Degenerate Triangles

[@jimmonroe5193]

Excellent explanation, to add to it, one of the used of degenerate cases is they can allow you to use a different set of rules for exploration. The idea of a degenerate triangle cascades forward. A degenerate quadrilateral with one side zero length for instance behaves like a triangle, with two sides zero, the degenerate quadrilateral behaves as a line segment. Why might this be useful? Think about computer animation, allowing for degenerate triangles allows the GPU to process everything as a triangle versus separate processes for segments. Some of y'all might want to read this: en.wikipedia.org/wiki/Degeneracy_(mathematics)

[@Radonatos]

In the primitive-assembly and rasterizer stages of the GPU you usually want to avoid degenerate triangles, because they only waste computing power with no output, and hence cull them as early as possible.

[@JosephWood1941-iz6mi]

Despite having a Mathematics A level GCE in the UK (in 1961), I had never heard of a degenerate triangle before. I'm pretty sure that it was never covered in the syllabus. Before I watched the video, I did figure out that 7-13-4 was impossible and that 2-3-1 was a straight line. That was just by drawing the triangles in my head. I love mathematics, but having spent 50-odd years as an Engineer, much of it in design, I never found the need to redefine a straight line as a degenerate triangle. "Degenerate" to me is something entirely different! I learn something new every day. At 82 years old, that's a must. Just found your channel and subscribed. Keep it up, mate. I'll be watching.

[@DeadsTBD]

not having a high level of mathematics but came to the same conclusion (only checked the smaller side) so yeah I had A was "technically" a triangle but didn't know it was called degenerate...

[@Slavic_Goblin]

I'm pretty sure the reason it's never mentioned is cause it's irrelevant, save for being a silly bit of trivia. ;) Following that same logic, a triangle is also a degenerate version of every other polygon Taking it to the extreme, a single point is degenerate from of all polygons. Btw, did you know that since two points are enough to define a side, a circle is also an... err, infinito-gon?

[@Scott-i9v2s]

@@Slavic_Goblin Excellent reasoning, sir! Because it shows that all polygons are based on the definition of a point. Now all that is needed is for educators to update themselves, rewrite the relevant curricula, & then teach up-to-date stuff right from the start! That leads to the concept of "degenerate" being no longer needed to explain obsoleted curricula. As to a circle being an infinito-gon: YUP!An infinite quantity of infinitely-short lines--which is right-there the definition of "limit"...

[@bubbahottep8644]

Strange. I learned the term in regard to polygons when I was in middle school. Of course that was 1975-ish, so the term might've only recently been applied.

[@enysuntra1347]

I doubt that. Regardless of which clade of engineering, you always come into a definition of forces/currents/potentials that splits it up into two orthogonal components, giving the components sin(a)F and cos(a)F. Every time this coincidentally only has one component with the other set to 0 (eg, a+j0 or three intersecting rods in a lattice where two of them point in the same direction), you have a degenerate triangle. Top dead centre and Bottom dead centre in a piston engine are degenerate triangles. sin(0) and sin(\pi) denote degenerate triangles. They are in fact incredibly common.

[@Aussie50InspiredDavidZ]

You know what's funny? I was taught all the way to Trigonometry and Calculus AB in High School but never once did I learn what a degenerate triangle was. I didn't even know what the requirements of making a triangle were until today. Thank you. Awesome explanation!!!

[@hannahrepollo]

literally same here, i did A level math and never even knew this

[@Kualinar]

Not ever touched at the collegial level UNLESS it's in a computer 3D graphics course.

[@Aussie50InspiredDavidZ]

@Kualinar actually, the only place that touches on Degenerate Triangles is Khan Academy's 7th grade Math. It only has one video and one exercise as far as I know. But, yeah, I never learned any of this in school.

[@shantanudhiman8194]

I just love your smile and the way you explain complex mathematical concepts. Your videos are absolutely calming and informative 😌👌🏼

[@Grecks75]

If you ask me, a line segment (or degenerate triangle as you called it), is usually NOT considered a triangle in geometry and other areas of mathematics. It is just a line segment. A triangle needs three distinct vertices, and the directions belonging to each pair of sides must be linearly independent vectors in the R^2 vector space. Degenerate cases are usually considered something special (an edge case, a limit case, outside the "normal" cases) and not lumped together with the regular cases. Similarly you would not call a pair of crossing lines a hyperbola, although it can be considered a degenerate hyperbola.

[@Ruddigore]

Fascinating. You say "never stop learning".... Well, this 70 year old has learnt something new today. Thank you.

[@Scott-i9v2s]

Am about the same age. I learned 4 things here. 1: The concept "degenerate triangles". 2: My teacher was not complete in his explanation of triangles. 3: The concept "degenerate" would never have existed (because it would have not been needed) if definitions had been taught -IN-FULL. 4: Even teachers outside the USA's educational system can be not-perfect--even though ever so much better!😄

[@igalbitan5096]

It's quicker (and enough) to compare only the longest side with the others. Here is why: Let's call C the longest side, we have automatically C >= A and C >= B, therefore A+C >= B and B+C >= A. We just have to check that A+B >= C.

