20 points are picked at random from the unit circle,how many would to lying inside the inscribed equilateral^?

Answer 1

Hi,

If the radius of the circle was 2, its area would be πr² = π(2)² = 12.566.

An inscribed equilateral triangle in this circle would have a radius of 2. If 3 radii were drawn to the vertices of the equilateral triangle, the triangle would be divided into 3 30-30-120° triangles where the 2 equal sides would both be 2. The area of a triangle can be found by the formula A = ½ab sin C using 2 sides and their included angle. So using the 120° angle and its 2 adjacent sides of length 2, the area of the 30-30-120° triangle is A = ½(2)(2) sin 120° = 1.732. Since the equilateral triangle is made up of 3 of these 30-30-120° triangles, the area of the equilateral triangle is 3*1.732 = 5.196.

Since the ratio of the area of the equilateral triangle to the area of the circle is 5.196/12.566 = .4135, the probability is that 41.35% of 20 randomly distributed points will be inside the triangle.

41.35% of 20 ≈ 8.27, so about 8 of the 20 points would be expected to be in the triangle.

I hope that helps!! :-)

Answer 2

Well, none of them, since only 3 of the points on the triangle are actually on the circle, and the probability that any individual point will be selected is zero.

Now, if (as I suspect), you meant to say 20 points are picked at random from the _inside_ of the unit circle (as opposed to the circle itself), then the problem is more interesting. Assuming a uniform distribution over the area of the circle, the probability of any given point being on the inside of the inscribed triangle is:

(area of inscribed equilateral triangle)/(area of circle)

The area of the unit circle is of course π. Let us assume WLOG that the triangle is oriented so that one vertex is at (1, 0). Then one of the other vertices is at (cos (2π/3), sin (2π/3)) = (-1/2, √3/2). So the length of one side of the triangle is √((1 - (-1/2))² + (0 - √3/2)²) = √(9/4 + 3/4) = √3. The height of an equilateral triangle is √3/2 times its base, which is 3/2. So the area is 1/2 * √3 * 3/2 = 3√3/4. Thus we have that the probability of any given point lying inside the triangle is:

3√3/(4π) ≈ 0.41349667

The expected number of points lying inside the triangle is then:

20 * 3√3/(4π) = 15√3/π ≈ 8.2699334

And we are done.

Answer 3

I think you are asking how many points would one expect to fall within an equilateral triangle inscribed within a unit circle.

The area of a unit circle is pi.
To find the area of an inscribed equilateral triangle, use the distance formula to find the length of one side that extends from (1,0) to (-1/2,[sqrt(3)]/2).
d = sqrt[3/4 + 9/4] = sqrt(12/4) = sqrt(3)
The height of the triangle is sqrt(3/4 + 3) = sqrt(15)/2
This is found using the Pythagorean Theorem and a right triangle consisting of the base, the height, and another side of the inscribed triangle (which is the hypotenuse).
The area of the inscribed triangle is [sqrt(3)/2]*sqrt(15)/2
or 3(sqrt(5))/4.
The number of points within the triangle would be
20*{3[sqrt(5)]/4}/pi.

Answer 4

The ratio of points inside the triangle will be the ratio of the area.
Assume circle is of unit radius, area of circle = pi =~ 3.142
Assume triangle ABC, center of circle O.
Area of triangle OAB = (1/2) * radius * radius * sin(120)
Since we assume unit radius ...
Area of triangle OAB = (1/2) sin(120) =~ 0.5 * 0.866 = 0.433.
3 times that = 1.299

So ratio of area = 1.299 / 3.142 =~ 0.4134
If you pick 20 points, then Expected points in triangle = 20 * 4.4134 = 8.268

Answer 5

8.27