Why are obtuse triangles more common?

[@sbares]

This problem is ill-posed from the very first sentence. "Three random points are taken in an infinite plane", well, what is the distribution? The solution given here assumes that the conditional distribution of C given "event E2" is uniform within the "eye", but it is not at all clear that this should be the case.

[@Falanwe]

There are no uniform distribution of points over the whole plane, the question is just nonsensical.

[@lord_of_love_and_thunder]

I think a restriction to the unit square should have been added to the problem specification.

[@Grammulka]

Yeah, it's pretty easy to prove that probability of the triangle to be obtuse is 1. Let's say we place the points in order. Then after we have A and B, we can move our axes such that A = (0, 0) and B = (1, 0). So now if C is such that x_C0, either angle BAC or angle ABC is obtuse. And the probability for x coordinate of C to get inside [0; 1] over probability to get outside of [0;1] is zero.

[@maxwelljennings4178]

I figure that the distribution is uniform across the plane (though it was *technically* never stated, it is generally presumed that "random" refers to an even distribution), and, regardless, whether the points are taken from a plane, a unit square, or a part of a circle should have no effect on the answer, as the triangles are not bound to any "smaller" set of numbers than the irrational numbers, and there are an infinite number of similar triangles that can be formed by dilating and rotating a single triangle. As for E2, I do agree, though I think it would have been better stated as defining AB to be the longest side of the triangle.

[@jimbolambo103]

@@Grammulka That was exactly my reasoning. It looks like many more triangles created by your method map to triangles that are inside the circle in the video than outside and so they are undercounted. Though having seen how making restrictions on the triangles that you consider gives different answers, as seen in the video, this method may not be a valid way to approach the problem either.

[@bsmith6276]

I guess I'd call this an overkill solution - you don't need an integral to find the area of the "eye" (which I often see called a lens). Call the intersections of the green and blue circles D and E. ABD and ABE are equilateral triangles. Then its easy to see angle DAE = angle DBE = 2*pi/3 radians. Now draw chord DE. This splits the lens into two equal halves by symmetry. Pick a circle, I'm going with blue. Then the area of left half-lens equals the area of the sector spanning angle DBE minus the area of triangle DBE. Those are easy to calculate as 4*pi/3 and sqrt(3), respectively. Then the area of the whole lens is just 2*(4*pi/3 - sqrt(3)) = 8*pi/3 - 2*sqrt(3), which is exactly what Michael got with the integral.

[@knutthompson7879]

Doing the integral seemed like a clunky way of finding the area. I like your way a lot better

[@JordHaj]

I did it as 4 times the area of a sector with radius 2 and angle 60deg (it would be sector ADB or any congruent one) minus the area of the rhombus ADBE with diagonals 2 and 2*sqrt(3) because we are double-counting the area of 2 triangles which form the rhombus, so the area of the eye is 4*pi*2^2*60deg/360deg - 1/2*2*2*sqrt(3) = 8pi/3 - 2*sqrt(3) As much as I like calculus, I would never calculate this area using it

[@Hiltok]

@@JordHaj Completely agree with your method being most geometrically intuitive, and that using calculus to find the area is adding too much seasoning to what is already a meal.

[@Yishay.S]

Did not read sorry. But it seems you reminded me the complement area that newton use to calculate pi. which is funny that other side for the equation what approx by an integral of an infinite series...

[@zanti4132]

Binge-watch Penn's videos, and you'll come to realize that sometimes he likes to find ridiculously complicated ways to solve problems, just for the fun of it. 😀

[@Bodyknock]

One thing missing is a justification for how you are randomly selecting the points. On the face of it it seems like the probability distribution of the points could change the answer and it’s not obvious how to select points where the probabilities align with something like a uniform distribution since it’s over an infinite plane.

[@ere4t4t4rrrrr4]

I think that there is a probability distribution where for each two triangles they are equally likely to be picked. My intuition is that this distribution is achieved by choosing two lengths and the angle between them randomly, but I'm not sure if this is even possible (what means to pick an arbitrary length randomly?)

[@JadeVanadiumResearch]

@@ere4t4t4rrrrr4 Picking a random length has the same problem as picking a random point. There is no uniform distribution over the entire number line.

[@metakaolin]

great point, I'd completely forgotten that there's no uniform distribution over the whole cartesian plane he should've either restricted it to an arbitrary but finite rectangle (which would give the obtained result), or selected a proper distribution (which would definitely change it)

[@JadeVanadiumResearch]

@@metakaolin The problem where we select an arbitrary rectangle and limit its size to infinity would be very interesting to see, but I think it won't give the answer that Michael finds, and I suspect it would depend on the proportions of the rectangle, and only on the proportions actually. The absolute size of the rectangle doesn't matter, because changing the scale preserves all angles and proportional distributions. I know for certain that if we compare between various non-rectangular shapes, you will get different answers, so the result definitely depends on the shape of the bounding set. I don't see a compelling reason for why rectangles would all give the same answers, given that non-rectangular shapes don't.

[@metakaolin]

@@JadeVanadiumResearch yeah, and there's also the fact that we don't really know if the limit actually converges (which I suspect it doesn't, otherwise we could define an infinite uniform distribution like that)

[@Krekkertje]

By the way, using calculus for that area of the “eye” is way overkill. You could have calculated a circle segment and substracted a triangle for half the eye and of course double it for the entire eye.

[@RafaelOliveira-cd9fz]

2:40 A note for the editor, I think that the right angle was ACB and not the height from C to AB. I do like the way the video is editted though

[@lorenpearson1230]

Agreed. All dropped lines from C will form a right angle with AB which is of no use to us, but only points on that circle will subtend an angle of 90deg at C, as that is the transition from all acute to obtuse.

[@jakobr_]

I feel like Bertrand’s paradox is relevant here

[@knutthompson7879]

That's what I was thinking the entire time. Our inability to strictly define what we mean by "uniformly distributed points on an infinite plane" makes the answer ambiguous.

[@solid4340]

@@knutthompson7879 I'd say uniform distribution in an infinite plane means every 2 angles in the triangle make 180 degree linear displacement. In 1 acute triangle system, each angle makes 1 displacement in at least 1 time unit. With an acceleration or weight flow of 180. kzbin.info/www/bejne/i2qXmaCPlNyZp6s S=1. D=1. T=1 w= 180 In my c polynomial when you plug these in, c = 181. The rate of acuteness. 181 x .1 = 18.1 % probability of an angle being acute. 18.1 x 2= 36.2 % prob. of 2 angles being acute. 100% obtuse triangle probability - 36.2 % acuteness prob. of 2 angles = 63.8% probability the triangle is obtuse

[@solid4340]

But then the assumption is that every angle displacement is taking place in 1 time unit on its path to acuteness or obtuseness

[@mrphlip]

5:55 - "Since we can assume that it's equally likely for any of these points to be landed on by C"... but can we just assume that? Even leaving aside the fact that we don't have a probability distribution to start from (and there is no uniform distribution for the entire plane)... the mapping from those three original points to the eventual location of C within the vesica piscis is non-trivial, being scaled by a varying amount depending on the locations of the selected points. It's not immediately obvious to me that the end result would be uniform, and handwaving away this part of the proof seems like a pretty major hole to me.

[@landsgevaer]

Hmm, not sure I agree with how points are just assumed to be randomly sampled from an infinite plane! (That is a friendly way of saying this solution is nonsense...) Suppose we rotate and scale the *shortest* side of the triangle to coincide with the line segment from (-1,0) to (+1,0) (as opposed to the longest side in the video). Then the third point must lie outside radius-2 disks centered at these two points in order for that to be the shortest. From all those points, all points in the strip -1

[@LouisOnAir]

Yes, I was also suspicious of this method and that is brilliant argument against it.

