MHB What Is the Probability That a Chord on a Unit Circle Is At Least 1 Unit Long?

[anemone]

Two points are picked at random on the unit circle $x^2+y^2=1$. Determine the probability that the chord joining the two points has length at least 1.

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[MarkFL]

Hello MHB Community,

anemone couldn't be with us this week, but she conscientiously made provisions for me to post this week's POTW for Secondary School/High School Students in her stead.

So, with no further ado...

Congratulations to the following members for their correct solutions::)

1. MarkFL
2. magneto
3. laura123
4. Ifdahl

Honorable mention goes also to chisigma as he simply misread the problem and computed the probability that the given chord is no greater than one unit in length.

Solution from magneto:

Spoiler

Consider a circle $O$ with radius $r = 1$. Let a point $A$ on the circumference be given.
Then there are two distinct points on the circumference $B$ and $B'$ where $AB = 1 = AB'$.
Note that $\triangle AOB$ and $\triangle AOB'$ are both equilateral, so $m \measuredangle BAB' = 120^\circ$,
and the major arc $BB'$ is $240^\circ$. If any point $C$ lies on the major arc $BB'$, the length of chord $AC \geq 1$.
So, the probability is $\frac{240}{360} = \frac 23$.

Solution from Ifdahl:

Spoiler

Two points picked at random on the unit circle spans an arc of length $\alpha$, where $0 \le \alpha \le \pi$. Thus picking two points at random corresponds to choosing a certain $\alpha$ - just like cutting out a piece of a circular birthday cake :). The pdf for a random choice of $\alpha$ is uniform continuous, because every possible $\alpha$-value has the same weight*. If the chord, that connects the points has length $c = 1$, we have the limit: $\alpha = \frac{\pi}{3}$. (Cf. attached figure).
Thus
$P(\alpha \le \frac{\pi}{3}) = P(c \le 1) = \frac{\frac{\pi}{3}-0}{\pi – 0}=\frac{1}{3}$

- and therefore the probability of choosing two points at random with a corresponding chord of length greater than $1$ is:

$ P(\alpha > \frac{\pi}{3}) = P(c > 1) = 1 - \frac{1}{3}= \frac{2}{3}$.

(*).This is not the case for the distribution of the cord length, $c$.

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