GEOMETRIC ART PROBLEM (Updated)

The problem solved may be stated thus:

Using only a drafting compass and straightedge (as did Euclid, the ancient Greek mathematician) draw a random horizontal line (“Line 1”) and locate a point (“Point 1”) some height above Line 1 randomly;  

Then draw another line (“Line 2”) parallel to Line 1 through Point 1;

Then draw a 30/60/90 right triangle with the 90-degree angle vertex on Point 1 and the hypotenuse lying on Line 1.

First, I had to make my own large compass, because no store-bought compass was big enough to achieve the necessary markings.  I made it out of a strip of lath, cut into two longer pieces, two much shorter pieces, and another strip to hold the compass in position and to also use as the “straightedge.”  I drilled 13/16” holes in both ends of the longer pieces and in one end of each short piece.  I drilled another hole some distance from the end of one of the longer pieces and in one end of the “straightedge” piece, then attached all of them with (4) #10-32 machine screws, washers and wing-nuts.  I placed a washer under the head of each screw, then one between the pieces of lath being screwed together, then another washer under each wing-nut.  I screwed one end of the two longer pieces together as a “pivot” for the compass.  I then attached the two shorter pieces to each of the two longer pieces with screws and wing-nuts, then I attached the “straightedge” to one of the longer pieces with the extra hole in it with a screw and wing-nut.  I secured the pencil to one of the shorter pieces with a binder clip and used another binder clip to hold the “straightedge” to the other compass arm.  Finally, I removed one of the other short pieces and drilled a shallow pilot hole in the end, drove a small brad securely into the pilot hole, then ground off the brad head with a grindstone to a fine point and reattached it to the compass.

Next, using the new, homemade compass, I cut two equidistant marks across Line 1 from and on either side of Point 1, then marked a central point (“Point 2”) some distance below the Line 1 with crossed arcs from those marks and equal in distance to the new Point 2 from the old Point 1.  Then I “dropped” a perpendicular (“Perp”) from Point 1 to Line 1 as extended through Point 2 below.  That Perp is thus to be one side of the equilateral triangle used to form the 30° angle on Line 1.  Using the compass, I then measured the distance from the intersection of Perp with Line 1 to one of those equidistant marks on it.  Again, using the compass, I then cut an equidistant arc horizontally from Point 1 over and above the equidistant mark.  Yet again, using the compass, I measured the distance up Perp from Line 1 to Point 1, then moved the compass over to the equidistant mark and cut another arc across the first, that being the measured point from which the parallel Line 2 would pass through Point 1.  And then I drew parallel Line 2.

To construct the triangle, I then used the compass to cut a mark on Line 1 (Point 3) from Point 1 equal to the distance down Perp from Point 1 above Line 1 to Point 2 below Line 1.  I then checked that length from Point 2 to Point 3 to be sure it was equal length.  That creates the equidistant triangle bisected by Line 1.  Every equidistant triangle has angles of 60 degrees at its vertices, so the bisected vertex angle would be 30 degrees.  I then drew a new line from Point 3 to Point 1, showing the 30-degree angle.

I then faintly extended that new line beyond Point 1 and “dropped” another perpendicular from it to Line 1, being on an angle with Line 1, thus forming both the 90-degree right angle at Point 1 and also the 60-degree angle at Line 1, since all triangles contain 180 degrees at their vertices.  I then drew in those respective lines to complete the 30/60/90 right triangle at Point 1.

I intend to paint this construction in full color as a piece of art.

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