# This note has been used to help create the Bertrand's Paradox wiki

Today we will try to answer a simple probability question:

A chord is selected at random inside a circle. What is the probability that the length of this chord is longer than the side length of an inscribed equilateral triangle in the circle?

We will attack this problem in three different ways.

**Solution 1**

First, we set a point to be stationary, and randomly select the other point. Clearly, when the other point is contained in the far \(120^{\circ}\) arc, the length of it is longer than the length of the side length of the triangle (shown in the picture as green), and elsewhere, it is shorter (shown as red). Thus, the probability is \(\dfrac{120^{\circ}}{360^{\circ}}=\boxed{\dfrac{1}{3}}\)

**Solution 2**

We randomly choose a point, then draw a horizontal line through it to form a chord in the circle. The probability that the chord is longer than the side length of the triangle is a little harder to figure out, but still easily done with some elementary geometry:

The side length of the equilateral triangle divides the radius of the circle into halves, as shown in the above diagram. Thus, the probability of the random chord being longer than the length of the side of the equilateral triangle is \(\boxed{\dfrac{1}{2}}\)

**Solution 3**

We pick a random point inside the circle, and draw a chord through it such that the point is the midpoint of the chord. Note that whenever the point picked is inside the circle in the middle, then the chord has a side length larger than the side of the triangle; otherwise, smaller.

Recall that the centroid of a triangle divides the medians into \(2:1\) pieces. Thus, \(R=2r\), or \(\dfrac{r}{R}=\dfrac{1}{2}\). Thus, the ratio of the two circles' areas is \(\dfrac{\pi r^2}{\pi R^2}=\left(\dfrac{1}{2}\right)^2=\boxed{\dfrac{1}{4}}\)

How can three different methods yield three different answers? Which one is the correct, and which ones are bogus? Post your thoughts in the comments below. Thanks for reading!

Daniel

For further information (spoilers!), see Bertrand's Paradox

## Comments

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TopNewestProblem with

Solution 2: All chords are not necessary parallel. They might be non-parallel.Problem with

Solution 3: Through one midpoint, infinite chords can pass, and the number of chords that pass through are relatively different for different midpoints. Hence,this solution is incorrect.First one is correct. If I fix one point , the probability that the second point is such that it satisfies the condition is \(\dfrac{1}{3}\) IRRESPECTIVE of where first point lies. I tried another method gives \(\dfrac{1}{3}\)

Choosing a chord at random is equivalent to choosing two points at random. I take \(x\) and \(y\) be the polar angles of two points, then

\(\dfrac{2 \pi}{3} < |x-y| < \dfrac{4 \pi}{3}\)

Draw the favorable region, and sample space, and get answer as \(\dfrac{1}{3}\) – Jatin Yadav · 3 years ago

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– Daniel Liu · 3 years ago

Your rebuttal for problem 3 is wrong. There is only one possible chord that can pass through a chosen point such that the point is the midpoint of the chord.Log in to reply

– Finn Hulse · 3 years ago

Excellent use of the word "rebuttal". The only word I know of that has a "butt" in it but is still used commonly in courtrooms. Watching courtroom movies is hilarious for me whenever I hear that word. :DLog in to reply

– David Lee · 3 years ago

Thought said english you bad..Log in to reply

– Finn Hulse · 3 years ago

Bad English but "butt" good so hear and say "yay!". :}Log in to reply

– Srujan Barai · 3 years ago

I didn't get the problem you specified with the Solution 2. It is perfectly correct from my perspective.Log in to reply

– Eric Hernandez · 2 years, 9 months ago

And butts fit well in this conversation because?Log in to reply

midpointof a chord, of which there is only 1 for that particular point. Daniel's given restriction makes it so that there aren't an infinite number of chords – Justin Wong · 3 years agoLog in to reply

arean infinite number of points! – Mursalin Habib · 3 years agoLog in to reply

None of these are

wrongin that sense. And none of them ismore rightthan the other. It boils down to what you mean by 'random'. The reason why there are 3 different answers is because the distributions are different in each case. – Mursalin Habib · 3 years agoLog in to reply

This was amongst the first examples our teacher discussed in probability class. – Rushi Jogdand · 1 year ago

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So what is the correct answer? Also can you tag all your coordinate geometry problems and notes with special tag which I can add to my profile? I didnt understand what wikipedia tried to explain. – Megh Parikh · 3 years ago

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I assumed it as 1/3 probability since it was an Equilateral one (sides have same length). I saw i was correct ! – Kevin Patel · 3 years ago

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I believe solution 2 is incorrect this is because all chords are not necessarily paralle – Mardokay Mosazghi · 3 years ago

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Solution 3 is wrong because you cannot say anything about the number of lines passing through when the point is at the center.

There is the difference in answer in Solution 1 and Solution 2 because of the difference in consideration of different types of lines. COMPLEX? Let me make it simpler.

In solution 1 we consider multiple lines different from each other on the basis of angle. For instance If we take a unit angle as 1° . (You can argue that we are considering ALL possible lines then the unit angle should be the smallest possible number but to get the answer we have to zoom in considering some particular angle. ) Now on drawing multiple lines 1° apart, after some instance the distance between two lines will increase, then what about the line that would fit in between? They are missed.

In solution 2 the number of lines are at equal distance from the initial stage to the final stage, therefore there is no chance in missing any of the lines.

Hence Solution 2, according to my perspectives, should be considered. – Srujan Barai · 3 years ago

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– Beakal Tiliksew · 3 years ago

Solution 2 won't work because he considered only parallel chordsLog in to reply

Would anyone mind explaining why the following solution is wrong?

It's obvious that this solution is incorrect by looking at Solution 1, which makes it intuitively obvious that the answer is \(\dfrac{1}{3}\), but I don't understand where I'm going wrong, or if this method is even applicable. Help! – Raj Magesh · 3 years ago

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– Raj Magesh · 3 years ago

A sudden question hit me just now: are there more chords of length \(1\) in a circle than there are chords of length \(2\)? Now that I think about it, it seems so. If I'm given an arc and I have to draw chords 1 unit apart from each other along the circumference, I'd get more chords if my chords were smaller. Could someone confirm this?Log in to reply

– Daniel Liu · 3 years ago

Infinities is a tricky concept to deal with... I believe that there are actually the same number, although don't quote me on that.Log in to reply

I think the answer would be (2-sqrt{3})/2. The length of a side of equilateral triangle is \sqrt{3}R. R is rhe radius of the circle. So 2R is the maximum length of a chord. Now we can say, we can draw infinite chords of length x where x is 2R max, and 0 min. And range of infinite number of chords with more length than \sqrt{3} is 2R<=x<\sqrt{3}R. Thus probablity is (2-\sqrt{3})/2. – Mosharraf Hossain · 3 years ago

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That first problem was on the AMC 10 B this year. – David Lee · 3 years ago

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I must say, creating diagrams really is a hassle! – Daniel Liu · 3 years ago

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Diagrams do help your explanation a lot. I agree that creating such diagrams can be time consuming, esp if you do not have good software (I hope you're not using paint!) – Calvin Lin Staff · 3 years ago

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I use the AoPS Wiki to render my ASY code. – Daniel Liu · 3 years ago

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