Galliv-Ant in Egypt

Let the apex vertex of the pyramid be $E$

Shortest distance will be the altitude from point $A$ to $BE$ + the altitude from point $C$ to $BE$ .

The side of the pyramid will be $100\sqrt{3}$ by the Pythagoras theorem. ( Side of the triangle formed by joining the point $E$ , point $B$ and the point M which is center of the $ABCD$ . Then the side BE is given by $\sqrt{BM^2 +ME^2} = (100\sqrt{2})^2 +100^2= 100\sqrt{3}$ )

Thus the answer will be twice the altitude shown in the figure

Solution diagram

Hence the answer is $\boxed{327}$

To solve this problem imagine that the pyramid is made of cardboard which you can unfold. Let the apex of the pyramid be E. Cut the pyramid open along the line AE. Unfold it and place it flat. Then $ABCE$ forms a quadrilateral where $EA=EB=EC=a$ and $AB=BC=200 m$ . The shortest path is therefore the straight line AC on the flat quadrilateral $ABCD$ .

To find $AC$ , we need to find $a$ . Let the center of the base be $F$ then $EF=100 m$ and $FA=\frac{1}{2}\sqrt{200^{2}+200^{2}}=100\sqrt{2}$ .

Then $a=EA=\sqrt{EF^{2}+FA^{2}}=\sqrt{100^{2}+100^{2}\times{2}}=100\sqrt{3}$ .

Let $\angle AEB=\theta$ . Then $sin\frac{\theta}{2}=\frac{\frac{1}{2}AB}{EA}=\frac{100}{100\sqrt{3}}=\frac{1}{\sqrt{3}}$ .

And $AC=2\times EA\times sin\theta=2\times 100\sqrt{3}\times \frac{2\sqrt{2}}{3}=326.5986\approx\boxed{327}$