[@helloweener2007]

Yes. I wanted to write the same. When it is true for the longest side, it must be true for all sides.

[@mikechappell4156]

What is the formula to find the longest side, and the two shorter sides? I assume you could do it if you wanted to use some set theory. Mathematically it seems difficult to describe. it is easier to express ( (A+B)>=C ) & ( (A+C)>=B ) & ( (B+C)>=A ) than to explain how to find the max and the other two sides to use. Keep in mind that 2 or 3 sides could be the same. There is no guarantee that you have one side of maximum length, though if you have multiple sides of max length you will always have a triangle. Just because you can do something intuitively doesn't mean it is easy to describe mathematically.

[@PhilosophicalNonsense-wy9gy]

But in case (A+B)=C the triangle has an area of 0 thus it ceases tl be a triangle, right?

[@mikechappell4156]

@@PhilosophicalNonsense-wy9gy While I agree with you, opinions differ. Degenerate triangles may have angles of 0°, 0°, 180°, but that''s a line in my book. Euclid's triangle inequality appears to indicate indicate (A+B)>C. Otherwise, you could argue that if the angle can be zero, the length can be zero, and we get octagons with three sides (five sides have lengths of zero.) IANAM.

[@Musicuber]

I was going to ask how a degenerate triangle was going to add up to 180 degrees because I thought both of the legs with the hypotenuse were 360 degrees each, but then I realized that a straight line is 180 degrees itself! Geometry can be so fascinating!!

[@sophiastern2719]

@@derwolf7810 you actually end up with two 0 degree angles along the longest length side, since a 0 degree angle in a triangle represents an immediate turnaround (two sides moving in the same direction away from a vertex). For another perspective consider the supplementary angle, which consists of two edges moving in opposite directions from a vertex, and thus meeting at a 180 degree angle. If an angle's supplement is 180 degrees, the angle itself must measure 0 degrees.

[@derwolf7810]

@@sophiastern2719 Yeah, i found my error. Thanks for your info, i deleted that message, round about the time you wrote your message, so i didn't notice you answered. Apologies for that.

[@Skank_and_Gutterboy]

My argument is that a polygon with zero area is no longer a polygon because it exists in

[@danquaylesitsspeltpotatoe8307]

Degenerate triangle = NOT triangle! you will notice theres no angles! a line does not have degrees!

[@Skank_and_Gutterboy]

@@danquaylesitsspeltpotatoe8307 I agree, your trig functions break down. You start getting complex numbers and divide-by-zero problems, too many things just don't make sense anymore.

[@johnjacobjinglehimerschmid3555]

You know at the ripe old age of 56, I really like these videos about math. What I realize about all of these is that the theorems and postulates of geometry are woefully not stressed enough. Math teachers should be putting a geometry question on every test to pound home how important knowing this basic information is.

[@TheDecagn]

There seems to be different interpretations on the triangle inequality theorem, specifically the difference between "2 sides must have a sum greater than the 3rd" and "2 sides must have a sum greater than OR equal to the 3rd". This is a bit frustrating but I think it should be important to consider what you want for a valid triangle. If you want a triangle that has an area > 0, then you would have to have the sums be greater than the 3rd side, but not equal.

[@jerry2357]

I was going to put exactly the same point. For an actual triangle, no angle can be equal to 0°. If A+B=C, then two angles equal 0° and one angle equals 180°. To me, that's not a triangle.

[@msskaggs3911]

@@jerry2357 I'm not a geometry expert, so I acknowledge that I could be incorrect, but I strongly agree with you here. I don't think how we are defining a triangle is relevant in this situation because if Sⁿ¹+Sⁿ² = Sⁿ³, then you have a line which is categorically not a triangle which is defined by having 3 sides (a/k/a LINES). The implications of a zero area triangle are pretty ridiculous, since you could just as easily show a triangle and claim it's a pentagon with 2 angles = 0º. Now, I would still answered C since this is the _most_ correct answer, and I could make a steel-man argument by claiming it's possible the length of A¹ equals 1.00000000001 and is reasonably reported as '1' which would create an actual triangle since technically Sⁿ¹+Sⁿ² > Sⁿ³. I know this isn't a great argument (or even a good argument), but it's better than any argument that can be made for the existence of Triangle C.

[@Scott-i9v2s]

@@jerry2357 But before you validly can use the word "actual", you must define its meaning. Saying "You know what I mean" is NOT a useful definition in a scientific field. You might require that a triangle must be distinguishable as such. So what must at least 1 angle minimally be? 0.0000000001 degree? Why not 1% of that value? Or even less? Would/should/could ANY value that is not exactly 0 be valid? What is still distinguishable?

