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

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## 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 \(\dfrac{1}{3}\)

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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.

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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. :D

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

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And butts fit well in this conversation because?

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Solution 3 stated to let the chosen point be the

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 chordsLog in to reply

arean infinite number of points!Log 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.Log in to reply

I must say, creating diagrams really is a hassle!

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Since "a picture says a thousand words", I'd think that the amount of hassle is the same.

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!)

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Nope, I'm using Asymptote Vector Graphic Language. It's really useful for creating math diagrams, although what you write is much like programming languages like JavaScript. So if you ever plan on using Asymptote, you need to first have a basic understanding of programming!

I use the AoPS Wiki to render my ASY code.

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That first problem was on the AMC 10 B this year.

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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.

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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!

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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?

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

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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.

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Solution 2 won't work because he considered only parallel chords

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I believe solution 2 is incorrect this is because all chords are not necessarily paralle

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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 !

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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.

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This was amongst the first examples our teacher discussed in probability class.

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