BrainDen.com - Brain Teasers

## Question

Last night at Morty's, Alex asked his buddies:

What is the probability that a random chord drawn through a circle

has length greater than the side of an inscribed equilateral triangle?

After scribbling a moment on the classy new, M - monogrammed napkins, each of them had an answer:

Ian quickly announced, it's 1/4.

Jamie was more optimistic with his result of 1/3.

Davie smiled and said, If ya do it truly randomly, it'll happen 1/2 the time.

Which, if any, of them is right?

This is one puzzle where ambiguity in the OP is not only OK, it's necessary. Give reasons, and please use spoilers.

## Recommended Posts

• 0 Choose any two points on the circle. If the angle between them is greater than 2pi/3 and 4pi/3, the chord will be longer. Therefore, the probability is 1/3.

If we choose from a set of chords that are parallel to one another, we have a probability of 1/2. If the chord chosen is within r/2 of the center of the circle, it will be longer than the chord defined by the triangle.

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• 0 i could see it would be perimeter/ circumference, which would give roughly 0.48. with this your basically randomly placing two points along the circumference, and then seeing if the line segment is longer than a side.

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• 0

Choose any two points on the circle. If the angle between them is greater than 2pi/3 and 4pi/3, the chord will be longer. Therefore, the probability is 1/3.

If we choose from a set of chords that are parallel to one another, we have a probability of 1/2. If the chord chosen is within r/2 of the center of the circle, it will be longer than the chord defined by the triangle.

As hall and bonanova mentioned, the key lies in how the 'random chord' was chosen.

If we choose a chord by randomly choosing a point in the circle, and draw a chord through the point such that it is perpendicular to the line connecting the point to the center, then the chord will be be larger than the side of an inscribed equilateral triangle if it falls within a circle of radius (1/2)*r centered at the same centre. that works out to be a probability of 1/4.

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• 0 As hall and bonanova mentioned, the key lies in how the 'random chord' was chosen.

If we choose a chord by randomly choosing a point in the circle, and draw a chord through the point such that it is perpendicular to the line connecting the point to the center, then the chord will be be larger than the side of an inscribed equilateral triangle if it falls within a circle of radius (1/2)*r centered at the same centre. that works out to be a probability of 1/4.

In doing so, the chord does not pass through the point that was randomly chosen. And the probability would be 1/2 not 1/4.

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• 0 A random point A is chosen as below. If we look at the equilateral triangle ABC, then the next point of the random chord should lie between (smaller) arc BC in order to for the chord to be longer than the side of the equilteral triangle.

The probability of this is = 120/360 = 1/3

Edited by DeeGee
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• 0 If you take a random point on the circle and draw a chord randomly this would give 1/3, but I don't think this would be a fairly random chord.

I suggest you should draw a chord without taking a starting point on the circle, for a random chord.

If OP had said that "you take a random point and then draw a chord" this would be the case.

But OP tells about only drawing a random chord.

I guess a random chord is drawn only if you draw all possible lines through the circle. But this is hard for me to calculate.

Instead, I took all lines parallel to each other through the circle. Then I got (sqrt(3)-1)/sqrt(3)=0.42 ???

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• 0 If you take a random point on the circle and draw a chord randomly this would give 1/3, but I don't think this would be a fairly random chord.

I suggest you should draw a chord without taking a starting point on the circle, for a random chord.

If OP had said that "you take a random point and then draw a chord" this would be the case.

But OP tells about only drawing a random chord.

I guess a random chord is drawn only if you draw all possible lines through the circle. But this is hard for me to calculate.

Instead, I took all lines parallel to each other through the circle. Then I got (sqrt(3)-1)/sqrt(3)=0.42 ???

How do you draw a random line in a given XY plane. Do you "draw" all possible lines of all possible lengths and then choose a random line? Or do you choose two random points and connect them? here's an example: you draw a random triangle in XY plane. What is the probability that origin (0,0) is contained in the triangle? Well, well... nice idea for a new topic on the forum!

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• 0

In doing so, the chord does not pass through the point that was randomly chosen. And the probability would be 1/2 not 1/4.

The chord, defined as the segment through the randomly chosen point and perpendicular to the line connecting the center to a randomly chosen point, should pass through the point. I include an image because a picture is worth 1000 words, plus or minus about 999 words or so.

The ambiquity in the original post is similar to asking, lets say we 'randomly choose' a point in the interval (0,1). What is its average value? There are an infinite number of density distribution available for sampling, such as the uniform distribution, gaussian, cauchy, poisson, and so on. And all of them are considered 'random sampling'. Thus we see that the answer depends on the random distribution we choose. As far as choosing a random chord goes, there are an infinite number of ways to sample a 'random chord' through the circle. Some additional ways to sample a random chord are

* Let the circle be drawn centered at (0,0), with radius 1. Sample the chord from the family of lines y = ax + b. The prior distribution on a and b will also change the answer accordingly.

* choose two random points within the circle's area. Connect the two points with a line. That is your chord.

* inscribe the circle within a square. Construct a chord by randomly choosing two points on different edges of the square and connect.

