Mascheroni Construction of a Regular Pentagon
What follows is a reproduction of a short note by Kurt Hofstetter giving a compass-only construction of a regular pentagon:
In 2002 we have given a simple 5-step compass-only (Mascheroni) construction of the golden section. Here we note that with two additional circles, it is possible to construct the vertices of a regular pentagon. As usual, we denote by P(Q) the circle with center P and passing through Q.
Construction 1. Given two points A and B,
- C_{1} = A(B),
- C_{2} = B(A) to intersect C_{1} at C and D,
- C_{3} = C(D) to intersect C_{1} at E and C_{2} at F,
- C_{4} = A(F),
- C_{5} = B(E) to intersect C_{4} at G and H.
- C_{6} = G(C) to intersect C_{3} at I and J,
- C_{7} = H(C) to intersect C_{3} at K and L.
Then DIKLJ is a regular pentagon.
Proof: In a 2002 paper, we have shown that the first five steps above lead to four collinear points C, D, G, H such that D divides CG, and C divides DH, in the golden section.
This means that in the isosceles triangle GCI, GC/IC = GC/DC = φ.. The base angles are 72° Therefore, ∠DCI = 72° By symmetry, ∠DCJ = 72°.
Also, in the isosceles triangle HCK, KC/CH = DC/CH = φ.. The base angles are ∠36°. It follows that ∠KCH = 36° By symmetry, ∠LCH = 36°, and ∠KCL = 72°.
Since C is on the line GH, ∠ICK = 180° - ∠GCI - &anf;KCH = 72°. By symmetry, ∠JCL = 72°.
Therefore, the five points D, I, K, L, J are equally spaced on the circle C_{3}.
They form the vertices of a regular pentagon.
Remark. The circle C_{7} can be replaced by C_{8} with center D and radius IJ. This intersects C_{3} at the same points K and L.
References
- K. Hofstetter, A Simple Compass-Only Construction of the Regular Pentagon, Forum Geometricorum Volume 8 (2008) 147-148.
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