Voila! Proofs With Iteratively Inscribed Similar Triangles

1. Voila! Proofs With Iteratively Inscribed Similar Triangles Christopher Thron Texas A&M University – Central Texas thron@ct.tamus.edu www.tarleton.edu/faculty/thron

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. Archimedes updated: parabola section =4/3 1 2 3 4 Break up 3 ’s: Evaluate area: (1 + ¼)  original  Iterate: perform skew transformations on each : 1 + ¼(1+¼ (1+¼ (… …))))  4/3 Skew transformation: doesn’t change areas: parabola parabola Original parabola section: tangent at  vertex is || to base

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. Puzzle pieces: Assemble: • 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

6. Filling in the details: • Why do the triangles fit in the holes? (and can you prove they do?) • Why do the same-colored segments line up? • How do we know that the segments all meet at a point? • SAS similarity, SSS similarity • Corresponding segments in similar pieces are || • ’s flipped by 1800still 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.

7. Summary: 2 ½  1 ½  ½  3 4 … • SAS & SSS similarity (to get central  to fit) • Corresponding ’s of || lines (converse) • ½  & 180o flip preserves || • Unique || line through a given point • Completeness of plane: and closure of line Details:

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. Filling in the details: • Why do the points all lie on the same line? • Why is the centroid on the line? • 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

11. Summary: 2 3 ½  1 ½  4… ½ 

12. 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: Clockwise Counterclockwise 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 + Cj,: