[Guest]

You draw an Equilateral Triangle inside a Circle, then at random you draw a Chord, what is the chance that the length of the Chord is smaller than or equal to the side of the triangle?

(try to solve this using different approaches)

[Guest]

2/3.

Take the triangle ABC, take the random chord and rotate it until A is one of it's end points. The other endpoint had 1/3 of a chance to be in each of the arcs AB, BC and CA. Too tired to keep writing...

[HoustonHokie]

A random chord subtends an arc between 0 and 180 degrees, with equal probability anywhere between 0 and 180. The chord of the equilateral triangle subtends an arc of 120 degrees. In order for the random chord to be smaller than or equal to the side of the equilateral triangle, it must subtend less than or equal to 120 degrees. So, I think the answer would be equivalent to 120 / 180 = 2/3.

[Guest]

You draw an Equilateral Triangle inside a Circle, then at random you draw a Chord, what is the chance that the length of the Chord is smaller than or equal to the side of the triangle?

(try to solve this using different approaches)

By random I'm going to assume that you mean two random points are chosen on the circumference and a chord is drawn joining them (but I'm sure other interpretations would be valid giving different answers)

Let two points on the the circumference by A and B. The clockwise arc angle between them is t with a probability function of: f(t) = 1/2pi {0<t<2pi).

Now the length of the chord is r*sqrt(2-2cos(t)). The length of the equilateral triangle chord is r*sqrt(3). Equating these gives values of t: 2pi/3 < t < 4pi/3 so that the chord is greater than the length of the triangle. This gives a probability of (4pi/3 - 2pi/3)/2pi

=1/3.

Edit: I was working out the chance the chord was bigger so take the complementary probability agrees with the above posts.

Edited April 28, 2010 by psychic_mind

[Guest]

You guys all have really math-y approaches but I thought of a real simple one

The chord defined by the side of an equilateral triangle is inevitably 1/3 the circumference of the circle. the probability that a chord would be shorter than that would seem to be 1/3, but that fails to account for the fact that a chord can only possibly be 1/2 the circumference of a circle, therefore the chord defined by the triangle would be 2/3 of the possible semicircular circumference. Thus, your answer is 2/3.

[Guest]

You draw an Equilateral Triangle inside a Circle, then at random you draw a Chord, what is the chance that the length of the Chord is smaller than or equal to the side of the triangle?

(try to solve this using different approaches)

That was rather easy as proved by the above posts. In order to make it more challenging, how about having any random equilateral triangle contained completely inside the circle (edges not necessarily on the circumference) and a random chord as before?

[Guest]

That was rather easy as proved by the above posts. In order to make it more challenging, how about having any random equilateral triangle contained completely inside the circle (edges not necessarily on the circumference) and a random chord as before?

Judging by the previous logic and extrapolation, I came up with this answer

The largest possible triangle would be one with all three points on the edge of the circle, therefore, the maximum likelihood would be 2/3, as proven before. The minimum side length for any equilateral triangle is 1x10^infinity, or the limit that approaches 0.

With all possibilities being equal between the two, the probability of any chord being as long as or shorter than the side length of any random triangle that fits within the same circle would be a limit that approaches 1/3.

[Guest]

That was rather easy as proved by the above posts. In order to make it more challenging, how about having any random equilateral triangle contained completely inside the circle (edges not necessarily on the circumference) and a random chord as before?

For any random equilateral triangle (of side length 'a'), contained completely inside the circle (of diameter 'd'), I get the probability as 1 - 2*cos^-1(a/d)/Pi.

Again, I move the triangle such that two of the vertices (say A and B) lie on the circumference and find the angle (say theta) b/w AB and AO (O being the center of the triangle). So any chord in the area covered within this angle will have length greater than 'a'. So the probability of having it smaller is 1 - theta/(Pi/2).

Note: for the original question this generic expression gives 2/3 (the correct answer) when we have a = d*sqrt(3)/2 <-- case where all three vertices are on the circle

Edited April 28, 2010 by johnogdenjunk

[peace*out]

this is my stupid brain asking, but what's a chord?

