Geometry/Math Question

[AWB]

Of the infinite number of possible triangles in geometry (flat co-planar ones, that is), what percentage are:
obtuse
acute
or right (one angle is 90o)

<marquee><font color="#FF0000">AWB</font></marquee>

[pluto]

There are an infinite number of acute triangles, an infinite number of obtuse triangles and an infinite number of right triangles. So let's see -- 3 * infinity = infinity. Hey wait! Is my calculator busted?

Any percentage of an infinite number is also infinite. So it's hard to talk about infinities without getting "wrapped around the axis", so to speak.

Your question, as it stands, really has no answer.


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"The inability of science to grasp Quality, as an object of enquiry, makes it impossible for science to provide a scale of values."
Robert Pirsig

[Boris B]

I didn't think you could have percentages of infinite numbers. I mean, there are an infinite number of possible right triangles, so, dividing that by infinity would get you, heck I don't know!

If you came up with an algorithm to generate random triangles, i.e. with random angles, and used it a bunch of times, I think almost none of them would be right triangles, since there is no reason to expect any of the angles would be exactly 90 degrees. So about half would be acute and half would be obtuse.

But I'm rusty on geometry so I might be all wet.

[Keeves]

The question is obtuse. Cecil is cute. And the posters are right.

[UncleBeer]

Well, since it takes at 3 points to define a plane, all triangles are co-planar (flat).

Unless I'm not understanding the question, there are an infinite number of each, even if you discard congruent triangles. That is to say, two legs of each triangle could have an infinite number of values as a length.

Okay, line up to take your potshots at me.

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"The problem with the world is that everyone is a few drinks behind." - Humphrey Bogart

[AWB]

Actually, there is an answer.

Here's a similar and simpler question:

What percentage of rectangles have short sides that are less than half the length of their long sides?

[UncleBeer]

Okay, damn it, there were no answers here when I started composing. Gotta stop answering the phone and concentrate on the important stuff.

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"The problem with the world is that everyone is a few drinks behind." - Humphrey Bogart

[AWB]

I stated the obvious (flat co-planar [triangles]) to eliminate spherical or terrestrial trianges from consideration.

[Hunsecker]

Well, the way i work it, if you generate a random triangle, you get:

50% obtuse
50% acute
0% right

So lets see if my math is right here:
To generate a random triangle, for our purposes all you need is two angles.
In any triangle, at least 2 of the angles are acute. Call 'em A and B, and the other angle C.

A + B > 90 => C < 90 => acute triangle
A + B < 90 => C > 90 => obtuse triangle
A + B = 90 => C = 90 => right triangle

So we plot, with 0 < A < 90 and 0 < B < 90, what a given pair of angles gets you.
The graph is just a square with a diagonal running from (0,90) to (90,0). Any point on the diagonal is a right triangle. Anything to the upper-right is acute, and anything to the lower-left is obtuse.

So for prob., you have to calculate the area of all the regions. That will get you the numbers above (area of the diagonal is 0)

[pluto]

Hunsecker is correct if you're asking what is the probability of a random triangle being obtuse, acute or right.

That's not what it asks in the OP, however.


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"The inability of science to grasp Quality, as an object of enquiry, makes it impossible for science to provide a scale of values."
Robert Pirsig

[Boris B]

What percentage of rectangles have short sides that are less than half the length of their long sides?

I guess half. The short sort side is anywhere between the length of the long side minus a negligible amount, to nothing at all plus a negligible amount. So it is effectively between 0.0 and 1.0 times the length of the long side. Half will be more than 5.0, half will be less.

I'm not sure I have any idea what I'm talking about though.

[Boris B]

Not 5.0! 0.5! Sorry.

[Mark Mal]

I guess it depends on how you randomly choose your triangles, but here is one way to answer the question:

Pick any two points in an infinite plane. It doesn't matter how far apart the points are, because the problem can be scaled up and down arbitrarily. Construct two parallel lines through the two points which are perpendicular to the line segment between the points (i.e., if you orient the plane so the points are side by side horizontally, construct two vertical lines through the points).

Now start to pick random third points in the plane. Any point you pick, plus the two original points, will define a triangle. If the third point falls between the two parallel lines, it will be an acute triangle. If it falls outside the lines, it will be an obtuse triangle, and if it falls directly on one of the lines, it will be a right triangle.

I would argue that percentage of randomly chosen points which fall between the lines is infinitely small, because the ratio of the area between the lines to the area outside the lines is infinitely small. (The area between the lines is infinite, of course, but it is only infinite in one direction. The area outside the lines is infinite in two directions - infinity squared, so to speak.)
The number of points which fall directly on one of the lines is infinitely small compared to both the ones between the lines, and the ones outside the lines.

So I would say that, in the limit, the percentage of random triangles which are obtuse is 100%, the percentage which are acute is 0%, and the percentage which are right is an even smaller 0%.


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"For what a man had rather were true, he more readily believes" - Francis Bacon

[Merper32]

i believe the answer is
ACUTE= 49.7222222222%
OBTUSE=49.72222222225
RIGHT= 00.55555555555%

[Merper32]

i believe the answer is
ACUTE= 49.7222222222%
OBTUSE=49.72222222225
RIGHT= 00.55555555555%
My logic on this was, if there are 180 degrees. 1 of them is right, and 89.5 are acute, and 89.5 are obtuse. therefore 89.5/180. 89.5/180 and 1/180. i think i'm wrong though



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I was walking down the street when something caught my eye... and dragged it 15 feet.--Emo Phillips

Someone said to me, "Make yourself a sandwich." Well, if I could make myself a sandwich, I wouldn't make myself a sandwich. I'd make myself a horny 18-year-old billionaire.

