Construction of a Pentagon

This construction may be done using only a compass—no ruler is needed. I have taken it from the excellent book What Is Mathematics? by Richard Courant and Herbert Robbins, recently reprinted by Oxford University Press with addenda by Ian Stewart.

Construction

(Use the check boxes to highlight steps in the construction.)

Start On a circle with center O, choose a point A.

Step 1 Construct points B, C, D on the circle so that OA = AB = BC = CD.

Step 2 Draw arcs with radius AC, centered at A and D, meeting at X outside the circle.

Step 3 Draw an arc with radius OX centered at A and meeting the circle at F.

Step 4 Draw arcs with radius FO, centered at F, meeting the circle at G and H.

Step 5 Draw arcs with radius OX, centered at G and H, meeting inside the circle at Y.

Finish Then AY is the length of an inscribed pentagon.

Proof

(Use the check boxes to highlight steps in the proof.)

P1 Let the radius of the circle be 1.

P2 The arcs AB, BC, and CD subtend angles of 60°, and triangles OAB, OBC, and OCD are equilateral.

P3 Therefore AC = √3.

P4 By Pythagoras, OX = √2.

P5 Therefore, arc AF subtends an angle of 90° , and F bisects arc BC.

P6 Triangles FGO and FHO are equilateral.

P7 GH bisects FO perpendicularly; let M be the intersection of GH and FO. Then GM = √(3/4).

P8 By Pythagoras, MY = √(5/4).

P9 But MO = 1/2. Therefore OY = (√5 - 1)/2 = τ - 1.

P10 Again by Pythagoras, AY² = 1 + (τ - 1)² = 1 + (2 - τ) = 3 - τ.

P11 Therefore AY = √(3 - τ). From the trigonometry of Golden-Ratio triangles we know that √(3 - τ) = 2 sin 36°.

Last modified on $Date: 2010-12-05 23:14:00 -0500 (Sun, 05 Dec 2010) $

Christopher J. Henrich