Probability that $|z_1+z_2|\geq \sqrt{2+\sqrt 3}$

Question

$z_1$ and $z_2$ are two distinct roots of $z^{101}=1$. Find the probability that $|z_1+z_2|\geq \sqrt{2+\sqrt 3}$

Answer: $\frac{4}{25}$

My attempt:

Let $z_1=\exp\left(i\cdot \frac{2m\pi}{101}\right)$ and $z_2=\exp\left(i\cdot \frac{2k\pi}{101}\right)$ with $m\neq k$ and $m,k\in \{0,1,\cdots ,100\}$.

Using this the given equation becomes $$\sqrt{2+2\cos\left(\frac{2\pi(m-k)}{101}\right)}\geq \sqrt{2+\sqrt 3}$$ From here I got, $$|m-k|\leq \frac{101}{12}$$ Since $m,k$ are integers, $$|m-k|\leq 8$$

My attempt matches with the given solution upto this point. But then, as I'm stuck with how to calculate the probability. The given solution gets $$P=\frac{101\cdot 16}{101\cdot 100}$$ in the very next step and I'm unable to understand why.

Any help would be great. Thanks!

Edit: My thoughts on calculating the probability

WLOG assume $m>k$.

• For $k=0$, values for $m$ are $1,2,\cdots 8$

• For $k=1$, values for $m$ are $2,3,\cdots 9$

And so on

• For $k=92$, values for $m$ are $93,94,\cdots 100$

Till here, each value of $k$ gives $8$ values of $m$. But from here onwards, that isn't the case.

• For $k=93$, values for $m$ are $94,95,\cdots 100$

And so on.

Thus, the "favourable" cases are $93\times 8 +7+6+5×4+3+2+1$ while the sample space has $100+99+\cdots +1$ elements. This gives the probability $\frac {772}{4950}$ which doesn't match.

Answer 1

Fix one vertex intially say $k=0$. Then $m$ can take the values $1,2, \ldots 8$, $-1,-2,\ldots,-8$(or $100,99,\ldots 93$) . Amounting to a total of $16$ ways.

Now the first vertex(k) can be shifted to the other $100$ vertices, and in each case there are $16$ corresponding values of $m$. Note that this approach includes the vertices as ordered pairs, i.e. making the choice of value $(m,k)$ different from $(k,m)$.

Hence the total number of outcomes, most also be made according to unordered pairs, which is ${101 \choose 2} \times 2 = 100 \times 101$.

Hence the required probability is $P= \frac{16 \times 101}{101 \times 100}= \frac{4}{25}$

Answer 2

Think of $m$ as fixed. Then since $m$ and $k$ are distinct, there are $100$ possibilities for $k$.

Now the condition $|m-k| \leq 8$ means $|m-k| = n$ for some integer $n \in [1,8]$. Each value of $n$ gives rise to two solutions for $k$, and there are $8$ possible values of $n$.

Can you finish it from here?

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Note also that technically the requirement on $m$ and $k$ is slightly different from $|m-k| \leq 8$. The actual requirement is that $|m-k| \equiv n \pmod{101}$ for some positive integer $n \leq 8$, and this takes care of the cases near the boundary.