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• 0 The chord, defined as the segment through the randomly chosen point and perpendicular to the line connecting the center to a randomly chosen point, should pass through the point. I include an image because a picture is worth 1000 words, plus or minus about 999 words or so.

I see what you are saying but in this case, you are only considering chords that are perpendicular to the line connecting the point with the center of the circle, while for a random chord passing through a point, it should have any random angle (with the line connecting the center of the circle)not just 90o

Edited by DeeGee
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• 0 I see what you are saying but in this case, you are only considering chords that are perpendicular to the line connecting the point with the center of the circle, while for a random chord passing through a point, it should have any random angle (with the line connecting the center of the circle)not just 90o

All chords contained in the circle have some point at which a line drawn from the circle's center intersects the chord at 90 degrees. So changing the angle at said random point only serves to move bushindo's random point.

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• 0 For any two random points on the circumfrence of the circle that form a chord, there is a point C which bisects the chord. The radial line from the center of the circle through C is perpendicular to this point.

The set of points where the chord length is equal to the side of the equalateral triangle is a circle of radius 1/2 r.

So. Anytime C falls outside this inner circle, the chord length of the random chord is less than the side of the triangle, and any time it is inside, the chord length is greater. So. The ratio of the times it falls in the outer circle annulus to the times it falls in the inner circle (chord is larger) is

(pi*r2 - pi*(r/2)2) / (pi*(r/2)2)

Which means that 3/4 of the time, C is in the annulus, and 1/4 of the time, C is in the inner circle. So the probability of being in the inner circle is 1/4.

Do i see that right?

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• 0 For any two random points on the circumfrence of the circle that form a chord, there is a point C which bisects the chord. The radial line from the center of the circle through C is perpendicular to this point.

The set of points where the chord length is equal to the side of the equalateral triangle is a circle of radius 1/2 r.

So. Anytime C falls outside this inner circle, the chord length of the random chord is less than the side of the triangle, and any time it is inside, the chord length is greater. So. The ratio of the times it falls in the outer circle annulus to the times it falls in the inner circle (chord is larger) is

(pi*r2 - pi*(r/2)2) / (pi*(r/2)2)

Which means that 3/4 of the time, C is in the annulus, and 1/4 of the time, C is in the inner circle. So the probability of being in the inner circle is 1/4.

Do i see that right?

I think this is the best answer. Without any idea of the probability density and as the OP and title suggest ambiguity to be key components of the answer I would tend to assume a uniform density.

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• 0 None of them are right. The probability is 2/9 that the chord length is greater.

Choose two random thetas, t1 and t2 between [0, 2PI). These represent the points on the circle. Take the absolute value of the difference between these two thetas, which is the angle formed between the two thetas. If that value is greater than PI, subtract PI (because you had an external angle, and the chord is drawn on the interior angle). If this value is greater than 120 degrees (2/3 PI), then the chord length formed between the two thetas is greater than the equilateral triangle base.

This happens 2/9 times.

(it should be noted that i did write a monte carlo program for this)

Edited by tpaxatb
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• 0

None of them are right. The probability is 2/9 that the chord length is greater.

Choose two random thetas, t1 and t2 between [0, 2PI). These represent the points on the circle. Take the absolute value of the difference between these two thetas, which is the angle formed between the two thetas. If that value is greater than PI, subtract PI (because you had an external angle, and the chord is drawn on the interior angle). If this value is greater than 120 degrees (2/3 PI), then the chord length formed between the two thetas is greater than the equilateral triangle base.

This happens 2/9 times.

(it should be noted that i did write a monte carlo program for this)

This is conceptually the same as picking two points randomly on the circle's circumfence, and computing the chord constructed therefrom. hall addressed this earlier in post #2. Shouldn't the probability of having a chord greater than the equilateral triangle base be 3/9?

Edited by bushindo
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• 0 This is conceptually the same as picking two points randomly on the circle's circumfence, and computing the chord constructed therefrom. hall addressed this earlier in post #2. Shouldn't the probability of having a chord greater than the equilateral triangle base be 3/9?

You're right. It was supposed to be if angle > 180, angle = 360-angle. Not if angle > 180, angle = angle - 180.

It is indeed, 1/3 in this case.  ##### Share on other sites
• 0 So they are all right, because the probability varies depending on how you define "draw a random chord".

Random chord 1: Pick one point inside the circle at random. Define the chord such that the point that is chosen is the bisector of the chord and the chord is perpendicular to the radial of the circle at that point. Chords drawn this way have a 1/4 chance being greater than the triangle.

Random chord 2: Pick two random points on the circumfrence of the circle (two random thetas). 1/3 of the time the chord is greater.

Random chord 3: Choose a random point between 0 and r along the (x axis) and define the chord as in explanation 1 above (it is parallel to the y axis). Then randomly rotate the chord about the origin. The point at which the chord is greater is when the point is < r/2; and the rotation has no affect on the probability.

Therefore the probability of being greater is 1/2.

Hmm. Now I'm randomly thinking there's gotta be another one somewhere... ##### Share on other sites
• 0 Hmm. Now I'm randomly thinking there's gotta be another one somewhere... If you are really interested in finding more possible options, here's one:

Lets say the circle of radius "r" is centered at the origin (0,0). Choose any random point in the plane such that its disctance from origin (0,0) is > r. The random chord is given by the line joining the two points of the cricle where the two tangents from this random point meet the circle.

