Since nobody has insulted me on this forum, and Torkain hasn't deleted any of my posts, I am going to post the full details of the Poussin "Binary Star" geometry, as well as the hexagon/pentagon which is required to produce it with full precision. I also like the date, as it is the 1st day of the 7th month, and the digits of the day/month/year total 13. Here is the article, it is directions for drawing the solution geometry into any good print of the Poussin painting. Of course, you would need some basic comprehension skils to actually carry it out. This article should devastate all naysayers and competetors........ To prove that my ‘binary star’ Poussin solution geometry is legitimate, I will now explain how to draw the geometry shown on my website, as well as the rest of the geometry that gives birth to the Binary Star. It consists of a combination of a perfect hexagon/hexagram combined (or ‘married’) with a pentagon/pentagram. The unusual thing about this ‘marriage’ is that in order to facilitate this particular combination, the pentagon must be every slightly shortened in one area. This results in a pentagon that still retains all five of its internal 108 degree angles, but when the pentagram is drawn within it, that pentagram has all of its angles altered from those of a true pentagram to varying extents.
We’ll begin by drawing a rectangle perfectly containing two overlapping equilateral triangles, one pointing up and the other down. The center of the top edge of this rectangle will be at the top of the left staff in the Poussin painting, and the upper right corner of the rectangle will be at the top of the right staff. Thus, parts of the rectangle will be outside of the painting’s left and lower edges.
Now we draw a horizontal hexagon within the rectangle, one point touching the center of the left edge and the other touching the center of the right edge. We must also draw a vertical line down the center of the rectangle, joining the top and bottom edges thereof.
We must now introduce the pentagon. The pentagon will be oriented horizontally, with one of its points oriented directly to the right. We do this by drawing lines from the two right corners of the rectangle, in toward its vertical centerline, at angles of 18 degrees from the top and bottom edges thereof (make the upper one go all the way across the rectangle), then drawing lines from the same two rectangle corners, out to the right, at angles of 36 degrees from the right edge, to meet and form a point. We now have a conjoined hexagon and pentagon, the pentagon being only slightly altered as I mentioned previously.
We could draw a full pentagram within this pentagon, but it won’t be required for our purposes. We will, however, need to draw a line from the exact center of the upper 36 degree line that we drew in the previous step, down and leftward to intersect the vertical centerline at the same point where the lower 18 degree line had intersected it. We may as well complete vertically oriented hexagram that fits inside the horizontal hexagon also, by drawing two horizontal lines across the rectangle ¼ way in from the top and bottom edges.
Now comes the part where the binary star is produced. Draw a line from the center of the top edge of the rectangle, down and rightward at 18 degrees from the vertical centerline. This will be the centerline of the large star, make it as long as you think you will need, to the bottom of the rectangle is convenient with plenty to spare. Now mark a point to the left of this 18 degree centerline, on the upper 18 degree pentagon line you drew across the rectangle, the same distance from the 18 degree vertical star centerline as the upper right corner is from the right of it. A compass is the easiest way to do this. That point you just marked will be the left arm point of the large star, and the upper right corner of the rectangle will its right arm point.
Now then, the line that runs diagonally from the top right corner of the rectangle, to its lower left corner, will be the lower arm line of the right arm of the star. In order to produce the left lower arm line, simply draw a line from the now established left arm point, down and rightward to pass through the intersection of the star’s centerline with the rectangle’s diagonal line which joins its upper right and lower left corners. Continue drawing this line until it touches the pentagon diagonal that we drew from the center of the upper 36 degree line. This touch point at the pentagon diagonal is where the right leg point of the star will be established. Now draw a line, from that now established right leg point, up to the center of the top edge of the rectangle (which is also the top point of the star). Now simply mirror this line, to the left side of the star’s centerline, to complete the left leg. The large star is now complete. The angles at the tips of the arms and legs will be long decimal numbers, and the angle of the top point of the star will be 66.0057724 degrees. Thus, by this convoluted procedure, we have obtained a star whose top angle is a perfect 66 degrees to two decimal places, a remarkable chance result of the marriage between the hexagon and pentagon. The inaccuracy is, therefore, only about 1/180th of one degree. The long decimal numbers of the arms and legs are of sufficient magnitude to actually be observable when taking precise angle readings from the features in the Poussin painting. This, then, provides a mathematical proof, if you will, of the Binary Star part of my proposed Poussin solution. I contend that no other possible solution can explain these anomalous readings. This is done by taking readings of the angles between the line connecting the tops of the left and right staves, and either the vertical tomb line (pointed at by the right shepherd), or the centerline of the right staff. The tomb line is obviously a more precise reference, as it is much finer. Both will be off from whole angle readings by 0.18189111505 degree. The tomb line is at exactly 6 degrees to the right staff, by the way.
The small star is exactly ¾ the size of the large star, its centerline passes through the right ‘shoulder’ intersection of the large star and is angled at exactly three degrees toward its centerline. To position its arm line, extend the right half of the top edge of the rectangle rightward for a distance Phi (1.618etc) times the length of said right half of rectangle edge. Then draw a line leftward from the right end of that extended line, at an angle of 3 degrees down from it. All the angles of the small star are identical to those of the large one except its vertical centerline is angled at 21degrees leftward, rather than 18.
Copyright Brian Ettinger July 1, 2003