[@Scott-i9v2s]

@@msskaggs3911 Re "The implications of a zero area triangle are pretty ridiculous, since you could just as easily show a triangle and claim it's a pentagon with 2 angles = 0º": Said claim is perfectly valid & not at all ridiculous. Were ridiculousness a valid argument here, then the concept of limits is absolutely ridiculous. How many of the zeroes in '1.00000000001' can one remove (or insert) and STILL "report it reasonably as '1' "? The idea of reasonableness being allowable is OK for engineers who must BUILD a thing, but it is a ridiculous concept in logic.

[@jerry2357]

@@Scott-i9v2s That rather depends on whether you are looking at a real, physical triangle, or a calculation. For a calculation, the largest angle must be less than 180°. The tolerance depends on the software you are using, or the resolution of the pocket calculator. Or if you're going old school, four figure trig tables worked to a precision of 1 minute of arc. For a real triangle, it rather depends on how you're doing the measurements. With a triangle drawn using a pencil on paper, with the angles measured using a school protractor, then you would be lucky to measure an angle of less than 0.5°. But you could measure a much smaller angle using a theodolite, down to seconds of arc.

[@sherylbegby]

Another way to state the triangle inequality is that the shortest path between any points is a straight line. The direct distance from two points will never be longer than the distance between the two points via a third point.

[@Grecks75]

A and C cannot be triangles (in the Euclidean plane) because the length of their sides do not satisfy the (strict) triangle inequality (for the Euclidean metric).

[@jamesmichuki5804]

I love the way this teacher makes things look so easy. Wishing all teachers were that gifted...😂😂

[@Madgearz]

The shortest distance between two points is a straight line. By this same definition, any path other than a straight line will be longer. Take 🔺️abc If ac is a straight line, then ab+bc must be longer. The sum of any two sides of a triangle will always be *greater* than the third. 1 +2 =3 7 +4

[@batavuskoga]

Excellent explanation . No way you can explain this better and more clearly

[@JSSTyger]

Well done sir, you are a great teacher. And this is coming from someone who minored in math applications.

[@SuperBartet]

If a flat triangle is a triangle, then a flat circle is a circle. So this explains flat earthers.

[@nymalous3428]

Nice! I love math, though I wouldn't say I'm particularly good at it. I don't know how I haven't stumbled upon your channel before now, but I'm glad I did. I am actually taking a math class this semester (going back to college after almost 20 years, oh boy) and I'm both excited and terrified. I appreciate your calm and methodical demeanor, and when you said, "A degenerate man is still a man," I knew I had to subscribe. Thanks for the great video!

[@PrimeNewtons]

You are making a good choice. Please search this channel for relevant videos. Or send me an email if you need specific topics addressed.

[@RoyTelling]

I have never heard of a Deteriorate Triangle thank you I learned something new

[@lukaskamin755]

Sorry, I can't agree, that a line segment is a triangle. No angle of triangle be either 0, or 180 degrees. And I'm absolutely positive, that the inequalities of triangle are strict, no equal sign is appropriate. By the way there's another part of the inequality |a-b|

[@XtreeM_FaiL]

Did he said that a triangle has to be on a flat surface? 180° is sum of all triangle angles only on flat surface.

[@harrymetu2746]

No, the triangle inequality in general holds for both the greater than or equal to case. It is only strictly greater than when considering non degenerate triangles.

[@lukaskamin755]

OK, I admit, I was wrong, but I'm sure in my school we never talked about degenerate triangles, nor have I seen this elsewhere. So that's a surprise for me. Didn't get what it has to flat or not flat space in comments, obviously triangles shown on a sheet of paper, they can't be any curvature? unless it's it mentioned unambiguously

[@wolvenedge6214]

​@@XtreeM_FaiL If you step out of Cartesian/Euclidean geometry, you can pretty easily make even the 4/7/13 tri work, but most of mathematics works in an assumed static, non-curving, neither hyperbolic nor hypobolic space unless explicitly stated otherwise. You're not strictly wrong, but the "gotcha!" carp is annoying.

[@jennifertate4397]

Thanks! Love these vids. And I love that you use a chalk board, specifically black. It all looks so great!

[@PrimeNewtons]

Thank you

[@biswambarpanda4468]

Wonderful sir you are a beautiful genius

[@shinypeter7]

What a great manner you have. Now you have to wade through all these thoughtful comments.

[@socraticmathtutor1869]

Another way of doing this uses the concept of semiperimeter, which is defined as s = (a+b+c)/2, where a,b and c are the side lengths. It turns out that we always have s >= a, s >= b, s >= c for any triangle. Thus, since the (7, 13, 4) triangle has semiperimeter s = 12, it should not have any sides longer than 12. So the side of length 13 gives a contradiction.

[@murirokcs5518]

this solution is actually very clever, I was worried I'd have to do lots of calculations! well done man!

[@itsandyagain]

I would argue that A + B > C (not equal), because if they are equal, you just have a straight line. Not a triangle

[@rjstegbauer]

I love all the "auxiliary" explanations and discussions!