[@JordHaj]

Sorry sir, are you saying that the "area of that strip is negligibly small compared to the entire plane?" Isn't this nonsensical itself? When dealing with infinities, including uncountable ones (which the plane definitely is), the simple argument of "well, this looks smaller so its area must also be smaller" doesn't apply. The set of natural numbers has the same power as the set of perfect squares. The sets of rational numbers and natural numbers have the same power. The set of real number between 0 and 1 has the same power as the set of all the real numbers. All pairs have "obvious" smaller ones. To be honest, I'm not sure how one would calculate the ratio of those infinite areas but that seems ridiculous anyways.

[@landsgevaer]

@@JordHaj It is a good point that you make, but it just underscores my own point that one needs to define the statistical distribution (/density function) of how likely points are to be picked. That holds for the video too. Just to be clear: my argument was not meant to be correct, as I already pointed out in the post, but to be just as wrong as the one in the video.

[@JordHaj]

@@landsgevaer Thanks for clarifying, I might have misunderstood you previously. Yes, your point is as valid as Michael's and mine(an other one, which I'll work on tomorrow) since the problem is ill-defined due to uniform distribution not applying to infinite regions.

[@cynodont7391]

That was also my first idea but fixing A=(-1,0), B=(+1,0) and C at random position also implies that that AB has a 0% probability of being the longest side. There is clearly a bias. In a truly random selection of points, each segment should have exactly a 1/3 probability of being the shortest, middle or longest side. Infinite is weird

[@kristianwichmann9996]

Without a probability distribution, this is a meaningless question.

[@TheEternalVortex42]

If we use a specific distribution we get a different result. For example let's say we use the uniform distribution over the unique square (very reasonable). Then with computer simulation I get that the chance of an obtuse triangle is 72.5% (which is different than what Michael got). I tried working it out exactly but you get some annoying integrals and it's kind of a pain. But it's doable with some more effort. Basically, given any two points inside the unit square A and B, consider the line segment AB. Then it forms a strip of the square if we take the two lines perpendicular to AB. This whole strip represents all acute triangles, except for those in the small circle with diameter AB. So given any two points the relevant probability is 1 - width of strip + d(A, B) * pi/4. Then you just do the quadruple integral of the values for A and B to get the overall probability. Maybe someone with more time can work it out :)

[@KD-onegaishimasu]

My numbers match yours -- while the problem seems scale-invariant, it also seems dependent on the shape of the support and the distribution within that shape. Since we can't normalize any uniform distn over the entire plane, different answers to the question seem possible

[@seneca983]

"For example let's say we use the uniform distribution over the unique square (very reasonable)." I'm not sure I agree that's reasonable. I think the distribution should at least have rotational symmetry.

[@KD-onegaishimasu]

@@seneca983 it should have several properties, but i think you pointed out the most important one. my difficulty is that the properties are incompatible with each other, so there is no uniquely correct answer. i'd like translational symmetry. perhaps then i can set the origin to be the center of mass. but then there is no uniform distribution on the plane. in practical terms, the question is poorly specified: we need to add additional information, and the answer changes depending on what we add. [enclosed soln] makes point 3 have a different distribution than the first two, but if that seems reasonable... >>> ci=lambda phat,n:phat+math.sqrt(phat*(1-phat)/n)*np.array([-1.96,1.96]) >>> def scC(): ... p1,p2=[[random.gauss(0,1) for i in [0,1]] for j in [0,1]] ... p3=list(-np.array(p1)-p2) ... a,b,c=sorted([rms_diff(p1,p2), rms_diff(p1,p3), rms_diff(p2,p3)]) ... return np.sign(a**2+b**2-c**2) ... >>> rms_diff=lambda a,b:math.sqrt(sum((np.array(a)-b)**2)) >>> collections.Counter(scC() for i in range(100000)) Counter({-1.0: 78490, 1.0: 21510}) >>> ci(.78490,100000) array([0.78235327, 0.78744673]) a growing square gave me 72.5%, he got 63.94%, and this gaussian answer is ~78.5% -- but replacing gauss with cauchy gives me ~88.7%. i think both my new answers have rotational symmetry

[@orionspur]

Without very careful definition of the distribution, there is a set theory violation when selecting the "longest side". For example: WLOG suppose the first point is the origin (0,0). Also WLOG (up to scaling) suppose the second point is (1,0). Finally, *apparently* WLOG, third point (x,y) is now "randomly chosen" in the plane. That seems fine, but it is not. If x1 the resulting triangle is now obtuse. That is effectively 100% of the plane, so the probability is 1.

[@JordHaj]

Edit: see other edits below I see a lot of people in the comments being confused about how we name the triangle and that it would change the answer etc. so I want to clarify that. I have already answered some of them, but I certainly cannot answer them all. At first we have an infinite plane. Then 3 random (yet nameless) points are generated. We find the lengths between them(doesn't change anything), and label them A, B, C such that AB is the longest. The we scale AB to be 2, rotate and translate such that A is at (-1, 0) and B is at (1, 0). Scaling, rotation and translation do not change the ratios of areas and thus the probability. We choose AB to be the longest, we don't rely on "C being inside the green and blue circles" or "C being inside the eye" etc., we SET AB such that C is ALWAYS inside the eye. If C were outside the green or blue circle(or both), then one of or both AC and BC would be longer than AB, and yet AB is the longest one, which is a contradiction. When AB is the longest, C MUST be inside the eye. Probability of C being outside of the eye is 0 by our labeling and construction. If AB is not the longest, re-label the point such that AB is the longest. Obviously, re-labeling points doesn't affect their distribution and thus probability of the triangle being obtuse. Edit 1: Now that I have really thought about the problem, it is very ill-defined since "random point on the plane" is also ill-defined. On a unit square - sure, on a unit disc - might have some problems(like Bertrand's paradox) but manageable, but whole planes - oh boy get ready for some trouble. Tomorrow I will try a method of defining A to be the point (0, 0) and B to be (1, 0) and C a random point such that both AC and BC are less than some R, and taking the limit R -> infinity and expressing the eye(that allowed region for C) area and the "obtuse triangle" area and their quotient and I will see if the answers match(probably not) and I will edit this comment once more. Edit 2: I have tried that method I mentioned in Edit 1, although I picked A to be (-1, 0). So I worked out the area of the region, which C had to be in for the triangle ABC to be obtuse, which is (8pi/3 - 2sqrt(3) + 2R^2 cos^-1(2/R) - 4R^2/sqrt(R^2-4)), and the total area of the lens where AC and BC are less than or equal to R (or the eye) is 2R^2cos^-1(1/R) The ratio of the areas is thus ((8pi/3 - 2sqrt(3))/[2R^2 cos(1/R)] - 2/[sqrt(R^2-4) cos^-1(1/R)] + cos^-1(2/R)/cos^-1(1/R)). The first two terms vanish as R -> infinity, and both 2/R and 1/R go to 0, and both cos^-1 terms go to pi/2. So the ratio of areas is 1 in the limit. This way is similar to picking AB to be the shortest side, except when C is in the semicircles centered at A and B that are "looking outwards" away from the origin, in which case, AB would be the middle one. Obviously, 1 is different from the result in the video.

