Solution to the Elliptical Rock Garden Wall Problem

by In Linux Find Truth
Solution to the Elliptical Rock Garden Wall Problem

Please take a moment to review the original problem that was posed earlier which challenged the reader to find a solution to reconstructing an elliptical garden rock wall from the partially-remaining wall that was discovered by a gardener who wished to rebuild the elliptical garden wall to its former glory.

After a few weeks, no one came forth to post a solution to the problem, so I wanted to share with you my proposed solution on how the elliptical garden rock wall can be reconstructed. This solution is an analytical one, which means that the wall was measured and transposed onto a Cartesian plane and graphically analyzed using an application call GeoGebra. The endpoint of the major axis of the elliptical wall was placed at the Origin (0, 0) and denoted by point E. Two additional points were identified and measured on the wall, which are points G, and H. Points G and H were chosen and were accurately measured to correspond to point (1, 2) and (3, 3), respectively on the Cartesian plane. See the diagram below:

Now, it can be proved conclusively in advanced Euclidean Geometry that if one can identify 5 non-collinear points on a 2D Euclidean plane, then a unique conic section of that plane can be drawn. In this particular case, the conic section in question is an ellipse, which is formed by the intersection of a cone passing through the Cartesian plane. This was shown in the Braikenridge-Maclaurin Theorem.

Since we have only 3 non-collinear points identified, we must find a way to discover at least 2 more points that lie on the ellipse before we can invoke the Braikenridge-Maclaurin Theorem and produce a unique conic section (ellipse) that represents the complete reconstructed wall. The process I employed to determine the 2 additional points is depicted in the following final solution graphic that was constructed using GeoGebra. The process is quite simple. If one rotates the image above counter-clockwise along the X-axis so that mirror images of points G and H, denoted by G' and H', respectively are constructed (since we know that the ellipse is symmetrical about the X-axis), then we have our 5 non-collinear points: H', G', E, G, and H. All that remains in solving this problem is to construct the ellipse from the 5 points on the Euclidean plane. This was performed as described in the following graphic:

This graphic represents my final solution to the elliptical garden wall problem. As you can see in the diagram, the 5 points and their Cartesian coordinates have been identified and a general equation for the ellipse is presented. If point G is represented by (p, q) and H by (r, s), then values for a and b as shown above can be determined as well. The value of a was calculated to be 4.5 and represents the distance from the center of the ellipse to a vertex of the ellipse along the major axis. The value of b was calculated to be 3.181980515 and represents the distance from the center of the ellipse to a vertex of the ellipse along the minor axis. Since a > b, the ellipse's major axis lies parallel to the X-axis. The precise equation representing our elliptical garden wall is shown directly underneath the ellipse in the diagram. With the equation calculated from our measurements, we can use the analytical method of identifying all the points around the ellipse and hence we are able to reconstruct the wall to its original glory.

Please leave comments below and if you have questions, I'll attempt to answer them at my earliest opportunity. If you worked on this problem and were unable to find a solution, I hope that you find this solution to be acceptable from an analytical standpoint. There may be another non-analytical or synthetic Geometric solution, but I was not able to discover it. If you do, please post it here in the comments section and I will certainly take a look at it.

May 10, 2019
by In Linux Find Truth
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