[@Xerock]

7,4,13. Technically 1,2,3 is two straight lines overlapping each other

[@taylorfinn1496]

Really enjoy your channel! You are a fantastic teacher.

[@decentsingersclub]

the way i like to think about it is: -take the two shortest sides -connect their ends -align them and start spreading them out -the biggest angle you can get is 180 degrees -therefore the length of the remaining side has to be between 0 and their sum

[@sammascreel]

okay, you just upped my level of understanding trig. THANK YOU!!!!!

[@digiacomtech5589]

One could argue that other 'Rules of Triangles' needs to be applied. The literal definition: it has to have three (3) non 0° angles between the lines, and the sum of those angles must = 180°. Yes 0° + 0° + 180° = 180°, but you have two angles that are exactly 0°. So applying the other rules for triangles, option 'A' drops out as well because there are angles of 0° between lines 1" & 3" and 2" & 3", and 1" & 2" is 180°, so all three lines literally collapse into a single line in 3D space (i.e. all lines are parallel), and a single line (or three parallel lines no matter the length) does NOT a triangle make.

[@stevewells5580]

Excellent! I truly enjoyed this. Math is so cool. Appreciate your talent for math instruction. Bless you.

[@TranquilSeaOfMath]

6:03 Triangle A is a degenerate triangle. We can exclude this from discussion of Plane Geometry since it is not unique to one plane. Nice video. You do a wonderful job, as always.

[@andreamiele5842]

Well if you look at the definition of a triangle, you can see that degenerate triangles are NOT triangles because they are not even polygons. A triangle is a polygon with 3 sides. A polygon contains a closed polygonal chain where every side of the chain is a side of the polygon. At least this is how it's defined in my country. The sides of a degenerate triangle do not form a closed polygonal chain therefore a degenerate triangle is not a triangle. The sin(theta) argument doesn't make any sense because 1) sin(theta) is defined using the unit circle , the theorem that allows to use sin(x) in right triangle assumes the existence of the triangle in the first place 2) even if we considered a degenerate triangle a triangle it would definitely not be a right one , as it would not have a 90° angle so the theorem would not apply.

[@Jeffman978]

Question: what if the triangle was 3,3,0 triangle, where two of the vertices lay on top of each other? Would that make it a right triangle? Or not because there’s no side length?

[@alienbydfault]

​@@Jeffman978uh no that would be just a line segment?

[@maximumhero9682]

Why is it called a degenerate triangle if it’s not a triangle?

[@enysuntra1347]

​@@Jeffman978There are operations that are only allowed on triangles, and a triangle can be between the real and imaginary coordinates of a complex number (a+jb). It's the whole Zero-discussion from the medieval ages all over again, when some people couldn't get around the concept that 0 was a number. If you exclude degenerate triangles from triangles, suddenly, half of engineering and maths breaks because there no longer is a 0; and worse, a sinus wave is no longer uninterrupted if \theta=0 suddenly is undefined, as well as every 0-transfer from positive to negative values. You can "define" a degenerate triangle to no longer hold in case of a specific context, but a degenerate triangle behaves like a triangle, can be constructed (using ruler & compass) like any other triangle, and it is necessary for many theoretical and practical engineering solutions.

[@enysuntra1347]

@andreamiele5842 as you write an answer on the internet, I am very sure that even in your country, a degenerate triangle is a triangle. Otherwise, you wouldn't have AC technology, so no networks, no radio, no AC power grids. What you postulate is in fact that sin and cos waves have undefined gaps whenever the wave transfers from the positive to the negative and vice versa. This obviously cannot be true.

[@bkailua1224]

What amazed me is, it took over 10 min to explain this. I also have met a few degenerate people in my time.

[@netritos7105]

I never thought I was going to relax seeing math videos

[@bubbahottep8644]

Informally, A is degenerate or a line segment, B is a pointy isoceles triangle, C is not even a degenerate triangle and its angles cannot be defined, and D is a very flat near-isoseles triangle.

[@gelbkehlchen]

Solution: For a triangle must be: a+b>c and a+c>b and b+c>a. The first (A): 1+2=3, no triangle; the second (B): o.k.; the third (C): 7+4

[@PatClevenger0709]

I always thought that the longest side should be smaller than the sum of the other two sides. If the sum was equal to the longest side, it would just lie right on top.

[@ondrejmatousek9723]

Great video, degenerative triangles absolutely blew my mind

[@annacerbara4257]

I think it is sufficient to verify that the greater of the sides is less than the sum of the other two. 😊

[@Gnerko123]

Yes! Tou can see this from the thumbnail already. Also, if the sum of the two minor sides is more than the greatest side, that is a sufficient condition that the triangle does exist.

[@n3l3sh]

As long as the triangle has an area thats greater than 0 its a triangle, so we can just use herons formula. Find the semi permeter of all three sides then compare if any side is above the semi perimeter , then we know that triangle has no area since we will get either 0 or a complex number as an answer.

