[i]String art[/i], or pin and thread art, according to [url=http://en.wikipedia.org/wiki/String_art]Wikipedia[/url], is characterized by an arrangement of colored thread strung between points to form abstract geometric patterns or representational designs such as a ship's sails, sometimes with other artist material comprising the remainder of the work.

In the figure above strings play the same role as tangents in the previous examples. As Markus Hohenwarter, inventor of GeoGebra refers in paper [url=http://www.fachgruppe-computeralgebra.de/data/JdM-2008/Sonderheft.pdf]GeoGebra: Vom Autodesign zur Computerschriftart[/url] (2008) and in the 8th chapter of the [url=www.geogebra.org/book/intro-en.pdf]GeoGebra 4.4 Introductory Book[/url] (2013), the segments we can see are tangents to a quadratic Bézier curve.[br][br]In the following [url=http://geogebratube.org/student/m135151]applet[/url] one can do a similar experiment by using GeoGebra, and eventually use its [b]Envelope[/b] command to check whether the resulted contour curve is quadratic.

By no mean the activity to draw the segments [color=#7d7dff]A[/color]B while [color=#7d7dff]A[/color] is dragged on grid points between (0,0) and (0,10)---and meanwhile B is moved between (10,0) and (0,0)---is an easy task for many types of school pupils. Also the result as getting the contour of the segments can be expected to be a straightforward way of the next step of understanding. However, obtaining the envelope equation is at a different step of difficulty level.[br][br]First of all, the obtained equation is of 5th grade, containing not only the curve itself, but its reflection to the x-axis, and also the x-axis itself. This equation is [math]x^4y-40x^3y-2x^2y^3+600x^2y-120xy^3-4000xy+y^5-200y^3+10000y=0[/math] which is an implicit equation, but it can be factorized into three factors. Unfortunately, factorization is not discussed at secondary level, so we need to find another approach to go into the very details.[br][br]Fortunately, all these problems can be managed in secondary school by changing the construction in some sense. On one hand, we will rotate the axes by 45 degrees to obtain an explicit equation: in this case one of the parabolas can be written in form [math]y=ax^2+bx+c[/math]. On the other hand, we will use a magnification of 10, so that we will obtain parabola [math]y=\frac{x^2}{2}+\frac{1}{2}[/math] which can be described with directrix [math]y=0[/math] and focus (0,1).

Now we prove that segment [color=#7d7dff]A[/color]B is always a tangent of the parabola described above. We will use only such methods which can be discussed in a secondary school as well. We would like to compute the equation of line [color=#7d7dff]A[/color]B to find the intersection point T of [color=#7d7dff]A[/color]B and the parabola.[br][br]So first we recognize that if point [color=#7d7dff]A[/color]=(-d,d), then point B=(1-d,1-d). Since line [color=#7d7dff]A[/color]B has an equation in form [math]y=ax+b[/math], we can set up equations for points [color=#7d7dff]A[/color] and B as follows: [math]d=a\cdot(-d)+b[/math] (1) and [math]1-d=a\cdot(1-d)+b[/math] (2). Now (1)-(2) results in [math]a=1-2d[/math] and thus, by using (1) again we get [math]b=2d-2d^2[/math].[br][br]Second, to obtain intersection point T we consider equation [math]ax+b=\frac{x^2}{2}+\frac{1}{2}[/math] which can be reformulated to search the roots of quadratic function [math]\frac{x^2}{2}-ax-b+\frac{1}{2}[/math]. If and only if the discriminant of this quadratic expression is zero, then [color=#7d7dff]A[/color]B is a tangent. Indeed, the determinant is [math](-a)^2-4\cdot\frac{1}{2}\cdot(-b+\frac{1}{2})=a^2+2b-1[/math] which is, after expanding a and b, obviously zero.[br][br]Despite this is an analytical proof, by computing the x-coordinate of T (which is [math]a=1-2d[/math]) the Reader may think of finding a synthetic proof as well.