Sources page biographical material

6.BR. WHAT IS A GENERAL TRIANGLE?
David & Geralda Singmaster, proposers; Norman Miller, solver. Problem E1705 -- Skewness of a triangle. AMM 71:6 (1964) 680 & 72:6 (1965) 669. Assume a  b  c. Define the skewness of the triangle as S = max {a/b, b/c, c/a} x min {a/b, b/c, c/a}. What triangles have maximum and minimum skewness? Minimum is S = 1 for any isosceles triangle. Maximum occurs for the degenerate triangle with sides 1, φ, 1+φ, where 1 + φ = φ², so φ = (1 + 5)/2 is the golden mean.

Baruch Schwarz & Maxim Bruckheimer. Let ABC be any triangle. MTr 81 (Nov 1988) 640-642. Assume AB < AC < BC and A < 90^o. Drawing BC and putting A above it leads to a small curvilinear triangular region where A can be. Making A equidistant from the three boundaries leads to a triangle with sides proportional to 33, 7, 8 and with angles 44.5^o, 58.5^o, 77^o. The sides are roughly in the proportion 6 : 7 : 8.

Gontran Ervynck. Drawing a 'general' triangle. Mathematics Review (Nov 1991). ??NYS -- cited by Anon., below. Notes that if we take an acute triangle with angles as different as possible, then we get the triangle with angles 45^o, 60^o, 75^o.

Anon. [possibly the editor, Tom Butts]. What is a 'general' triangle? Mathematical Log 37:3 (Oct 1993) 1 & 6. Describes above two results and mentions Guy's article in 8.C. Gives an argument which would show the probability of an acute triangle is 0.

Anon. [probably the editor, Arthur Dodd]. A very scalene triangle. Plus 30 (Summer 1995) 18-19 & 23. (Content says this is repeated from a 1987 issue -- ??NYS.) Uses the same region as Schwarz & Bruckheimer, below. Looks for a point as far away from the boundaries as possible and takes the point which gives the angles 45^o, 60^o, 75^o.

In 1995?, I experimented with variations on the definition of skewness given in the first item above, but have not gotten much. However, taking a = 1, we have 1  b  c  b + 1. Plotting this in the b, c plane gives us a narrow strip extending to infinity. For generality, it would seem that we want c = b + ½, but there is no other obvious condition to select a central point in this region. As fairly random points, I have looked at the case where c = b², which gives b = (1 + 3)/2 = 1.366.., c = 1.866.. -- this triangle has angles about 31.47^o, 45.50^o, 103.03^o -- and at the case where b = 3/2, which gives a triangle with sides proportional to 2, 3, 4 with angles about 28.96^o, 46.57^o, 104.46^o. In Mar 1996, I realised that the portion of the strip corresponding to an acute triangle tends to 0 !!

I have now (Mar 1996) realised that the situation is not very symmetric. Taking c = 1, we have 0  a  b  1  a + b and plotting this in the a, b plane gives us a bounded triangle with vertices at (0, 1), (½, ½), (1, 1). There are various possible central points of this triangle. The centroid is at (1/2, 5/6), giving a triangle with sides 1/2, 5/6, 1 which is similar to 3, 5, 6, with angles 29.93^o, 56.25^o, 93.82^o. An alternative point in this region is the incentre, which is at (½, ½{-1 + 22}), giving a triangle similar to 1, -1 + 22, 2 with angles 29.85^o, 65.53^o, 84.62^o. The probability of an acute triangle in this context is 2 - π/2 = .429.