Here’s the answer. I don’t have a server to post the picture that I used, so I’ll have to describe it very carefully.

I’ll deal first with all triangles whose longest side is a given length, C.

[list=1][li]Orient side C so it is horizontal. Let its enpoints be p[sub]1[/sub] and p[sub]2[/sub].[/li]

[li]Draw a circle around p[sub]1[/sub] with radius C. Draw another circle around p[sub]2[/sub] with radius C. The intersection of these two circles is an American-football-shaped area.[/li]

*All the points contained in this area are the third points of the triangles whose longest side is C.* Any points outside of this area are a distance greater than C from either p[sub]1[/sub] or p[sub]2[/sub], which means they describe triangles whose longest side is *not* C.

[li]Within the football, label one of the apexes (pointy ends) p[sub]3[/sub] and the other p[sub]4[/sub]. Draw a line between p[sub]1[/sub] and p[sub]3[/sub], p[sub]1[/sub] and p[sub]4[/sub], p[sub]2[/sub] and p[sub]3[/sub], and p[sub]2[/sub] and p[sub]4[/sub]. These lines and line p[sub]1[/sub]p[sub]2[/sub] should outline two equilateral triangles of length C.[/li]

[li]Draw a cirlce whose diameter is C. It should be bisected by line p[sub]1[/sub]p[sub]2[/sub].[/li]

From trigonometry, we know that any triangle inscribed in a semicircle is a right triangle. So, every triangle described by p[sub]1[/sub], p[sub]2[/sub] and a point on this circle is a right triangle. *(The percentage of right triangles, therefore, is the area of the circumference of the circle [not the circle itself] divided by the area of the football. Since the area of a line is infinitesimally small, this percentage is essentially 0.)*

Also from trig, any triangle described by p[sub]1[/sub], p[sub]2[/sub], and a point *inside* the circle is obtuse.

So, the percent of obtuse triangles is the area of the circle divided by the area of the football.[/list=1]

Now, let:[ul][li]R be the area of the circle = π(C/2)[sup]2[/sup][/li][li]T be the aera of one of the equilateral triangles = C[sup]2[/sup] X sqrt(3)/4[/li][li]W be the area of a wedge of the large circle drawn around p[sub]1[/sub], described by one of the equilateral triangles plus one of the lunar-shaped (lune) areas between the large circle and the line p[sub]2[/sub]p[sub]3[/sub][/li]This area is 1/6 of the large circle, which equals πC[sup]2[/sup]/6.

[li]L be the area of a lune, which equals W-T.[/li][li]Z be the ratio of obtuse to all triangles, which equals R/(4L + 2T)[/ul][/li]Substituting:

Z = R/(4(W-T) +2T) = C/(4W - 2T)

Z = (π(C/2)[sup]2[/sup])/(4(πC[sup]2[/sup]/6) - 2(C[sup]2[/sup]sqrt(3)/4))

Z = (C[sup]2/sup)/(C[sup]2/sup)

Cancelling C[sup]2[/sup] gives

Z = (π/4)/((2π/3)-(sqrt(3)/2)) = 0.63982… or 63.982%

And since this answer is independent of the length of side C, this answer applies to all triangles.