NMTC 2K15 #29 Kaprekar Contest

$AE$ is the arc of the circle centered $D$

$\Rightarrow$ $AD$ is the radius of the circle.

Now, we know that the distance from the centre of the circle to the boundary is always uniform. Hence,

$\Rightarrow$ $DE\space=\space AD\space =\space \text{radius of the circle}\space =\space 2$

Now,see the figure. $\Rightarrow\space \bigtriangleup DEC$ is right-angled at $C$ .

Now, according to pythogorean theorem,

$\Rightarrow(DE)^2\space =\space (EC)^2\space +\space (DC)^2$

$\Rightarrow\space 4\space =\space (EC)^2\space +\space 1$

$\Rightarrow\space EC\space =\space \sqrt{3}$

$\boxed{\therefore\space BE\space =\space BC\space -\space EC\space =\space 2\space -\space \sqrt{3}}$

$\boxed{\therefore\space BE\space =\space 0.268}$

I used the unit circle to solve this one. Since we have an arc with radius 2 and CD = 1 we have coordinates corresponding to 2 x (1/2,1/2\sqrt(3)). This means BE = 2-\sqrt(3).