[NanoByte]

what percentage are:

obtuse
acute
or right

<blink><font color="#FF0000" size=10>100%.</font></blink>

Ray

[Hunsecker]

pluto writes:
Hunsecker is correct if you're asking what is the probability of a random triangle being obtuse, acute or right.
That's not what it asks in the OP, however.


Yeah, well, that was the only way I could make sense of the whole "percentage of an infinite number" thing.

[TheDude]

The correct answer is 50% acute, 50% obtuse and 0% right. This is how we get there.

We know from geometry that if two sides and the included angle of any triangle are the same as another triangle, the two triangles are congruent(SAS). Therefore, a list of all possible ordered pairs of numbers (S1, A, S2) is a list of all possible triangles with no repeats (switching S1 and S2 is a repeat i.e. (3,45,5)=(5,45,3)), where S1,S2>=0 and 0<=A<=180.

Since, S1, S2, and A are all independent variables, they make up a portion of a three dimensional space, namely that portion of the first octant with 0<=y<=180.

We know that the percentage of ordered pairs in a particular region is equal to the percentage of the total volume enclosed in the region in question. And we also know that the region 0<=y<90 is the region consisting of only acute triangles, the region y=90 is the region consisting of only right triangles and 90<y<=180 is the region consisting of only obtuse angles.

Since the acute triangle region is equal to the obtuse triangle region and the right triangle region only consists of the infinitely thin border between the two, the relative numbers must be 50% acute, 0% right, 50% obtuse.

One problem with your example, Mark, is that it counts many triangles multiple times. Otherwise, your method of counting is just as valid as mine because the fundamental problem with this question comes from the attempt to count infinities. The one thing we can say for sure is that the infinity of acute and obtuse triangles is larger than the infinity of right triangles, but that the infinities of acute and obtuse triangles can be directly correlated 1 to 1. (Try correlating (S1, A, S2) to (S1, 180-A, S2)). This also reemphasizes the 50%, 0%, 50% conclusion.

TheDude

[Mark Mal]

I guess it really does depend on how you choose your triangles. If you randomly pick S1, A, and S2, like theDude, then you will get 50% acute, 50% obtuse. If you start with two fixed points in a plane and randomly choose a third, like I did, you will get 100% obtuse, 0% acute (I'm not sure what theDude means when he says I'm counting triangles multiple times, but I think you can avoid that by limiting the third point to one quadrant of the plane. The four quadrants are symmetric and will give you four identical sets of triangles.)

I think the way that makes the most sense is to choose three random points in three-dimensional space: (x1, y1, z1), (x2, y2, z2), and (x3, y3, z3), where xi, yi, and zi can range from -infinity to +infinity; and then determine how many of the resulting triangles are obtuse and how many are acute. I have no idea how to do that, though.


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"For what a man had rather were true, he more readily believes" - Francis Bacon

[AWB]

Mark Mal:
Pick any two points in an infinite plane. It doesn't matter how far apart the points are, because the problem can be scaled up and down arbitrarily. Construct two parallel lines through the two points which are perpendicular to the line segment between the points (i.e., if you orient the plane so the points are side by side horizontally, construct two vertical lines through the points).
Now start to pick random third points in the plane. Any point you pick, plus the two original points, will define a triangle. If the third point falls between the two parallel lines, it will be an acute triangle. If it falls outside the lines, it will be an obtuse triangle, and if it falls directly on one of the lines, it will be a right triangle.

You're on the right track, but you missed.

In your method, for example, if the first two random points are (0,0) and (10,0), the parallel lines would be x=0 and x=10. Now, say the third point is (5,0.5). This clearly forms an obtuse triangle, contrary to your statement "If the third point falls between the two parallel lines, it will be an acute triangle"

[TheIncredibleHolg]

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.

[AWB]

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.
Orient side C so it is horizontal. Let its enpoints be p1 and p2.

Draw a circle around p1 with radius C. Draw another circle around p2 with radius C. The intersection of these two circles is an American-football-shaped area.

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 p1 or p2, which means they describe triangles whose longest side is not C.

Within the football, label one of the apexes (pointy ends) p3 and the other p4. Draw a line between p1 and p3, p1 and p4, p2 and p3, and p2 and p4. These lines and line p1p2 should outline two equilateral triangles of length C.

Draw a cirlce whose diameter is C. It should be bisected by line p1p2.

From trigonometry, we know that any triangle inscribed in a semicircle is a right triangle. So, every triangle described by p1, p2 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 p1, p2, 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:[list] R be the area of the circle = &pi;(C/2)2
T be the aera of one of the equilateral triangles = C2 X sqrt(3)/4
W be the area of a wedge of the large circle drawn around p1, described by one of the equilateral triangles plus one of the lunar-shaped (lune) areas between the large circle and the line p2p3
This area is 1/6 of the large circle, which equals &pi;C2/6.
L be the area of a lune, which equals W-T.
Z be the ratio of obtuse to all triangles, which equals R/(4L + 2T)
Substituting:

Z = R/(4(W-T) +2T) = C/(4W - 2T)
Z = (&pi;(C/2)2)/(4(&pi;C2/6) - 2(C2sqrt(3)/4))
Z = (C2(&pi/4))/(C2((2&pi;/3)-(sqrt(3)/2)))

Cancelling C2 gives

Z = (&pi;/4)/((2&pi;/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.

[TheIncredibleHolg]

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?

[TheDude]

Holg,
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.

TheDude

[AWB]

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.

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

[TheIncredibleHolg]

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.

[voguevixen]

ZZZZZZZZZZZZZZZZZZZZZZZZZZZZZZ. Huh? What?