[@Hiltok]

You are correct that the "arbitrary points on an infinite plane" leads to impossibilities and the problem isn't well defined. The expression Without Loss Of Generality is not safe to use here because the choices made have consequences - or perhaps it is better to say antecedent conditions - built in. Try this: From the three arbitrary points, choose AB such that AB is the mid-length side with AC the shortest side and BC the longest side, except in the (zero probability case) that our three points are truly equidistant and we have an equilateral triangle. Now, for AB to be mid-length, construct the zone where C can be. Draw a circle of radius |AB| centred at A and exclude the 'lens' that intersects with a similar circle centred at B. For an obtuse angle BAC, C must in the half of the circle beyond a diameter perpendicular to AB. For an acute angle, C lies on the two remaining parts of the circle on B's side of that perpendicular diameter outside the lens of intersection. Now, the area that produces an obtuse angle is (1/2)*pi*|AB|^2 = half-circle and the area that produces an acute angle is (1/2)*pi*|AB|^2 - ( (2/3)*pi*|AB|^2 - sqrt(3)/2 )*|AB|^2 = half-circle - lens This gives a probability of an obtuse triangle of approximately 0.821. So, fixing AB to be the longest side gives about 0.639 and fixing AB to be the mid-length side gives about 0.821. If those choices could be made "Without Loss OF Generality" the probabilities would be equal. We are losing some aspect of the general problem when we fix the length of AB. For completeness, let us consider what happens when we fix AB to be the shortest side. This produces another kind of problem as the region for C to give an acute angled triangle is infinite in area (within perpendiculars to A and B but outside the 'lens') but seemingly smaller than the infinite area of the region that produces an obtuse triangle (outside the perpendiculars to A and B). Can we divide the infinities to get a probability? No. We might try looking at it in an analogous way to the Cauchy Principle Value that we apply to divergent integrals and say the probability of an acute angled triangle is the length of AB divided by the infinity of the length of the real number line, which goes to zero, leaving the probability of an obtuse angled triangle almost certain. This really isn't valid as it attempts to sum an infinite number ( = |R^2| ) of zeroes and compare that sum to another equally infinite sum of zeroes. Uniform distribution on infinite space doesn't work. . The problem here is that when we make a choice about the length of AB, it cannot be done without losing generality because the choice encodes a scaling and orientation to what comes after. Each of the three choices for fixing the length of AB is equally valid. One produces a nice number approx. 0.693. One produces a nice number approx. 0.821. The third really exposes the problem as one of competing infinities and what it means to be trying to measure probabilities on an infinite space without a sub-space of non-zero probabilities.

[@lexyeevee]

@@Hiltok i can't explain why the answers are different, but "we can't pick the two of three points that are furthest apart because that imposes a condition on them" is not very satisfying. that seems not fundamentally different from imposing an arbitrary order on an equation with symmetric variables, which this channel does all the time with number theory problems - where the number of choices is also infinite.

[@deinauge7894]

@@lexyeevee the problem arises when assuming, that the probabilities are proportional to the areas after making the choice... i will try some finite cases and post the results...

[@deinauge7894]

ok, the calculations are quite messi. did it numerically now. in a rectangle, the chance of getting an obtuse triangle is about 0.725 in a circle, it is ~0.745 and with gaussian distribution ~0.75 all three test-cases are scale invariant. the probability does not depend on r (r being sidelength, radius or standard deviation)

[@KD-onegaishimasu]

@@deinauge7894 Okay I just did the square case and my numbers agree with yours. I'll probably try the circle next. Edited to add: Intuitively I feel like a thin rectangle would have an increased Pr of yielding obtuse triangles, was your 72.5% shape-invariant? I get scale invariance, but was confused on this point

[@ilyasakhundzada6604]

Interestingly, that when one randomly picks one angle of triangle to be more than 90 degrees(i.e. obtuse), keeping in mind that the sum is always 180 degrees, the probability of obtuse triangle appears to be 3/4. To achieve this result consider the surface of x+y+z=pi plane in a space as area of all possible events. Then construct areas corresponding to x+ypi/2, x+zpi/2, z+ypi/2 - these are favorable events. The ratio of favorable events to all events, i.e. probability equals to 3/4.

[@apprentice-k5e]

I approached the problem differently and (of course) got a different answer. I have three angles that sum to 180, so I can work on the simplex, scaled to add up to 180. Think about the positive quadrant of the x-y plane where the x coordinate represents the angle at corner A, the y coordinate represents the angle at corner B, and the angle at corner C is 180 - x - y. Then any point inside the triangle with corners (0,0), (180,0) and (0,180) represents a valid triangle (or rather a family of valid triangles). It is then simple geometry to calculate that the proportion of the area that represents obtuse triangles is 3/4. The answers are different because the underlying probability distribution over triangles is different. The random construction process for the video consists of selecting arbitrary points A and B and then picking a point C such that the distances |AC| and |BC| are less than or equal to |AB|. Doing that avoids the badly ill-defined problem of selecting points on an infinite plane. Clearly my approach has a different underlying construction method (and thus a different underlying probability distribution over triangles), but my geometry is not good enough to come up with what the construction method would be.

[@goodplacetostop2973]

12:30 Alice in Triangleland

[@orbik_fin]

I used brute Monte Carlo to find the answer for normal-distributed points (~0.750), and for uniform points within a square (~0.725). Michael's interpretation leads to a special distribution for one point, which doesn't match the initial problem description. And his answer is quite different: ~0.639. Choosing a longest side AB to restrict the location of C seems similar to the Monty Hall paradox. When some possibilities are eliminated, the distribution of remaining options changes. One door has 1/3, the other 2/3. I thought of a simpler question with a similar paradox: Choose 2 random real numbers x, y from [0, +inf[. What is the probability that y is on the interval [x, 2x]? Well, it's a finite interval out of an infinite one, so its probability should be 0. But let's apply Michael's method and notice that if y > x, then x should be uniformly distributed between [0, y[, and the condition is satisfied on half of that interval. And by symmetry there's a 50% chance that y > x, so the answer is 0.25. How is this possible?!

[@lexyeevee]

the monty hall paradox involves an oracle introducing new information into the problem, contingent on a decision we've already made. if we have three points, we can easily choose the two that are furthest apart ourselves.

[@seneca983]

@@lexyeevee "if we have three points, we can easily choose the two that are furthest apart ourselves" That we can do ourselves is not the point in this case.

[@natepolidoro4565]

So that's about a 63.94% chance.

[@artsmith1347]

I, too, wanted a numerical value for the given answer: 3 * pi / (8 * pi - 6 * sqrt(3)) = 0.639382560712

[@samueldevulder]

Let's round this to 64%.

[@artsmith1347]

@@samueldevulder A numerical value was needed before any rounding could be done. How much rounding to do is influenced by practical needs and personal preference. I often don't want others to do "too much" rounding "for" me. I will round to the precision I desire when there are more digits than I need. We could round 60%. My first guess was that the probability would be 50%. Whether we round to 64% , 60% (the nearest multiple of 10%), or 75% (the nearest multiple of 25%), the conclusion is the same: the answer is not 50%. Without a text form of the answer, I couldn't verify the answer is 63.94%. It was easiest for me to copy and paste what my calculator displayed on both the left and right side of the '=' sign, so that is what I did.

[@jacoboribilik3253]

I did the following and got a different anwer. Since you can´t define a uniform probability distribution in the plane I assumed any three angles adding up to pi radians are equally likely. That means the plane given by x+y+z=pi restricted to the first octant has all the relevant information we are interested in and we implicitly defined a triangle with vertices (pi,0,0), (0,pi,0) and (0,0,pi). A triangle with angles x,y,z ( where x, y and z are defined as the angles appearing from left to right in order when the triangle is picked at random in the Cartesian plane) is obtuse if and only if there exists a pair of angles {z,w} belonging to {x,y,z} such that 0

[@darkshoxx]

I really like the animation at the beginning. But maybe double check it. The right angle drawn at 2:40 is a right angle, but it definitley is not the angle he is talking about 😅

[@MushookieMan]

Isn't it implied there is a uniform distribution on an infinite plane? I thought that was impossible.

[@theimmux3034]

Aren't the lattice points of the Cartesian plane distributed uniformly?

[@dudz1978]

I thought the same thing. If we choose points with some other distribution, could the result be different?

[@skylardeslypere9909]

We aren't dealing with an infinite plane though. WLOG we can set the length AB = 2 and then achieve this eye shape. Now our universe is just this eye shape, because we only consider points within this eye shape.

[@MushookieMan]

@@skylardeslypere9909 Changing the problem statement to an easier, not equivalent problem is a silly way to solve a problem. I think the only sensible way to find the "true" answer is to define a sequence of smaller distributions that has a uniform distribution over the entire plane as a limit.