[@princearthur5532]

You are a great 😃 teacher. I like your style

[@joanmackie1735]

I hope nobody tries to explain degenerate triangles to kids just starting to learn geometry. They will easily see that the sum of any two sides must be greater than the third. They often like to use compasses, so they could prove (or even discover) this rule for themselves.This is about as much knowledge as the average (non-mathematician) person needs to go about their daily life. Still, it’s fascinating to know that so much exists beyond what I learned in school in the 1950’s.

[@Ahmad-ik6vz]

Answer is A&C. The prime condition for construction of triangle is that the sum of any two sides must be greater than 3rd side. The triangle 4,7,13 can not exist. Also the triangle 1,2,3 can not exist.

[@nickmcginley4570]

Can answer by inspection. A 1, 2, 3 triangle is a line segment of length 3. Nothing left over for area.

[@richd6362]

I never heard of degenerate triangles. Very interesting. Thanks for the video.

[@brianfeuerman1732]

I know of their existence but it always seemed a bit more trivial than something I’d ever have to consider in math. It might be a culture difference.

[@Galaxy-ww2og]

Thank you for the explanation. The solution and explanation well explained.

[@FuriousMaximum]

This channel is growing by the day! Congrats on 100K!

[@kilroy987]

The total lengths any two sides of a triangle must be greater than the length of the third side in order to form a triangle. So the 1-2-3 triangle isn't really a triangle because the 1-2 sides form a straight line in order to connect with the 3 side. But that all depends on the accepted definition of a triangle. The 4-7-13 triangle can't exist at all because 4-7 adds up to 11, which is too short to connect with the 13 side. The other two triangles are fine.

[@jimmonroe5193]

You are using the standard High School version of the triangle inequality theorem. They lied to you for the sake of simplicity. This is one of the reasons I hated teaching HS geometry.

[@kevinkasp]

My exact same thoughts. But after watching the video and seeing him draw the unit circle and showing the trig functions on it, I had to admit there must exist Sin (0) and Cos (90°) and accept the consequences of them. In other words in order for me to accept all the ideas and rules of trigonometry, I must also accept the idea that a triangle with an angle of zero degrees exists. Obviously if it has an angle of zero degrees the two “short” sides must be drawn on top of the long side, so it just looks like a line.

[@christianemden7637]

I agree, not sure that this video is going to convince me of anything different

[@christianemden7637]

He also uses greater equal instead if greater than,which is what the definition requires.

[@jimmonroe5193]

@@christianemden7637under pure Euclidean geometry perhaps, but maths have moved past this. Using >= connects this definition with other metric spaces using the generalized rule |a|+|b| >= |a+b|, e.g., vector spaces. Moreover as he mentioned degenerate triangles are critical to applications of computer graphics. We may not like degenerate triangles, but we need them.

[@captainlengthwidth6692]

My answer to the 'is the 1,2,3 triangle a triangle?' was 'yes' - on the grounds that all triangles would look like a straight line with a point on it if viewed from *the plane of the triangle*. No flatlander would be able to see all three vertices at the same time. It only looks like a three sided shape from our third dimensional, perpendicular to the 2D plane, viewpoint. Triangles on a sphere have interior angles of more than 180 degrees - up to, I just realised, 900 degrees... if you put a degenerate triangle on a sphere with two vertices at the poles and the third on the equator. 360 at each pole and 180 at the equator = 900 - I'm sure there is a HUGE flaw in my thinking here...

[@phunkydroid]

7:50 going to dispute this, the probability is not zero. If you can intentionally pick numbers to make it zero, then those same numbers can come from random selection as well. The odds may be small but they can't be zero unless it's impossible to do intentionally as well.

[@MD-vs9ff]

Oh boy is there a video for you. 3Blue1Brown made a video with a title like "0% chance doesn't mean impossible".

[@serbanudrea9429]

I did not participate in the pool exactly because of the 1-2-3 case. In my opinion it's a matter of definition, if one considers this case a triangle or not. When I learned about triangles it was considered a special kind of triangle (don't remember exactly if "degenerate" was used to name it). However, I mentioned that presently some definitions seem to differ from how I learned/remember them.

[@ericklugaro730]

Also all the angles of a triangle is 180. Assign letters to each point and you can show it as well. So even with the degenerate triangle is two zeros and one 180 degrees. Just a different proof it is a triangle.

[@PS-mh8ts]

another method is to explore the terms that make up heron's formula A=√[s(s-a)(s-b)(s-c)] assuming a, b, and c are all positive, if any of (s-a), (s-b) or (s-c) is zero, the triangle is degenerate. if any of (s-a), (s-b), or (s-c) is negative, a triangle cannot be formed with side-lengths a, b, and c. for example, in a=1, b=2, and c=3, s=(a+b+c)/2=(1+2+3)/2=3. thus, we find that s-c=0, and hence the triangle is degenerate in case where a=7, b=4, and c=13, s=(7+4+13)/2=12. in this case we find that s-c is negative, and hence it's impossible to form a triangle

[@alain5726]

Excellente explication, bravo et merci.

