Three inside Ten

The closest decagon vertex to $E$ is either $D$ or the other vertex from the decagon side. Let $ED=s$ . $$ It's not hard to find that the central angle of a regular decagon is $\frac{2\pi}{10}=\frac{\pi}{5}$ , from where we find the internal angle to be $\pi-\frac{\pi}{5}=\frac{4\pi}{5}=144^{\text{o}}$ . From the symmetry at vertex $A$ we have $2\angle BAE+60^{\text{o}}=144^{\text{o}}$ and so $\angle BAE=42^{\text{o}}$ . The sum of the angles of the pentagon $ABCDE$ is $540^{\text{o}}$ , from which we find $\angle AED=66^{\text{o}}$ . Now let's connect points $A$ and $D$ . Obviously $ABCD$ is a trapezoid with equal legs of a unit length, so it's an isosceles trapezoid. Then $\angle BAD=\angle CDA = 36^{\text{o}}$ . Also $\angle EAD$ is $6^{\text{o}}$ . Drawing heights from $B$ and $C$ to the base $AD$ splits the trapezoid to two congruent right triangles and a rectangle. Then it's easily found that $AD=1+2\text{cos}36^{\text{o}}$ Consider $\triangle ADE$ . The Law of Sines gives: $\frac{AD}{\text{sin}66^{\text{o}}}=\frac{s}{\text{sin}6^{\text{o}}}$ Substituting for $AD$ and rearranging results in: $s=\frac{\text{sin}6^{\text{o}}\left(1+2\text{cos}36^{\text{o}}\right)}{\text{sin}66^{\text{o}}}\approx 0.29955714555539187$ $s<0.5$ so it will be the desired distance. Note that the original drawing gives the intuition that $s=DE$ is indeed the desired length but sometimes this can be misleading. Finally: $\lfloor{10^8s}\rfloor=\fbox{29955714}$