I'm a professor of mathematics at the University of North Carolina Asheville. I've also done a fair amount of consulting work over the years focusing on scientific and data visualization.
I'm happy to be part of the first Observable Ambassador's cohort.
IFS QuickstartIterated Function SystemsConstruction of the Serpinski triangleThe Sierpinski pedal triangleDeterministic IFS algorithmsSelf-similar spiralsStochastic IFS algorithmsGeneralized Sierpinski trianglesThe Z-CurvePoly-gasketsThe Koch snowflakeBarnsley's fernIteratedFunctionSystem ClassAffineFunction ClassSimilarity transformationsThe TwindragonBox-counting dimension examplesMcMullen carpetsThe Twindragon (with boundary)Self-affine tilesTutorial on self-affine tilesA self-affine tile with holesSelf-affine tiles via polygon mergeThe Eisenstein fractionsGolden rectangle fractalsDeterministic digraph IFS algorithmsA stochastic digraph IFS algorithmMixed self-similarityFractalized squareHilbert's coordinate functionsRauzy Fractals
Similarity and the Pedal Triangle
comments = [
html`The Sierpinski pedal triangle is an interesting modification of the simpler and more well-known Sierpinski triangle that is a bit more challenging to analyze mathematically and to generate algorithmically. The fact that it's a well-defined, self-similar set at all hinges on a not so obvious construction of Euclidean geometry. This page presents that consruction.`,
html`The construction starts with any <em>acute</em> triangle with vertices ${tex`A`}, ${tex`B`}, and ${tex`C`}. As is common in these types of constructions, we will also refer to the measure of each of those angles as ${tex`A`}, ${tex`B`}, and ${tex`C`}.`,
html`From each vertex, we project a line perpendicularly to the opposite side determining three new points ${tex`A_1`}, ${tex`B_1`}, and ${tex`C_1`}. Note that the three projected lines intersect at a common point called the <em>orthocenter</em> of the triangle, that we denote ${tex`H`}.`,
html`Now, the new points ${tex`A_1`}, ${tex`B_1`}, and ${tex`C_1`} together determine a new triangle called the <em>Pedal triangle</em> inscribed inside the original triangle. Furthemore, the complement of the Pedal triangle relative to the original forms three more triangles:
${tex.block`\triangle AB_1C_1, \: \triangle A_1BC_1, \: \text{and } \triangle A_1B_1C.`}
The key observation we need to prove is that each of those three new triangles are geometrically similar to the original.`,
html`In the course of our proof, we'll use two results taken right from <a target="_blank" href="http://aleph0.clarku.edu/~djoyce/elements/toc.html">Euclid's Elements</a>. The first of these is <a target="_blank" href="http://aleph0.clarku.edu/~djoyce/elements/bookIII/propIII21.html">Proposition 21 of Book III</a> that states
<blockquote>If two angles in a circle subtend the same arc of that circle, then they are congruent.</blockquote>
For example, the theorem implies that the two angles shown in the figure are congruent.`,
html`The second is <a target="_blank" href="http://aleph0.clarku.edu/~djoyce/elements/bookIII/propIII31.html">Proposition 31 of Book III</a> that states
<blockquote>If an angle in a circle subtends the diameter of that circle, then that angle is right.</blockquote>
The angle shown in the figure now, for example, is a right angle.`,
html`Returning to our proof, note that ${tex`\angle B_1BC`} is complementary to ${tex`\angle C`} (i.e. the sum of their measures must be ${tex`90^{\circ}`}). Thus, the measure of ${tex`\angle B_1BC`} is ${tex`90^{\circ}-C`}.`,
html`As a result, ${tex`\angle A_1HB`} is congruent to ${tex`\angle C`}, since it too is complementary to ${tex`\angle B_1BC`}.`,
html`<div>Now, we bring Euclid's theorems to bear. First, we draw a circle whose diameter is ${tex`\overline{BH}`}. Note that ${tex`\angle HC_1B`} is right by construction. Thus, ${tex`C_1`} lies on the circle by Euclid III 31. Similarly, ${tex`A_1`} lies on the circle.</div>
<div>Next, we see that ${tex`\angle A_1HB`} and ${tex`\angle A_1C_1B`} subtend the same arc in that circle. Thus, they are congruent by Euclid III 21. As a result, ${tex`\angle A_1C_1B`} is congruent to ${tex`\angle C`} as well.</div>`,
html`Of course, ${tex`\angle BA_1C_1`} must be congruent to ${tex`\angle A`} since the angles in ${tex`\triangle BA_1C_1`} must sum to ${tex`180^{\circ}`}. This imples that ${tex`\triangle BA_1C_1`} is similar to the original ${tex`\triangle ABC`} since all the angles are the same.`,
html`Naturally, the same argument applies to the other triangles as well.`,
html`Thus, the three light-gray triangles are all similar to the original triangle as claimed.`,
html`Of course, a major point is that the process can be iterated over...`,
html`and over...`,
html`Until the limit is clear.`
]
// The images were all generated using Mathematica code I have from years ago.
// Observable is just used to coordinate the presentation.
images = Promise.all([
FileAttachment("pedalProof00.png").image(),
FileAttachment("pedalProof01.svg").image(),
FileAttachment("pedalProof02@1.svg").image(),
FileAttachment("pedalProof03.svg").image(),
FileAttachment("pedalProof04.svg").image(),
FileAttachment("pedalProof05.svg").image(),
FileAttachment("pedalProof06.svg").image(),
FileAttachment("pedalProof07.svg").image(),
FileAttachment("pedalProof08.svg").image(),
FileAttachment("pedalProof09@1.svg").image(),
FileAttachment("pedalProof10.svg").image(),
FileAttachment("pedalProof11.svg").image(),
FileAttachment("pedalProof12.svg").image(),
FileAttachment("pedalProof13.svg").image(),
FileAttachment("pedalProof14.png").image()
])
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