Geometry/Math Question

What is this, a quiz show? Are you asking a question because you actually want to know something, AWB? Or do you know the solution and just want to see others sweat and tell them “ha-ha, you’re wrong”?

For anyone who still cares: The fault that AWB finds in Mark’s solution can easily be resolved by defining a circle between the first two points (the smallest possible, the one that just touches the two points on its opposite ends, if you know what I mean). Then:

  • If the third point falls within this circle or outside the two lines, the triangle is obtuse.
  • If the third point falls on the circle or on one of the lines, the triangle is right.
  • Otherwise, the triangle is acute.
    Since the area of the circle is finite, the result is not affected.

Dude, you’re wrong when you say “we also know that the region 0<=y<90 is the region consisting of only acute triangles”. If one of the sides is significantly longer than the other, then an obtuse angle will occur at the far end of the shorter side. (This is easiest to imagine if y is very small.) In fact, you also count each possible triangle multiple times, i.e. at least three times for each of its angles that you could choose to name “y”.

As you say, the basic problem is in trying to enumerate the (non-enumerable) set of all triangles. I’m not sure there is any solution at all, but Mark’s approach seems to be the best so far.

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.

So, the game show host has succeeded in showing the audience how ignorant the candidates are. (Or so he thinks.)

Now what was all that for?

You’re right, of course, as our kindly professor has demonstrated. Although I do enjoy thinking about these types of problems, this is exactly the sort of behavior that would get one labeled as a troll around these parts. If I were AWB, I would be more careful from now on.


To Nickrz (and anyone else concerned),

I didn’t mean for this topic to be troll-type. I just thought this was an interesting geometry puzzle and was offering it as such.

If this was the wrong forum to post it, please excuse me.

<marquee><font bgcolor="#000000" color="#FF0000">A</font><font bgcolor="#000000" color="#FFFFFF">W</font><font bgcolor="#000000" color="#0000FF">B</font></marquee>

AWB, the forum was okay (IMHO), but next time you pose a solved problem just for play, you should clearly label it as such.

Otherwise, people may no longer like to spend time and effort on any of your unsolved problems.