Probability of decomposing an interval Let $U_1$ and $U_2$ be independent, uniformly distributed random variables on the interval $ I := [0,1] $. We decompose $I$ into three parts $I_1$, $I_2$ and $I_3$.
First I am supposed to give a proof, that the probability, that all lengths of $I_1$, $I_2$, $I_3$ are less than $\frac{1}{2}$, is $\frac{1}{4}$. 
Then we consider three random points $V_1, V_2, V_3 \in \mathbb{R}^2$, that are independent, uniformly distributed on the unit circle.
Give the probability that these three points are the corners of an acute triangle (every angle less than 90°).
I have done a few approaches to this but they all ended in nothing.. I assume, I have to use the first part with the intervals to show the second part with the triangle, but I failed at that, too..
 A: The probability that $I_3$ is greater than $\frac12$ is clearly $\frac14$. Now convince yourself that the three interval lengths all have the same distribution by starting with a circle, uniformly choosing $3$ points on it, uniformly picking one of them to cut the circle, producing the unit interval, and noting that the resulting distribution is the same as in your situation; thus all cyclic permutations of the interval lengths are equiprobable.
So every interval has probability $\frac14$ to be greater than $\frac12$; that's three disjoint events whose probabilities add up to $\frac34$; that leaves $\frac14$ for the complementary event that none of them are greater than $\frac12$.
For the second part, use the fact that for points on a circle, a triangle over a diameter has a right angle, so that's the border case between obtuse and acute triangles.
A: Another way to solve this is.
Let $u_1$ and $u_2$ be the two cuts in a 0-1 line.  
You have the condition
$u_1\lt \frac{1}{2}$
$u_2 - u_1\lt \frac{1}{2}\implies u_2\le \frac{1}{2} + u_1$
$u_2\gt u_1$
Putting all together
we get
The required probability $= \int_{0}^{\frac{1}{2}}\int_{u_1}^{\frac{1}{2}+u_1} du_2du_1 = \frac{1}{4}$
