Two isosceles right triangles, one larger ($\triangle ABC$) than the other ($\triangle DEF$), superimposed such that the smaller fits completely within the boundary of the larger.

However, $\triangle DEF$ should be sized and placed such that the following constraints are met for a given $\triangle ABC$:

  • $DE$ is a fixed distance, $h$ from $AB$
  • $EF$ is a fixed distance, $g$ from $BC$
  • $DF$ is a fixed distance, $i$ from $AC$

It is simple to calculate $E$ but I'm struggling to calculate the coordinates for $D$ and $F$.

  • $D(h, ?)$
  • $E(h, g)$
  • $F(?, g)$

Alternatively, given the offset $E$, what should be the lengths of $DE$ and $EF$ to satisfy the constraint for distance $i$.

enter image description here


For a given outer isosceles right triangle, $\triangle ABC$, I wish to define an inner isosceles right triangle, $\triangle DEF$, where the distance between all parallel sides are also given.

Geometric dilation works where the distances are all equal but I haven't found an analytical solution for where they are all different.


I am trying to produce a parameterized CAD drawing of a hollow triangular shaped box, where the thickness of each wall can be different and defined for any size box (using OpenSCAD).

  • $\begingroup$ You did not give any coordinates! $\endgroup$ – QuIcKmAtHs Jan 20 '18 at 10:37

Take a look at the figure below:

enter image description here

Here, I used the OP's notations. The key to the solution is that $\alpha=45^{\circ}.$ Since $\cos(\alpha)=\frac1{\sqrt2}$, $$m=\sqrt2\ i.$$ That is the equation of $e$ is

$$y=-x+10-\sqrt2\ i.$$

It is easy, then, to find the intersection points sought for. For example, in the case of point $D$, we have $x=1$ and $y=-x+10-\sqrt 2\ i$, so $E=\big(1,9-\sqrt 2\ i\big)$.

  • $\begingroup$ This answers my question succinctly and clearly shows how you derived the answer which helped me think about how to approach the problem. Thank you very much. $\endgroup$ – groenewaldd Jan 22 '18 at 6:56

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