[@JadeVanadiumResearch]

@@dudz1978 Yes, you and @interdimensional are both correct. There is no "uniform distribution" over the entire plane. If you had such a distribution, the probability a point lands in any particular bounded set would be zero, and by countable sub-additivity you can prove the measure of the entire plane is zero, contradicting the requirement that the total probability is 1. The only way to formulate this sort of problem is to restrict the points to some bounded set, and then somehow take the limit as the size of that set tends to infinity. However, as you point out, if we take this limit in different ways you will get different answers, so the question has no unambiguous answer. Through some clever reasoning, I was able to find a way to take the limit in such a way that the triangles are obtuse at least 8/9ths of the time. However, I also found a different way to take the limit which forces the triangles to be obtuse at most 7/9ths of the time. These two figures are clearly irreconcilable, so the answer depends on how exactly you take the limit, which the problem hasn't specified.

[@pwmiles56]

Doesn't selecting AB as the longest side place a condition on C, which may or may not be uniformly distributed on the region so defined? Suppose AB is any side and C is "anywhere". The triangle is obtuse if C is inside the circle with diameter AB, or outside the perpendiculars through A and B. Since this region is indefinitely "larger", the answer is P(obtuse) = 1.

[@rtheben]

Right on spot, in fact you can say a B is the middle side or you can see a B is the shortest side you get different answers. So indeed there is loss Generality in making the choice

[@rtheben]

Basically it’s the axiom of choice at work, an ill work

[@rtheben]

On the other hand if you set AB to be the middle side you get 3pi/(2pi+3sqrt(3)), yet different

[@JordHaj]

I don't see a problem with "loss of generality." We have three arbitrary points. We calculate the lengths between them and only then do we label them as A, B and C such that AB is the longest. This does not violate the "WLOG" statement and does not "place a condition on C" as far as I'm aware. Then we scale AB to be length of 2 and rotate and translate so that A is (-1, 0) and B is (1, 0). These transformations(homothety, rotation and translation) do not change ratios of areas. By the choice of labeling, C cannot be outside of the "eye", since if it were, AC or BC(or both) would we larger than AB, which contradicts our construction and labeling.

[@pwmiles56]

@@JordHaj "Arbitrary" is the tricky concept here. Arbitrary on what? As others have pointed out, the problem is ill-defined. I tried a Monte Carlo simulation on a square. After 1000's of samples it returned P(obtuse) ~ 76% vs. Michael's answer, 63.9% I'm guessing Lewis Carroll (C.L.Dodgson) laid a trap here

[@kevskevs]

Do we really *have* to define AB to be the longest side? If we take AB to be the *shortest* side and scale the triangle so that A=( - 1|0 ), B=( 1|0 ), then the triangle is obtuse if C is outside of the strip between x-coordinates - 1 and 1. In other words, the entire plane excluding this 2-unit wide vertical strip (and the x-axis) gives us obtuse triangles ... which would make the probability aribitrarily close to 1, wouldn't it?

[@solid4340]

1 acute triangle system. Each angle is making 1 displacement. With an acceleration or weight flow of 180. We are observing this system in at least 1 time unit. kzbin.info/www/bejne/i2qXmaCPlNyZp6s d=1. s = 1. T = 1. w = 180. When you plug these into my c polynomial, you get 181 = c. Rate of an acute angle. 181 x .1 = 18.1 % probability for an angle to be acute. 18.1 x 2 = 36.2 % probability of two angles being acute. If we take the 100% probability of an obtuse triangle and subtract the 36.2 % acuteness and you get a 63.8% probability of the triangle being obtuse.

[@KD-onegaishimasu]

We do not have to, but if the assumption is truly wolog it'll make the problem easier without changing the answer. But I think you're right! Bc it does seem to change the answer

[@JohnWilliams-iy2vx]

For the curious, please check out Gilbert Strang's video from 7 years ago: " Are Random Triangles Acute or Obtuse"

[@egillandersson1780]

Very interesting ! So, as I felt, Michael's approach is one of the possible approches but is not "without lost of generality".

[@kevinmartin7760]

Some people have brought up two objections to this answer: One is perhaps related to Bertrand's Paradox any time you try considering points picked randomly on the infinite plane. If you want to pick a random point anywhere on an infinite plane with even distribution, the probability of that point being within any finite boundary will be zero, thus Michael's evaluation of the answer involves dividing zero by zero (E_1 and E_2 are both zero). Also, the choice that AB be the longest side of the triangle seems arbitrary. One could equally choose AB to be the shortest edge, or even not choose at all, in which case getting a non-obtuse triangle would seem to be vanishingly small because to get one, the third point would have to land in a strip of finite width perpendicular to AB. Although this strip has finite width, it does have infinite length, so we can't just say that there is zero probability for the third point to land there (as mentioned above) because it is not a finite area. To get out of this conundrum, consider the projection P of the third point onto the infinite line AB: In this case we have a finite length AB that the projection must land in to avoid an obtuse triangle, even though the probability of P being at any particular point on this infinite line is evenly distributed. This is again the probability of a point being in a finite region of an infinitely-sized distribution and so it zero. I wonder if there is a nitpick-resistant solution if we instead confine our points to be taken from an even distribution within a finite square. There will be many tiny triangles for which the stripe argument in the above paragraph would make them almost assuredly obtuse, but would this be balanced out by larger triangles? At least in this case one could do a computer simulation to find an approximation to the answer...

[@kevinmartin7760]

As to the latter, I just tried such a program, and found that about 72.52% of the triangles are obtuse (after testing 1.6 billion triangles). The points were chosen using the random number generator, asking for a value between 0 (inclusive) and 1 (exclusive) for each of the 6 coordinates. The triangle was checked for being obtuse by calculating the squares of the three side lengths and seeing if any was larger than the sum of the squares of the other two side lengths (viz. law of cosines). Of course, this result could be affected by the finite precision inherent in computer numbers. I expect the result for a finite square would be the same as in the infinite plane; the probability for the finite square is constant regardless of the size of the square, and if you take the limit of this as the size of the square goes to infinity, you still get the same probability.

[@kevinmartin7760]

Just on a lark I changed the program to only test the triangle if all three points lie in the circle of radius 0.5 centered at (0.5, 0.5). It gives the correct reject rate (51.55% = 1-(pi/4)^3) but now 71.97% of the triangles are obtuse! It seems I could make the same limit argument, enlarging the radius of the circle to infinity...

[@solid4340]

kzbin.info/www/bejne/i2qXmaCPlNyZp6s 1 = d. 1 = s. 1 = t. 180 = w. I got 181 = c. 181 x .1 = 18.1 % acuteness 18.2 x 2 = 36.2% acuteness of 2 angles A and B. 100% obtuseness - 18.1 % = 63.8 % obtuseness

[@TheEternalVortex42]

@@kevinmartin7760 There is a boundary effect because the strip of acuteness gets cut off at the edge of the region you are sampling. with a circle that region is slightly larger because it's rounded so you get slightly more acute triangles. This does give a good explanation of why there is no single answer to the question posed.

[@kevinmartin7760]

@@TheEternalVortex42 At first I thought this might explain it (though it took me a minute to figure out what you were saying), but by the same argument, when using a square sample space, the zone outside the "strip of acuteness" is much larger (by a factor of about 4/pi) for any particular size of triangle which would appear to bias things in the other direction. I still suspect the difference is an artifact of the finite precision arithmetic being used in the program. It cannot check the infinite number of triangles smaller than the precision, so the result is biased a tiny bit by an excess of large triangles which, as they approach the size of the sample shape, tend to be more likely to be acute. This bias could depend on the shape of the sample space.

[@Utesfan100]

If we proceed as shown, but assume that AB is the smallest side, then C must be outside the union of the interiors of the two ircles draw. We get an acute angle if c is in the strip orthogonal to the segment AB, and an obtuse tringle if it is in either half planes to either side. Relative to the infinite area, the finite areas near the two circles is negligible. But the measure of the strip is also negligible to the half planes, as they extend in an extra dimension. Thus P(obtuse)=1. This is yet another interpretation of what it means to have a random distribution of points in a plane.

[@QuantumHistorian]

I have a different answer and I think the reason is I'm assuming a different distribution in a subtle way, but I can't pin point _exactly_ what the difference is, any insight would be appreciated. I generate 3 random points, sequentially, on the infinite plane, and label them A, B and C in that order. I set up my co-ordinate system such that A is at (-1, 0) and B at (+1, 0), and leave C unconstrained. Using a similar construction as to the video, we have that the triangle is acute if and only if C lies in an oculus formed by the overlap of two circles of radius 2 centred at A and B respectively, but *not* in the inscribed circle with diameter AB. If C lies anywhere else on the infinite plane, the triangle is obtuse. As one area is finite and the other infinite, we have that with probability 1 the triangle is obtuse. This is basically identical to the video, except that I choose A and B by the order in which the points were generated, rather than by being the longest line. Yet the answer is drastically different. Have I done something wrong, or is the question just ill posed with a bad (implicit) distribution?

[@TheEternalVortex42]

Since there's no uniform distribution on the infinite plane, I suspect you can get any answer you want by picking different distributions.

[@QuantumHistorian]

@@TheEternalVortex42 I suspect you're right. But I don't even think I've picked a different distribution to the video, I've only assumed an ordering. As we're only dealing with finitely many points (even if they are on the infinite plane), I'm surprised this makes a difference.

[@cynodont7391]

@@QuantumHistorian Think about that : What is the probability that AB is the shortest, middle or longest side of the distribution? With your method, AB has a 100% probability of being the shortest and 0% to be the middle or the longest. There is clearly a bias. The prob should be 1/3 for each

[@KD-onegaishimasu]

@@QuantumHistorian I'm wondering about this as well -- how did you generate the random points? Perhaps I can help find some subtle difference

[@ethandavis7310]

@@cynodont7391 To modify the situation, if you take AB to be any side you reach the exact same conclusion.

[@bsmith6276]

An alternate solution approach: I will start with the same simplifying idea that Michael did: scale the triangle so that its longest side is a fixed length. But I will set that length to 1 in this solution. Let A and B be the lengths of the other two sides. Then we have 0

[@solid4340]

Well we have 3 angles. Each angle is making 1 displacement. With a acceleration or weight flow of 180 in 1 system. We are observing this in 1 time unit. kzbin.info/www/bejne/i2qXmaCPlNyZp6s If s=1 system, d=1 displacement, t= 1 time unit, and w = 180. C = 181. This is the rate of an acute angle. 181 x .1 is the 18.1 % probability of an angle being 90. 18.1 x 2 = 36.2 % chance of angle A and B being acute. The 100% probability of a triangle being obtuse. Subtract the 36.2 % probability of Angles A and B, and you get 63.8% chance the triangle is obtuse. This is how I got my answer close to the 64%

[@solid4340]

The c polynomial thing and link is my P= NP proof. Could be right could be wrong. I'm currently trying to solve problems with it to see if it's legitimate

[@ThainaYu]

Remind me of 3b1b video about probability. Changing the way to measure probability can change result Such as this problem. If we change the assumption from longest side to shortest side. The result might be infinitely probability to be acute triangle

[@ReCaptchaHeinz]

I think I disagree with the solution. I thought about this problem long ago, and my approach "showed" that in fact, there are infinitely many more obtuse triangles than acute triangles... Let A B C 3 points in the plane that form the triangle T. Let 2r be the distance AB. WLOG, let A land in (-r,0), and B in (r,0). (just rotate and translate the triangle, with no scaling, which is different from what you showed in the video) This defines regions: 1) the "outside" of the interval in the x axis, namely ((-∞,-r)U(r, +∞))xR 2) the circle with center (0,0) and radius r 3) the "rest", namely, the region (-r, r)×R - the circle discribed in 2) Now, the point C can land in ANYWHERE on the plane, every point being equally likely. - If C lands in regions 1) or 2) then T is obtuse. - If C lands in the region 3), T is acute - If C lands in the borders of these regions, T is rectangle. Being the region 2) finite, while 1) and 3) are infinite, we can ignore it and simplify the calculations and proceed as follows: P(T acute) = P(C lands on (-r, r)×R) = P(the x coordinate is between -r and r) = (length of (-r, r)) / (length of R))= 2/∞ = 0 then P(obtuse) = 1-0 = 1 At this point: I would love to know if I am wrong and where did i commit the mistake (because i feel i didn't). On the other hand, what doesn't sound exactly right to me in the video is that C is not equally likely to land in the points of the regions defined (the circle and the eye), and that, due to the scaling, there might be a non-uniform density function over the regions.. In the end, you're squishing arbitrarily big triangles in a finite region, so that transformation of the plane may imply such non-uniformity (again, that's my opinion, I'm may be wrong)

[@TaleshicMatera]

we did very similar things and I got the same answer (there are infinitely more obtuse triangles than acute triangles), but since I was playing around with drawing those regions, it is fun to plot the various types of triangles (equilateral, isosceles, scalene, acute, right, obtuse) and you can (almost) rank them (scalene, obtuse, acute, right/isosceles, equilateral) but can't (meaningfully) define ratios of their populations.

[@alexfekken7599]

The title of this video asks for the cardinality, not a probability, so I almost skipped this as too trivial: the cardinalities are the same.

[@kevinjohnson4531]

I showed this to my 13 year old and his first question was “what does this say about what percentage of triangles are right triangles?” Very proud moment for me.

[@londonalicante]

I bet he was shocked that the answer is 0%.

[@borisjerkunica4442]

at 2:49 the right angle should be drawn on the angle ACB. It's interesting to watch as the channel grows and multiple people are involved how these artifacts start appearing 🙂

[@paologat]

The problem is not well defined, as it assumes a uniform probability density over an infinite plane - which can’t exist as the probability per unit area would be uniformly zero. A rigorous way to prove the result would be : 1) pick the three points from a uniform distribution over a finite convex set (e.g. a square or a circle, to have an easy start) and calculate the probability; 2) then take the limit as the set scales linearly to infinite size (easy, as the calculation is scale independent); 3) and finally prove the result is independent from the shape you started from. Which is most definitely false (for example the probability of an obtuse triangle is almost 100% if you start with a thin rectangle as your domain). Or you may fix the problem by stating explicitly the process by which you pick the three points.

[@orbik_fin]

Technically Lewis doesn't say uniform distribution, so we could interpret he must've meant normal distribution, and gotten a well defined answer of about 0.75.

[@bollyfan1330]

There is a much easier way to prove. Consider the first two points of the triangle, say A and B. Draw the line perpendicular to AB at B. Any point C on other side of the perpendicular line than the side with line segment AB will form an obtuse triangle because ABC will be an obtuse angle, creating an obtuse angle triangle. Hence just by this we know that there are exactly as many obtuse angled triangles as all acute angled triangles. Now doing the same on the other side, by drawing the perpendicular line at A instead of B, choosing C on that side will produce obtuse triangles with BAC being an obtuse angle. Since the half is completely a part of the half that had the all acute angles from before. If you choose A to have coordinate (0,0) and B to be (b, 0) without loss of generality, the Y-coordinate of C is irrelevant, for C being (c, y), if c is in (- infinity, 0) U (b, + infinity), it will be an obtuse triangle and it will not be obtuse angled triangle when c is in [0, b]. Clearly assuming a uniform distribution, there are more possible values of C that will create obtuse triangles than the ones that will create non-obtuse triangles.

[@dirkvanmaercke8469]

This is an ill-posed problem because a uniform (or even an informative) distribution over an infinite plane does not exist. The problem can only be solved by taking the limit over some real distribution (that has finite values in at least some part of the plane) with some scaling parameter A and then taking the limit as A goes to zero (or infinity). But because you have used a length measure (L = max of lengths AB, AC, BC) your solution will be function of the ration L/A and that means you could get any value you want ! E.g one can easily prove that p(acute) = 0. Consider two random points A and B on the infinite plane. Construct a Cartesian coordinate system with O = midpoint between A and B, X axis aligned with AB, Y axis perpendicular. Set L = length of AB/2 ; then A has coordinates (-L,0) and B similarly (+L,0). Now take another random point C in the infinite plane. The triangle ABC will be acute if and only if C has Y coordinate in the interval (-L, +L) and falls outside the circle with center 0 and radius L. Consider a uniform distribution for C on the square of size R centered at O. One can easily calculate the areas corresponding to acute and obtuse triangles. Taking the limit as R goes to infinity, you get p(acute) = 0 ! You'll get the same result for ANY real probability density because when extending the plane to infinity, the area of the strip of width (-L,+L) becomes infinitely small compared to the total area of the plane. Note this does NOT mean that acute triangles are impossible, only that their probability has zero measure which is not rare when taking probabilities over uncountable sets (e.g. the probability that 3 random points form a single straight line is also zero, still Euclide tells us that there are infinitely many points that align with any two distinct points of the plane).

[@Czeckie]

I'm sure you can Bertrand's Paradox this problem into any solution

[@sairamganna350]

I'm trying to understand this with differenet way I use X+Y+Z = 180* and drawn it in 3Dspace, resulting in a triangle And Now I pointed out 90* points and plotted points taking X as 90* and Y as 90* and Z as 90*, I now got another triangle with in the bigger one with each side lengths halved. So take any point inside that smaller triangle represent the combination of (X, Y, Z) whose sum is 180* and each values of angles are

[@gottfriedschuss5999]

"From Lewis Caroll..." Is this the mathematician who wrote, "Alice's Adventures in Wonderland"?

[@VideoFusco]

I do not believe at all that the reported construction is equivalent to considering all three points uniformly distributed in the plane.

[@aditaggarwal3486]

Before I see the solution: I think, the largest angle can have a measure from 60 to (juuuuust less than) 180 degrees. 30 degrees for acute and right angles, 90 degrees for obtuse angles. The answer should be 3/4.

[@williamangelogonzales148]

Just stumbled upon this video because I'm looking for a similar problem that I could not solve "Suppose the perimeter of a triangle is 8cm. Find the probability that it is obtuse".

[@Krekkertje]

Let A and B be points in the plane. Let a and b be lines through A and B that are perpendicular to line segment AB. The only way to get an obtuse triangle ABC is if C is within the circle with diameter AB or if C is not between a and b. The area between a and b is infinite in only one direction whereas the rest of the plane is infinite in all directions. So the plane segment that that produces acute triangles is “smaller” than the plane segment that produces obtuse triangles. Ergo, there are more obtuse triangles. Of course I’m aware that I’m being sketchy by comparing two infinities.

[@jamesfortune243]

I liked how the circles were set up. That taught me more than the solution did.

[@JohnSmith-ux3xm]

We can do a similar calculation but choose AB to be the side with middle length. The steps are similar, and involve areas of circles and equilateral triangles. Following this approach I calculate the probability of the triangle being obtuse as 3pi/(2pi+3 sqrt(3)) = 0.821021 Is this just further evidence that the question is rather dodgy?

[@ΠαναγιώτηςΓιόφτσος]

So, what is a rough approximation of that final answer? Pratically, I mean, how likely is it that a random triangle will be obtuse, with a useful approximation?

[@panagiotisapostolidis6424]

it's close to 64%, you can find a few similar problems on math stack exchange

[@rogerkearns8094]

Good question. I tried to approximate it in my head, calling pi 22/7 and root three 7/4, but I'm getting too old for doing that sort of head work it seems...

[@rogerkearns8094]

@@panagiotisapostolidis6424 ...so, thanks for the answer. ;)

[@solid4340]

3 angles in 1 system. Each angle makes a linear displacement every 1 time unit. s=1. d=1. t=1. The acceleration of each LINEAR displacement is 180. kzbin.info/www/bejne/i2qXmaCPlNyZp6s If you plug these into my c polynomial, you get c=181. This is the rate each angle is going to achieve acuteness. 181 x .1 = 18.1% probability each angle is acute. 18.1 x 2 = 36.2% probability angles A and B are acute. If we take the 100% probability of a triangle being obtuse, and subtract the 36.2% of acuteness, we have a 63.8% probablility the triangle is obtuse

[@solid4340]

I made a c polynomial to make any P=NP and it seems to verify alot of problems. Based on the comments if the answer is close to 64% and I got 63.8 do you think my math checks out?

[@richardfarrer5616]

Isn't calculus overkill here? If you call one of the intersections of the blue and green circles D and the other E then it's obvious triangle ABD is an equilateral triangle so angle BAC is 60 degrees. Then DAE is 120 degrees. The area of the segment bounded by the y axis and the green circle to the right of it is thus area of sector ADE - area of triangle ADE = pi/3 (2^2) - sqrt(3) = 4pi/3 - sqrt(3). The area of the eye is twice that, or 8pi/3 - 2sqrt(3)

[@notEphim]

As already stated in comments, this is not a well formulated problem, bc no distribution is specified and there is no uniform distribution on the plane. However i know a beautiful proof that the probability is 3/4, assuming that for any finite set of points the distribution is uniform within condition that point is chosen from that finite set. Granted, i actually don't know if such distribution exists For any orthocentric quadruple there are exactly one triple that contains vertices of acute triangle and three triples that contain vertices of obtuse triangle. Therefore the probability is 3/4

[@dirkvanmaercke8469]

After taking A and B so that |AB| is either the shortest or the longest of the segments CHANGES the conditional probability distribution of C ! And that would be hard to work out (because of the infamous 0 divide by 0). I guess this is also true when we first select any 2 points A and B, than add the point C. After some deeper thinking, I guess the right answer is p(acute) = 1/4 and p(obtuse) = 3/4. As was suggested by Ilyas and close to the numerical result found by The EternalVortex. Here is my "proof". Any 3 points A,B and C in the plane form a circle (fact). If the pdf of A, B and C are (almost) uniform in sufficiently large area covering this circle, than the positions of the points on the circle will also be uniform (proof using measurement theory ?). . The question can therefore be rephrased as : given any 3 random points on a circle, what is the probability they will form an acute/obtuse triangle ? Answer : 1/4 and 3/4 respectively.

[@stlemur]

This fact might actually have implications in cosmology...there's a kind of general cosmological model called Bianchi IX where the combination of forces along three different axes lie within some triangle. The naive assumption people make is that if the triangle is chosen at random then triangle is probably close to equilateral...

[@ArpegiusWhooves]

2:43 The mentioned right angle is on ACB corner, but someone drew it on a height XD

[@bosorot]

Can anyone explain that with this area/probability comparison? The probability to get a right triangle is 0 because there is no area , just a line . But that is not true .

[@andrewparker8636]

I agree that the problem is basically nonsensical. The assumption and the transformation at the start is effectively defining the problem, not transforming it. I can use a similar argument to state that the probability of an obtuse triangle is infinite. Let's order the random points 1,2,3. Do a transformation similar to Michael's so that 1 and 2 lie on the x-axis at -1 and 1. Now, in this case the 3rd point can be anywhere. Similar to before, the triangle is obtuse if point 3 is inside the unit circle. If point 3 has x coordinate < -1 or > 1 then the triangle is also obtuse. In other words, to be acute, point 3 must be in the strip between x=-1 and x=1 and not in the unit circle. However, it's pretty easy to see that the probability of being in that strip is zero. You can in fact be a bit more concrete here and take a limit. So the triangle as prob=1 of being obtuse. Again, this argument also doesn't make sense. I'm basically formulating a different argument that says: Imagine you form a triangle using (-1,0), (1,0) and some randomly chosen point in the plane. What's the prob of the triangle having an obtuse angle. It's a different problem. Note that as mentioned above, you can make sense of choosing the 3rd point by taking a limits, i.e. start with a fixed rectangle, use uniform distributions, let the y-bounds tend to infinity then the x-bounds. The prob of an obtuse angle with tend to 1.

[@aerospace1472]

I dont know why you chose to constraint the triangle in a circle. Can't we choose just place two points randomly on the infinite plane. In order for the triangle to be obtuse, the 3rd point should be placed in the half-plane delimited by the line perpendicular to the line passing through the two points, at one of them. We get a 1/2 probability in this case. Is this reasoning false?

[@tolkienfan1972]

Exactly what distribution are we assuming? It's well known that there is no uniform distribution over the reals.

[@s4623]

Well, you could have solved for the area of the eye by computing the area of the segment (sector[fractional circle] - triangle[law of cosine]) in order to stick to geometry.

[@joaosouza9914]

hey, i tried this problem and come with a solution that p%=0 and I wanted to know where i missed, so... I started picking both "A" and "B" randomly in the plane since "C" is the one who decides the angles, then i noticed that if you draw AB, and draw two lines perpendicular to AB both in A and B and draw the region bounded to these 2 lines, such that AB is inside this region, you can notice that C must be in that region to be an acute triangle, then the probability is the "area" of that region divided by the entrire plane minus that region, and hmmm, well, visually the region is infinitly smaller than the plane, so, p%=0.

[@ethandavis7310]

The problem here is that both areas are infinite, even though the strip seems smaller. I agree with you and believe the probability of an obtuse triangle is 1, but I don't know how to do the formal math for this situation. Also sidenote the magenta circle Michael made in the video exists here too between AB, but doesn't really impact anything

[@LeetMath]

yeah i mean the distribution is poorly defined but ok so suppose you got your first two points, ok then lets scale the whole plane so that those points lie on (0,0) and (0,1). for the triangle to be acute the third point has to lie in 0

[@LeetMath]

but we are dealing with infinities here so it’s kind of weird. an easier problem would be to pick three points randomly on a circle, and see if the triangle formed is acute or obtuse.

[@LarkyLuna]

I bought a book of works of Lewis Caroll with Alice/through the looking glass Was really surprised when I found it was full of math problems as well lol

[@Grizzly01]

2:43 could have done with illustrating the actual right angle being talked about by Michael: ∠ACB

[@g10royalle]

What about for when the points turn out collinear? Then there are no triangles formed

[@hevado01]

Statistically i guess that probability approaches zero, being the area of the line divided over that of the infinite plane

[@_Xeto]

What about the probability of the triangle being a right angled one? Because I feel like taking the area gave us the probability of the triangle having a 90 degree or higher angle, 90 degrees included. If you found the rest of the eye’s area, it would be the probability of the triangle having a 90 degree or lower angle including 90. So the probability of exactly 90 would turn out to be zero. It would also turn out to be zero because the “area” that belongs to exactly 90 is just the perimeter of the circle, and as an area it’s 0. But obviously the 90 degree case doesn’t have 0 probability. So what gives?

[@egillandersson1780]

It's not a contradiction. Same if I say "Pick randomly a real number. What is the probability that your number is 𝛑 ? Or even a multiple of 𝛑 ?" This probability is 0, even if 𝛑 and its multiples exist. If you compare

[@JesusP7]

It may be worth mentioning that scaling AB to size 2 doesn't change the probability

[@LouisOnAir]

That doesn't, but always taking AB to be the longest side might (and will for many probability distributions over R^2)

[@iljas275]

Take every obtuse triangle and draw a bisector from an obtuse angle dividing it into two triangles. We can get two acute, one acute and one not acute, and two right triangles. If we get a right triangle, repeat this procedure with the right triangle. Thus on every obtuse triangle we have more than one acute triangle. So the quantity of acute triangles is bigger than obtuse triangles. Is my proof right?

[@AndrewWeimholt]

The spherical triangle version of this might be interesting. Can have 0, 1, 2, or 3 obtuse angles.

[@stkhan1945]

.could u do the right angled triangle probability case..Michael.. loved this as always..joy to watch.. aside: never stop watching your concept rich math posts...

[@ethandavis7310]

Probability of a right triangle is most likely zero

[@darreljones8645]

The probability, rounded off to four decimal places, is about 63.94%. So there are actually more obtuse triangles.

[@TheEternalVortex42]

If we instead let AB be the *second longest* side and proceed in the same way we will get a different answer, namely 3pi / (2 pi + 3 sqrt(3)) which is ~82% and thus a totally different answer. This suggests that the approach taken in the video is not really reasonable.

[@holyshit922]

This Lewis Carroll , less known as Charles L Dodgeson He also was a mathematician

[@allykid4720]

What is the probability that those 3 points will all be collinear?

[@zunaidparker]

Mathematically zero.

[@neilgerace355]

Every triangle that has an obtuse angle also has an acute one, but the converse is not true.

[@SaveSoilSaveSoil]

In case this has not been mentioned: check out Richard Guy's paper "There Are Three Times as Many Obtuse-Angled Triangles as There Are Acute-Angled Ones" published in Mathematics Magazine Vol. 66, No. 3 (Jun., 1993), pp. 175-179

[@zanti4132]

Actually, based on this video, only about 64% of all triangles are obtuse. 😀

[@seneca983]

@@zanti4132 But that doesn't seem to match numerical results from simulations that some commenters have tried out.

[@zanti4132]

I don't see a problem with the exact mathematics, so why trust a simulation? If you can demonstrate a flaw in the math, have at it.

[@seneca983]

@@zanti4132 The mathematics here don't specify what kind of distribution these random points are drawn from and you can't have a uniform distribution over an infinite plane. Thus you can't just assume the third point is uniformly distributed over the "eye". See also e.g. Bertrand's paradox.

[@KD-onegaishimasu]

Instead of 63.94%, I get 72.0% - 72.8% by simulating random points uniformly on a random square, and limiting that square's size to infinity changes nothing. I'll try a circular simulation next, but I don't see why point C would be distributed uniformly across the area of the eye?

[@koenth2359]

Hi, Although I have doubts that confining the triangle to a finite square is the way to go, I more or less duplicated your results. Based on a sample of 10 million observations, and an estimated standard deviation using the formula σ_e = p(1-p)/sqrt(N), I get (72.5202 +/- 0.0063 ) % Edit: this should be (72.520 +/- 0.014) %, see discussion below, so that the 95%- confidence interval is 72.492 to 72.548 For completeness I give you my code and numerical output. import random #makes a triangle out of three points in the unit square def Obtuse(): x = []; y = []; d2 = [] for i in range(3): x.append(random.random()) y.append(random.random()) for i in range(3): v1=i%3 v2=(i+1)%3 # three squared side lengths: d2.append( (x[v1]-x[v2])**2 + (y[v1]-y[v2])**2 ) d2.sort() return d2[2]>d2[1]+d2[0] trials=1000000 count=0 for trial in range(trials): if Obtuse(): count+=1 print(f'fraction obtuse = {count/trials}') This gave me (with 10 runs of a million trials each): 0.724753, 0.725597, 0.725421, 0.724567 , 0.725508, 0.724561, 0.725273, 0.724964, 0.726151, 0.725225

[@koenth2359]

In Michael's method, the eye comes from the assumption that (without loss of generality) AB is chosen to be the longest side of the triangle. Therefore the distance from the third point to either A or B cannot be more than the distance from A to B. So these are the large circles.

[@stewartcopeland4950]

@@koenth2359 Note that with a disc of unit radius, I obtained a probability of 0.730, and I also found your result of 0.725 with a unit square.

[@koenth2359]

@@stewartcopeland4950 I Modified my code into uniform a distribution over a disk of radius 1. With a million trials I get 0.7196 +/- 0.0002. Are u sure your distribution was uniform? Edit: error margin should have been. 0.7196 +/- 0.0005

[@KD-onegaishimasu]

@@koenth2359 Thank you for the source code! With one million (ten times) it doesn't change the the conclusions at all but I got σ_e = sqrt(p(1-p)/N) instead for a 95% confidence interval of 0.72489 to 0.725514 (for the square). So I'm trying to get (72.5202 +/- 0.0063 ) % and so far haven't been able to

[@alre9766]

Meaning the probability of the 3 random points making an obtuse triangle is 63.938... %.

[@Ghaz002]

just from looking at the eye shape: does this mean that the chance of a "random" triangle being a right triangle is 0%?

[@zanti4132]

The probability that the three points line up to form a right angle approaches zero, so yes. Of course, that's just how the calculation works out and does not imply that right triangles don't exist. The probability a positive integer is a prime number approaches zero, but prime numbers exist.

[@herbertschulz4713]

Do you get the same result in a polar coordinate system?

[@bflopolska]

What if we formulate the creation of a triangle by taking a straight angle and dividing it randomly into 3 pieces?

[@TheEternalVortex42]

Then you will get 3/4 as the answer :)

[@rwex1]

I didn't watch the full video, but I can answare the title simply: It's because having a random triangle whose all angles are less than 90 degrees is less probable than having just 2 of its angles less than 90 degrees.

[@voyageur8001]

It is not the probability - it is the math_expectation the point C to be into the defined area.

[@mrosskne]

what allows you to replace x+1 with 2 sin theta?

[@BarryRowlingsonBaz]

You can pretty much replace any variable with any function of another variable in an integral, as long as you change the bounds and change the "dx" to "d-theta". The skill is finding a substitution that takes you from an integral you can't do to one you can! He could have replaced x+1 with arctan(sqrt(theta)), or exp(theta), but those wouldn't help much! You never see the wrong paths in these proofs!

[@bob53135]

I'm pretty sure you cannot say that the distribution of C is uniform on its restricted area as soon as you decide AB to be the longest size. If you first choose A and B in the plane, then there's a infinity times bigger area where putting C in it would create an obtuse triangle. So, for me, it will almost surely be an obtuse triangle (probability of acute triangle is zero, though not impossible).

[@JordHaj]

We label the points such that AB is the longest side. No matter the (initial) location of three points, we can always label them in a way that AB in the longest side. For example, if C lied outside the green and blue circles, would AB still be the longest? Certainly not, since both AC and BC would be longer than the radii of those circles, which is AB. However we chose AB to be the longest, didn't we? Then we must re-label them so that the new AB is the longest.

[@bob53135]

@@JordHaj I'm not denying that you can label AB the longest side as you will. I'm saying that it changes where you would mostly find C inside the allowed area.

[@JordHaj]

@@bob53135 Could you please elaborate further? I don't quite get how it would change it. Are you saying that the density inside the eye is not uniform anymore?

[@bob53135]

@@JordHaj I say that I'm not convinced (yet) that we can assume that. I thought it was not the case, because I found a contradictory result, but after some further inspection, it seems my way leads to some paradoxes 😅 (uniform density on the plane is quite problematic…) And as I was trying to find some other more intuitive situation where the uniform density of C would change, I couldn't find any. And so, now, I'm thorn between thinking the assumption is all right, and between thinking it may be more like the Bertrand paradox, where this assumption gives just one result among others possible ones.

[@bob53135]

Let's simplify all this and think about just three random real numbers, x, y, z, labelled such as x

[@ethandavis7310]

I don't think this solution is correct. I believe the answer to be 1. If instead of taking the two points corresponding to the longest side, we instead take the first two points to be given as placed, we can then partition the plane into areas that give obtuse and acute triangles respectively if the third point is placed there. To qualify, I don't know the rules of dealing with infinities here so I could be wrong, but the area in which an acute triangle can be produced, while still infinite, seems much more "restricted" than the area where an obtuse is produced. Would love a discussion. Edit: So I've read further in the comments section and seen a couple napkin proofs on why you can't have a uniform distribution over the infinite plane, making the premise of the question impossible, so my answer cannot be sound. However, Michael has made a reasonable interpretation of the problem but he has lost generality (but the general problem has no solution anyway so whatever)

[@koenth2359]

My immediate intuition is that by rotation, translation and scaling (preserving both angles and uniform probability distribution for selecting points), every triangle ABC with angles α,β,γ can be mapped into A' =(0,0), B'=(1,0), C' =(x,y). Now α>90° when x90° when x>1 and γ>90° in the circular area of which AB is a diameter. Clearly the remaining area is an infinitesimal small part of the plane. Edit: on second thought, and also after watching the video, I think that line of thought has to be erroneous. Maybe I agree with those that see Bertrands paradox, but infinity plays a crucial role as well. Initially I thought: the most natural distribution is uniform, meaning that the probability that vertex A (or any other) is within a certain area is just portional to that area. But for finite areas that probability is 0. Also, the expected sides of our triangle are infinite then. Can we still talk about 'the longest side'?

[@KD-onegaishimasu]

I see no way to disagree with your reasoning, yet I get 72.5% -- but your answer seems correct as well!

[@VanNguyen-kx6gx]

No resolution.

[@mohamedfarouk9654]

So the probability of it being a right triangle is zero?

[@JayTemple]

The title of the video invites a much simpler answer: There are aleph-sub-C of each.

[@egillandersson1780]

I'm not comfortable with your "without lost of generality" when you choose first AB as the longest side. Maybe you are wright but it's not so obvious for me.

[@hoangnguyennguyen6445]

how about C not in the green either, it could be obtuse too

[@hoangnguyennguyen6445]

oh, then AB not the longest

[@RexxSchneider]

It's hard work, trying to pick points on an infinite plane so perhaps we should examine something a bit smaller to see if we can gain any insight. I'll make use of the "strip" method suggested by other commentators. Let's pick two points to start with, and set our coordinate system so that A is at (-1,0) and B is at (+1,0). Now, I'm too old and tired to go very far in order to pick C, so I'll confine myself to picking C from the square region between the lines x = ±a and y = ±a, where a is somewhat bigger than 1. Now given a uniform distribution of C in that square, we find that triangle ABC is acute if C lies the region between the lines x = ±1 and y = ±a, so that area is 2 x 2a = 4a. The area of the whole square, side 2a, is 4a^2, so the region where triangle ABC is obtuse has an area (4a^2 - 4a) and therefore the probability that the triangle is obtuse is (4a^2 - 4a) / 4a^2 = 1 - 1/a. Since a isn't much bigger than 1 in my case, the probability is close to 0. Now, some of my friends are a lot younger and vigorous than I, so they may be able to pick from a much larger area than I can, and for them the probability of the triangle being obtuse clearly increases towards 1 (I was going to say "in the limit", but that's too much of a strong conclusion for my little thought experiment). So, the result seems to be that the further away from AB that one is willing to go in picking C, the closer the probability of an obtuse triangle gets to 1. Everybody can have their own personal result, which only depends on how far they are prepared to go. Even Michael Penn will get the result in the video with his 'a' being close to Euler's number, e.

[@pseudolullus]

Nice little review problem

[@archimidis]

There must be more obtuse triangles. If you start with 2 sides that form an acute angle, you can end up with either an acute or obtuse triangle, but if the 2 sides form an obtuse angle you can only get an obtuse triangle.

[@ingobojak5666]

Fun, but to answer the question in a comprehensible form: probability is 63.94%.

[@gary.h.turner]

So, to answer the title question, there are more obtuse triangles than acute triangles by a factor of 63.94/(100-63.94) = 1.773

[@schweinmachtbree1013]

​@@gary.h.turner I'm nitpicking, but the answer depends on what you mean by "more" - if one means cardinality then there are as many obtuse triangles as there are acute triangles; there are uncountably many of each, and in fact there are also uncountably many right triangles. If however by "more" one means "more likely when chosen at random" then the video shows that there are more obtuse triangles than acute triangles, and there are far fewer right triangles (which have probability 0 of being chosen at random, but -- since the sample space is infinite -- this of course doesn't mean that there are no right triangles)

[@TheEternalVortex42]

@@schweinmachtbree1013 The video doesn't really explain what process we are using to choose "at random" so it doesn't even answer that question