Inscribing square in circle in just seven compass-and-straightedge steps

Problem Here is one of the challenges posed on Euclidea, a mobile app for Euclidean constructions: Given a O centered on point O with a point A on it, inscribe ABCD within the circle — in just seven elementary steps. Euclidea hints that the first two steps use the compass, the third uses the straightedge, and the last four use the straightedge to draw the sides themselves of ABCD.

Definitions The problem is not considered solved until each line containing one of the four sides of the desired square are drawn; merely finding the vertices of the square does not suffice. Naming and creating points are of course allowed and fortunately do not count toward the seven elementary steps allowed in this problem. Other than these ‘zero-step’ steps, Euclidea permits only two elementary steps, each costing one step:

  1. Create an infinite line connecting two points using an unmarked, one-dimensional, infinitely long straightedge. (Even merely extending a given line segment costs one step.)
  2. Create a circle using a compass that collapses immediately thereafter.

Research of previous Mathematics Stackexchange questions I am aware that there is a seven-step process previously described at How can I construct a square using a compass and straight edge in only 8 moves?. Notwithstanding the post’s title, it actually has just seven steps since its first corresponds to constructing the given O. This solution fails, however, since the resulting inscribed square is neither inscribed in the given O nor inclusive of the given point A As a vertex.

Attempt 1: 8-step solution using perpendicular bisectors

  1. Take one step to extend ¯AO to the other side of O.
    • Take as point C the new intersection between said line and circle.
  2. Take three steps to define L, the perpendicular bisector of diameter ¯AC.
    • Take as points B and D the intersections of L with O.
  3. Take four steps to draw the sides themselves of ABCD.

Attempt 2: 8-step solution using a 15-75-90 triangle It turns out that @Blue’s successful 7-step solution uses much the same circles and 15-75-90 triangle as the one proposed here.

8-step process

  1. Take one step to create A with radius AO.
    • Take as point E1 the ‘left’ resulting point of intersection.
    • Take as point E2 the ‘right’ resulting point of intersection.
  2. Take one step to create P with radius E1O.
  3. Take one step to create Q with radius E1E2.
    • Take as point C the intersection point between circle Q and circle O.
    • Take as point F the intersection point between circle Q and circle A.
  4. Take one step to create E1F.
    • Take as point G1 the resulting ‘top’ intersection point with P.
    • Take as point G2 the resulting ‘bottom’ intersection point with P.
  5. Take one step to create AG1 to effectively draw ¯AB.
    • Take as point B the resulting intersection between said line and O.
  6. Take one step to create ¯BC.
  7. Take one step to create AG2 to effectively draw ¯AD.
    • Take as point D the resulting intersection between said line and O.
  8. Take one step to create ¯CD.
    • This completes desired ABCD, albeit in one too many steps.

Answer

image description here

  1. Construct A through O.
  • Let P1 and P2 be the points where A meets O.
  1. Construct P1 through P2.
  • Let C be the (other) point where P1 meets O.
  1. Construct OP1.
  • Let Q1 and Q2 be the points where OP1 meets P1.
  1. Construct CQ1.
  • Let B be the point where CQ1 meets O.
  • Note that we have constructed BC.
  1. Construct CQ2.
  • Let D be the point where CQ2 meets O.
  • Note that we have constructed CD.
  1. Construct AB.
  2. Construct AD.

Square ABCD, with constructed edge-lines, is inscribed in O. (Proof that the quadrilateral is, in fact, a square, is left as an exercise to the reader.)


Edit. Having been asked to elaborate on the square

  • As of Step 2, we know P1P2C is equilateral and that ¯OA is on the perpendicualr bisector of side ¯P1P2. Therefore, ¯AC is a diameter of O, and we have that ABC and ADC (for point D constructed later) are right angles by Thales’ Theorem.

  • As of Step 4, as observed by Jan and Tristan in the comments, ¯Q1Q2 is a diameter of P1, so Q1CQ2 is a right angle. Therefore, ABCD is at least a rectangle.

  • Now, define a:=|¯OA|, so that |¯P1P2|=a3 and |¯OQ1|=a(1+3). Since AOQ1=60, if we let R be the foot of the perpendicular from Q1 to OA, then |¯OR|=a2(1+3) and
    |¯Q1R|=a32(1+3)=a2(3+3)=a+|¯OR|=|¯CR| Thus, Q1CR=45 and we may conclude that ABCD is a square.

Attribution
Source : Link , Question Author : PDE , Answer Author : PranshuKhandal

Leave a Comment