I want to calculate the intersection points of the following image:


Assume that the three points of the triangle could be located anywhere.

How would I do this?

  • $\begingroup$ I think the formula for the quadratic spline can be adapted for my needs. $\endgroup$ – posfan12 Aug 28 '13 at 22:15
  • $\begingroup$ The image illustrates a parabola as the "envelope" of lines of the form $$x(a-t)+yt=t(a-t)$$ (where $a$ is the (constant) length of the "legs" of the figure, and $t$ is a parameter that varies from $0$ to $a$). The way envelopes work, the parabola doesn't contain points of intersection of the lines; the parabola is tangent to each line at some point (usually not the point of intersection with another line). Do you really want the points-of-intersection? Or do you want the points(-of-tangency) actually on the parabola? $\endgroup$ – Blue Aug 28 '13 at 23:10
  • $\begingroup$ I need the points of intersection. I already know how to generate the spline. $\endgroup$ – posfan12 Sep 4 '13 at 0:55

Let's first solve the problem where the figures right-corner is at the origin, and its legs (of unit length) align with the axes.

The intercept-intercept form of the line equation gives this parameterization of the various lines: $$\frac{x}{t} + \frac{y}{1-t} = 1 \quad \text{or, in fraction-free form,} \quad x(1-t) + y t = t (1-t)$$ for $t$ a parameter between $0$ and $1$ (the extreme values being valid only in the fraction-free version).

Now, if we divide each leg into $n$ pieces (with $n+1$ equally-spaced points), then the intersection of the lines corresponding to $t=\frac{i}{n}$ and $t=\frac{j}{n}$, with $i, j \in \{0, 1, \dots, n\}$, is the point $$P_{ij} := \frac{1}{n^2}\large(\;ij\;,\;(n-i)(n-j)\;\large)$$

When the figure is located "anywhere", we need to apply a simple transformation. Note that, because the intersection points are defined via ratios of lengths of parallel segments, an affine transformation (that is, a linear transformation followed by a translation) will preserve the pattern.

Let's say that the resulting figure should have its (not-necessarily-right) corner at $C$ and its (not-necessarily-unit-length) legs ending at points $A$ and $B$.

We can write our original intersection points $P_{ij}$ as $$P_{ij} = \frac{ij}{n^2} u + \frac{(n-i)(n-j)}{n^2} v$$ where $u$ and $v$ are the unit vectors in the positive $x$ and $y$ directions.

Replacing $u$ with $A-C$ and $v$ with $B-C$ transforms the figure into the correct shape, though its corner remains at the origin; adding $C$ translates the figure into place. So, our transformed intersection points are given by

$$\begin{align} P^\prime_{ij} &= \frac{ij}{n^2} (A-C) + \frac{(n-i)(n-j)}{n^2} (B-C) + C \\[6pt] &= \frac{ij}{n^2} A \;+\; \frac{( n-i )( n - j )}{n^2} B \;+\; \frac{i (n-j) + j(n-i)}{n^2} C \end{align}$$

  • $\begingroup$ As I said in my original post, the three points of the triangle formed by the two outer segments can be located anywhere, in any configuration. I don't think your formula works when that is the case. $\endgroup$ – posfan12 Sep 4 '13 at 20:19
  • $\begingroup$ Ah, right. Well, we just need to apply an appropriate transformation. I'll adjust my answer. $\endgroup$ – Blue Sep 4 '13 at 20:37
  • $\begingroup$ Sorry. How do I solve this type of equation? I am accustomed to using parametric formulas where there are two equations - one for the x value, one for the y value. $\endgroup$ – posfan12 Sep 4 '13 at 23:14
  • $\begingroup$ Give the triangle's vertices coordinates, $A(x_a,y_a)$, $B(x_b,y_b)$, $C(x_c,y_c)$. For brevity here, let $\alpha$, $\beta$, $\gamma$ be the multipliers on $A$, $B$, $C$. Then, writing $(x,y)$ for $P^\prime_{ij}$, the formula $$P^\prime_{ij} = \alpha \; A + \beta \; B + \gamma \; C$$ simply means $$\begin{align}x &= \alpha \; x_a + \beta \; x_b + \gamma \; x_c \\ y &= \alpha \; y_a + \beta \; y_b + \gamma \; y_c\end{align}$$ $\endgroup$ – Blue Sep 4 '13 at 23:40

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