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Voila! Proofs With Iteratively Inscribed Similar Triangles Christopher Thron Texas A&M University – Central Texas

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1 Voila! Proofs With Iteratively Inscribed Similar Triangles Christopher Thron Texas A&M University – Central Texas

2 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

3 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

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

5 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:

6 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 similarity B.Corresponding segments in similar pieces are ||  ’s flipped by 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 :

7 ½  …  SAS & SSS similarity (to get central  to fit)  Corresponding  ’s of || lines (converse)  ½  & 180 o flip preserves ||  Unique || line through a given point  Completeness of plane: and closure of line Details: ½  Summary:

8 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.

9 What does this (appear to) show?  The points all lie on a single segment  The 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

10 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 point B. Cauchy sequence, completeness, closure of line C. Lengths are obtained as alternating +/- sum of segment lengths Filling in the details :

11 ½  … ½  Summary:

12  The above example iterates the operation of inscribing 180 o -rotated similar  ’s.  Try inscribing similar  ’s at other  ’s  180 o.  Depending on , there are three cases: Counterclockwise Clockwise Inverted Figures drawn with: C.a.R. (Compass and Ruler), zirkel.sourceforge.net/JavaWebStart/zirkel.jnlp

13 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.

14 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  + C j,:

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