\sqrt{2}}{2} = 2\sqrt{2}$.

I am stuck here.The answer for $a+b+c+d=10$

Answer 1

The triangle has area $$ \frac12 \times \sqrt2 \times (1- \sqrt2/2) = \frac{\sqrt2 - 1}{2}, $$ so the total polygon has area $$ 2\sqrt{2} - \frac{\sqrt2 - 1}{2} = \frac{3}{2}\sqrt{2} + \frac12 = \frac{3\sqrt{2} + 1}{2}, $$ at least in lowest terms. The result is $a + b + c + d = 8$, contrary to the 10 you claimed.

Answer 2

$$\frac{a\sqrt{b}+c}{d}~=~2\sqrt{2}~~~~\text{with}~~~~a+b+c+d~=~10$$

The numerator of the $abcd$-fraction contains one square root plus a number. Hence the sum of the four variable equals an integer the easiest way to get rid of $c$ is by setting it $0$. From thereone we not got $a+b+d=10$. Now $b$ has to be $2$ so that we can arrive at the square root of $2$. Now we got $$a+d=8~~\text{and}~~\frac{a\sqrt{2}}{d}=2\sqrt{2}$$ So $a$ has to be a multiply of $d$, to be exact $a=2d$. Plugging this into $a+d=8$ leads us to $a=\frac{16}3$ and $d=\frac83$. Verifying by setting this values in the first equations:

$$\begin{align} \frac{\frac{16}{3}\sqrt{2}+0}{\frac83}~=~2\sqrt{2}~~~~&\text{and}~~~~\frac{16}3+2+0+\frac83~=~10\\ \frac{16\sqrt{2}}{8}~=~2\sqrt{2}~~~~&\text{and}~~~~\frac{16+8+6}{3}~=~10\\ 2\sqrt{2}~=~2\sqrt{2}~~~~&\text{and}~~~~10~=~10 \end{align}$$

Therefore $a=\frac{16}3,b=2,c=0,\frac83$ but this only holds for $a,b,c,d \in \mathbb{Q}$.

Answer 3

You got stuck at a very odd point. You basically solved the hard part of the problem. So compute the are of the triangle, subtract it from the area of the octagon, and express the result in the desired form.

Area of triangle: $\frac{1}{2}\sqrt{2}(1-\frac{\sqrt{2}}{2}) = \frac{\sqrt{2}-1}{2}$.

So the area of the polygon is $2\sqrt{2}- \frac{\sqrt{2}-1}{2}= \frac{3\sqrt{2}+1}{2}$.

The result is not unique though. Given any number $t$, $a=3t, b=2, c=t, d=2t$ is a solution (and there are many others). So $a+b+c+d= 6t+2$ can be any number.

Some condition must be missing, check the problem again.

Answer 4

\begin{align} p_1&=(\tfrac{\sqrt2}2,\tfrac{\sqrt2}2) ,\\ p_2&=(0,1) ,\\ p_3&=(-\tfrac{\sqrt2}2,\tfrac{\sqrt2}2) ,\\\ \dots \end{align}

\begin{align} S&= (p_{1x}-p_{3x})^2+\tfrac32(p_{1x}-p_{3x})(p_{2y}-p_{1y}) =\frac{3\sqrt{2}+1}{2} ,\\ &= \frac{3k\sqrt{2}+k}{2k} ,\\ a&=3k,\quad b=2,\quad c=k,\quad d=2k ,\\ a+b+c+d&=6k+2, \quad k\in\mathbb{R} . \end{align}

In order to get $a+b+c+d=10$, we need to have $k=\tfrac43$.

Answer 5

Let us use Roots of unity!

Here's a hint (by the way this question is from the ARML and I recommend you try it out because it is quite nice). We know that if $z\neq1$ is an $z$th root of unity, that $1+z+z^2+z^3+z^4+\cdots+z^{n-1}=0$. Also, note that the $n$th roots of unity form a regular $n$-gon with a vertex at $(1,0)$.

ANSWER (SPOILER):

After all this you will get an answer of $\frac{6\sqrt{2}+1}{4}$, so that $a+b+c+d=13$, so our answer is $\boxed{13}$