voila! proofs with iteratively inscribed similar triangles christopher thron texas a&m...

Voila! Proofs With Iteratively Inscribed Similar Triangles

Christopher ThronTexas A&M University – Central Texas

thron@ct.tamus.eduwww.tarleton.edu/faculty/thron

Why’s it so great to iterate?• Ancient: “method of exhaustion” was

used by Archimedes to find areas.

• Modern: Fractals (now part of the standard high school geometry curriculum)

• Visually appealing, and amenable to modern software.

• Hugely important technique in modern analysis

• Can lead to proofs that are visually immediate (“Voila!”)*

* Although technical details can be nasty

Original parabola section: tangent at vertex is || to base

1

Archimedes updated: parabola section =4/3

Skew transformation:

doesn’t change areas: parabola parabola

2

Break up 3 ’s:

Evaluate area: (1 + ¼) original

3

Iterate: perform skew transformations on each :

1 + ¼(1+¼ (1+¼ (… …)))) 4/3

4

1. The Centroid Theorem

• The three medians of a triangle meet at a single point (called the centroid)

• The centroid divides each median in the ratio of 2:1

What does this (appear to) show? Blue, red, and green lines all meet at a

single point. Dark-colored segments are 2 as long

as light-colored segments

Puzzle pieces:

Assemble:

A. Why do the triangles fit in the holes? (and can you prove they do?)B. Why do the same-colored segments line up?C. How do we know that the segments all meet at a point?

A. SAS similarity, SSS similarityB. Corresponding segments in similar pieces are ||

’s flipped by 1800 still have corresponding segments ||.C. Completeness property --Cauchy seq. in the plane converges to a unique point. Closure property: a line in a plane contains all its limit points.

Filling in the details:

½ 1 2

3 4 …

SAS & SSS similarity (to get central to fit)Corresponding ’s of || lines (converse)½ & 180o flip preserves ||Unique || line through a given pointCompleteness of plane: and closure of line

Details:

½

½

Summary:

2. The Euler Segment

• The circumcenter, centroid, and orthocenter of a triangle are collinear

• The centroid divides the segment from orthocenter to circumcenter in the ratio 2:1.

What does this (appear to) show?The points all lie on a single segmentThe line must contain the centroid,

because the triangles shrink down to the centroid.

By considering lengths of segments, the centroid splits the segment as 2:1

A. Why do the points all lie on the same line?B. Why is the centroid on the line?C. Why is the |Ortho-Centroid|: |Circum – Centroid| = 2:1?

A. Similar reasoning to last time: unique parallel line through a given pointB. Cauchy sequence, completeness, closure of lineC. Lengths are obtained as alternating +/- sum of segment lengths

Filling in the details:

½ 1 2

3 4…

½

½

Summary:

The above example iterates the operation of inscribing 180o-rotated similar ’s.

Try inscribing similar ’s at other ’s 180o. Depending on , there are three cases:

CounterclockwiseClockwise Inverted

Figures drawn with: C.a.R. (Compass and Ruler), zirkel.sourceforge.net/JavaWebStart/zirkel.jnlp

Given ABC (clockwise). Successively inscribe similar ’s at any clockwise angle .

The inscribed ’s converge to a point P with the property: PAB = PBC = PCA.

Summary:

• The “equal angle” point P is unique (Proof: “3 impossible regions”)• P is called a Brocard point• Any sequence of clockwise-inscribed similar ’s will converge to the

Brocard point, as long as the size 0 (the 3 fan shapes are always similar)

• The vertices of the fan shapes lie on three logarithmic spirals: of the form: ln(r) = k + Cj,: