That’s a lot more complicated, I suspect ðŸ™‚ A possible approach might be to eliminate all cases whee c^2 > a^2 + b^2, but even that needs to be handled carefully. Shall think about it and reply again if I find a reasonable answer, but from the point of view of aptitude tests, you may safely assume it is out of scope. If you’re writing an olympiad, however…

I tried doing it in a different way and landed on 1/4. Can you please let me know if it is legit?

Let the pieces be A,B and C. Let us choose A first. Clearly it cannot be greater than 1/2 of the total length. So probability of choosing an appropriate A which can lead to a triangle is 1/2.

Now for B.
Let the length of A be x. So 1-x is left. Suppose we take half of it for B. So length of B is (1-x)/2. Then length of C will be 1-(x+(1-x)/2) = 1/2 + x/2. Sum of lengths of A and B = x + (1-x)/2 = 1/2+x/2, which is same as that of C and so violates the triangle rule. So length of B has to be less than half of what is left after A. So we have 1/2 probability of choosing B which leads to a triangle. So total probability = 1/2 * 1/2 = 1/4.

I know it looks rather clumsy, but do you think it is correct? Please let me know

No, I think you have made an error in this statement “Then length of C will be 1-(x+(1-x)/2) = 1/2 + x/2”. It will in fact be (1-x)/2 only. You can use this kind of approach too, but to prove its correctness requires integration.

if the stick is broken into 3 parts, what is the probability that the parts form acute angled triangle?

That’s a lot more complicated, I suspect ðŸ™‚ A possible approach might be to eliminate all cases whee c^2 > a^2 + b^2, but even that needs to be handled carefully. Shall think about it and reply again if I find a reasonable answer, but from the point of view of aptitude tests, you may safely assume it is out of scope.

If you’re writing an olympiad, however…regards

J

The graphical method is as always very elegant.

I tried doing it in a different way and landed on 1/4. Can you please let me know if it is legit?

Let the pieces be A,B and C. Let us choose A first. Clearly it cannot be greater than 1/2 of the total length. So probability of choosing an appropriate A which can lead to a triangle is 1/2.

Now for B.

Let the length of A be x. So 1-x is left. Suppose we take half of it for B. So length of B is (1-x)/2. Then length of C will be 1-(x+(1-x)/2) = 1/2 + x/2. Sum of lengths of A and B = x + (1-x)/2 = 1/2+x/2, which is same as that of C and so violates the triangle rule. So length of B has to be less than half of what is left after A. So we have 1/2 probability of choosing B which leads to a triangle. So total probability = 1/2 * 1/2 = 1/4.

I know it looks rather clumsy, but do you think it is correct? Please let me know

No, I think you have made an error in this statement “Then length of C will be 1-(x+(1-x)/2) = 1/2 + x/2”. It will in fact be (1-x)/2 only. You can use this kind of approach too, but to prove its correctness requires integration.

regards

J

Oh no! I was so thrilled to get the correct answer.

Thanks a lot for the swift reply.