[@PhilosophicalNonsense-wy9gy]

In my opinion, claiming that a+b may be equal to c means like saying every line is a triangle, and it also applying to other polygons would mean a line is every kind of shape in existence, give you put a set of ∞ number of vertices. Remember ∞={1,2,3,4,...}

[@emanisrirangasai2438]

The area of triangle in the first triangle having sides as 1, 2 & 3 will be 0 This satisfies the other axiom : "the angle of a straight line is 180 degrees" Hence this is triangle

[@hriturajverma4449]

It is very simple to explain by heron's formula

[@lucasdetex8703]

I found The Victor Wooten of Mathematics. Enjoyable to listen and learning is a bonus.

[@Xebtria]

very easy to solve. add the smallest side (a) and the second smallest side (b) together. In order to form any triangle, they must be bigger than the longest side (c). theoretically, if the angle between the smallest and the second smallest side is 180°, the other two angles are 0°, therefore it is technically a line and not a triangle, but assuming we would still define this shape as a triangle, then a + b = c. but if a+b < c, then the triangle does not exist. therefore triangle C does not exist, as 7 + 4 = 11 < 13. triangle A would be the example which I brought up earlier, which is 1 + 2 = 3 = 3, therefore it is a line and not a triangle per se. triangles B (7 + 1 = 8 > 7) and D (10 + 13 = 23 > 21) do exist, they are just very "flat" or "slim" triangles. so the question of whether triangle A is actually a triangle or just a line is a matter of how you define a triangle. if one is happy with one 180° angle and two 0° angles, then A is a triangle as well. edit: after having watched the video, well I guess that was the whole point of the video :D but why do I have to check all 3 formulas? the largest + anything is always bigger than the second largest by definition. and the largest + the second largest is always bigger than the smallest, also by definition. so you have to determine what the largest is and check if it is smaller than the other two added together

[@Scott-i9v2s]

@Xebtria Re "why do I have to check all 3 formulas?": Because the definition for triangle generalizes (defines ALL triangles), ie states no specific value for any of the sides.

[@Xebtria]

@@Scott-i9v2s I am also generalizing here. if a

[@Scott-i9v2s]

@@Xebtria The equation: a+b >= c is NOT the same equation as: a+c >= b NOR is it the same equation as: b+c >= a So one must be given all 3 lengths before one can determine relative sizes &c. Your step (a) is in reality already 3 calculations: is a >=b? T/F is a >=c? T/F is b >=c? T/F Then you add step (b), a 4th calculation: is a+b >= c? This is the same quantity of calculations as in the formal generalization, because the 4th calculation is determining the sum of the 3 initial truth-states. So both methods have the same level of efficiency. Just (slightly) different methods. I see no reason to think that your method might lead to errors.

[@Xebtria]

@@Scott-i9v2s I yield, I understand it now, you are correct.

[@Aderon]

*flashes back to Cyberchase* Jokes aside, your content is really good, and while I don't think they covered degenerate triangles, I fondly remember the triangle inequality from watching it all those years ago.

[@Hogscraper]

Perhaps in other languages the usage is different but I was taught that the word triangle used alone is always assumed to be a non-degenerate polygon because a triangle is a polygon, polygons are two dimensional objects by definition, (an enclosed area), and a line segment can't be said to be a two-dimensional object.

[@TomCee53]

Can I get comments then on the extrapolation of this. When all three sides are zero (a point), is it still a triangle?

[@nimabhutia6690]

Video is awesome 😎

[@krielsavino5368]

Love the videos. Intro music is hot! where is from?

[@ProHolmes]

If we assume than any angle of a triange is more than 0 degrees, then any side should be less than the sum of 2 other sides. Which says that A and C are not triangles. This all is correct if we are in euclidean geometry.

[@maxforsberg8852]

You can evaluate this problem by using the law of cosines where you take the longest side as the opposite side. When applying this on the 1-2-3 triangle we find that cos of the angle between the 1 and the 2 is -1 which means 180 degrees which becomes the collapsed degenerate triangle. When applying the same logic on the 7-4-13 triangle we get that cos of the angle between the 4 and the 7 becomes -13/7 but this is impossible because the values of cos range from -1 to 1 when the angle is between 0 and 180 degrees. Hence the 7-4-13 triangle doesn’t exist

[@nekomatafuyu]

I would say that A is mathematically a triangle, but in common parlance it is not. The trick is to know why the question is being asked so you can answer appropriately.

[@JSSTyger]

Ill say the 7-4-13 triangle. If I have sides 7 and 4, the maximum for the other side is 11.

[@akarsh240]

"Never stop learning. Those who stop learning, stop living"

[@GOLDman4856]

Its not a triangle

[@allenhonaker4107]

" In the word of mathematics some things are not normal " should be enshrined in stone.😂😂😂

[@iscs_]

If the lines a and b are equal to c, would the "shape" then still be considered a triangle? wouldn't it just be a straight line?