[Guest]

The triangle is a fixed size touching the circle with each of it's corner...

You are all correct on the 2/3 answer and when first read the riddle I answered the same, but to tell you the truth this wasn't a riddle but rather a paradox cause there was another answer:

Draw a diameter in the circle that is perpendicular to one of the triangles sides, the point of meeting with the diameter and the triangle shall be called D, and the point between the diamteter nad the circle (as in the edge of it) shall be called E, center of circle is O, now simple geometry rules tell that OD=DE, you make a chord that is perpendicular to the diameter, such chord will be larger than the triangle's side if it cuts it somewhere between point D and point E (look at illustration) since that area is half the length of the diameter, according to this solution, the chance is 1/2!

This is the paradox, there was a third solution but I don't remember it and I didn't understand it completely but it offers even another answer, the book claimed that this is a paradox cause by the fact that there are infinity chances for chords and infinity is undefined, though I believe in the first solution (the one you all answered) and I doubt this one a bit cause if you spin the diameter here the points on OD travel half the distance than the points on DE, so what do you guys think? (I'll try to get the third solution later)

this is my stupid brain asking, but what's a chord?

It's any straight line inside a circle connecting two points, the Diameter is a kind of chord...(and don't feel bad cause I had to look it up as well before posting this riddle, English isn't my first language...)

[Guest]

For any random equilateral triangle (of side length 'a'), contained completely inside the circle (of diameter 'd'), I get the probability as 1 - 2*cos^-1(a/d)/Pi.

Again, I move the triangle such that two of the vertices (say A and B) lie on the circumference and find the angle (say theta) b/w AB and AO (O being the center of the triangle). So any chord in the area covered within this angle will have length greater than 'a'. So the probability of having it smaller is 1 - theta/(Pi/2).

Note: for the original question this generic expression gives 2/3 (the correct answer) when we have a = d*sqrt(3)/2 <-- case where all three vertices are on the circle

Does 1-theta/(pi/2) have a number associated with it? is it close to 1/3? Is theta 60 degrees?

[peace*out]

It's any straight line inside a circle connecting two points, the Diameter is a kind of chord...(and don't feel bad cause I had to look it up as well before posting this riddle, English isn't my first language...)

kk, thanks Anza!!

[Guest]

Are we assuming this equilateral triangle is inscribed in the circle and not randomly placed? If it is inscribed like everyone is thinking then I think the answer is pretty simple.

Starting the chord at one of the triangle points B and moving the endpoint R around the circle the chord will be longer than a side when the endpoint is between the opposite points A and C. So chance of the chord being longer than a triangle side is 1/3 or 33%.

[Guest]

Draw a diameter in the circle that is perpendicular to one of the triangles sides, the point of meeting with the diameter and the triangle shall be called D, and the point between the diamteter nad the circle (as in the edge of it) shall be called E, center of circle is O, now simple geometry rules tell that OD=DE, you make a chord that is perpendicular to the diameter, such chord will be larger than the triangle's side if it cuts it somewhere between point D and point E (look at illustration) since that area is half the length of the diameter, according to this solution, the chance is 1/2!

The reason for this so called paradox is ambiguity in the way a random chord is defined.

One way to define a random chord is to pick two points randomly on the circumference of the circle and connect them by a line.

Another way to make a random chord would be to take a random point on a diameter of the circle and draw a perpendicular line through that point.

The important thing to realize is that the two methods are not equivalent. Most people (including Braindeners) would assume the first method and correctly come to an answer of 2/3. Some people may assume the second and correctly come to an answer of 1/2. Why the difference? Let's take a look at your diagram again.

While it's true that OD=DE, we're actually covering 1/3 of the circumference of the circle with the chords in DE and only 1/6 with the chords in OD. If we adjust the weight of the probabilities accordingly, you'll end up with the familiar answer of 2/3.

