Given two circles about the same center, to inscribe in the greater circle an equilateral polygon with an even number of sides which does not touch the lesser circle.

Let *ABCD* and *EFGH* be the two given circles about the same center *K.*

It is required to inscribe in the greater circle *ABCD* an equilateral polygon with an even number of sides which does not touch the circle *EFGH.*

Draw the straight line *BKD* through the center *K,*
and draw *GA* from the point *G* at right angles to the straight line *BD,* and carry it through to *C.*

Therefore *AC* touches the circle *EFGH.*

Then, bisecting the circumference *BAD,* bisecting the half of it, and doing this repeatedly, we shall leave a circumference less than *AD.*

Let such be left, and let it be *LD.*

Draw *LM* from *L* perpendicular to *BD,* and carry it through to *N.* Join *LD* and *DN.*

Therefore *LD* equals *DN.*

Now, since *LN* is parallel to *AC,* and *AC* touches the circle *EFGH,* therefore *LN* does not touch the circle *EFGH.* Therefore *LD* and *DN* are far from touching the circle *EFGH.*

If, then, we fit into the circle *ABCD* straight lines equal to the straight line *LD* and placed repeatedly, then there is inscribed in the circle *ABCD* an equilateral polygon with an even number of sides which does not touch the lesser circle *EFGH.*

Q.E.F.

This construction will actually generate a polygon whose number of sides is a power of 2 such as 8, 16, 32, etc. The next proposition requires a polygon where the number of sides is not just even, but a multiple of 4, which conveniently this construction generates.

Furthermore, the next proposition requires not just that the polygon not touch the inner circle, but the chords joining alternate vertices also not touch the inner circle, which again this construction satisfies.