[@kazedcat]

Straight line is not a shape it is a line. But mathematically a triangle with 0 area will still work it does not become undefined. Since the system does not blow up then it is part of the solution set.

[@plentyofpaper]

*This has been corrected thanks to an error pointed out in later comments. I had originally misidentified the failure condition as the final answer* I used roughly this principle in a college statistics course I was taking (although I didn't get into degenerate triangles.) I think the problem was: If you take a line segment, and randomly cut it in 2 places, what's the probability you can make a triangle out of the remaining pieces? I solved this, and expanded on it for any n-gon using the principle that all cuts must fall on the same half of the line segment. Looking at the failure condition (all cuts are on the same side) that's 1/2 to make a triangle. 3 cuts, 1/2^2 to make a quadrilateral. 4 cuts, 1/2^3 to make a pentagon, etc. The success condition is 1 those amounts. So the generic formula was the probability of creating a succesful n-gon with this method would be 1-1/(2^(n-2)) While I was doing this, my instructor was showing the class some completely different method that I had trouble following, then assigned solving for a quadrilateral as extra credit. I showed him what I came up with, and verified that my answers were consistent with his. But he wasn't convinced my reasoning was sufficient. He thought all cuts falling on the same half seemed intuitive, but wasn't completely convinced and wanted a proof. Which wasn't something I felt up to.

[@SuperTheNobody]

actually the cuts have to be on opposite halves and the distance between them must be at least half the length, the chance for a triangle ends up being 1/4, very interesting problem though

[@plentyofpaper]

@@SuperTheNobody Indeed, I did this incorrectly just now. I had established a failure condition in my original comment. So the correct answer is 1-[1/(2^(n-2))] I'll go add an edit. Thank you for the catch!

[@Rai_Te]

If you have the 3 angles and the length of one side of a triangle, you usually can calculate all other properties (like the missing sidelenghts). This is NOT the case for a degenerate triangle as you cannot calculate the missing sides from the following set of values: 0deg, 0deg, 180deg, 10units ... the only conclusion you can have is that a+b=10. So, one should be very careful with degenerated triangles, as some operations are not possible with them. (Which is why I never liked this definition).

[@enysuntra1347]

I give you one word why we need it, though: Sine waves. And all the bells and whistles that come with them, like sound, AC, complex numbers. A degenerate triangle is to geometry as the number 0 is to arithmetics. There are good reasons to exclude it from |N (creating a new subset |N_0 for the Natural numbers plus the number 0: Peano's definition of rational numbers |Q is any integer |Z divided by any natural number |N, which only holds if 0 is not a natural number), but you need it nevertheless the moment you go beyond natural numbers.

[@nerze3157]

​@@enysuntra1347Well, I just discovered that the definition of |N isn't universal. When I learned it (and in the ISO standard), 0 IS considered a natural number and part of |N. To consider the natural numbers excluding 0, we used |N*

[@nerze3157]

One interesting side effect is that with your definition, (Z,+) isn't a group

[@enysuntra1347]

@@nerze3157 This changes all the time. |N and |N_0 is what I first learned, and it's not uncommon. However, the only difference is that you have to write |N_\{0} (N without {0}) if you define |N as beginning with 0, you need both sets.

[@enysuntra1347]

@@nerze3157 Help me out here. (|Z,+) isn't changed by using Peano's |N, it's still the same set. Why does it lose its group status?

[@jpopelish]

I like the example of a degenerate circle. A circle is the locus of points that are all the same distance from a central point. A degenerate circle is the locus of points that are all zero distance from the central point. Zero is a mathematical distance.

[@chinareds54]

A point is the ultimate degenerate form of almost every geometric figure.

[@amritanshshrivastava-i7b]

thanks sir I really appricate your work this helps a lot

[@N8570E]

The 'Triangle' 1 2 3, is not a triangle, it is a straight-line segment. I.E., 1 + 2 = 3. Also, it is not a LINE! It is a line segment. Mathematicians should care. Entertaining. Thank you for your efforts. May you and yours stay well and prosper.

[@PrimeNewtons]

Thanks for the info!

[@JohnGalt0902]

Yes a straight line is technically a triangle, and technically correct is the best correct.

[@anacreon212]

I think in school we never were taught about degenerate triagles and in fact the test did not have the or equals part it was just A + B > C, B + C > A, and A + C > B for it to be a triangle that exists.

[@Scott-i9v2s]

So there you see an issue with the educational system--in NOT teaching the correct definition(s).

[@jrboxi2301]

Sum of length of two sides is greater than 3 rd side.Ans-A

[@jrboxi2301]

A,C,D

[@Weirdtopia]

Thanks for the very simple explanation. I have a question with warning that I'm not a math expert and have very basic math education. I came across degenerate triangles when finding out if "in theory" that three points are in the same coordinates, does that count as a triange but also if also counts as an "equilateral triangle"? I found out from Wiki (gain of salt) that three points is a degenerate triangles however each angle is undefined. Would we consider that we can't determine that it's an equilateral triangle because the angles are undefined or do we say that it's possiable but not sure. I'm curious what people think and I'm sorry if my questions doesn't make sense :).

