Three points are chosen uniformly at random from the perimeter of circle. The probability that the triangle formed by these is acute can be expressed as ab where a and b are coprime positive integers. What is the value of a+b?
Place point A anywhere on the circle.
Draw a diameter from point A.
Now place point B. It lies on one side of that diameter.
There is a 50% chance that point C lies on the same side of the diameter as point B.
Now, if C is at the end of the diameter, ABC is a right triangle.
ABC is acute if C lies on the opposite side of the diameter from B.
The chance of that is 1/2, so a+b=3
its wrong
its five...
let the points be abc,
put "a" anywhere. you can place b at diametrically opposite or not (1/2) then if they are not diametrical opposite then there are two sides where we can place c.and one gives acute and other gives obtuse... so 1/2
so 1/2*1/2=1/4 so 5..:P :)
To find the probability that the triangle formed by the three randomly chosen points on the perimeter of a circle is acute, we need to consider the different cases.
Case 1: All three points are close to each other on the circle.
If all three points are close to each other on the circle, then the triangle formed will always be acute. This case contributes to the probability.
Case 2: Two of the points are close to each other, and the third point is far away.
If two of the points are close to each other and the third point is far away, then the triangle formed will always be obtuse. This case does not contribute to the probability.
Case 3: All three points are evenly spaced on the circle.
If all three points are evenly spaced on the circle, then the triangle formed will be equilateral and therefore acute. This case contributes to the probability.
Considering these cases, we can see that all acute triangles are formed in Case 1 and Case 3 scenarios.
In Case 1, we can choose any initial point on the circle as the starting point, and the probability of choosing two consecutive points close to each other is 2/3.
In Case 3, since all three points are evenly spaced on the circle, we have only one possible configuration. The probability of this case occurring is 1/3.
Therefore, the probability of forming an acute triangle is 2/3 + 1/3 = 1.
The value of a+b is 1+1 = 2.