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 geometryteachers?

$\begingroup$ Related to this question is the observation that even if you succeed to draw a "general triangle" as described above, you can still "prove" that it has to equal sides (hence two equal angles). Of course, you cheat somewhere, but it is very subtle. This was shown to me by my math teacher when I was 12, and I never forget the argument. This teacher claimed that "Geometry is the art of making correct reasoning from wrong pictures"; in French "la Géométrie est l'art de raisonner juste sur des figures fausses". $\endgroup$ – Denis Serre Apr 25 '13 at 14:15

7$\begingroup$ The moment you have optimized the triangle, it ceases to be general. $\endgroup$ – Lee Mosher Apr 25 '13 at 14:58

1$\begingroup$ Well, if you maximise the differences you mention, the answer is 456075. But in the attempt to generate the most general triangle we have arrived at one that is very special... $\endgroup$ – user25199 Apr 25 '13 at 14:58

$\begingroup$ Thanks for comments and answers. As Wille says: The student must not identify the general triangle as a special one. So it has to be drawn very unobtrusively. The next question would be then how to draw this triangle freehands. $\endgroup$ – Rhett Butler Apr 25 '13 at 19:53

5$\begingroup$ Discussion at tea.mathoverflow.net/discussion/1582/… $\endgroup$ – Daniel Moskovich May 3 '13 at 10:10
The book by "Humor in der Mathematik" by Friedrich Wille (from the 1970s or 1980s) contains the tongueincheek 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.

$\begingroup$ There is a formal, machineverified proof of this theorem at afp.sourceforge.net/entries/GeneralTriangle.shtml $\endgroup$ – Joachim Breitner May 13 '14 at 12:08

$\begingroup$ The stated theorem is clearly false. General triangles (as defined here) include, among many others, 105/70/5, 105/65/10, and 75/60/45. $\endgroup$ – Steven Landsburg Aug 5 '16 at 23:12

$\begingroup$ They only defined the notion of "general" for acute triangles. You must remember that this proof is from the 20th century, when suffrage was limited to cis90 triangles. $\endgroup$ – Goldstern Aug 29 '16 at 16:13
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 "psychomathematical" 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.

$\begingroup$ Thanks for the answer, 1+. But why are no general triangles on the sphere possible? There you have the chance to subdivide more than 180° into three angles. $\endgroup$ – Rhett Butler May 3 '13 at 20:08

$\begingroup$ If the largest angle is $\le 75$ (Conditions 1, 3), the next one is $\le 60$ (Condition 2) and the next one is $\le 45$ (Condition 2), then the total angle is $\le 180$ which is impossible on a sphere. $\endgroup$ – user6976 May 3 '13 at 21:34

2$\begingroup$ @Mark: I see. I overlooked the notobtuse condition. Personally, in a 270° triangle I would admit an obtuse angle. $\endgroup$ – Rhett Butler May 4 '13 at 15:22

6$\begingroup$ Perhaps you are right: in the spherical world obtuseness may not be considered abnormal. $\endgroup$ – user6976 May 5 '13 at 3:14