Construct a pentagon avoiding compass use
2 circles, 13 lines, 17 points



Try it on GeoGebra


Let circle(A, B) intersect circle(B, A) at C and D.
Let AB intersect circle(A, B) again at E.
Let AB intersect circle(B, A) again at F.
Let AD intersect circle(A, B) again at G.
Let AD intersect CF at H.
Let BG intersect DF at I.
Let HI intersect circle(A, B) at J and K.
Let BG intersect EJ at L.
Let BJ intersect EG at M.
Let BG intersect EK at N.
Let BK intersect EG at O.
Let LM intersect circle(A, B) at P and S.
Let NO intersect circle(A, B) at Q and R.


Then EPQRS is a regular pentagon.

Why it works

Let BE intersect GJ at T, and let BE intersect GK at U.  The complete quadrilateral BEGJ shows that T is the polar of LM, which is the intersection of the tangents at P and S.  Similarly, the complete quadrilateral BEGK shows that U is the polar of NO, which is the intersection of the tangents at Q and R.

Let FG intersect HI at V.  The diagonals DV and GI of the complete quadrilateral DGVI intersect FH at harmonic conjugates with respect to F and H; since the first is at ∞, the second is the midpoint C of FH, which is to say that C, D, V are collinear.

Let CG intersect HI at W.



Now for the fun part.  Line FUBAT is perspective about G to line VKIHJ, which is perspective about D to circle CKDGJ, which is perspective about C to line HKVWJ, which is perspective about G to line AUF∞T.  Composing these four perspecitivities yields a projectivity FUBAT ⌅ AUF∞T.  Since a one-dimensional projectivity is determined by three points, T and U are determined as the two fixed points of FBA ⌅ AF∞.

Assigning coordinates with A = 0, B = −1, F = −2, this projectivity is defined by x ↦ 4/x + 2, and its fixed points T = 1 + √5 = sec(2π/5) and U = 1 − √5 = −sec(2π/10), exactly as required to make EPQRS a regular pentagon.