[@helloweener2007]

Imagine constructing the triangle with a ruler and a compass. When you draw the longest line with the ruler and try to find the 3rd corner with the compass you have to draw circles that are in sum bigger than the longest side or the circles will never intersect.

[@mjmeans7983]

In the beginning you didn't specify that you were using a flat coordinate system. Now consider polar and spherical coordinates. Then consider the triangle as vectors on a spacetime manifold.

[@MotorMcMatt]

Very nice! Thank you for this tutorial. Clear and concise!

[@josephmathes]

If you want to get into special cases, they can all be triangles in a non-euclidian space

[@PrimeNewtons]

This is still Euclidean plane.

[@crosswingrobots]

The triangle inequality holds in any space where movement has equal cost in opposite directions and edges are defined as the globally lowest cost path between two points.

[@namewithheld8115]

I have to agree with you. If we're willing to accept that a straight line is a triangle in this context, then we could also accept that the 4-7-13 is a triangle because the sides meet in an unspecified Z plane.

[@nicholasharvey1232]

Triangle C immediately jumped out at me because its longest side is longer than its other two sides combined, which is impossible. Also, if Triangle A is a real triangle then it has a height of zero, since its longest side (which I'll designate as the base) is the same length as the two others combined. Triangle B is a perfectly possible isosceles triangle, and I see nothing wrong with Triangle D (which is scalene).

[@billh5923]

None A and C because the sum of two sides is

[@NichelleSerena]

Mid point theorem states that the line joining two opposite sides of the triangle is half of the third side

[@tinybabybread]

It still applies to this degenerate triangle Half of the 1 unit segment is 0.5. Half of the 2u segment is 1 They sum up to 1.5, which is half of the third side being 3u

[@QuippersUnited]

First of all, great video. I'd never heard of degenerate geometry, so this was very interesting to me. I'm still confused on the concept, however. A triangle is defined as having three sides, vertices, and angles, however, if one of those angles is 180 degrees, then it's not conducive to the placement of a vertex. If I can place a vertex on a 180 degree angle, I can place that same vertex anywhere on a straight line with no discernable alteration to the geometry. I could even place multiple vertices on the line and say it's a degenerate square, or hexagon, or n-gon. In short, the provided values produce an exact geometry, but the geometry in isolation does NOT provide exact values. Is there a justification for the inclusion of ambiguous geometries?

[@cwldoc4958]

Very nicely done video, thank you for posting. The usefulness of triangles with one angle of 0 degrees is apparent. However, two such triangles with angles of (0,90,90) and (0, 45,135), respectfully, would have the same vertices and appear indistinguishable. In fact, if one specifies such a triangle by listing the vertices, then the only angle that is defined is the one that is zero, so we would in effect have a triangle with only one angle.

[@nerze3157]

I fail to see how you could construct a triangle with angles of (0,45,135) as it would expand into the 2nd dimension. It think the only angles possible for a flat triangle are of (0,90,90) and (0,0,180).

[@nerze3157]

Well, except if all sides are of length 0, which is another triangle that is quite peculiar lol.

[@spicymickfool]

Note a quick test is half the perimeter >= any side.

[@mmmk6322]

6:11 Yes. Three vertices exist (triangle definition does not state that the vertices cannot exist within the edges). Sum of angles is 180: 0,0,180. Three edges. This triangle is undefined as there are infinite number of triangles with the longest side being 10 inches tho. A different classification of triangles that look exactly like straight lines?

[@sebastiengross7849]

3:49 Nice one. And wouldn't it be more visual too also use circles? At each end of the line 10 you draw a circle or radius 1 and 2. If the the circles do not intersect, obviously the triangle cannot be drawn. Hence the circles must either intersect in 2 points or be tangent.

[@Kualinar]

The 1-2-3 triangle is a degenerate triangle. The sides of length 1 and 2 have a combined length of 3, the same as the third side. The 4-7-13 triangle do not exist. (the shorter sides lengths add up to 11, witch is shorter than the longer side, so, it's not a closed shape.) I like your explanation and illustration.

[@MrSeezero]

It has to be greater than only, because if that sum is equal to the 3rd side, then you only have a segment which is not a triangle.

[@MrSeezero]

I looked at your video further and looked up the idea of a degenerate triangle and actually found it. I learn something new everyday.

[@DanDart]

Very clear and helpful. Hilarious too.

[@reidpattis3127]

If any two sides add up to a third side, then the three points must be colinear. Hence, cannot be an triangle. I feel like there might be some mixup with the so-called triangle inequality in vector calculus, which does deal with equality.

[@jimmonroe5193]

The problem with many high school geometry courses is they sacrifice depth for simplicity. The any three noncolinear points requirement is a simplification to avoid discussing degenerate cases.