Probability that the length of chord AB is greater than the radius of a circle.

Relevant wiki: Circles Problem Solving - Basic

Consider the case where the chord is equal to the radius. Then the measure of the arc between the two points is $60^{\circ}$ (it creates the equilateral triangle shown above). Now randomly select the first point. If the second point falls within $60^{\circ}$ of this point, then the chord would be smaller than the radius. This is $\frac{120^{\circ}}{360^{\circ}} = \frac{1}{3}$ of the circumference. For the chord to be longer than the radius, the second point needs to land on the other $\boxed{\frac{2}{3}}$ of the circumference.

Relevant wiki: Properties of Equilateral Triangles

The red portion on the circumference is representative of where the points A and B can be placed for the chord to be greater than the radius of the circle.

Fix one point

Now draw a equilateral triangle having side length r vertex = centre of circle and the fixed point, chord is between the fixed point and the remaining vertex

Now rotate the chord along centre as the fixed point(towards the centre of circle) a point will come when it's length again equal to r

So the angle between them is 120

So answer is 120/180 = 0.667 $\textbf{APPROXIMATELY}$

Relevant wiki: Cosine Rule (Law of Cosines)

Let $s, r,$ and $\theta$ be the length of the chord, radius, and the measure of the included angle respectively.

By law of cosine, we get

$s = \sqrt{2r^2(1-cos\theta)} > r$

$2r^2(1-cos\theta) > r^2$

$1-cos\theta > \dfrac{1}{2}$

$cos\theta < \dfrac{1}{2}$

Notice that $0 < \theta \leq 180$ and the only values that satisfy are $60< \theta \leq 180.$

$\therefore$ The probablility is $\dfrac{120}{180} = \boxed{\dfrac{2}{3}}.$

The chord would equal the radius if an equilateral triangle were formed, in which case the central angle would be 60 degrees = 1/3 of the total possible of 180. therefore, a cental angle > 60 results in a chord > radius, so the answer is 2/3. Ed Gray

This is for people that are familiar with multiple choice questions and do a lot of them regularly. Not for people who don't appreciate humor.

This is not intended to be a formal solution for anyone to take too seriously. You'll understand what I mean about this $after$ you properly read it.

The answer choices are $\frac{1}{3}$ , $\frac{1}{2}$ , $\frac{2}{3}$ , $\frac{5}{6}$ . $2$ of these answers are above half, and $1$ of them is below.

When a person is creating a multiple choice question, they usually try to make it so that the correct answer is hard to guess based on comparison to the other answer choices. They can achieve this in many different ways. For example, they generally make all the answers around the same interval. And also some other ways I don't know how to put into words. (i know you probably think i'm stupid now but continue reading anyway. please.

Going back to the beginning, more answers are concentrated above 1/2, so you can assume that the answer is probably above 1/2. *idk how to explain this, i never really tried to explain it before until now, so bear with me and my cheesy explanations. The answer probably won't be exactly 1/2 because half is such a perfect number and it's too good to be true.

eg. if the correct answer of a random question is 3/4. The creator of the question will know the general thinking process of a person trying to solve it. They will probably list the other answer choices at around the same interval. (they won't make the choices of eg 1/4, 3/2, 3/4, and 384 because 384 is obviously wrong) And since it is a fractions question, they will probably list another choice with the same numerator, and another choice with the same denominator. They will also have more answer similar to 3/4, just so it's harder to estimate the answer without actually doing calculations.

There are 2 numbers with the denominator of 3. There aren't any other 2 numbers with a common denominator. So you can assume that the answer probably has a denominator of 3. They listed 1/2 because it is a common and central number, and it is an obvious choice for what to list, because a lot of people (like noobs) will automatically assume any two random events have a 50/50 chance of happening. They probably listed 5/6 because since the correct answer is above 1/2, it is more likely for someone to guess wrong 5/6, than a number like 1/4. And 6 is a multiple of 3, so it is easy(er) for someone to make a mistake and accidentally think the denominator is 6 instead of 3.

**This is me being too lazy to calculate the answer.

or maybe just too stupid to figure out how to calculate it

maybe I'm too good at psycology.