[Guest]

^ That's not my explaination that's the book's explaination and I already said before if you rotate the diameter let's say 30° you skip over twice the points in DE than in OD...

BUT, I just looked at the 3rd solution and this one is a bit harder to disprove:

If you define a random chord by a randomly picked point inside the circle (not on the circle or the center itself) there is only one chord that will cross that point with it being exactly in the middle.

Let's call that point Point P, I've just doodled a simulator in Flash and this is what it looks like:

Illustrator

Now if you calculate the green area is 1/4th the big circle and green is 3/4th, so according to this theory the chances are 3/4...

This is the one boggling me and I can't find what's wrong with it besides only the fact that if Point P resided on the center there are infinite possibilities for a chord (diameter) there...

[Guest]

I think you guys all missed it. That or I made a mistake. See Spoiler:

The answer is 1/2

or

50% of chords are shorter than the side of the triangle

Math and an illustration to follow when I get home from work.

Hints: Consider a half-circle with a chord equal to the length of the side of our equilateral triangle parallel to the flat side of the half-circle. Draw any chord parallel to these lines. Consider that any chord on a full circle could be represented on your parallel-chords-half-circle by rotating the half-circle. What fraction of chords go above and below your original triangle side line?

[Guest]

I think you guys all missed it. That or I made a mistake. See Spoiler:

The answer is 1/2

or

50% of chords are shorter than the side of the triangle

Math and an illustration to follow when I get home from work.

Hints: Consider a half-circle with a chord equal to the length of the side of our equilateral triangle parallel to the flat side of the half-circle. Draw any chord parallel to these lines. Consider that any chord on a full circle could be represented on your parallel-chords-half-circle by rotating the half-circle. What fraction of chords go above and below your original triangle side line?

I think I see what's going on here. With my work above my chords had one endpoint fixed and the other moved around the circle. My solution was 1/3. As with the second solution, both endpoints are free to move about and the solution there is 1/2. Which approach is the correct one I wonder.

[Guest]

This is the one boggling me and I can't find what's wrong with it besides only the fact that if Point P resided on the center there are infinite possibilities for a chord (diameter) there...

There is nothing wrong with the approach. All three answers are correct. The flaw is actually in the question itself since it does not specify what a 'random chord' is. This allows for different interpretations and therefore different answers.

Which approach is the correct one I wonder.

I don't know if there is a generally accepted method of generating a random chord, but I think I like the third one Anza described the best.

[Guest]

I think I see what's going on here. With my work above my chords had one endpoint fixed and the other moved around the circle. My solution was 1/3. As with the second solution, both endpoints are free to move about and the solution there is 1/2. Which approach is the correct one I wonder.

For the first approach, it is entirely valid cause the first point is not "fixed" cause it doesn't matter what first point you choose they all make the same type of chances for the second point, for example let's say they ask you to randomly choose two numbers from 1-9 and now what's the chance of them summing up to 10? the first number doesn't matter what you choose, let's say it's n, now the second number needs to be n-1 which is 1 number from the ten that you have so the chance is 1/10...

The second approach is the one I'm skeptical about, if you rotate the whole thing let's say by 30°, the centers and end points of all the short chords move twice the distance of the center and end points of the long chords (draw a short chord and a long chord and see what happens when you rotate the entire circle, measure the distances that each chord goes)...

The third approach (the one with the P Point in illustration above) the problem is that if P was exactly in the center, there are infinity probabilities for chords (which are diameters) there...

[Guest]

Found this on Wikipedia, it's pretty cool actually. Particularly this part. Apparently the second method gives the best answer, which I would have thought was the worst of the bunch

Edited April 30, 2010 by Tuckleton

[Guest]

^ Hey nice find Tuckleton, didn't know this paradox was that famous...

I also thought the second method is the worst of the bunch and I kinda still do, that Jaynes dude has some pretty good points though I'm kinda still not convinced that the first method isn't a good answer, the real problem here I suppose, is the fact that there is no use whatsoever to the solution of this problem other than to entertain bored mathematicians...