Neuberg's Theorem (1892):
From point P construct perpendicular lines to the sides (or their extensions) of ΔA0B0C0. Repeat the same process from P to the Miquel ΔA1B1C1, and again to Miquel ΔA2B2C2. Then ΔA3B3C3 is similar to ΔA0B0C0.
(Below is a dynamic geometry sketch illustrating the theorem - also see A generalization of Neuberg's Theorem and the Simson line, using equi-inclined lines to the sides).
Investigate
1) Use the sketch below and drag P to investigate where the third pedal ΔA3B3C3 has maximum area.
2) Ensure that you drag P to coincide with I, the incentre of ΔA0B0C0. What do you notice?
3) Drag any of A0, B0, or C0. Then repeat steps 1) and 2).
4) Formulate a conjecture about the location of P in the interior of the initial triangle so that ΔA3B3C3 has maximum area (with respect to a fixed ΔA0B0C0).
Maximising the Area of the 3rd Pedal Triangle in Neuberg's theorem
Challenge
Can you explain why (prove that) that your conjecture in 4) above is true?
Explore More
What happens if P is dragged outside ΔA0B0C0?
5) Click on the 'Link to Neuberg triangle with excentres' button. This sketch is a 'zoomed-in' version of the preceding one showing the excentres E1, E2 and E3.
6) Drag P to each of the excentres E1, E2 and E3. What do you notice?
7) For a fixed ΔA0B0C0, write down the area of ΔA3B3C3 at each of I, E1, E2 and E3. Can you find a relationship between all four measurements?
8) Challenge: Can you explain why (prove that) your observations in 6) and 7) are true?
Submitted Paper
A paper "Optimizing Triangle Areas: Miquel Circles Center
Triangle and 3rd Pedal Triangle" by Hans Humenberger & myself, with proofs of these results, has been submitted for publication. All Rights Reserved.
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Created 23 July 2023 by Michael de Villiers, using WebSketchpad.