I have a feeling it will probably lead to the probability of chord being greater than the side of eq triangle of 1/2...

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• 0

If you are really interested in finding more possible options, here's one:

Lets say the circle of radius "r" is centered at the origin (0,0). Choose any random point in the plane such that its disctance from origin (0,0) is > r. The random chord is given by the line joining the two points of the cricle where the two tangents from this random point meet the circle.

I have a feeling it will probably lead to the probability of chord being greater than the side of eq triangle of 1/2...

Good one, and I agree there must be others. This one, owing to the preponderance of points far from the circle, will create a lot of near-diameters.

Nonetheless, it's random in the sense the generating point lies at a random point in the plane.

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• 0 Which, if any, of them is right?

They are all correct depending on the probability density generated by the way a random chord is chosen. Basically, you could generate any probability from 0 to 1 inclusive.

Edited by Semper Rideo
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• 0 Good one, and I agree there must be others. This one, owing to the preponderance of points far from the circle, will create a lot of near-diameters.

Nonetheless, it's random in the sense the generating point lies at a random point in the plane.

One thing...I don't think this works, because the set of chords defined along the exact diameter cannot be defined...by using this method, you could never define a chord that is the exact diameter of the circle, the two tangents are parallel and could never converge to a single point outside the circle...

ugh my head hurts from too many of these things ##### Share on other sites
• 0

One thing...I don't think this works, because the set of chords defined along the exact diameter cannot be defined...by using this method, you could never define a chord that is the exact diameter of the circle, the two tangents are parallel and could never converge to a single point outside the circle...

ugh my head hurts from too many of these things Interesting method here. Despite the apparent shortcoming of not being able to define a chord with length 2r, this method has the following properties

* The probability that the produced chord is larger than the length of the incribed triangle is 1.

* The expected chord length produced is the diameter of the circle, 2r. (actually, it is the limit, but close enough).

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• 0 I havent read yet wanted to put my two cents in before i read everything but i would think that from the center of the circles perspective the chords points are 120deg apart. so any points less the 120deg apart are smaller lines. So the first point of the chord doesnt matter as its a circle so one point is as good as any other.

so anything <120deg in the positive direction is too short

anything >-120deg in the negative direction is too short

that leaves 120 deg of being longer

so 1/3

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• 0 quick program a little messy but it does the trick

```public class circ

{

public static void main (String args[])

{

double point1[]={0,0};

double point2[]={0,0};

double r=100;

double ran=0;

double ran2=0;

double si,co,si2,co2,dis,dis2;

dis2=173.205;

int counter2=0;

double ans=0;

for(int counter=0;counter<10000;counter++)

{

ran=Math.random()*2*3.141592654;

ran2=Math.random()*2*3.141592654;

si=Math.sin(ran);

si2=Math.sin(ran2);

co2=Math.cos(ran2);

co=Math.cos(ran);

point1=co*r;

point1=si*r;

point2=co2*r;

point2=si2*r;

dis=(point1-point2)*(point1-point2)+(point1-point2)*(point1-point2);

dis=Math.sqrt(dis);

if(dis>dis2)

{

counter2++;

}

}

ans=((double)counter2)/10000;

}

}

```

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• 0 final - this is for you ```

#define _USE_MATH_DEFINES 1

#include <iostream>

#include <cmath>

#include <cstdlib>

#include <ctime>

#define itercount 1000000

// The chord length of the equilateral triangle is:  2*cos(30)/r = 2*root(3)/2*1/r = root(3)/r

static double equalateral_triangle_leg_length = sqrt(3.0) / circleRadius;

static void RandomChordHalf()

{

std::cout << "Performing Test P=0.5" << std::endl;

int greater = 0;

for (int i = 0; i < itercount; i++)

{

// I know the distribution is small for rand() but..this is qad...

double dist_to_centerpoint = (double)rand()/(RAND_MAX+1) * circleRadius;

// Define the chord as where the line y=dist_to_centerpoint intersects the circle at the origin of radius r..

// Then rotate the circle about the origin a random amount.

// The length of the chord is 2* root(radius^2 - distance to centerpoint^2).

// The length is unchanged by the rotation (so i didn't include it in the calculations).

double testlength = 2 * sqrt(circleRadius * circleRadius - dist_to_centerpoint * dist_to_centerpoint);

if (testlength > equalateral_triangle_leg_length)

++greater;

}

std::cout << greater << " out of " << itercount << " are greater (" << (double)greater / itercount << "%)" << std::endl;

}

int main(int argc, char* argv[])

{

srand((unsigned)(time(0)));

RandomChordHalf();

return 0;

}

```

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• 0 well if thats true then I cant say much except, whats my mistake? I cant find one in either of ours. I think mine is more reliable(who would have guessed) because I choose two random points then figure it out. Where you take a subset of chords and generalize. Oh and you need sqrt(3)*circleRadius not divide

but anyway I guess ill program this and see because I don't get where a mistake is in either

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