During courses on geometry it is sometimes necessary to draw a triangle on the blackboard that can easily be recognized as a general triangle. It must not be rectangular and must not have two or more equal angles. Further all angles should be less than $\pi/2$. Has anybody optimized this old problem of geometry-teachers?
The book by "Humor in der Mathematik" by Friedrich Wille (from the 1970s or 1980s) contains the tongue-in-cheek theorem "Up to similarity, there is a unique general triangle". (Google book search for "Friedrich Wille" "Allgemeines Dreieck")
"General" is defined as "all angles must differ from each other, and from 90 degress, by at least 15 degrees".
Given some further axioms (like: diagonals of acute angles have to be visually distinct from line of symmetry) it is also shown that there is a unique general quadrilateral, and that this quadrilateral is "teeming" (another technical term) with general triangles.
Since the question is reopened, I convert my comments into an answer.
The "general triangle" must have angles 75,60,45 degrees (as in Carl Dettmann's comment above), and so there is only one such triangle up to similarity. Here is a "psycho-mathematical" proof. 1) The triangle should not be obtuse (otherwise it is not a general triangle but an obtuse one). 2) The difference between any two angles should be at least 15 degrees (otherwise for an untrained eye of a student the triangle will look too isosceles). 3) The biggest angle should be at most 75 degrees (at least 15 degrees apart from 90), otherwise the triangle looks "too right". 4) The sum of angles must be 180 degrees (otherwise the 5th postulate would be broken). Parts 1),2),3), 4) immediately imply the claim.
PS There are many general triangles on the hyperbolic plane and none on the sphere.
protected by Lucia May 14 '15 at 0:32
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