I'm not sure

**idk stands for "I don't know"

btw, in case you were wondering, i used this method of guessing to guess the answer and i actually guessed it right.

i'm proud of me now

EDIT: THIS WEBSITE DOESN'T LET ME DO DOUBLE SPACING I'M TRIGGERED

Choose any point on the circle, then draw a circle of radius r from that point as in the figure. Now any point on the pink line drawn will be at a distance > r , so we need to find out the perimeter covered by the pink line.

Pink Line = Perimeter of half circle i.e. (Pi r ) + 2 (Perimeter of the blue line).

Notice the green triangle is equilateral

Perimeter of the blue line = The blue line spans 30 degree of the circle so would be Pi r /6

Therefore Pink Line Perimeter = Pi r + 2*Pi r /6

Total circumference = 2*Pi r

Dividing them we get our answer as 2/3

What you must consider in this problem is at what point is S = R. Well, that’s when △SRR is equilateral. This is at the point where the angle between the two R sides is 60°. Given the symmetry of circles this occurs twice. Once on once half and once on the other meaning in total there are 120° of this circle where R ≤ S. This means there is 240° where S > R. Assuming the two points that the radii go to are picked completely randomly. Then given in 240° or $\frac{2}{3}$ of the circle S is larger the probability of S being larger is when the points are random is $\frac{2}{3}$

So yeah im stopping to think why (it's intuitive but there could possibly be something interesting...probably not) picking two points at random is equivalent probabilistically to choosing thr inner angle.

Yeah it's just the chance that the inner angle is bigger than 60 degrees (which corresponds to an aqulateral triangle which is the sorts border g event between the two outcome spaces. )

Anyway any angle bigger than sixty up to max (180) so 120 over 180....2(60)/3(60)

(pi)r = 180
r = 57(appx.)
so ratio becomes.....57 : 180
or...............................1 ; 3
so...1 - 1/3 = 2/3

Fix one point (p1) in place. Now pick a random second point(p2) on the circumference of the circle. Draw a triangle between the centre of the circle(M) , the fixed and second points. If the angle casted at M between the lines (p1M) & (p2M) is equal to 60 then the triangle formed would be an equilateral triage, thus the line (p1p2) would be exactly equal to the radius. Any lower than 60 degrees and said segment would be smaller, any larger it would be bigger. More than 180 degrees simply creates a new line, reverted from the one draw until now. So we have a case of degrees from 0 - 60 results in negative and 60-max(180) results in positive. 120/180=2/3

If we only consider half of a circle, picking randomly two points on the circle is like choosing an angle right there :

Therefore the chord is equal to the radius if the angle is 60°. If the angle is smaller than 60° then the chord will be smaller than the radius and because we can choose any angle between 0° and 180° , this situation occurs 1/3 of the time. Then we extend this solution to a full circle by adding 1/3 for the other half. Conclusion, the probabilty is 1/3 + 1/3 = 2/3

2 Solutions

1.If angle between the two radii (of the corresponding points)is greater than 60 then the above condition will be satisfied

Since the points are random the maximum angle is 180 and any angle from more than zero until 180 is possible

SO THE probability is 180-60/180 = 2/3

Second. Since the square of 3rd side if angle is ¤ is given by a^2 + b^2 - 2ab cos¤ Given that

2R^2 - 2R^2 cos¤ > R^2

Solving we get cos¤>1/2 ;i.e ¤>60

Maximum angle is 180 SO THE probability is 180-60/180 = 2/3

6 equilateral triangles, with 2 sides being radius and 1 being the chord, can fit inside of a circle (forming an hexagon shape). Remove all but 2 opposing triangles, and any isosceles triangle with a chord size bigger than the radius fits in between them, and fits the criteria.

The angle when line S is not longer than the radius starts exactly or below pi/3 rad. since there are 2 parts having pi/3 rad (opposite sides), and pi/3 rad is 1/6 of tau, that means the probability of line S being longer than the radius is 2/3 I'm 10 years old by the way..

None of these solutions are correct. However, the probability that the chord drawn between two points placed randomly along the circumference of the circle is greater than or equal to the radius of the circle is 2/3.